Optimal. Leaf size=113 \[ -\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {12 a^4 \cos (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {31 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {95 a^4 x}{8} \]
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Rubi [A] time = 0.16, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2709, 2648, 2638, 2635, 8, 2633} \[ -\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {12 a^4 \cos (c+d x)}{d}+\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac {31 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {95 a^4 x}{8} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2633
Rule 2635
Rule 2638
Rule 2648
Rule 2709
Rubi steps
\begin {align*} \int (a+a \sin (c+d x))^4 \tan ^2(c+d x) \, dx &=a^2 \int \left (-8 a^2-\frac {8 a^2}{-1+\sin (c+d x)}-8 a^2 \sin (c+d x)-7 a^2 \sin ^2(c+d x)-4 a^2 \sin ^3(c+d x)-a^2 \sin ^4(c+d x)\right ) \, dx\\ &=-8 a^4 x-a^4 \int \sin ^4(c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (7 a^4\right ) \int \sin ^2(c+d x) \, dx-\left (8 a^4\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\left (8 a^4\right ) \int \sin (c+d x) \, dx\\ &=-8 a^4 x+\frac {8 a^4 \cos (c+d x)}{d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {7 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{2} \left (7 a^4\right ) \int 1 \, dx+\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {23 a^4 x}{2}+\frac {12 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {31 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}-\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=-\frac {95 a^4 x}{8}+\frac {12 a^4 \cos (c+d x)}{d}-\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {31 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 1.10, size = 125, normalized size = 1.11 \[ \frac {(a \sin (c+d x)+a)^4 \left (-1140 (c+d x)+192 \sin (2 (c+d x))-3 \sin (4 (c+d x))+1056 \cos (c+d x)-32 \cos (3 (c+d x))+\frac {1536 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{96 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^8} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 179, normalized size = 1.58 \[ -\frac {6 \, a^{4} \cos \left (d x + c\right )^{5} + 32 \, a^{4} \cos \left (d x + c\right )^{4} - 73 \, a^{4} \cos \left (d x + c\right )^{3} + 285 \, a^{4} d x - 288 \, a^{4} \cos \left (d x + c\right )^{2} - 192 \, a^{4} + 3 \, {\left (95 \, a^{4} d x - 127 \, a^{4}\right )} \cos \left (d x + c\right ) + {\left (6 \, a^{4} \cos \left (d x + c\right )^{4} - 26 \, a^{4} \cos \left (d x + c\right )^{3} - 285 \, a^{4} d x - 99 \, a^{4} \cos \left (d x + c\right )^{2} + 189 \, a^{4} \cos \left (d x + c\right ) - 192 \, a^{4}\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.31, size = 231, normalized size = 2.04 \[ \frac {a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+4 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+6 a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{4} \left (\tan \left (d x +c \right )-d x -c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 181, normalized size = 1.60 \[ -\frac {32 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{4} + 3 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{4} + 72 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{4} + 24 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{4} - 96 \, a^{4} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.26, size = 363, normalized size = 3.21 \[ -\frac {95\,a^4\,x}{8}-\frac {\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (285\,c+285\,d\,x-326\right )}{24}\right )-\frac {a^4\,\left (285\,c+285\,d\,x-896\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (285\,c+285\,d\,x-570\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-570\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-1430\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-2154\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {95\,a^4\,\left (c+d\,x\right )}{2}-\frac {a^4\,\left (1140\,c+1140\,d\,x-3014\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {285\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (1710\,c+1710\,d\,x-1770\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {285\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (1710\,c+1710\,d\,x-3606\right )}{24}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int 4 \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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