Optimal. Leaf size=70 \[ -\frac{15 \csc (a+b x)}{8 b}+\frac{15 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}+\frac{5 \csc (a+b x) \sec ^2(a+b x)}{8 b} \]
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Rubi [A] time = 0.0467545, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2621, 288, 321, 207} \[ -\frac{15 \csc (a+b x)}{8 b}+\frac{15 \tanh ^{-1}(\sin (a+b x))}{8 b}+\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}+\frac{5 \csc (a+b x) \sec ^2(a+b x)}{8 b} \]
Antiderivative was successfully verified.
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Rule 2621
Rule 288
Rule 321
Rule 207
Rubi steps
\begin{align*} \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=\frac{5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=-\frac{15 \csc (a+b x)}{8 b}+\frac{5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b}\\ &=\frac{15 \tanh ^{-1}(\sin (a+b x))}{8 b}-\frac{15 \csc (a+b x)}{8 b}+\frac{5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac{\csc (a+b x) \sec ^4(a+b x)}{4 b}\\ \end{align*}
Mathematica [C] time = 0.0135408, size = 27, normalized size = 0.39 \[ -\frac{\csc (a+b x) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\sin ^2(a+b x)\right )}{b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.029, size = 76, normalized size = 1.1 \begin{align*}{\frac{1}{4\,b\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{4}}}+{\frac{5}{8\,b\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{15}{8\,b\sin \left ( bx+a \right ) }}+{\frac{15\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00028, size = 107, normalized size = 1.53 \begin{align*} -\frac{\frac{2 \,{\left (15 \, \sin \left (b x + a\right )^{4} - 25 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{5} - 2 \, \sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.09936, size = 261, normalized size = 3.73 \begin{align*} \frac{15 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 30 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 4}{16 \, b \cos \left (b x + a\right )^{4} \sin \left (b x + a\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 = 1.22019, size = 99, normalized size = 1.41 \begin{align*} -\frac{\frac{2 \,{\left (7 \, \sin \left (b x + a\right )^{3} - 9 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac{16}{\sin \left (b x + a\right )} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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