Optimal. Leaf size=188 \[ -\frac {2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d \sqrt {a^2-b^2}}+\frac {a x \left (4 a^2-b^2\right )}{b^5}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \sin (c+d x) \cos (c+d x)}{b^3 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {4 \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d} \]
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Rubi [A] time = 0.74, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2889, 3048, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ \frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d \sqrt {a^2-b^2}}+\frac {a x \left (4 a^2-b^2\right )}{b^5}-\frac {2 a \sin (c+d x) \cos (c+d x)}{b^3 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {4 \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2735
Rule 2889
Rule 3023
Rule 3048
Rule 3049
Rule 3050
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \frac {\sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx\\ &=-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\int \frac {\sin ^2(c+d x) \left (-3 \left (a^2-b^2\right )+4 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\int \frac {\sin (c+d x) \left (8 a \left (a^2-b^2\right )-b \left (a^2-b^2\right ) \sin (c+d x)-12 a \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\int \frac {-12 a^2 \left (a^2-b^2\right )+4 a b \left (a^2-b^2\right ) \sin (c+d x)+2 \left (a^2-b^2\right ) \left (12 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\int \frac {-12 a^2 b \left (a^2-b^2\right )-6 a \left (a^2-b^2\right ) \left (4 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=\frac {a \left (4 a^2-b^2\right ) x}{b^5}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\left (a^2 \left (4 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^5}\\ &=\frac {a \left (4 a^2-b^2\right ) x}{b^5}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\left (2 a^2 \left (4 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=\frac {a \left (4 a^2-b^2\right ) x}{b^5}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}+\frac {\left (4 a^2 \left (4 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=\frac {a \left (4 a^2-b^2\right ) x}{b^5}-\frac {2 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 \sqrt {a^2-b^2} d}+\frac {\left (12 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}-\frac {2 a \cos (c+d x) \sin (c+d x)}{b^3 d}+\frac {4 \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{b d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 2.53, size = 246, normalized size = 1.31 \[ \frac {\frac {96 a^4 c+96 a^4 d x+96 a^3 b c \sin (c+d x)+96 a^3 b d x \sin (c+d x)+24 a^2 b^2 \sin (2 (c+d x))+12 a b \left (8 a^2-b^2\right ) \cos (c+d x)-24 a^2 b^2 c-24 a^2 b^2 d x-24 a b^3 c \sin (c+d x)-24 a b^3 d x \sin (c+d x)+4 a b^3 \cos (3 (c+d x))-2 b^4 \sin (2 (c+d x))-b^4 \sin (4 (c+d x))}{a+b \sin (c+d x)}-\frac {48 a^2 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{24 b^5 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.90, size = 643, normalized size = 3.42 \[ \left [\frac {4 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 6 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - 2 \, {\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}, \frac {2 \, {\left (a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, a^{6} - 5 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + 3 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2} + {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, a^{5} b - 5 \, a^{3} b^{3} + a b^{5}\right )} d x - 6 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{5} - a b^{7}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 261, normalized size = 1.39 \[ \frac {\frac {3 \, {\left (4 \, a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {6 \, {\left (4 \, a^{4} - 3 \, a^{2} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {6 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} - b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{4}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.55, size = 460, normalized size = 2.45 \[ \frac {2 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {12 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 a^{2}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{5}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3}}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a^{3}}{d \,b^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {8 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{5} \sqrt {a^{2}-b^{2}}}+\frac {6 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{3} \sqrt {a^{2}-b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.85, size = 1688, normalized size = 8.98 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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