Optimal. Leaf size=184 \[ -\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{a^2 \sinh (a+b x)}{4 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 \sinh (a+b x)}{2 b^4}+\frac{a \cosh (a+b x)}{2 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{x^3 \cosh (a+b x)}{4 b} \]
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Rubi [A] time = 0.380206, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6532, 6742, 2638, 3296, 2637, 3298} \[ -\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{a^2 \sinh (a+b x)}{4 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 \sinh (a+b x)}{2 b^4}+\frac{a \cosh (a+b x)}{2 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{x^3 \cosh (a+b x)}{4 b} \]
Antiderivative was successfully verified.
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Rule 6532
Rule 6742
Rule 2638
Rule 3296
Rule 2637
Rule 3298
Rubi steps
\begin{align*} \int x^3 \text{Shi}(a+b x) \, dx &=\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{1}{4} b \int \frac{x^4 \sinh (a+b x)}{a+b x} \, dx\\ &=\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{1}{4} b \int \left (-\frac{a^3 \sinh (a+b x)}{b^4}+\frac{a^2 x \sinh (a+b x)}{b^3}-\frac{a x^2 \sinh (a+b x)}{b^2}+\frac{x^3 \sinh (a+b x)}{b}+\frac{a^4 \sinh (a+b x)}{b^4 (a+b x)}\right ) \, dx\\ &=\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{1}{4} \int x^3 \sinh (a+b x) \, dx+\frac{a^3 \int \sinh (a+b x) \, dx}{4 b^3}-\frac{a^4 \int \frac{\sinh (a+b x)}{a+b x} \, dx}{4 b^3}-\frac{a^2 \int x \sinh (a+b x) \, dx}{4 b^2}+\frac{a \int x^2 \sinh (a+b x) \, dx}{4 b}\\ &=\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{x^3 \cosh (a+b x)}{4 b}-\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{Shi}(a+b x)+\frac{a^2 \int \cosh (a+b x) \, dx}{4 b^3}-\frac{a \int x \cosh (a+b x) \, dx}{2 b^2}+\frac{3 \int x^2 \cosh (a+b x) \, dx}{4 b}\\ &=\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{x^3 \cosh (a+b x)}{4 b}+\frac{a^2 \sinh (a+b x)}{4 b^4}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}-\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{Shi}(a+b x)+\frac{a \int \sinh (a+b x) \, dx}{2 b^3}-\frac{3 \int x \sinh (a+b x) \, dx}{2 b^2}\\ &=\frac{a \cosh (a+b x)}{2 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{x^3 \cosh (a+b x)}{4 b}+\frac{a^2 \sinh (a+b x)}{4 b^4}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}-\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{Shi}(a+b x)+\frac{3 \int \cosh (a+b x) \, dx}{2 b^3}\\ &=\frac{a \cosh (a+b x)}{2 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{x^3 \cosh (a+b x)}{4 b}+\frac{3 \sinh (a+b x)}{2 b^4}+\frac{a^2 \sinh (a+b x)}{4 b^4}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}-\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{Shi}(a+b x)\\ \end{align*}
Mathematica [A] time = 0.199126, size = 94, normalized size = 0.51 \[ \frac{\left (b^4 x^4-a^4\right ) \text{Shi}(a+b x)+\left (a^2-2 a b x+3 b^2 x^2+6\right ) \sinh (a+b x)+\left (-a^2 b x+a^3+a b^2 x^2+2 a-b^3 x^3-6 b x\right ) \cosh (a+b x)}{4 b^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.051, size = 156, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\it Shi} \left ( bx+a \right ){b}^{4}{x}^{4}}{4}}-{\frac{ \left ( bx+a \right ) ^{3}\cosh \left ( bx+a \right ) }{4}}+{\frac{3\, \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) }{4}}-{\frac{ \left ( 3\,bx+3\,a \right ) \cosh \left ( bx+a \right ) }{2}}+{\frac{3\,\sinh \left ( bx+a \right ) }{2}}+a \left ( \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) -2\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) +2\,\cosh \left ( bx+a \right ) \right ) -{\frac{3\,{a}^{2} \left ( \left ( bx+a \right ) \cosh \left ( bx+a \right ) -\sinh \left ( bx+a \right ) \right ) }{2}}+{a}^{3}\cosh \left ( bx+a \right ) -{\frac{{a}^{4}{\it Shi} \left ( bx+a \right ) }{4}} \right ) } \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^{3}{\rm Shi}\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^{3} \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^{3} \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^{3}{\rm Shi}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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