Optimal. Leaf size=268 \[ -\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {6 b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}-\frac {6 b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)^3}{x} \]
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Rubi [A] time = 0.58, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5865, 5801, 5831, 3322, 2264, 2190, 2531, 2282, 6589} \[ -\frac {6 b \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {6 b \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}+\frac {6 b \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}-\frac {6 b \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )}{\sqrt {a^2+1}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )}{\sqrt {a^2+1}}-\frac {\sinh ^{-1}(a+b x)^3}{x} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2264
Rule 2282
Rule 2531
Rule 3322
Rule 5801
Rule 5831
Rule 5865
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^3}{x^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^3}{\left (-\frac {a}{b}+\frac {x}{b}\right )^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}+3 \operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^2}{\left (-\frac {a}{b}+\frac {x}{b}\right ) \sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}+3 \operatorname {Subst}\left (\int \frac {x^2}{-\frac {a}{b}+\frac {\sinh (x)}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}+6 \operatorname {Subst}\left (\int \frac {e^x x^2}{-\frac {1}{b}-\frac {2 a e^x}{b}+\frac {e^{2 x}}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}+\frac {6 \operatorname {Subst}\left (\int \frac {e^x x^2}{-\frac {2 a}{b}-\frac {2 \sqrt {1+a^2}}{b}+\frac {2 e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}-\frac {6 \operatorname {Subst}\left (\int \frac {e^x x^2}{-\frac {2 a}{b}+\frac {2 \sqrt {1+a^2}}{b}+\frac {2 e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {(6 b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 e^x}{\left (-\frac {2 a}{b}-\frac {2 \sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}+\frac {(6 b) \operatorname {Subst}\left (\int x \log \left (1+\frac {2 e^x}{\left (-\frac {2 a}{b}+\frac {2 \sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 e^x}{\left (-\frac {2 a}{b}-\frac {2 \sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}+\frac {(6 b) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {2 e^x}{\left (-\frac {2 a}{b}+\frac {2 \sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\sqrt {1+a^2}}\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {(6 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{a-\sqrt {1+a^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{\sqrt {1+a^2}}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{a+\sqrt {1+a^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{\sqrt {1+a^2}}\\ &=-\frac {\sinh ^{-1}(a+b x)^3}{x}-\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {3 b \sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {6 b \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}+\frac {6 b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}-\frac {6 b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )}{\sqrt {1+a^2}}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 259, normalized size = 0.97 \[ -\frac {6 b x \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-6 b x \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )-6 b x \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+6 b x \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+\sqrt {a^2+1} \sinh ^{-1}(a+b x)^3-3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac {\sqrt {a^2+1}-e^{\sinh ^{-1}(a+b x)}+a}{\sqrt {a^2+1}+a}\right )+3 b x \sinh ^{-1}(a+b x)^2 \log \left (\frac {\sqrt {a^2+1}+e^{\sinh ^{-1}(a+b x)}-a}{\sqrt {a^2+1}-a}\right )}{\sqrt {a^2+1} x} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{x^{2}}, x\right ) \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {\arcsinh \left (b x +a \right )^{3}}{x^{2}}\, dx \]
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 \[ -\frac {\log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{3}}{x} + \int \frac {3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2}}{b^{3} x^{4} + 3 \, a b^{2} x^{3} + {\left (3 \, a^{2} b + b\right )} x^{2} + {\left (a^{3} + a\right )} x + {\left (b^{2} x^{3} + 2 \, a b x^{2} + {\left (a^{2} + 1\right )} x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^3}{x^2} \,d x \]
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 \[ \int \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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