Optimal. Leaf size=98 \[ \frac {x^2}{4 b}-\frac {\text {Chi}(2 b x)}{b^3}+\frac {\log (x)}{b^3}-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {5 \sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b} \]
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Rubi [A]
time = 0.09, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6683, 12,
3391, 30, 6677, 2644, 6681, 3393, 3382} \begin {gather*} -\frac {\text {Chi}(2 b x)}{b^3}+\frac {2 \text {Shi}(b x) \sinh (b x)}{b^3}+\frac {\log (x)}{b^3}+\frac {5 \sinh ^2(b x)}{4 b^3}-\frac {2 x \text {Shi}(b x) \cosh (b x)}{b^2}-\frac {x \sinh (b x) \cosh (b x)}{2 b^2}+\frac {x^2 \text {Shi}(b x) \sinh (b x)}{b}+\frac {x^2}{4 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2644
Rule 3382
Rule 3391
Rule 3393
Rule 6677
Rule 6681
Rule 6683
Rubi steps
\begin {align*} \int x^2 \cosh (b x) \text {Shi}(b x) \, dx &=\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}-\frac {2 \int x \sinh (b x) \text {Shi}(b x) \, dx}{b}-\int \frac {x \sinh ^2(b x)}{b} \, dx\\ &=-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}+\frac {2 \int \cosh (b x) \text {Shi}(b x) \, dx}{b^2}-\frac {\int x \sinh ^2(b x) \, dx}{b}+\frac {2 \int \frac {\cosh (b x) \sinh (b x)}{b} \, dx}{b}\\ &=-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {\sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}+\frac {2 \int \cosh (b x) \sinh (b x) \, dx}{b^2}-\frac {2 \int \frac {\sinh ^2(b x)}{b x} \, dx}{b^2}+\frac {\int x \, dx}{2 b}\\ &=\frac {x^2}{4 b}-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {\sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}-\frac {2 \int \frac {\sinh ^2(b x)}{x} \, dx}{b^3}-\frac {2 \text {Subst}(\int x \, dx,x,i \sinh (b x))}{b^3}\\ &=\frac {x^2}{4 b}-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {5 \sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}+\frac {2 \int \left (\frac {1}{2 x}-\frac {\cosh (2 b x)}{2 x}\right ) \, dx}{b^3}\\ &=\frac {x^2}{4 b}+\frac {\log (x)}{b^3}-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {5 \sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}-\frac {\int \frac {\cosh (2 b x)}{x} \, dx}{b^3}\\ &=\frac {x^2}{4 b}-\frac {\text {Chi}(2 b x)}{b^3}+\frac {\log (x)}{b^3}-\frac {x \cosh (b x) \sinh (b x)}{2 b^2}+\frac {5 \sinh ^2(b x)}{4 b^3}-\frac {2 x \cosh (b x) \text {Shi}(b x)}{b^2}+\frac {2 \sinh (b x) \text {Shi}(b x)}{b^3}+\frac {x^2 \sinh (b x) \text {Shi}(b x)}{b}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 72, normalized size = 0.73 \begin {gather*} \frac {2 b^2 x^2+5 \cosh (2 b x)-8 \text {Chi}(2 b x)+8 \log (x)-2 b x \sinh (2 b x)+8 \left (-2 b x \cosh (b x)+\left (2+b^2 x^2\right ) \sinh (b x)\right ) \text {Shi}(b x)}{8 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.37, size = 76, normalized size = 0.78
method | result | size |
derivativedivides | \(\frac {\hyperbolicSineIntegral \left (b x \right ) \left (b^{2} x^{2} \sinh \left (b x \right )-2 b x \cosh \left (b x \right )+2 \sinh \left (b x \right )\right )-\frac {b x \cosh \left (b x \right ) \sinh \left (b x \right )}{2}+\frac {b^{2} x^{2}}{4}+\frac {5 \left (\cosh ^{2}\left (b x \right )\right )}{4}+\ln \left (b x \right )-\hyperbolicCosineIntegral \left (2 b x \right )}{b^{3}}\) | \(76\) |
default | \(\frac {\hyperbolicSineIntegral \left (b x \right ) \left (b^{2} x^{2} \sinh \left (b x \right )-2 b x \cosh \left (b x \right )+2 \sinh \left (b x \right )\right )-\frac {b x \cosh \left (b x \right ) \sinh \left (b x \right )}{2}+\frac {b^{2} x^{2}}{4}+\frac {5 \left (\cosh ^{2}\left (b x \right )\right )}{4}+\ln \left (b x \right )-\hyperbolicCosineIntegral \left (2 b x \right )}{b^{3}}\) | \(76\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{2} \cosh {\left (b x \right )} \operatorname {Shi}{\left (b x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^2\,\mathrm {sinhint}\left (b\,x\right )\,\mathrm {cosh}\left (b\,x\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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