Optimal. Leaf size=206 \[ \frac{(1-m) (3-m) \text{Unintegrable}\left (\frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d x^2+d},x\right )}{8 d^2}-\frac{b c (3-m) x^{m+2} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{8 d^3 (m+2)}-\frac{b c x^{m+2} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{m+2}{2},\frac{m+4}{2},-c^2 x^2\right )}{4 d^3 (m+2)}+\frac{(3-m) x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (c^2 x^2+1\right )}+\frac{x^{m+1} \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (c^2 x^2+1\right )^2} \]
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Rubi [A] time = 0.253305, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin{align*} \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx &=\frac{x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{(b c) \int \frac{x^{1+m}}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac{(3-m) \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac{x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{(3-m) x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{b c x^{2+m} \, _2F_1\left (\frac{5}{2},\frac{2+m}{2};\frac{4+m}{2};-c^2 x^2\right )}{4 d^3 (2+m)}-\frac{(b c (3-m)) \int \frac{x^{1+m}}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac{((1-m) (3-m)) \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{(3-m) x^{1+m} \left (a+b \sinh ^{-1}(c x)\right )}{8 d^3 \left (1+c^2 x^2\right )}-\frac{b c (3-m) x^{2+m} \, _2F_1\left (\frac{3}{2},\frac{2+m}{2};\frac{4+m}{2};-c^2 x^2\right )}{8 d^3 (2+m)}-\frac{b c x^{2+m} \, _2F_1\left (\frac{5}{2},\frac{2+m}{2};\frac{4+m}{2};-c^2 x^2\right )}{4 d^3 (2+m)}+\frac{((1-m) (3-m)) \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{d+c^2 d x^2} \, dx}{8 d^2}\\ \end{align*}
Mathematica [A] time = 5.66301, size = 0, normalized size = 0. \[ \int \frac{x^m \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.47, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) }{ \left ({c}^{2}d{x}^{2}+d \right ) ^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{m}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\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 = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{m}}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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