Integrand size = 23, antiderivative size = 23 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\frac {2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d}+\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\text {Int}\left (\frac {\tan (a+b x)}{c+d x},x\right ) \]
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Not integrable
Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 \cos (a+b x) \sin (a+b x)}{c+d x}-\frac {\sin ^2(a+b x) \tan (a+b x)}{c+d x}\right ) \, dx \\ & = 3 \int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx-\int \frac {\sin ^2(a+b x) \tan (a+b x)}{c+d x} \, dx \\ & = 3 \int \frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx+\int \frac {\cos (a+b x) \sin (a+b x)}{c+d x} \, dx-\int \frac {\tan (a+b x)}{c+d x} \, dx \\ & = \frac {3}{2} \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx+\int \frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx-\int \frac {\tan (a+b x)}{c+d x} \, dx \\ & = \frac {1}{2} \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx+\frac {1}{2} \left (3 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx+\frac {1}{2} \left (3 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx-\int \frac {\tan (a+b x)}{c+d x} \, dx \\ & = \frac {3 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{2 d}+\frac {3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {1}{2} \cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx+\frac {1}{2} \sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx-\int \frac {\tan (a+b x)}{c+d x} \, dx \\ & = \frac {2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d}+\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\int \frac {\tan (a+b x)}{c+d x} \, dx \\ \end{align*}
Not integrable
Time = 2.89 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx \]
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Not integrable
Time = 0.54 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
\[\int \frac {\sec \left (x b +a \right ) \sin \left (3 x b +3 a \right )}{d x +c}d x\]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{d x + c} \,d x } \]
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Exception generated. \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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Not integrable
Time = 0.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 7.91 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{d x + c} \,d x } \]
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Not integrable
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{d x + c} \,d x } \]
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Not integrable
Time = 29.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )}{\cos \left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]
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