Integrand size = 14, antiderivative size = 195 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=-\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d} \]
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Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5858, 5773, 5818, 5819, 3389, 2211, 2236, 2235} \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=-\frac {4 \sqrt {\pi } e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 \sqrt {\pi } e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {8 \sqrt {(c+d x)^2+1}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {2 \sqrt {(c+d x)^2+1}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}} \]
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Rule 2211
Rule 2235
Rule 2236
Rule 3389
Rule 5773
Rule 5818
Rule 5819
Rule 5858
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+b \text {arcsinh}(x))^{7/2}} \, dx,x,c+d x\right )}{d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}+\frac {2 \text {Subst}\left (\int \frac {x}{\sqrt {1+x^2} (a+b \text {arcsinh}(x))^{5/2}} \, dx,x,c+d x\right )}{5 b d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}+\frac {4 \text {Subst}\left (\int \frac {1}{(a+b \text {arcsinh}(x))^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}+\frac {8 \text {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sqrt {a+b \text {arcsinh}(x)}} \, dx,x,c+d x\right )}{15 b^3 d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {8 \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{15 b^4 d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {4 \text {Subst}\left (\int \frac {e^{-i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{15 b^4 d}+\frac {4 \text {Subst}\left (\int \frac {e^{i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{15 b^4 d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {8 \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{15 b^4 d}+\frac {8 \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{15 b^4 d} \\ & = -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d (a+b \text {arcsinh}(c+d x))^{5/2}}-\frac {4 (c+d x)}{15 b^2 d (a+b \text {arcsinh}(c+d x))^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \text {arcsinh}(c+d x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\frac {-6 b^2 e^{\text {arcsinh}(c+d x)}-2 e^{-\text {arcsinh}(c+d x)} \left (4 a^2+2 a b (-1+4 \text {arcsinh}(c+d x))+b^2 \left (3-2 \text {arcsinh}(c+d x)+4 \text {arcsinh}(c+d x)^2\right )\right )+8 e^{a/b} \sqrt {\frac {a}{b}+\text {arcsinh}(c+d x)} (a+b \text {arcsinh}(c+d x))^2 \Gamma \left (\frac {1}{2},\frac {a}{b}+\text {arcsinh}(c+d x)\right )-4 e^{-\frac {a}{b}} (a+b \text {arcsinh}(c+d x)) \left (e^{\frac {a}{b}+\text {arcsinh}(c+d x)} (2 a+b+2 b \text {arcsinh}(c+d x))+2 b \left (-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )\right )}{30 b^3 d (a+b \text {arcsinh}(c+d x))^{5/2}} \]
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\[\int \frac {1}{\left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right )^{\frac {7}{2}}}d x\]
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Exception generated. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \]
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\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
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\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
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\[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
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Timed out. \[ \int \frac {1}{(a+b \text {arcsinh}(c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]
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