Optimal. Leaf size=205 \[ 2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-\frac {1}{3} \sinh ^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.35, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5865, 5799, 5561, 2190, 2531, 2282, 6589} \[ 2 \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+2 \sinh ^{-1}(a+b x) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-2 \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-2 \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-\frac {1}{3} \sinh ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 5561
Rule 5799
Rule 5865
Rule 6589
Rubi steps
\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^2}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^2}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \cosh (x)}{-\frac {a}{b}+\frac {\sinh (x)}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{3} \sinh ^{-1}(a+b x)^3+\frac {\operatorname {Subst}\left (\int \frac {e^x x^2}{-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}+\frac {e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {e^x x^2}{-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}+\frac {e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{3} \sinh ^{-1}(a+b x)^3+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-2 \operatorname {Subst}\left (\int x \log \left (1+\frac {e^x}{\left (-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )-2 \operatorname {Subst}\left (\int x \log \left (1+\frac {e^x}{\left (-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac {1}{3} \sinh ^{-1}(a+b x)^3+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-2 \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {e^x}{\left (-\frac {a}{b}-\frac {\sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )-2 \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {e^x}{\left (-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac {1}{3} \sinh ^{-1}(a+b x)^3+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-2 \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 )-2 \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 )\\ &=-\frac {1}{3} \sinh ^{-1}(a+b x)^3+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )\\ \end {align*}
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Mathematica [A] time = 0.03, size = 251, normalized size = 1.22 \[ 2 \sinh ^{-1}(a+b x) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}-\frac {\sqrt {a^2+1}}{b}\right ) b}\right )+2 \sinh ^{-1}(a+b x) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right ) b}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-2 \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^2 \log \left (\frac {e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right )}+1\right )+\sinh ^{-1}(a+b x)^2 \log \left (\frac {e^{\sinh ^{-1}(a+b x)}}{b \left (\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right )}+1\right )-\frac {1}{3} \sinh ^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{x}, x\right ) \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\arcsinh \left (b x +a \right )^{2}}{x}\, dx \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{2}}{x}\,{d x} \]
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 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^2}{x} \,d x \]
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 \[ \int \frac {\operatorname {asinh}^{2}{\left (a + b x \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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