Optimal. Leaf size=135 \[ \frac {a \tan ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}-\frac {a \cot ^2(c+d x)}{2 d}+\frac {3 a \log (\tan (c+d x))}{d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {15 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d} \]
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Rubi [A] time = 0.16, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2834, 2620, 266, 43, 2621, 288, 321, 207} \[ \frac {a \tan ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}-\frac {a \cot ^2(c+d x)}{2 d}+\frac {3 a \log (\tan (c+d x))}{d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {15 b \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d} \]
Antiderivative was successfully verified.
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Rule 43
Rule 207
Rule 266
Rule 288
Rule 321
Rule 2620
Rule 2621
Rule 2834
Rubi steps
\begin {align*} \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \csc ^3(c+d x) \sec ^5(c+d x) \, dx+b \int \csc ^2(c+d x) \sec ^5(c+d x) \, dx\\ &=\frac {a \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,\tan (c+d x)\right )}{d}-\frac {b \operatorname {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}\\ &=\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {a \operatorname {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}\\ &=\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {a \operatorname {Subst}\left (\int \left (3+\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=-\frac {a \cot ^2(c+d x)}{2 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}+\frac {a \tan ^4(c+d x)}{4 d}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=\frac {15 b \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {a \cot ^2(c+d x)}{2 d}-\frac {15 b \csc (c+d x)}{8 d}+\frac {3 a \log (\tan (c+d x))}{d}+\frac {5 b \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac {b \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac {3 a \tan ^2(c+d x)}{2 d}+\frac {a \tan ^4(c+d x)}{4 d}\\ \end {align*}
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Mathematica [C] time = 0.61, size = 86, normalized size = 0.64 \[ -\frac {a \left (2 \csc ^2(c+d x)-\sec ^4(c+d x)-4 \sec ^2(c+d x)-12 \log (\sin (c+d x))+12 \log (\cos (c+d x))\right )}{4 d}-\frac {b \csc (c+d x) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\sin ^2(c+d x)\right )}{d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 211, normalized size = 1.56 \[ \frac {24 \, a \cos \left (d x + c\right )^{4} - 12 \, a \cos \left (d x + c\right )^{2} + 48 \, {\left (a \cos \left (d x + c\right )^{6} - a \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 3 \, {\left ({\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a - 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a + 5 \, b\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, b \cos \left (d x + c\right )^{4} - 5 \, b \cos \left (d x + c\right )^{2} - 2 \, b\right )} \sin \left (d x + c\right ) - 4 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 133, normalized size = 0.99 \[ -\frac {3 \, {\left (8 \, a - 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (8 \, a + 5 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 48 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 151, normalized size = 1.12 \[ \frac {a}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3 a}{4 d \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3 a}{2 d \sin \left (d x +c \right )^{2}}+\frac {3 a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 b}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 b}{8 d \sin \left (d x +c \right )}+\frac {15 b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 140, normalized size = 1.04 \[ -\frac {3 \, {\left (8 \, a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a + 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, b \sin \left (d x + c\right )^{5} + 12 \, a \sin \left (d x + c\right )^{4} - 25 \, b \sin \left (d x + c\right )^{3} - 18 \, a \sin \left (d x + c\right )^{2} + 8 \, b \sin \left (d x + c\right ) + 4 \, a\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.11, size = 146, normalized size = 1.08 \[ \frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {3\,a}{2}-\frac {15\,b}{16}\right )}{d}-\frac {\frac {15\,b\,{\sin \left (c+d\,x\right )}^5}{8}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {25\,b\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {9\,a\,{\sin \left (c+d\,x\right )}^2}{4}+b\,\sin \left (c+d\,x\right )+\frac {a}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {3\,a}{2}+\frac {15\,b}{16}\right )}{d} \]
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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