Integrand size = 24, antiderivative size = 206 \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c} \]
[Out]
Time = 0.22 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5791, 3393, 3384, 3379, 3382} \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c} \]
[In]
[Out]
Rule 3379
Rule 3382
Rule 3384
Rule 3393
Rule 5791
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\cosh ^6\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c} \\ & = \frac {\text {Subst}\left (\int \left (\frac {5}{16 x}+\frac {\cosh \left (\frac {6 a}{b}-\frac {6 x}{b}\right )}{32 x}+\frac {3 \cosh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{16 x}+\frac {15 \cosh \left (\frac {2 a}{b}-\frac {2 x}{b}\right )}{32 x}\right ) \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c} \\ & = \frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c}+\frac {\text {Subst}\left (\int \frac {\cosh \left (\frac {6 a}{b}-\frac {6 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{32 b c}+\frac {3 \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 )}{16 b c}+\frac {15 \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 )}{32 b c} \\ & = \frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c}+\frac {\left (15 \cosh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{32 b c}+\frac {\left (3 \cosh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {6 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{32 b c}-\frac {\left (15 \sinh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {2 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{32 b c}-\frac {\left (3 \sinh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {6 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{32 b c} \\ & = \frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}+\frac {5 \log (a+b \text {arcsinh}(c x))}{16 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 (a+b \text {arcsinh}(c x))}{b}\right )}{32 b c} \\ \end{align*}
Time = 0.56 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.74 \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+6 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )+10 \log (a+b \text {arcsinh}(c x))-15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-6 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{32 b c} \]
[In]
[Out]
Time = 0.46 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {{\mathrm e}^{\frac {6 a}{b}} \operatorname {Ei}_{1}\left (6 \,\operatorname {arcsinh}\left (c x \right )+\frac {6 a}{b}\right )+6 \,{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right )+15 \,{\mathrm e}^{\frac {2 a}{b}} \operatorname {Ei}_{1}\left (2 \,\operatorname {arcsinh}\left (c x \right )+\frac {2 a}{b}\right )+15 \,{\mathrm e}^{-\frac {2 a}{b}} \operatorname {Ei}_{1}\left (-2 \,\operatorname {arcsinh}\left (c x \right )-\frac {2 a}{b}\right )+6 \,{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right )+{\mathrm e}^{-\frac {6 a}{b}} \operatorname {Ei}_{1}\left (-6 \,\operatorname {arcsinh}\left (c x \right )-\frac {6 a}{b}\right )-20 \ln \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{64 b c}\) | \(163\) |
[In]
[Out]
\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \]
[In]
[Out]
\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{a + b \operatorname {asinh}{\left (c x \right )}}\, dx \]
[In]
[Out]
\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \]
[In]
[Out]
\[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int { \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \operatorname {arsinh}\left (c x\right ) + a} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \text {arcsinh}(c x)} \, dx=\int \frac {{\left (c^2\,x^2+1\right )}^{5/2}}{a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \]
[In]
[Out]