Optimal. Leaf size=135 \[ -\frac {15 a \sin (c+d x)}{8 d}+\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac {5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac {15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \cos ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {3 b \sec ^2(c+d x)}{2 d}-\frac {3 b \log (\cos (c+d x))}{d} \]
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Rubi [A] time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2834, 2592, 288, 321, 206, 2590, 266, 43} \[ -\frac {15 a \sin (c+d x)}{8 d}+\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac {5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac {15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \cos ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {3 b \sec ^2(c+d x)}{2 d}-\frac {3 b \log (\cos (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 43
Rule 206
Rule 266
Rule 288
Rule 321
Rule 2590
Rule 2592
Rule 2834
Rubi steps
\begin {align*} \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx &=a \int \sin (c+d x) \tan ^5(c+d x) \, dx+b \int \sin ^2(c+d x) \tan ^5(c+d x) \, dx\\ &=\frac {a \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^3} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^5} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{4 d}-\frac {b \operatorname {Subst}\left (\int \frac {(1-x)^3}{x^3} \, dx,x,\cos ^2(c+d x)\right )}{2 d}\\ &=-\frac {5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}+\frac {(15 a) \operatorname {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}-\frac {b \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^3}-\frac {3}{x^2}+\frac {3}{x}\right ) \, dx,x,\cos ^2(c+d x)\right )}{2 d}\\ &=\frac {b \cos ^2(c+d x)}{2 d}-\frac {3 b \log (\cos (c+d x))}{d}-\frac {3 b \sec ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {15 a \sin (c+d x)}{8 d}-\frac {5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}+\frac {(15 a) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}\\ &=\frac {15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \cos ^2(c+d x)}{2 d}-\frac {3 b \log (\cos (c+d x))}{d}-\frac {3 b \sec ^2(c+d x)}{2 d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {15 a \sin (c+d x)}{8 d}-\frac {5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac {a \sin (c+d x) \tan ^4(c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.36, size = 133, normalized size = 0.99 \[ -\frac {a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {5 a \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d}-\frac {b \left (2 \sin ^2(c+d x)-\sec ^4(c+d x)+6 \sec ^2(c+d x)+12 \log (\cos (c+d x))\right )}{4 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 138, normalized size = 1.02 \[ \frac {8 \, b \cos \left (d x + c\right )^{6} + 3 \, {\left (5 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, b \cos \left (d x + c\right )^{4} - 24 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (8 \, a \cos \left (d x + c\right )^{4} + 9 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.28, size = 124, normalized size = 0.92 \[ -\frac {8 \, b \sin \left (d x + c\right )^{2} - 3 \, {\left (5 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (5 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (18 \, b \sin \left (d x + c\right )^{4} + 9 \, a \sin \left (d x + c\right )^{3} - 24 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) + 8 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 205, normalized size = 1.52 \[ \frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a \sin \left (d x +c \right )}{8 d}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {b \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {3 b \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 b \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 b \ln \left (\cos \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 121, normalized size = 0.90 \[ -\frac {8 \, b \sin \left (d x + c\right )^{2} - 3 \, {\left (5 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (5 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{3} + 12 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) - 10 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.16, size = 304, normalized size = 2.25 \[ \frac {3\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15\,a}{8}+3\,b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {15\,a}{8}-3\,b\right )}{d}-\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{4}-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\frac {25\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+12\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}+12\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {25\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-6\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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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