Optimal. Leaf size=149 \[ -\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d} \]
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Rubi [A] time = 0.20, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ -\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 894
Rule 2837
Rubi steps
\begin {align*} \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (b^2-x^2\right )}{b^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \left (b^2-x^2\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a^4 \left (1-\frac {b^2}{a^2}\right )+a \left (a^2-b^2\right ) x-\left (a^2-b^2\right ) x^2+a x^3-x^4+\frac {a^5-a^3 b^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 127, normalized size = 0.85 \[ \frac {\frac {60 a^3 (a-b) (a+b) \log (a+b \sin (c+d x))}{b^6}-\frac {60 a^2 (a-b) (a+b) \sin (c+d x)}{b^5}+\frac {30 a (a-b) (a+b) \sin ^2(c+d x)}{b^4}-\frac {20 (a-b) (a+b) \sin ^3(c+d x)}{b^3}+\frac {15 a \sin ^4(c+d x)}{b^2}-\frac {12 \sin ^5(c+d x)}{b}}{60 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.79, size = 127, normalized size = 0.85 \[ \frac {15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, a^{3} b^{2} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 10 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 149, normalized size = 1.00 \[ -\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 20 \, b^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b \sin \left (d x + c\right )^{2} + 30 \, a b^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right ) - 60 \, a^{2} b^{2} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.30, size = 182, normalized size = 1.22 \[ -\frac {\sin ^{5}\left (d x +c \right )}{5 b d}+\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 b^{2} d}-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{3 d \,b^{3}}+\frac {\sin ^{3}\left (d x +c \right )}{3 b d}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{2 d \,b^{4}}-\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{2 b^{2} d}-\frac {a^{4} \sin \left (d x +c \right )}{d \,b^{5}}+\frac {a^{2} \sin \left (d x +c \right )}{b^{3} d}+\frac {a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{6}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 131, normalized size = 0.88 \[ -\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.08, size = 133, normalized size = 0.89 \[ \frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {1}{3\,b}-\frac {a^2}{3\,b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^5}{5\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^5-a^3\,b^2\right )}{b^6}-\frac {a\,{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{2\,b}+\frac {a^2\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b^2}}{d} \]
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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