Optimal. Leaf size=130 \[ \frac{2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right ) \text{Hypergeometric2F1}\left (\frac{5}{2},-\frac{-5 b n+2 i m+2 i}{4 b n},-\frac{-9 b n+2 i m+2 i}{4 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{5 i b n+2 m+2} \]
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Rubi [A] time = 0.0991482, antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4509, 4507, 364} \[ \frac{2 x^{m+1} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac{5}{2},\frac{1}{4} \left (5-\frac{2 i (m+1)}{b n}\right );-\frac{2 i m-9 b n+2 i}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right )}{5 i b n+2 m+2} \]
Antiderivative was successfully verified.
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Rule 4509
Rule 4507
Rule 364
Rubi steps
\begin{align*} \int x^m \sec ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\left (x^{1+m} \left (c x^n\right )^{-\frac{1+m}{n}}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1+m}{n}} \sec ^{\frac{5}{2}}(a+b \log (x)) \, dx,x,c x^n\right )}{n}\\ &=\frac{\left (x^{1+m} \left (c x^n\right )^{-\frac{5 i b}{2}-\frac{1+m}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \sec ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^{-1+\frac{5 i b}{2}+\frac{1+m}{n}}}{\left (1+e^{2 i a} x^{2 i b}\right )^{5/2}} \, dx,x,c x^n\right )}{n}\\ &=\frac{2 x^{1+m} \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right )^{5/2} \, _2F_1\left (\frac{5}{2},\frac{1}{4} \left (5-\frac{2 i (1+m)}{b n}\right );-\frac{2 i+2 i m-9 b n}{4 b n};-e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \sec ^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right )}{2+2 m+5 i b n}\\ \end{align*}
Mathematica [A] time = 2.1366, size = 182, normalized size = 1.4 \[ \frac{2 x^{m+1} \sqrt{\sec \left (a+b \log \left (c x^n\right )\right )} \left (\left (b^2 n^2+4 m^2+8 m+4\right ) \left (1+e^{2 i a} \left (c x^n\right )^{2 i b}\right ) \text{Hypergeometric2F1}\left (1,-\frac{-3 b n+2 i m+2 i}{4 b n},-\frac{-5 b n+2 i m+2 i}{4 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )-(i b n+2 m+2) \left (-b n \tan \left (a+b \log \left (c x^n\right )\right )+2 m+2\right )\right )}{3 b^2 n^2 (i b n+2 m+2)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.287, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sec \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \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 [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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