Optimal. Leaf size=186 \[ -\frac{a^2 \text{Shi}(2 a+2 b x)}{2 b^3}-\frac{a \text{Chi}(2 a+2 b x)}{b^3}-\frac{\text{Shi}(2 a+2 b x)}{b^3}-\frac{2 x \text{Shi}(a+b x) \sinh (a+b x)}{b^2}+\frac{2 \text{Shi}(a+b x) \cosh (a+b x)}{b^3}+\frac{a \log (a+b x)}{b^3}+\frac{\sinh (2 a+2 b x)}{8 b^3}+\frac{a \cosh (2 a+2 b x)}{4 b^3}-\frac{x \cosh (2 a+2 b x)}{4 b^2}+\frac{\sinh (a+b x) \cosh (a+b x)}{b^3}+\frac{x^2 \text{Shi}(a+b x) \cosh (a+b x)}{b}-\frac{x}{b^2} \]
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Rubi [A] time = 0.584283, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 16, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6542, 5617, 6741, 6742, 2638, 3296, 2637, 3298, 6548, 2635, 8, 3312, 3301, 6540, 5448, 12} \[ -\frac{a^2 \text{Shi}(2 a+2 b x)}{2 b^3}-\frac{a \text{Chi}(2 a+2 b x)}{b^3}-\frac{\text{Shi}(2 a+2 b x)}{b^3}-\frac{2 x \text{Shi}(a+b x) \sinh (a+b x)}{b^2}+\frac{2 \text{Shi}(a+b x) \cosh (a+b x)}{b^3}+\frac{a \log (a+b x)}{b^3}+\frac{\sinh (2 a+2 b x)}{8 b^3}+\frac{a \cosh (2 a+2 b x)}{4 b^3}-\frac{x \cosh (2 a+2 b x)}{4 b^2}+\frac{\sinh (a+b x) \cosh (a+b x)}{b^3}+\frac{x^2 \text{Shi}(a+b x) \cosh (a+b x)}{b}-\frac{x}{b^2} \]
Antiderivative was successfully verified.
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Rule 6542
Rule 5617
Rule 6741
Rule 6742
Rule 2638
Rule 3296
Rule 2637
Rule 3298
Rule 6548
Rule 2635
Rule 8
Rule 3312
Rule 3301
Rule 6540
Rule 5448
Rule 12
Rubi steps
\begin{align*} \int x^2 \sinh (a+b x) \text{Shi}(a+b x) \, dx &=\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 \int x \cosh (a+b x) \text{Shi}(a+b x) \, dx}{b}-\int \frac{x^2 \cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx\\ &=\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \frac{x^2 \sinh (2 (a+b x))}{a+b x} \, dx+\frac{2 \int \sinh (a+b x) \text{Shi}(a+b x) \, dx}{b^2}+\frac{2 \int \frac{x \sinh ^2(a+b x)}{a+b x} \, dx}{b}\\ &=\frac{2 \cosh (a+b x) \text{Shi}(a+b x)}{b^3}+\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \frac{x^2 \sinh (2 a+2 b x)}{a+b x} \, dx-\frac{2 \int \frac{\cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx}{b^2}+\frac{2 \int \left (\frac{\sinh ^2(a+b x)}{b}-\frac{a \sinh ^2(a+b x)}{b (a+b x)}\right ) \, dx}{b}\\ &=\frac{2 \cosh (a+b x) \text{Shi}(a+b x)}{b^3}+\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \left (-\frac{a \sinh (2 a+2 b x)}{b^2}+\frac{x \sinh (2 a+2 b x)}{b}+\frac{a^2 \sinh (2 a+2 b x)}{b^2 (a+b x)}\right ) \, dx+\frac{2 \int \sinh ^2(a+b x) \, dx}{b^2}-\frac{2 \int \frac{\sinh (2 a+2 b x)}{2 (a+b x)} \, dx}{b^2}-\frac{(2 a) \int \frac{\sinh ^2(a+b x)}{a+b x} \, dx}{b^2}\\ &=\frac{\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac{2 \cosh (a+b x) \text{Shi}(a+b x)}{b^3}+\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{\int 1 \, dx}{b^2}-\frac{\int \frac{\sinh (2 a+2 b x)}{a+b x} \, dx}{b^2}+\frac{a \int \sinh (2 a+2 b x) \, dx}{2 b^2}+\frac{(2 a) \int \left (\frac{1}{2 (a+b x)}-\frac{\cosh (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b^2}-\frac{a^2 \int \frac{\sinh (2 a+2 b x)}{a+b x} \, dx}{2 b^2}-\frac{\int x \sinh (2 a+2 b x) \, dx}{2 b}\\ &=-\frac{x}{b^2}+\frac{a \cosh (2 a+2 b x)}{4 b^3}-\frac{x \cosh (2 a+2 b x)}{4 b^2}+\frac{a \log (a+b x)}{b^3}+\frac{\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac{2 \cosh (a+b x) \text{Shi}(a+b x)}{b^3}+\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{\text{Shi}(2 a+2 b x)}{b^3}-\frac{a^2 \text{Shi}(2 a+2 b x)}{2 b^3}+\frac{\int \cosh (2 a+2 b x) \, dx}{4 b^2}-\frac{a \int \frac{\cosh (2 a+2 b x)}{a+b x} \, dx}{b^2}\\ &=-\frac{x}{b^2}+\frac{a \cosh (2 a+2 b x)}{4 b^3}-\frac{x \cosh (2 a+2 b x)}{4 b^2}-\frac{a \text{Chi}(2 a+2 b x)}{b^3}+\frac{a \log (a+b x)}{b^3}+\frac{\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac{\sinh (2 a+2 b x)}{8 b^3}+\frac{2 \cosh (a+b x) \text{Shi}(a+b x)}{b^3}+\frac{x^2 \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{2 x \sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{\text{Shi}(2 a+2 b x)}{b^3}-\frac{a^2 \text{Shi}(2 a+2 b x)}{2 b^3}\\ \end{align*}
Mathematica [A] time = 0.307408, size = 123, normalized size = 0.66 \[ \frac{-4 a^2 \text{Shi}(2 (a+b x))+8 \text{Shi}(a+b x) \left (\left (b^2 x^2+2\right ) \cosh (a+b x)-2 b x \sinh (a+b x)\right )-8 a \text{Chi}(2 (a+b x))-8 \text{Shi}(2 (a+b x))+8 a \log (a+b x)+5 \sinh (2 (a+b x))+2 a \cosh (2 (a+b x))-2 b x \cosh (2 (a+b x))-8 b x}{8 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.061, size = 170, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}\cosh \left ( bx+a \right ){\it Shi} \left ( bx+a \right ) }{b}}-2\,{\frac{x{\it Shi} \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{{b}^{2}}}+2\,{\frac{\cosh \left ( bx+a \right ){\it Shi} \left ( bx+a \right ) }{{b}^{3}}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}x}{2\,{b}^{2}}}+{\frac{a \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2\,{b}^{3}}}+{\frac{5\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{4\,{b}^{3}}}-{\frac{3\,x}{4\,{b}^{2}}}-{\frac{3\,a}{4\,{b}^{3}}}-{\frac{{\it Shi} \left ( 2\,bx+2\,a \right ) }{{b}^{3}}}-{\frac{a{\it Chi} \left ( 2\,bx+2\,a \right ) }{{b}^{3}}}+{\frac{a\ln \left ( bx+a \right ) }{{b}^{3}}}-{\frac{{a}^{2}{\it Shi} \left ( 2\,bx+2\,a \right ) }{2\,{b}^{3}}} \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^{2}{\rm Shi}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \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 (x^{2} \sinh \left (b x + a\right ) \operatorname{Shi}\left (b x + a\right ), 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*} \int x^{2} \sinh{\left (a + b x \right )} \operatorname{Shi}{\left (a + b x \right )}\, dx \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 x^{2}{\rm Shi}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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