Optimal. Leaf size=257 \[ \frac{5 b \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{16 d^2}+\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{16 d^2}-\frac{b \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{8 d^2}-\frac{b \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^2}+\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{16 d^2}+\frac{5 b \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{16 d^2}-\frac{\cos (a+b x)}{8 d (c+d x)}+\frac{\cos (3 a+3 b x)}{16 d (c+d x)}+\frac{\cos (5 a+5 b x)}{16 d (c+d x)} \]
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Rubi [A] time = 0.344944, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac{5 b \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b c}{d}+5 b x\right )}{16 d^2}+\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{16 d^2}-\frac{b \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{8 d^2}-\frac{b \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^2}+\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{16 d^2}+\frac{5 b \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{16 d^2}-\frac{\cos (a+b x)}{8 d (c+d x)}+\frac{\cos (3 a+3 b x)}{16 d (c+d x)}+\frac{\cos (5 a+5 b x)}{16 d (c+d x)} \]
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 ^2(a+b x)}{(c+d x)^2} \, dx &=\int \left (\frac{\cos (a+b x)}{8 (c+d x)^2}-\frac{\cos (3 a+3 b x)}{16 (c+d x)^2}-\frac{\cos (5 a+5 b x)}{16 (c+d x)^2}\right ) \, dx\\ &=-\left (\frac{1}{16} \int \frac{\cos (3 a+3 b x)}{(c+d x)^2} \, dx\right )-\frac{1}{16} \int \frac{\cos (5 a+5 b x)}{(c+d x)^2} \, dx+\frac{1}{8} \int \frac{\cos (a+b x)}{(c+d x)^2} \, dx\\ &=-\frac{\cos (a+b x)}{8 d (c+d x)}+\frac{\cos (3 a+3 b x)}{16 d (c+d x)}+\frac{\cos (5 a+5 b x)}{16 d (c+d x)}-\frac{b \int \frac{\sin (a+b x)}{c+d x} \, dx}{8 d}+\frac{(3 b) \int \frac{\sin (3 a+3 b x)}{c+d x} \, dx}{16 d}+\frac{(5 b) \int \frac{\sin (5 a+5 b x)}{c+d x} \, dx}{16 d}\\ &=-\frac{\cos (a+b x)}{8 d (c+d x)}+\frac{\cos (3 a+3 b x)}{16 d (c+d x)}+\frac{\cos (5 a+5 b x)}{16 d (c+d x)}+\frac{\left (5 b \cos \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{5 b c}{d}+5 b x\right )}{c+d x} \, dx}{16 d}+\frac{\left (3 b \cos \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\sin \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{16 d}-\frac{\left (b \cos \left (a-\frac{b c}{d}\right )\right ) \int \frac{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{8 d}+\frac{\left (5 b \sin \left (5 a-\frac{5 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{5 b c}{d}+5 b x\right )}{c+d x} \, dx}{16 d}+\frac{\left (3 b \sin \left (3 a-\frac{3 b c}{d}\right )\right ) \int \frac{\cos \left (\frac{3 b c}{d}+3 b x\right )}{c+d x} \, dx}{16 d}-\frac{\left (b \sin \left (a-\frac{b c}{d}\right )\right ) \int \frac{\cos \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{8 d}\\ &=-\frac{\cos (a+b x)}{8 d (c+d x)}+\frac{\cos (3 a+3 b x)}{16 d (c+d x)}+\frac{\cos (5 a+5 b x)}{16 d (c+d x)}+\frac{5 b \text{Ci}\left (\frac{5 b c}{d}+5 b x\right ) \sin \left (5 a-\frac{5 b c}{d}\right )}{16 d^2}+\frac{3 b \text{Ci}\left (\frac{3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac{3 b c}{d}\right )}{16 d^2}-\frac{b \text{Ci}\left (\frac{b c}{d}+b x\right ) \sin \left (a-\frac{b c}{d}\right )}{8 d^2}-\frac{b \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{8 d^2}+\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{16 d^2}+\frac{5 b \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b c}{d}+5 b x\right )}{16 d^2}\\ \end{align*}
Mathematica [A] time = 2.13917, size = 212, normalized size = 0.82 \[ \frac{-2 \left (b \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )+b \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )+\frac{d \cos (a+b x)}{c+d x}\right )+5 b \sin \left (5 a-\frac{5 b c}{d}\right ) \text{CosIntegral}\left (\frac{5 b (c+d x)}{d}\right )+3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )+3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )+5 b \cos \left (5 a-\frac{5 b c}{d}\right ) \text{Si}\left (\frac{5 b (c+d x)}{d}\right )+\frac{d \cos (3 (a+b x))}{c+d x}+\frac{d \cos (5 (a+b x))}{c+d x}}{16 d^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 367, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{2}}{8} \left ( -{\frac{\cos \left ( bx+a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\frac{-ad+bc}{d}} \right ) }-{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \sin \left ({\frac{-ad+bc}{d}} \right ) } \right ) } \right ) }-{\frac{{b}^{2}}{80} \left ( -5\,{\frac{\cos \left ( 5\,bx+5\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-5\,{\frac{1}{d} \left ( 5\,{\frac{1}{d}{\it Si} \left ( 5\,bx+5\,a+5\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 5\,{\frac{-ad+bc}{d}} \right ) }-5\,{\frac{1}{d}{\it Ci} \left ( 5\,bx+5\,a+5\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 5\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) }-{\frac{{b}^{2}}{48} \left ( -3\,{\frac{\cos \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-3\,{\frac{1}{d} \left ( 3\,{\frac{1}{d}{\it Si} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) }-3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 3\,{\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 = 1.96772, size = 593, normalized size = 2.31 \begin{align*} -\frac{1073741824 \, b^{2}{\left (E_{2}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac{b c - a d}{d}\right ) - 536870912 \, b^{2}{\left (E_{2}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{2}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) - 536870912 \, b^{2}{\left (E_{2}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{2}\left (-\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) + b^{2}{\left (-1073741824 i \, E_{2}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) + 1073741824 i \, E_{2}\left (-\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) + b^{2}{\left (536870912 i \, E_{2}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - 536870912 i \, E_{2}\left (-\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + b^{2}{\left (536870912 i \, E_{2}\left (\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - 536870912 i \, E_{2}\left (-\frac{5 i \, b c + 5 i \,{\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right )}{17179869184 \,{\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 = 0.63162, size = 867, normalized size = 3.37 \begin{align*} \frac{32 \, d \cos \left (b x + a\right )^{5} - 32 \, d \cos \left (b x + a\right )^{3} + 10 \,{\left (b d x + b c\right )} \cos \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{5 \,{\left (b d x + b c\right )}}{d}\right ) + 6 \,{\left (b d x + b c\right )} \cos \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) \operatorname{Si}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) - 4 \,{\left (b d x + b c\right )} \cos \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\right ) - 2 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{b d x + b c}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{b d x + b c}{d}\right )\right )} \sin \left (-\frac{b c - a d}{d}\right ) + 3 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{3 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{3 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right ) + 5 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{5 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{5 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{5 \,{\left (b c - a d\right )}}{d}\right )}{32 \,{\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 ^{2}{\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 )^{2}}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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