Integrand size = 27, antiderivative size = 144 \[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \text {arcsinh}(c x))}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5} \]
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Time = 0.21 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5819, 3393, 3384, 3379, 3382} \[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \text {arcsinh}(c x))}{8 b c^5} \]
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Rule 3379
Rule 3382
Rule 3384
Rule 3393
Rule 5819
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sinh ^4\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c^5} \\ & = \frac {\text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cosh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{8 x}-\frac {\cosh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{2 x}\right ) \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c^5} \\ & = \frac {3 \log (a+b \text {arcsinh}(c x))}{8 b c^5}+\frac {\text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^5}-\frac {\text {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{2 b c^5} \\ & = \frac {3 \log (a+b \text {arcsinh}(c x))}{8 b c^5}-\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^5} \\ & = -\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \text {arcsinh}(c x))}{8 b c^5}+\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b c^5}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^5} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.76 \[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {4 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-3 \log (a+b \text {arcsinh}(c x))-4 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{8 b c^5} \]
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Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right )+{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right )-4 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right )-4 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right )-6 \ln \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{5} b}\) | \(115\) |
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\[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x^{4}}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
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\[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x^{4}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \sqrt {c^{2} x^{2} + 1}}\, dx \]
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\[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x^{4}}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
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\[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x^{4}}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^4}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x^4}{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {c^2\,x^2+1}} \,d x \]
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