Optimal. Leaf size=111 \[ -\frac {a^3 \cos ^3(c+d x)}{d}+\frac {7 a^3 \cos (c+d x)}{d}+\frac {a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {19 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {51 a^3 x}{8} \]
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Rubi [A] time = 0.17, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2872, 2648, 2638, 2635, 8, 2633} \[ -\frac {a^3 \cos ^3(c+d x)}{d}+\frac {7 a^3 \cos (c+d x)}{d}+\frac {a^3 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {19 a^3 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {51 a^3 x}{8} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 2638
Rule 2648
Rule 2872
Rubi steps
\begin {align*} \int \sin (c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=a^2 \int \left (-4 a-\frac {4 a}{-1+\sin (c+d x)}-4 a \sin (c+d x)-4 a \sin ^2(c+d x)-3 a \sin ^3(c+d x)-a \sin ^4(c+d x)\right ) \, dx\\ &=-4 a^3 x-a^3 \int \sin ^4(c+d x) \, dx-\left (3 a^3\right ) \int \sin ^3(c+d x) \, dx-\left (4 a^3\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\left (4 a^3\right ) \int \sin (c+d x) \, dx-\left (4 a^3\right ) \int \sin ^2(c+d x) \, dx\\ &=-4 a^3 x+\frac {4 a^3 \cos (c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {2 a^3 \cos (c+d x) \sin (c+d x)}{d}+\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{4} \left (3 a^3\right ) \int \sin ^2(c+d x) \, dx-\left (2 a^3\right ) \int 1 \, dx+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-6 a^3 x+\frac {7 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {19 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{8} \left (3 a^3\right ) \int 1 \, dx\\ &=-\frac {51 a^3 x}{8}+\frac {7 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {19 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.79, size = 125, normalized size = 1.13 \[ \frac {(a \sin (c+d x)+a)^3 \left (-204 (c+d x)+40 \sin (2 (c+d x))-\sin (4 (c+d x))+200 \cos (c+d x)-8 \cos (3 (c+d x))+\frac {256 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{32 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 178, normalized size = 1.60 \[ -\frac {2 \, a^{3} \cos \left (d x + c\right )^{5} + 8 \, a^{3} \cos \left (d x + c\right )^{4} - 15 \, a^{3} \cos \left (d x + c\right )^{3} + 51 \, a^{3} d x - 56 \, a^{3} \cos \left (d x + c\right )^{2} - 32 \, a^{3} + {\left (51 \, a^{3} d x - 67 \, a^{3}\right )} \cos \left (d x + c\right ) + {\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{3} \cos \left (d x + c\right )^{3} - 51 \, a^{3} d x - 21 \, a^{3} \cos \left (d x + c\right )^{2} + 35 \, a^{3} \cos \left (d x + c\right ) - 32 \, a^{3}\right )} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 167, normalized size = 1.50 \[ -\frac {51 \, {\left (d x + c\right )} a^{3} + \frac {64 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 32 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 144 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 19 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.60, size = 212, normalized size = 1.91 \[ \frac {a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+3 a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 162, normalized size = 1.46 \[ -\frac {8 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{3} + {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{3} + 12 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} - 8 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 14.70, size = 363, normalized size = 3.27 \[ -\frac {51\,a^3\,x}{8}-\frac {\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\frac {a^3\,\left (51\,c+51\,d\,x-58\right )}{8}\right )-\frac {a^3\,\left (51\,c+51\,d\,x-160\right )}{8}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{8}-\frac {a^3\,\left (51\,c+51\,d\,x-102\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-102\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-266\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-374\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {51\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (204\,c+204\,d\,x-538\right )}{8}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{4}-\frac {a^3\,\left (306\,c+306\,d\,x-342\right )}{8}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {153\,a^3\,\left (c+d\,x\right )}{4}-\frac {a^3\,\left (306\,c+306\,d\,x-618\right )}{8}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
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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