Optimal. Leaf size=115 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]
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Rubi [A] time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2707, 88} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]
Antiderivative was successfully verified.
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Rule 88
Rule 2707
Rubi steps
\begin {align*} \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+\frac {a^3}{4 (a-x)^3}-\frac {a^2}{(a-x)^2}+\frac {23 a}{16 (a-x)}-\frac {a^2}{8 (a+x)^2}+\frac {7 a}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 123, normalized size = 1.07 \[ -\frac {a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}-\frac {5 a \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 159, normalized size = 1.38 \[ \frac {16 \, a \cos \left (d x + c\right )^{4} + 2 \, a \cos \left (d x + c\right )^{2} + 7 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]
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 [A] time = 0.15, size = 147, normalized size = 1.28 \[ \frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a \sin \left (d x +c \right )}{8 d}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 95, normalized size = 0.83 \[ \frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \sin \left (d x + c\right ) + \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.63, size = 235, normalized size = 2.04 \[ \frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {23\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {7\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]
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 \[ a \left (\int \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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