Optimal. Leaf size=246 \[ -\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^3 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b^3 d}+\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{8 b^3 d}-\frac{9 e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )}{8 b^3 d}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]
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Rubi [A] time = 0.611828, antiderivative size = 305, normalized size of antiderivative = 1.24, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3303, 3298, 3301, 5657} \[ -\frac{9 e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}+\frac{9 e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{9 e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 5865
Rule 12
Rule 5667
Rule 5774
Rule 5669
Rule 5448
Rule 3303
Rule 3298
Rule 3301
Rule 5657
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^2}{\left (a+b \sinh ^{-1}(c+d x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sinh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}+\frac{e^2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{1}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}-\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{b^3 d}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (e^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{b^3 d}-\frac{\left (e^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{x}{b}\right )}{x} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{b^3 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}+\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}-\frac{\left (9 e^2 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}+\frac{\left (9 e^2 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}-\frac{\left (9 e^2 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b^2 d}\\ &=-\frac{e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{2 b d \left (a+b \sinh ^{-1}(c+d x)\right )^2}-\frac{e^2 (c+d x)}{b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{3 e^2 (c+d x)^3}{2 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )}-\frac{9 e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{9 e^2 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^3 d}+\frac{e^2 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}+\frac{9 e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{9 e^2 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c+d x)\right )}{8 b^3 d}-\frac{e^2 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )}{b^3 d}\\ \end{align*}
Mathematica [A] time = 0.729361, size = 216, normalized size = 0.88 \[ \frac{e^2 \left (-\frac{4 b^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{\left (a+b \sinh ^{-1}(c+d x)\right )^2}+9 \left (-\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )+\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )\right )\right )+8 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-8 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\frac{4 b \left (-3 (c+d x)^3-2 (c+d x)\right )}{a+b \sinh ^{-1}(c+d x)}\right )}{8 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.114, size = 507, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}{b^{3} \operatorname{arsinh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{arsinh}\left (d x + c\right ) + a^{3}}, x\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*} e^{2} \left (\int \frac{c^{2}}{a^{3} + 3 a^{2} b \operatorname{asinh}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}\, dx + \int \frac{d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname{asinh}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}\, dx + \int \frac{2 c d x}{a^{3} + 3 a^{2} b \operatorname{asinh}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asinh}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asinh}^{3}{\left (c + d x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{2}}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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