Integrand size = 23, antiderivative size = 23 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {2 b \cos (2 a+2 b x)}{d^2 (c+d x)}-\frac {4 b^2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^3}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\text {Int}\left (\frac {\tan (a+b x)}{(c+d x)^3},x\right ) \]
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Not integrable
Time = 0.49 (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 = {} \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]
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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)^3}-\frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^3}\right ) \, dx \\ & = 3 \int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx-\int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ & = 3 \int \frac {\sin (2 a+2 b x)}{2 (c+d x)^3} \, dx+\int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^3} \, dx-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = \frac {3}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3} \, dx+\int \frac {\sin (2 a+2 b x)}{2 (c+d x)^3} \, dx-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {3 \sin (2 a+2 b x)}{4 d (c+d x)^2}+\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3} \, dx+\frac {(3 b) \int \frac {\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{2 d}-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {3 b \cos (2 a+2 b x)}{2 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2}-\frac {\left (3 b^2\right ) \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d^2}+\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{2 d}-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (2 a+2 b x)}{d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2}-\frac {b^2 \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d^2}-\frac {\left (3 b^2 \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}{d^2}-\frac {\left (3 b^2 \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}{d^2}-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (2 a+2 b x)}{d^2 (c+d x)}-\frac {3 b^2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^3}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2}-\frac {3 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {\left (b^2 \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}{d^2}-\frac {\left (b^2 \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}{d^2}-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (2 a+2 b x)}{d^2 (c+d x)}-\frac {4 b^2 \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^3}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\int \frac {\tan (a+b x)}{(c+d x)^3} \, dx \\ \end{align*}
Not integrable
Time = 4.68 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]
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Not integrable
Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
\[\int \frac {\sec \left (x b +a \right ) \sin \left (3 x b +3 a \right )}{\left (d x +c \right )^{3}}d x\]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\text {Timed out} \]
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Not integrable
Time = 0.43 (sec) , antiderivative size = 308, normalized size of antiderivative = 13.39 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]
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Not integrable
Time = 0.40 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]
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Not integrable
Time = 30.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )}{\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^3} \,d x \]
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