Optimal. Leaf size=115 \[ -\frac{15 a \csc (c+d x)}{8 d}+\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}+\frac{b \tan ^2(c+d x)}{d}+\frac{b \log (\tan (c+d x))}{d} \]
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Rubi [A] time = 0.142762, antiderivative size = 115, normalized size of antiderivative = 1., 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, 2621, 288, 321, 207, 2620, 266, 43} \[ -\frac{15 a \csc (c+d x)}{8 d}+\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{b \tan ^4(c+d x)}{4 d}+\frac{b \tan ^2(c+d x)}{d}+\frac{b \log (\tan (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 2834
Rule 2621
Rule 288
Rule 321
Rule 207
Rule 2620
Rule 266
Rule 43
Rubi steps
\begin{align*} \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x)) \, dx &=a \int \csc ^2(c+d x) \sec ^5(c+d x) \, dx+b \int \csc (c+d x) \sec ^5(c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (c+d x)\right )}{d}+\frac{b \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (c+d x)\right )}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{(1+x)^2}{x} \, dx,x,\tan ^2(c+d x)\right )}{2 d}\\ &=\frac{5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}+\frac{b \operatorname{Subst}\left (\int \left (2+\frac{1}{x}+x\right ) \, dx,x,\tan ^2(c+d x)\right )}{2 d}\\ &=-\frac{15 a \csc (c+d x)}{8 d}+\frac{b \log (\tan (c+d x))}{d}+\frac{5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{b \tan ^2(c+d x)}{d}+\frac{b \tan ^4(c+d x)}{4 d}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (c+d x)\right )}{8 d}\\ &=\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{15 a \csc (c+d x)}{8 d}+\frac{b \log (\tan (c+d x))}{d}+\frac{5 a \csc (c+d x) \sec ^2(c+d x)}{8 d}+\frac{a \csc (c+d x) \sec ^4(c+d x)}{4 d}+\frac{b \tan ^2(c+d x)}{d}+\frac{b \tan ^4(c+d x)}{4 d}\\ \end{align*}
Mathematica [C] time = 0.377658, size = 76, normalized size = 0.66 \[ -\frac{a \csc (c+d x) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\sin ^2(c+d x)\right )}{d}-\frac{b \left (-\sec ^4(c+d x)-2 \sec ^2(c+d x)-4 \log (\sin (c+d x))+4 \log (\cos (c+d x))\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.062, size = 120, normalized size = 1. \begin{align*}{\frac{a}{4\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{5\,a}{8\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{15\,a}{8\,d\sin \left ( dx+c \right ) }}+{\frac{15\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{b}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{b}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00408, size = 170, normalized size = 1.48 \begin{align*} \frac{{\left (15 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (15 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, b \log \left (\sin \left (d x + c\right )\right ) - \frac{2 \,{\left (15 \, a \sin \left (d x + c\right )^{4} + 4 \, b \sin \left (d x + c\right )^{3} - 25 \, a \sin \left (d x + c\right )^{2} - 6 \, b \sin \left (d x + c\right ) + 8 \, a\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.14112, size = 429, normalized size = 3.73 \begin{align*} \frac{16 \, b \cos \left (d x + c\right )^{4} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) +{\left (15 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) -{\left (15 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 30 \, a \cos \left (d x + c\right )^{4} + 10 \, a \cos \left (d x + c\right )^{2} + 4 \,{\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 4 \, a}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.25322, size = 181, normalized size = 1.57 \begin{align*} \frac{{\left (15 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (15 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac{16 \,{\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )} + \frac{2 \,{\left (6 \, b \sin \left (d x + c\right )^{4} - 7 \, a \sin \left (d x + c\right )^{3} - 16 \, b \sin \left (d x + c\right )^{2} + 9 \, a \sin \left (d x + c\right ) + 12 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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