Optimal. Leaf size=360 \[ \frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \cosh ^{-1}(c+d x)\right )}{b}\right )}{3 b^4 d}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{8 e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3} \]
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Rubi [A] time = 0.887219, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {5866, 12, 5668, 5775, 5666, 3303, 3298, 3301} \[ \frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{8 e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3} \]
Antiderivative was successfully verified.
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Rule 5866
Rule 12
Rule 5668
Rule 5775
Rule 5666
Rule 3303
Rule 3298
Rule 3301
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \cosh ^{-1}(c+d x)\right )^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \cosh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \cosh ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}-\frac{e^3 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-1+x} \sqrt{1+x} \left (a+b \cosh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{b d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{-1+x} \sqrt{1+x} \left (a+b \cosh ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{e^3 \operatorname{Subst}\left (\int \frac{x}{\left (a+b \cosh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \cosh ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{8 e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^3 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{b^3 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (2 x)}{2 (a+b x)}-\frac{\cosh (4 x)}{2 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{8 e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{b^3 d}+\frac{\left (e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{b^3 d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{8 e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{b^4 d}+\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{b^4 d}+\frac{\left (4 e^3 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^3 \cosh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (4 e^3 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}-\frac{\left (4 e^3 \sinh \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c+d x)\right )}{3 b^3 d}\\ &=-\frac{e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b d \left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{e^3 (c+d x)^2}{2 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}-\frac{2 e^3 (c+d x)^4}{3 b^2 d \left (a+b \cosh ^{-1}(c+d x)\right )^2}+\frac{e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}-\frac{8 e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{3 b^3 d \left (a+b \cosh ^{-1}(c+d x)\right )}+\frac{e^3 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}+\frac{4 e^3 \cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{e^3 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}-\frac{4 e^3 \sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \cosh ^{-1}(c+d x)\right )}{3 b^4 d}\\ \end{align*}
Mathematica [A] time = 1.31323, size = 330, normalized size = 0.92 \[ \frac{e^3 \left (-\frac{2 b^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{\left (a+b \cosh ^{-1}(c+d x)\right )^3}+\frac{b^2 \left (3 (c+d x)^2-4 (c+d x)^4\right )}{\left (a+b \cosh ^{-1}(c+d x)\right )^2}-30 \left (\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )+\log \left (a+b \cosh ^{-1}(c+d x)\right )\right )+8 \left (4 \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )+\cosh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )-4 \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )-\sinh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\cosh ^{-1}(c+d x)\right )\right )+3 \log \left (a+b \cosh ^{-1}(c+d x)\right )\right )-\frac{2 b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (8 (c+d x)^3-3 (c+d x)\right )}{a+b \cosh ^{-1}(c+d x)}+6 \log \left (a+b \cosh ^{-1}(c+d x)\right )\right )}{6 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.164, size = 860, normalized size = 2.4 \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^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b^{4} \operatorname{arcosh}\left (d x + c\right )^{4} + 4 \, a b^{3} \operatorname{arcosh}\left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \operatorname{arcosh}\left (d x + c\right )^{2} + 4 \, a^{3} b \operatorname{arcosh}\left (d x + c\right ) + a^{4}}, 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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{3}}{{\left (b \operatorname{arcosh}\left (d x + c\right ) + a\right )}^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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