Optimal. Leaf size=291 \[ \frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}-\frac {2 b^2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 b^2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e} \]
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Rubi [A] time = 0.47, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5799, 5561, 2190, 2531, 2282, 6589} \[ \frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e}-\frac {2 b^2 \text {PolyLog}\left (3,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 b^2 \text {PolyLog}\left (3,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 5561
Rule 5799
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+e x} \, dx &=\operatorname {Subst}\left (\int \frac {(a+b x)^2 \cosh (x)}{c d+e \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\operatorname {Subst}\left (\int \frac {e^x (a+b x)^2}{c d-\sqrt {c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )+\operatorname {Subst}\left (\int \frac {e^x (a+b x)^2}{c d+\sqrt {c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+\frac {e e^x}{c d-\sqrt {c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}-\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+\frac {e e^x}{c d+\sqrt {c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {e e^x}{c d-\sqrt {c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-\frac {e e^x}{c d+\sqrt {c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {e x}{-c d+\sqrt {c^2 d^2+e^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{e}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {e x}{c d+\sqrt {c^2 d^2+e^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}+\frac {2 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 b^2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e}-\frac {2 b^2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e}\\ \end {align*}
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Mathematica [A] time = 0.22, size = 273, normalized size = 0.94 \[ \frac {6 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )+6 b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )+3 \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )+3 \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )-\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{b}-6 b^2 \text {Li}_3\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )-6 b^2 \text {Li}_3\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{3 e} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{e x + d}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{e x +d}\, dx \]
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 \[ \frac {a^{2} \log \left (e x + d\right )}{e} + \int \frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{e x + d} + \frac {2 \, a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{e x + d}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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