Optimal. Leaf size=275 \[ 3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-\frac {1}{4} \sinh ^{-1}(a+b x)^4 \]
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Rubi [A] time = 0.40, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5865, 5799, 5561, 2190, 2531, 6609, 2282, 6589} \[ 3 \sinh ^{-1}(a+b x)^2 \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-6 \sinh ^{-1}(a+b x) \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )-6 \sinh ^{-1}(a+b x) \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )+6 \text {PolyLog}\left (4,\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+6 \text {PolyLog}\left (4,\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{\sqrt {a^2+1}+a}\right )-\frac {1}{4} \sinh ^{-1}(a+b x)^4 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 5561
Rule 5799
Rule 5865
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\sinh ^{-1}(a+b x)^3}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)^3}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \cosh (x)}{-\frac {a}{b}+\frac {\sinh (x)}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{4} \sinh ^{-1}(a+b x)^4+\frac {\operatorname {Subst}\left (\int \frac {e^x x^3}{-\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^3}{-\frac {a}{b}+\frac {\sqrt {1+a^2}}{b}+\frac {e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{4} \sinh ^{-1}(a+b x)^4+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-3 \operatorname {Subst}\left (\int x^2 \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 )-3 \operatorname {Subst}\left (\int x^2 \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}{4} \sinh ^{-1}(a+b x)^4+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-6 \operatorname {Subst}\left (\int x \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 )-6 \operatorname {Subst}\left (\int x \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}{4} \sinh ^{-1}(a+b x)^4+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+6 \operatorname {Subst}\left (\int \text {Li}_3\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 )+6 \operatorname {Subst}\left (\int \text {Li}_3\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}{4} \sinh ^{-1}(a+b x)^4+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{a-\sqrt {1+a^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )+6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{a+\sqrt {1+a^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )\\ &=-\frac {1}{4} \sinh ^{-1}(a+b x)^4+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (1-\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {1+a^2}}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {1+a^2}}\right )+6 \text {Li}_4\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 = 346, normalized size = 1.26 \[ 3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}-\frac {\sqrt {a^2+1}}{b}\right ) b}\right )+3 \sinh ^{-1}(a+b x)^2 \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right ) b}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (-\frac {a}{b}-\frac {\sqrt {a^2+1}}{b}\right ) b}\right )-6 \sinh ^{-1}(a+b x) \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(a+b x)}}{\left (\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right ) b}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a-\sqrt {a^2+1}}\right )+6 \text {Li}_4\left (\frac {e^{\sinh ^{-1}(a+b x)}}{a+\sqrt {a^2+1}}\right )+\sinh ^{-1}(a+b x)^3 \log \left (\frac {e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right )}+1\right )+\sinh ^{-1}(a+b x)^3 \log \left (\frac {e^{\sinh ^{-1}(a+b x)}}{b \left (\frac {\sqrt {a^2+1}}{b}-\frac {a}{b}\right )}+1\right )-\frac {1}{4} \sinh ^{-1}(a+b x)^4 \]
Antiderivative was successfully verified.
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fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{x}, x\right ) \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\arcsinh \left (b x +a \right )^{3}}{x}\, dx \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{x}\,{d x} \]
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 \[ \int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^3}{x} \,d x \]
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 \[ \int \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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