Optimal. Leaf size=112 \[ \frac {2 \text {Shi}(2 b x)}{3 b^3}-\frac {4 \text {Shi}(b x) \cosh (b x)}{3 b^3}-\frac {5 \sinh (b x) \cosh (b x)}{6 b^3}+\frac {4 x \text {Shi}(b x) \sinh (b x)}{3 b^2}+\frac {5 x}{6 b^2}+\frac {x \sinh ^2(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {2 x^2 \text {Shi}(b x) \cosh (b x)}{3 b} \]
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Rubi [A] time = 0.15, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6536, 6542, 12, 5372, 2635, 8, 6548, 6540, 5448, 3298} \[ \frac {2 \text {Shi}(2 b x)}{3 b^3}+\frac {4 x \text {Shi}(b x) \sinh (b x)}{3 b^2}-\frac {4 \text {Shi}(b x) \cosh (b x)}{3 b^3}+\frac {5 x}{6 b^2}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {5 \sinh (b x) \cosh (b x)}{6 b^3}+\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {2 x^2 \text {Shi}(b x) \cosh (b x)}{3 b} \]
Antiderivative was successfully verified.
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Rule 8
Rule 12
Rule 2635
Rule 3298
Rule 5372
Rule 5448
Rule 6536
Rule 6540
Rule 6542
Rule 6548
Rubi steps
\begin {align*} \int x^2 \text {Shi}(b x)^2 \, dx &=\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {2}{3} \int x^2 \sinh (b x) \text {Shi}(b x) \, dx\\ &=-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {2}{3} \int \frac {x \cosh (b x) \sinh (b x)}{b} \, dx+\frac {4 \int x \cosh (b x) \text {Shi}(b x) \, dx}{3 b}\\ &=-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {4 \int \sinh (b x) \text {Shi}(b x) \, dx}{3 b^2}+\frac {2 \int x \cosh (b x) \sinh (b x) \, dx}{3 b}-\frac {4 \int \frac {\sinh ^2(b x)}{b} \, dx}{3 b}\\ &=\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {\int \sinh ^2(b x) \, dx}{3 b^2}+\frac {4 \int \frac {\cosh (b x) \sinh (b x)}{b x} \, dx}{3 b^2}-\frac {4 \int \sinh ^2(b x) \, dx}{3 b^2}\\ &=-\frac {5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {4 \int \frac {\cosh (b x) \sinh (b x)}{x} \, dx}{3 b^3}+\frac {\int 1 \, dx}{6 b^2}+\frac {2 \int 1 \, dx}{3 b^2}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {4 \int \frac {\sinh (2 b x)}{2 x} \, dx}{3 b^3}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {2 \int \frac {\sinh (2 b x)}{x} \, dx}{3 b^3}\\ &=\frac {5 x}{6 b^2}-\frac {5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {2 \text {Shi}(2 b x)}{3 b^3}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 78, normalized size = 0.70 \[ \frac {4 b^3 x^3 \text {Shi}(b x)^2-8 \text {Shi}(b x) \left (\left (b^2 x^2+2\right ) \cosh (b x)-2 b x \sinh (b x)\right )+8 \text {Shi}(2 b x)+8 b x-5 \sinh (2 b x)+2 b x \cosh (2 b x)}{12 b^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Shi}\left (b x\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Shi}\left (b x\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 84, normalized size = 0.75 \[ \frac {\frac {b^{3} x^{3} \Shi \left (b x \right )^{2}}{3}-2 \Shi \left (b x \right ) \left (\frac {b^{2} x^{2} \cosh \left (b x \right )}{3}-\frac {2 b x \sinh \left (b x \right )}{3}+\frac {2 \cosh \left (b x \right )}{3}\right )+\frac {b x \left (\cosh ^{2}\left (b x \right )\right )}{3}-\frac {5 \sinh \left (b x \right ) \cosh \left (b x \right )}{6}+\frac {b x}{2}+\frac {2 \Shi \left (2 b x \right )}{3}}{b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Shi}\left (b x\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {sinhint}\left (b\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {Shi}^{2}{\left (b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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