Optimal. Leaf size=86 \[ \frac {a^4 \sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {4 a^2 \cos (c+d x)}{3 d}-\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+2 a^2 x \]
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Rubi [A] time = 0.25, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2869, 2765, 2968, 3023, 12, 2735, 2648} \[ -\frac {4 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+2 a^2 x \]
Antiderivative was successfully verified.
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Rule 12
Rule 2648
Rule 2735
Rule 2765
Rule 2869
Rule 2968
Rule 3023
Rubi steps
\begin {align*} \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx &=a^4 \int \frac {\sin ^3(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\sin (c+d x) (-2 a-4 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {-2 a \sin (c+d x)-4 a \sin ^2(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac {4 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {1}{3} a \int \frac {6 a^2 \sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac {4 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\left (2 a^3\right ) \int \frac {\sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=2 a^2 x-\frac {4 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\left (2 a^3\right ) \int \frac {1}{a-a \sin (c+d x)} \, dx\\ &=2 a^2 x-\frac {4 a^2 \cos (c+d x)}{3 d}+\frac {a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {2 a^3 \cos (c+d x)}{d (a-a \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.02, size = 131, normalized size = 1.52 \[ \frac {a^2 (\sin (c+d x)+1)^2 \left (-3 \cos (c+d x)+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (8 \sin (c+d x)-7)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {1}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+6 c+6 d x\right )}{3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 168, normalized size = 1.95 \[ -\frac {3 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} d x - {\left (6 \, a^{2} d x + 11 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + {\left (6 \, a^{2} d x - 13 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (12 \, a^{2} d x - 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} + 2 \, {\left (3 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 86, normalized size = 1.00 \[ \frac {2 \, {\left (3 \, {\left (d x + c\right )} a^{2} - \frac {3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}\right )}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.48, size = 162, normalized size = 1.88 \[ \frac {a^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 95, normalized size = 1.10 \[ \frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - a^{2} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.16, size = 182, normalized size = 2.12 \[ 2\,a^2\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,a^2\,\left (9\,d\,x-24\right )}{3}-6\,a^2\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {2\,a^2\,\left (9\,d\,x-6\right )}{3}-6\,a^2\,d\,x\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2\,a^2\,\left (12\,d\,x-18\right )}{3}-8\,a^2\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {2\,a^2\,\left (12\,d\,x-22\right )}{3}-8\,a^2\,d\,x\right )-\frac {2\,a^2\,\left (3\,d\,x-10\right )}{3}+2\,a^2\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
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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