Optimal. Leaf size=111 \[ \frac {\sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {5}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {10 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]
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Rubi [A] time = 0.11, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2833, 12, 43} \[ \frac {\sin ^3(c+d x)}{3 a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {5}{d \left (a^3 \sin (c+d x)+a^3\right )}-\frac {10 \log (\sin (c+d x)+1)}{a^3 d}+\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 2833
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \sin ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{a^5 (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (6 a^2-3 a x+x^2-\frac {a^5}{(a+x)^3}+\frac {5 a^4}{(a+x)^2}-\frac {10 a^3}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^6 d}\\ &=-\frac {10 \log (1+\sin (c+d x))}{a^3 d}+\frac {6 \sin (c+d x)}{a^3 d}-\frac {3 \sin ^2(c+d x)}{2 a^3 d}+\frac {\sin ^3(c+d x)}{3 a^3 d}+\frac {1}{2 a d (a+a \sin (c+d x))^2}-\frac {5}{d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.70, size = 106, normalized size = 0.95 \[ \frac {32 \sin ^5(c+d x)-80 \sin ^4(c+d x)+320 \sin ^3(c+d x)+\sin ^2(c+d x) (1023-960 \log (\sin (c+d x)+1))-6 \sin (c+d x) (320 \log (\sin (c+d x)+1)-21)-960 \log (\sin (c+d x)+1)-417}{96 a^3 d (\sin (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 117, normalized size = 1.05 \[ \frac {10 \, \cos \left (d x + c\right )^{4} + 115 \, \cos \left (d x + c\right )^{2} - 120 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (d x + c\right )^{4} - 24 \, \cos \left (d x + c\right )^{2} + 37\right )} \sin \left (d x + c\right ) - 80}{12 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 89, normalized size = 0.80 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {3 \, {\left (10 \, \sin \left (d x + c\right ) + 9\right )}}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a^{6} \sin \left (d x + c\right )^{3} - 9 \, a^{6} \sin \left (d x + c\right )^{2} + 36 \, a^{6} \sin \left (d x + c\right )}{a^{9}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 101, normalized size = 0.91 \[ \frac {\sin ^{3}\left (d x +c \right )}{3 a^{3} d}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2 a^{3} d}+\frac {6 \sin \left (d x +c \right )}{a^{3} d}+\frac {1}{2 d \,a^{3} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {10 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {5}{d \,a^{3} \left (1+\sin \left (d x +c \right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 95, normalized size = 0.86 \[ -\frac {\frac {3 \, {\left (10 \, \sin \left (d x + c\right ) + 9\right )}}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} - \frac {2 \, \sin \left (d x + c\right )^{3} - 9 \, \sin \left (d x + c\right )^{2} + 36 \, \sin \left (d x + c\right )}{a^{3}} + \frac {60 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{6 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.13, size = 108, normalized size = 0.97 \[ \frac {6\,\sin \left (c+d\,x\right )}{a^3\,d}-\frac {5\,\sin \left (c+d\,x\right )+\frac {9}{2}}{d\,\left (a^3\,{\sin \left (c+d\,x\right )}^2+2\,a^3\,\sin \left (c+d\,x\right )+a^3\right )}-\frac {10\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}-\frac {3\,{\sin \left (c+d\,x\right )}^2}{2\,a^3\,d}+\frac {{\sin \left (c+d\,x\right )}^3}{3\,a^3\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 8.06, size = 394, normalized size = 3.55 \[ \begin {cases} - \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {120 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {60 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} + \frac {2 \sin ^{5}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {5 \sin ^{4}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} + \frac {20 \sin ^{3}{\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {120 \sin {\left (c + d x \right )}}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} - \frac {90}{6 a^{3} d \sin ^{2}{\left (c + d x \right )} + 12 a^{3} d \sin {\left (c + d x \right )} + 6 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{5}{\relax (c )} \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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