Como Calcular essa integral

PAra encontrarmos a integral dada, realizaremos os cálculos abaixo:

\(\begin{align} & \int_{{}}^{{}}{f(t)}=\int_{{}}^{{}}{{{e}^{(3x)}}tan(x)} \\ & \int_{{}}^{{}}{{{e}^{(3x)}}tan(x)}=\int_{{}}^{{}}{\frac{{{e}^{3x}}\sin x}{\cos x}} \\ & \int_{{}}^{{}}{\frac{{{e}^{3x}}\sin x}{\cos x}}=\left( \frac{2}{39}+\frac{i}{13} \right){{e}^{3x}}\left[ \left( 3+2i \right)\left( \frac{-3i}{2}-\frac{3i}{2}-{{e}^{2ix}} \right) \right]- \\ & -3{{e}^{2ix}}\left( 1,1-\frac{3i}{2};2-\frac{3i}{2};-{{e}^{2ix}} \right) \\ & \int_{{}}^{{}}{\frac{{{e}^{3x}}\sin x}{\cos x}}=\left( \frac{2}{39}+\frac{i}{13} \right){{e}^{3x}}\left[ \left( 3+2i \right)\left( \frac{-3i}{2}-\frac{3i}{2}-{{e}^{2ix}} \right) \right]- \\ & -3{{e}^{2ix}}\left( 1,1-\frac{3i}{2};2-\frac{3i}{2};-{{e}^{2ix}} \right)+C \\ \end{align}\ \)

Portanto, a integral será \(\begin{align} & \int_{{}}^{{}}{\frac{{{e}^{3x}}\sin x}{\cos x}}=\left( \frac{2}{39}+\frac{i}{13} \right){{e}^{3x}}\left[ \left( 3+2i \right)\left( \frac{-3i}{2}-\frac{3i}{2}-{{e}^{2ix}} \right) \right]- \\ & -3{{e}^{2ix}}\left( 1,1-\frac{3i}{2};2-\frac{3i}{2};-{{e}^{2ix}} \right)+C \\ \end{align}\ \).