Optimal. Leaf size=115 \[ -\frac {3 \sinh ^{-1}(a+b x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {Li}_3\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{2 b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {(a+b x)^2+1}}+\frac {\sinh ^{-1}(a+b x)^3}{b}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (e^{2 \sinh ^{-1}(a+b x)}+1\right )}{b} \]
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Rubi [A] time = 0.21, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5867, 5687, 5714, 3718, 2190, 2531, 2282, 6589} \[ -\frac {3 \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(a+b x)}\right )}{2 b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {(a+b x)^2+1}}+\frac {\sinh ^{-1}(a+b x)^3}{b}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (e^{2 \sinh ^{-1}(a+b x)}+1\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 5687
Rule 5714
Rule 5867
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^3}{\left (1+x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \operatorname {Subst}\left (\int \frac {x \sinh ^{-1}(x)^2}{1+x^2} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \operatorname {Subst}\left (\int x^2 \tanh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {6 \operatorname {Subst}\left (\int \frac {e^{2 x} x^2}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {6 \operatorname {Subst}\left (\int x \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}-\frac {3 \sinh ^{-1}(a+b x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {3 \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\sinh ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}-\frac {3 \sinh ^{-1}(a+b x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 b}\\ &=\frac {\sinh ^{-1}(a+b x)^3}{b}+\frac {(a+b x) \sinh ^{-1}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \sinh ^{-1}(a+b x)^2 \log \left (1+e^{2 \sinh ^{-1}(a+b x)}\right )}{b}-\frac {3 \sinh ^{-1}(a+b x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {Li}_3\left (-e^{2 \sinh ^{-1}(a+b x)}\right )}{2 b}\\ \end {align*}
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Mathematica [A] time = 0.63, size = 128, normalized size = 1.11 \[ \frac {2 \sinh ^{-1}(a+b x)^2 \left (\frac {\left (-\sqrt {a^2+2 a b x+b^2 x^2+1}+a+b x\right ) \sinh ^{-1}(a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2+1}}-3 \log \left (e^{-2 \sinh ^{-1}(a+b x)}+1\right )\right )+6 \sinh ^{-1}(a+b x) \text {Li}_2\left (-e^{-2 \sinh ^{-1}(a+b x)}\right )+3 \text {Li}_3\left (-e^{-2 \sinh ^{-1}(a+b x)}\right )}{2 b} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \operatorname {arsinh}\left (b x + a\right )^{3}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \, {\left (3 \, a^{2} + 1\right )} b^{2} x^{2} + a^{4} + 4 \, {\left (a^{3} + a\right )} b x + 2 \, a^{2} + 1}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 203, normalized size = 1.77 \[ -\frac {\left (b^{2} x^{2}-\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x b +2 a b x -\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a +a^{2}+1\right ) \arcsinh \left (b x +a \right )^{3}}{b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}+\frac {2 \arcsinh \left (b x +a \right )^{3}}{b}-\frac {3 \arcsinh \left (b x +a \right )^{2} \ln \left (1+\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}-\frac {3 \arcsinh \left (b x +a \right ) \polylog \left (2, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}+\frac {3 \polylog \left (3, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{2 b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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