Optimal. Leaf size=260 \[ -\frac{10 b x^2 \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{PolyLog}\left (4,-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{PolyLog}\left (4,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 b \sqrt{x} \text{PolyLog}\left (5,-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \sqrt{x} \text{PolyLog}\left (5,e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \text{PolyLog}\left (6,-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{240 b \text{PolyLog}\left (6,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d} \]
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Rubi [A] time = 0.273744, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {14, 5437, 4182, 2531, 6609, 2282, 6589} \[ -\frac{10 b x^2 \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{PolyLog}\left (4,-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{PolyLog}\left (4,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 b \sqrt{x} \text{PolyLog}\left (5,-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \sqrt{x} \text{PolyLog}\left (5,e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \text{PolyLog}\left (6,-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{240 b \text{PolyLog}\left (6,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 5437
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a x^2+b x^2 \text{csch}\left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{a x^3}{3}+b \int x^2 \text{csch}\left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{a x^3}{3}+(2 b) \operatorname{Subst}\left (\int x^5 \text{csch}(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{(10 b) \operatorname{Subst}\left (\int x^4 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(10 b) \operatorname{Subst}\left (\int x^4 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{(40 b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(40 b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(120 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{(120 b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{(240 b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(240 b) \operatorname{Subst}\left (\int x \text{Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{(240 b) \operatorname{Subst}\left (\int \text{Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(240 b) \operatorname{Subst}\left (\int \text{Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{(240 b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{(240 b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^6}\\ &=\frac{a x^3}{3}-\frac{4 b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{10 b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{40 b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{40 b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{120 b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{120 b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{240 b \text{Li}_6\left (-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{240 b \text{Li}_6\left (e^{c+d \sqrt{x}}\right )}{d^6}\\ \end{align*}
Mathematica [A] time = 2.61733, size = 273, normalized size = 1.05 \[ \frac{2 b \left (-5 d^4 x^2 \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )+5 d^4 x^2 \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )+20 d^3 x^{3/2} \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )-20 d^3 x^{3/2} \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )-60 d^2 x \text{PolyLog}\left (4,-e^{c+d \sqrt{x}}\right )+60 d^2 x \text{PolyLog}\left (4,e^{c+d \sqrt{x}}\right )+120 d \sqrt{x} \text{PolyLog}\left (5,-e^{c+d \sqrt{x}}\right )-120 d \sqrt{x} \text{PolyLog}\left (5,e^{c+d \sqrt{x}}\right )-120 \text{PolyLog}\left (6,-e^{c+d \sqrt{x}}\right )+120 \text{PolyLog}\left (6,e^{c+d \sqrt{x}}\right )+d^5 x^{5/2} \log \left (1-e^{c+d \sqrt{x}}\right )-d^5 x^{5/2} \log \left (e^{c+d \sqrt{x}}+1\right )\right )}{d^6}+\frac{a x^3}{3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.073, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b{\rm csch} \left (c+d\sqrt{x}\right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.13899, size = 352, normalized size = 1.35 \begin{align*} \frac{1}{3} \, a x^{3} - \frac{2 \,{\left (\log \left (e^{\left (d \sqrt{x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt{x}\right )}\right )^{5} + 5 \,{\rm Li}_2\left (-e^{\left (d \sqrt{x} + c\right )}\right ) \log \left (e^{\left (d \sqrt{x}\right )}\right )^{4} - 20 \, \log \left (e^{\left (d \sqrt{x}\right )}\right )^{3}{\rm Li}_{3}(-e^{\left (d \sqrt{x} + c\right )}) + 60 \, \log \left (e^{\left (d \sqrt{x}\right )}\right )^{2}{\rm Li}_{4}(-e^{\left (d \sqrt{x} + c\right )}) - 120 \, \log \left (e^{\left (d \sqrt{x}\right )}\right ){\rm Li}_{5}(-e^{\left (d \sqrt{x} + c\right )}) + 120 \,{\rm Li}_{6}(-e^{\left (d \sqrt{x} + c\right )})\right )} b}{d^{6}} + \frac{2 \,{\left (\log \left (-e^{\left (d \sqrt{x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt{x}\right )}\right )^{5} + 5 \,{\rm Li}_2\left (e^{\left (d \sqrt{x} + c\right )}\right ) \log \left (e^{\left (d \sqrt{x}\right )}\right )^{4} - 20 \, \log \left (e^{\left (d \sqrt{x}\right )}\right )^{3}{\rm Li}_{3}(e^{\left (d \sqrt{x} + c\right )}) + 60 \, \log \left (e^{\left (d \sqrt{x}\right )}\right )^{2}{\rm Li}_{4}(e^{\left (d \sqrt{x} + c\right )}) - 120 \, \log \left (e^{\left (d \sqrt{x}\right )}\right ){\rm Li}_{5}(e^{\left (d \sqrt{x} + c\right )}) + 120 \,{\rm Li}_{6}(e^{\left (d \sqrt{x} + c\right )})\right )} b}{d^{6}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{2} \operatorname{csch}\left (d \sqrt{x} + c\right ) + a x^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \operatorname{csch}{\left (c + d \sqrt{x} \right )}\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{csch}\left (d \sqrt{x} + c\right ) + a\right )} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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