Optimal. Leaf size=195 \[ -\frac{4 \sqrt{\pi } e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{4 \sqrt{\pi } e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{(c+d x)^2+1}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 \sqrt{(c+d x)^2+1}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]
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Rubi [A] time = 0.465366, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5863, 5655, 5774, 5779, 3308, 2180, 2204, 2205} \[ -\frac{4 \sqrt{\pi } e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{4 \sqrt{\pi } e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{(c+d x)^2+1}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 \sqrt{(c+d x)^2+1}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 5863
Rule 5655
Rule 5774
Rule 5779
Rule 3308
Rule 2180
Rule 2204
Rule 2205
Rubi steps
\begin{align*} \int \frac{1}{\left (a+b \sinh ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{8 \operatorname{Subst}\left (\int \frac{\sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{4 \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{8 \operatorname{Subst}\left (\int e^{\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}+\frac{8 \operatorname{Subst}\left (\int e^{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac{2 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(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 \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}\\ \end{align*}
Mathematica [A] time = 0.189301, size = 238, normalized size = 1.22 \[ \frac{8 e^{a/b} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Gamma}\left (\frac{1}{2},\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-4 e^{-\frac{a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right ) \left (2 b \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )+e^{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (2 a+2 b \sinh ^{-1}(c+d x)+b\right )\right )-2 e^{-\sinh ^{-1}(c+d x)} \left (4 a^2+2 a b \left (4 \sinh ^{-1}(c+d x)-1\right )+b^2 \left (4 \sinh ^{-1}(c+d x)^2-2 \sinh ^{-1}(c+d x)+3\right )\right )-6 b^2 e^{\sinh ^{-1}(c+d x)}}{30 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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