Optimal. Leaf size=89 \[ -\frac {11 a^3 x}{2}+\frac {5 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d} \]
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Rubi [A]
time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2788, 2727,
2718, 2715, 8, 2713} \begin {gather*} -\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {5 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {11 a^3 x}{2} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 2713
Rule 2715
Rule 2718
Rule 2727
Rule 2788
Rubi steps
\begin {align*} \int (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)-3 a \sin ^2(c+d x)-a \sin ^3(c+d x)\right ) \, dx\\ &=-4 a^3 x-a^3 \int \sin ^3(c+d x) \, dx-\left (3 a^3\right ) \int \sin ^2(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\\ &=-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 {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx+\frac {a^3 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {11 a^3 x}{2}+\frac {5 a^3 \cos (c+d x)}{d}-\frac {a^3 \cos ^3(c+d x)}{3 d}+\frac {4 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A]
time = 0.32, size = 115, normalized size = 1.29 \begin {gather*} \frac {(a+a \sin (c+d x))^3 \left (-66 (c+d x)+57 \cos (c+d x)-\cos (3 (c+d x))+\frac {96 \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 )}+9 \sin (2 (c+d x))\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.23, size = 167, normalized size = 1.88
method | result | size |
risch | \(-\frac {11 a^{3} x}{2}+\frac {19 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {19 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {8 a^{3}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}-\frac {a^{3} \cos \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\) | \(98\) |
derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(167\) |
default | \(\frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(167\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 117, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\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} + 9 \, {\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} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - 18 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{6 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 154, normalized size = 1.73 \begin {gather*} -\frac {2 \, a^{3} \cos \left (d x + c\right )^{4} - 7 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} d x - 30 \, a^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} + 3 \, {\left (11 \, a^{3} d x - 15 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 33 \, a^{3} d x + 9 \, a^{3} \cos \left (d x + c\right )^{2} - 21 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \end {gather*}
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 \begin {gather*} a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 10.33, size = 288, normalized size = 3.24 \begin {gather*} -\frac {11\,a^3\,x}{2}-\frac {\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (33\,c+33\,d\,x-38\right )}{6}\right )-\frac {a^3\,\left (33\,c+33\,d\,x-104\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {11\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (33\,c+33\,d\,x-66\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-66\right )}{6}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-120\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-192\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {33\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (99\,c+99\,d\,x-246\right )}{6}\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 )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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