Optimal. Leaf size=218 \[ \frac{9 e^{\frac{a \sqrt{-(m+1)^2}}{m+1}} x^{m+1} \left (c x^2\right )^{\frac{1}{6} (-m-1)}}{16 \sqrt{-(m+1)^2}}-\frac{9 e^{\frac{a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{6}}}{32 \sqrt{-(m+1)^2}}+\frac{e^{\frac{3 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{16 \sqrt{-(m+1)^2}}-\frac{(m+1) e^{-\frac{3 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}}{8 \sqrt{-(m+1)^2}} \]
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Rubi [A] time = 0.304595, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {4493, 4489} \[ \frac{9 e^{\frac{a \sqrt{-(m+1)^2}}{m+1}} x^{m+1} \left (c x^2\right )^{\frac{1}{6} (-m-1)}}{16 \sqrt{-(m+1)^2}}-\frac{9 e^{\frac{a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{6}}}{32 \sqrt{-(m+1)^2}}+\frac{e^{\frac{3 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \left (c x^2\right )^{\frac{m+1}{2}}}{16 \sqrt{-(m+1)^2}}-\frac{(m+1) e^{-\frac{3 a (m+1)}{\sqrt{-(m+1)^2}}} x^{m+1} \log (x) \left (c x^2\right )^{\frac{1}{2} (-m-1)}}{8 \sqrt{-(m+1)^2}} \]
Antiderivative was successfully verified.
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Rule 4493
Rule 4489
Rubi steps
\begin{align*} \int x^m \sin ^3\left (a+\frac{1}{6} \sqrt{-(1+m)^2} \log \left (c x^2\right )\right ) \, dx &=\frac{1}{2} \left (x^{1+m} \left (c x^2\right )^{\frac{1}{2} (-1-m)}\right ) \operatorname{Subst}\left (\int x^{-1+\frac{1+m}{2}} \sin ^3\left (a+\frac{1}{6} \sqrt{-(1+m)^2} \log (x)\right ) \, dx,x,c x^2\right )\\ &=\frac{\left (\sqrt{-(1+m)^2} x^{1+m} \left (c x^2\right )^{\frac{1}{2} (-1-m)}\right ) \operatorname{Subst}\left (\int \left (\frac{e^{-\frac{3 a (1+m)}{\sqrt{-(1+m)^2}}}}{x}-3 e^{\frac{a \sqrt{-(1+m)^2}}{1+m}} x^{\frac{1}{3} (-2+m)}-e^{\frac{3 a (1+m)}{\sqrt{-(1+m)^2}}} x^m+3 e^{\frac{a (1+m)}{\sqrt{-(1+m)^2}}} x^{\frac{1}{3} (-1+2 m)}\right ) \, dx,x,c x^2\right )}{16 (1+m)}\\ &=\frac{9 e^{\frac{a \sqrt{-(1+m)^2}}{1+m}} x^{1+m} \left (c x^2\right )^{\frac{1}{6} (-1-m)}}{16 \sqrt{-(1+m)^2}}-\frac{9 e^{\frac{a (1+m)}{\sqrt{-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac{1+m}{6}}}{32 \sqrt{-(1+m)^2}}+\frac{e^{\frac{3 a (1+m)}{\sqrt{-(1+m)^2}}} x^{1+m} \left (c x^2\right )^{\frac{1+m}{2}}}{16 \sqrt{-(1+m)^2}}+\frac{e^{-\frac{3 a (1+m)}{\sqrt{-(1+m)^2}}} \sqrt{-(1+m)^2} x^{1+m} \left (c x^2\right )^{\frac{1}{2} (-1-m)} \log (x)}{8 (1+m)}\\ \end{align*}
Mathematica [F] time = 0.429437, size = 0, normalized size = 0. \[ \int x^m \sin ^3\left (a+\frac{1}{6} \sqrt{-(1+m)^2} \log \left (c x^2\right )\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sin \left ( a+{\frac{\ln \left ( c{x}^{2} \right ) }{6}\sqrt{- \left ( 1+m \right ) ^{2}}} \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.08219, size = 278, normalized size = 1.28 \begin{align*} \frac{9 \,{\left (\cos \left (2 \, a\right ) \sin \left (3 \, a\right ) - \cos \left (3 \, a\right ) \sin \left (2 \, a\right )\right )} c^{\frac{5}{6} \, m + \frac{5}{6}} x^{\frac{5}{3}} x^{\frac{4}{3} \, m} + 18 \,{\left (\cos \left (3 \, a\right ) \sin \left (4 \, a\right ) - \cos \left (4 \, a\right ) \sin \left (3 \, a\right )\right )} c^{\frac{1}{2} \, m + \frac{1}{2}} x x^{\frac{2}{3} \, m} - 2 \,{\left (c^{\frac{7}{6} \, m + 1} x^{2} x^{2 \, m} \sin \left (3 \, a\right ) + 2 \,{\left ({\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac{1}{6} \, m} m +{\left (\cos \left (3 \, a\right )^{2} \sin \left (3 \, a\right ) + \sin \left (3 \, a\right )^{3}\right )} c^{\frac{1}{6} \, m}\right )} \log \left (x\right )\right )} c^{\frac{1}{6}} x^{\frac{1}{3}}}{32 \,{\left ({\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac{2}{3} \, m} m +{\left (\cos \left (3 \, a\right )^{2} + \sin \left (3 \, a\right )^{2}\right )} c^{\frac{2}{3} \, m}\right )} c^{\frac{2}{3}} x^{\frac{1}{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 0.503488, size = 350, normalized size = 1.61 \begin{align*} \frac{{\left ({\left (4 i \, m + 4 i\right )} e^{\left (-{\left (m + 1\right )} \log \left (c\right ) - 2 \,{\left (m + 1\right )} \log \left (x\right ) + 6 i \, a\right )} \log \left (x\right ) + 9 i \, e^{\left (-\frac{1}{3} \,{\left (m + 1\right )} \log \left (c\right ) - \frac{2}{3} \,{\left (m + 1\right )} \log \left (x\right ) + 2 i \, a\right )} - 18 i \, e^{\left (-\frac{2}{3} \,{\left (m + 1\right )} \log \left (c\right ) - \frac{4}{3} \,{\left (m + 1\right )} \log \left (x\right ) + 4 i \, a\right )} - 2 i\right )} e^{\left (\frac{1}{2} \,{\left (m + 1\right )} \log \left (c\right ) + 2 \,{\left (m + 1\right )} \log \left (x\right ) - 3 i \, a\right )}}{32 \,{\left (m + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sin ^{3}{\left (a + \frac{\sqrt{- m^{2} - 2 m - 1} \log{\left (c x^{2} \right )}}{6} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [C] time = 2.45635, size = 1751, normalized size = 8.03 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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