Integrand size = 24, antiderivative size = 149 \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {\text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {2 a}{b}\right )}{b^2 c}-\frac {\text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {4 a}{b}\right )}{2 b^2 c}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c} \]
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Time = 0.21 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5790, 5819, 5556, 3384, 3379, 3382} \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {\sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{b^2 c}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c}-\frac {\left (c^2 x^2+1\right )^2}{b c (a+b \text {arcsinh}(c x))} \]
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Rule 3379
Rule 3382
Rule 3384
Rule 5556
Rule 5790
Rule 5819
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}+\frac {(4 c) \int \frac {x \left (1+c^2 x^2\right )}{a+b \text {arcsinh}(c x)} \, dx}{b} \\ & = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {4 \text {Subst}\left (\int \frac {\cosh ^3\left (\frac {a}{b}-\frac {x}{b}\right ) \sinh \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{b^2 c} \\ & = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {4 \text {Subst}\left (\int \left (\frac {\sinh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{8 x}+\frac {\sinh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{4 x}\right ) \, dx,x,a+b \text {arcsinh}(c x)\right )}{b^2 c} \\ & = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {\text {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{2 b^2 c}-\frac {\text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{b^2 c} \\ & = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}+\frac {\cosh \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 )}{b^2 c}+\frac {\cosh \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 )}{2 b^2 c}-\frac {\sinh \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 )}{b^2 c}-\frac {\sinh \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 )}{2 b^2 c} \\ & = -\frac {\left (1+c^2 x^2\right )^2}{b c (a+b \text {arcsinh}(c x))}-\frac {\text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {2 a}{b}\right )}{b^2 c}-\frac {\text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {4 a}{b}\right )}{2 b^2 c}+\frac {\cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{b^2 c}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\frac {-\frac {2 b \left (1+c^2 x^2\right )^2}{a+b \text {arcsinh}(c x)}-2 \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {2 a}{b}\right )-\text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {4 a}{b}\right )+2 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{2 b^2 c} \]
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Time = 0.49 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.71
method | result | size |
default | \(-\frac {4 b \,c^{4} x^{4}+8 b \,c^{2} x^{2}+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )+2 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )-{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )-2 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) a +2 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right ) a -{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) a -2 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right ) a +4 b}{4 c \,b^{2} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}\) | \(255\) |
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\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \]
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\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
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\[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\left (1+c^2 x^2\right )^{3/2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\int \frac {{\left (c^2\,x^2+1\right )}^{3/2}}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \]
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