Optimal. Leaf size=179 \[ \frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}+\frac{b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}-\frac{b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}-\frac{\sin (4 a+4 b x)}{8 d (c+d x)} \]
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Rubi [A] time = 0.268476, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}+\frac{b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}-\frac{b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}-\frac{\sin (4 a+4 b x)}{8 d (c+d x)} \]
Antiderivative was successfully verified.
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Rule 4406
Rule 3297
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx &=\int \left (\frac{\sin (2 a+2 b x)}{4 (c+d x)^2}+\frac{\sin (4 a+4 b x)}{8 (c+d x)^2}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\sin (4 a+4 b x)}{(c+d x)^2} \, dx+\frac{1}{4} \int \frac{\sin (2 a+2 b x)}{(c+d x)^2} \, dx\\ &=-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}-\frac{\sin (4 a+4 b x)}{8 d (c+d x)}+\frac{b \int \frac{\cos (2 a+2 b x)}{c+d x} \, dx}{2 d}+\frac{b \int \frac{\cos (4 a+4 b x)}{c+d x} \, dx}{2 d}\\ &=-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}-\frac{\sin (4 a+4 b x)}{8 d (c+d x)}+\frac{\left (b \cos \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{4 b c}{d}+4 b x\right )}{c+d x} \, dx}{2 d}+\frac{\left (b \cos \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}{2 d}-\frac{\left (b \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{4 b c}{d}+4 b x\right )}{c+d x} \, dx}{2 d}-\frac{\left (b \sin \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}{2 d}\\ &=\frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Ci}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}+\frac{b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{Ci}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}-\frac{\sin (2 a+2 b x)}{4 d (c+d x)}-\frac{\sin (4 a+4 b x)}{8 d (c+d x)}-\frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{2 d^2}-\frac{b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b c}{d}+4 b x\right )}{2 d^2}\\ \end{align*}
Mathematica [A] time = 1.67309, size = 151, normalized size = 0.84 \[ -\frac{-4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )-4 b \cos \left (4 a-\frac{4 b c}{d}\right ) \text{CosIntegral}\left (\frac{4 b (c+d x)}{d}\right )+4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )+4 b \sin \left (4 a-\frac{4 b c}{d}\right ) \text{Si}\left (\frac{4 b (c+d x)}{d}\right )+\frac{2 d \sin (2 (a+b x))}{c+d x}+\frac{d \sin (4 (a+b x))}{c+d x}}{8 d^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.023, size = 256, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{2}}{8} \left ( -2\,{\frac{\sin \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}+2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 2\,{\frac{-ad+bc}{d}} \right ) }+2\,{\frac{1}{d}{\it Ci} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 2\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) }+{\frac{{b}^{2}}{32} \left ( -4\,{\frac{\sin \left ( 4\,bx+4\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}+4\,{\frac{1}{d} \left ( 4\,{\frac{1}{d}{\it Si} \left ( 4\,bx+4\,a+4\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 4\,{\frac{-ad+bc}{d}} \right ) }+4\,{\frac{1}{d}{\it Ci} \left ( 4\,bx+4\,a+4\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 4\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [C] time = 1.74053, size = 406, normalized size = 2.27 \begin{align*} -\frac{b^{2}{\left (2 i \, E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 i \, E_{2}\left (-\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + b^{2}{\left (i \, E_{2}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{2}\left (-\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) + 2 \, b^{2}{\left (E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\left (-\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + b^{2}{\left (E_{2}\left (\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{2}\left (-\frac{4 i \, b c + 4 i \,{\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right )}{16 \,{\left (b c d +{\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.557949, size = 597, normalized size = 3.34 \begin{align*} -\frac{4 \, d \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) + 2 \,{\left (b d x + b c\right )} \sin \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{4 \,{\left (b d x + b c\right )}}{d}\right ) + 2 \,{\left (b d x + b c\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) -{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) -{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{4 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{4 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac{4 \,{\left (b c - a d\right )}}{d}\right )}{4 \,{\left (d^{3} x + c d^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{3} \sin \left (b x + a\right )}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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