Integrand size = 10, antiderivative size = 154 \[ \int x \text {Shi}(a+b x)^2 \, dx=\frac {\cosh (2 a+2 b x)}{4 b^2}-\frac {\text {Chi}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}+\frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}-\frac {a \text {Shi}(2 a+2 b x)}{b^2} \]
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Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.400, Rules used = {6673, 6677, 5736, 6873, 6874, 2718, 3379, 6681, 3393, 3382, 6669, 6675, 5556, 12} \[ \int x \text {Shi}(a+b x)^2 \, dx=-\frac {\text {Chi}(2 a+2 b x)}{2 b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}-\frac {a \text {Shi}(2 a+2 b x)}{b^2}+\frac {\text {Shi}(a+b x) \sinh (a+b x)}{b^2}+\frac {a \text {Shi}(a+b x) \cosh (a+b x)}{b^2}+\frac {\log (a+b x)}{2 b^2}+\frac {\cosh (2 a+2 b x)}{4 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}-\frac {x \text {Shi}(a+b x) \cosh (a+b x)}{b} \]
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Rule 12
Rule 2718
Rule 3379
Rule 3382
Rule 3393
Rule 5556
Rule 5736
Rule 6669
Rule 6673
Rule 6675
Rule 6677
Rule 6681
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}-\frac {a \int \text {Shi}(a+b x)^2 \, dx}{2 b}-\int x \sinh (a+b x) \text {Shi}(a+b x) \, dx \\ & = -\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}+\frac {\int \cosh (a+b x) \text {Shi}(a+b x) \, dx}{b}+\frac {a \int \sinh (a+b x) \text {Shi}(a+b x) \, dx}{b}+\int \frac {x \cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx \\ & = \frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}+\frac {1}{2} \int \frac {x \sinh (2 (a+b x))}{a+b x} \, dx-\frac {\int \frac {\sinh ^2(a+b x)}{a+b x} \, dx}{b}-\frac {a \int \frac {\cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx}{b} \\ & = \frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}+\frac {1}{2} \int \frac {x \sinh (2 a+2 b x)}{a+b x} \, dx+\frac {\int \left (\frac {1}{2 (a+b x)}-\frac {\cosh (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b}-\frac {a \int \frac {\sinh (2 a+2 b x)}{2 (a+b x)} \, dx}{b} \\ & = \frac {\log (a+b x)}{2 b^2}+\frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}+\frac {1}{2} \int \left (\frac {\sinh (2 a+2 b x)}{b}+\frac {a \sinh (2 a+2 b x)}{b (-a-b x)}\right ) \, dx-\frac {\int \frac {\cosh (2 a+2 b x)}{a+b x} \, dx}{2 b}-\frac {a \int \frac {\sinh (2 a+2 b x)}{a+b x} \, dx}{2 b} \\ & = -\frac {\text {Chi}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}+\frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}-\frac {a \text {Shi}(2 a+2 b x)}{2 b^2}+\frac {\int \sinh (2 a+2 b x) \, dx}{2 b}+\frac {a \int \frac {\sinh (2 a+2 b x)}{-a-b x} \, dx}{2 b} \\ & = \frac {\cosh (2 a+2 b x)}{4 b^2}-\frac {\text {Chi}(2 a+2 b x)}{2 b^2}+\frac {\log (a+b x)}{2 b^2}+\frac {a \cosh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {x \cosh (a+b x) \text {Shi}(a+b x)}{b}+\frac {\sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {a (a+b x) \text {Shi}(a+b x)^2}{2 b^2}+\frac {x (a+b x) \text {Shi}(a+b x)^2}{2 b}-\frac {a \text {Shi}(2 a+2 b x)}{b^2} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.62 \[ \int x \text {Shi}(a+b x)^2 \, dx=\frac {\cosh (2 (a+b x))-2 \text {Chi}(2 (a+b x))+2 \log (a+b x)+4 ((a-b x) \cosh (a+b x)+\sinh (a+b x)) \text {Shi}(a+b x)-2 \left (a^2-b^2 x^2\right ) \text {Shi}(a+b x)^2-4 a \text {Shi}(2 (a+b x))}{4 b^2} \]
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Time = 0.97 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {\operatorname {Shi}\left (b x +a \right )^{2} \left (-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{2}\right )-2 \,\operatorname {Shi}\left (b x +a \right ) \left (-a \cosh \left (b x +a \right )+\frac {\left (b x +a \right ) \cosh \left (b x +a \right )}{2}-\frac {\sinh \left (b x +a \right )}{2}\right )-a \,\operatorname {Shi}\left (2 b x +2 a \right )+\frac {\cosh \left (b x +a \right )^{2}}{2}+\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Chi}\left (2 b x +2 a \right )}{2}}{b^{2}}\) | \(113\) |
default | \(\frac {\operatorname {Shi}\left (b x +a \right )^{2} \left (-\left (b x +a \right ) a +\frac {\left (b x +a \right )^{2}}{2}\right )-2 \,\operatorname {Shi}\left (b x +a \right ) \left (-a \cosh \left (b x +a \right )+\frac {\left (b x +a \right ) \cosh \left (b x +a \right )}{2}-\frac {\sinh \left (b x +a \right )}{2}\right )-a \,\operatorname {Shi}\left (2 b x +2 a \right )+\frac {\cosh \left (b x +a \right )^{2}}{2}+\frac {\ln \left (b x +a \right )}{2}-\frac {\operatorname {Chi}\left (2 b x +2 a \right )}{2}}{b^{2}}\) | \(113\) |
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\[ \int x \text {Shi}(a+b x)^2 \, dx=\int { x {\rm Shi}\left (b x + a\right )^{2} \,d x } \]
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\[ \int x \text {Shi}(a+b x)^2 \, dx=\int x \operatorname {Shi}^{2}{\left (a + b x \right )}\, dx \]
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\[ \int x \text {Shi}(a+b x)^2 \, dx=\int { x {\rm Shi}\left (b x + a\right )^{2} \,d x } \]
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\[ \int x \text {Shi}(a+b x)^2 \, dx=\int { x {\rm Shi}\left (b x + a\right )^{2} \,d x } \]
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Timed out. \[ \int x \text {Shi}(a+b x)^2 \, dx=\int x\,{\mathrm {sinhint}\left (a+b\,x\right )}^2 \,d x \]
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