Optimal. Leaf size=441 \[ -\frac{20 a b x^2 \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 a b x^{3/2} \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 a b x \text{PolyLog}\left (4,-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{PolyLog}\left (4,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \text{PolyLog}\left (6,-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 a b \text{PolyLog}\left (6,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{20 b^2 x^{3/2} \text{PolyLog}\left (2,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 b^2 x \text{PolyLog}\left (3,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 b^2 \sqrt{x} \text{PolyLog}\left (4,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{PolyLog}\left (5,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}-\frac{2 b^2 x^{5/2}}{d} \]
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Rubi [A] time = 0.63134, antiderivative size = 441, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5437, 4190, 4182, 2531, 6609, 2282, 6589, 4184, 3716, 2190} \[ -\frac{20 a b x^2 \text{PolyLog}\left (2,-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{PolyLog}\left (2,e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 a b x^{3/2} \text{PolyLog}\left (3,-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{PolyLog}\left (3,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 a b x \text{PolyLog}\left (4,-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{PolyLog}\left (4,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \text{PolyLog}\left (6,-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 a b \text{PolyLog}\left (6,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{20 b^2 x^{3/2} \text{PolyLog}\left (2,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 b^2 x \text{PolyLog}\left (3,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 b^2 \sqrt{x} \text{PolyLog}\left (4,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{PolyLog}\left (5,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}-\frac{2 b^2 x^{5/2}}{d} \]
Antiderivative was successfully verified.
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Rule 5437
Rule 4190
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4184
Rule 3716
Rule 2190
Rubi steps
\begin{align*} \int x^2 \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^5 (a+b \text{csch}(c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^5+2 a b x^5 \text{csch}(c+d x)+b^2 x^5 \text{csch}^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}+(4 a b) \operatorname{Subst}\left (\int x^5 \text{csch}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^5 \text{csch}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}-\frac{(20 a b) \operatorname{Subst}\left (\int x^4 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(20 a b) \operatorname{Subst}\left (\int x^4 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (10 b^2\right ) \operatorname{Subst}\left (\int x^4 \coth (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{(80 a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(80 a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (20 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x^4}{1-e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(240 a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{(240 a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{\left (40 b^2\right ) \operatorname{Subst}\left (\int x^3 \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 b^2 x^{3/2} \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 a b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{(480 a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(480 a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{\left (60 b^2\right ) \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 b^2 x^{3/2} \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{30 b^2 x \text{Li}_3\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{240 a b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{(480 a b) \operatorname{Subst}\left (\int \text{Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(480 a b) \operatorname{Subst}\left (\int \text{Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{\left (60 b^2\right ) \operatorname{Subst}\left (\int x \text{Li}_3\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 b^2 x^{3/2} \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{30 b^2 x \text{Li}_3\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{240 a b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{30 b^2 \sqrt{x} \text{Li}_4\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 a b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{(480 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{(480 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^6}-\frac{\left (30 b^2\right ) \operatorname{Subst}\left (\int \text{Li}_4\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 b^2 x^{3/2} \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{30 b^2 x \text{Li}_3\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{240 a b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{30 b^2 \sqrt{x} \text{Li}_4\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 a b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \text{Li}_6\left (-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 a b \text{Li}_6\left (e^{c+d \sqrt{x}}\right )}{d^6}-\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(x)}{x} \, dx,x,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}\\ &=-\frac{2 b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}-\frac{8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{2 b^2 x^{5/2} \coth \left (c+d \sqrt{x}\right )}{d}+\frac{10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^2 \text{Li}_2\left (-e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 a b x^2 \text{Li}_2\left (e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 b^2 x^{3/2} \text{Li}_2\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (-e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 a b x^{3/2} \text{Li}_3\left (e^{c+d \sqrt{x}}\right )}{d^3}-\frac{30 b^2 x \text{Li}_3\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{240 a b x \text{Li}_4\left (-e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 a b x \text{Li}_4\left (e^{c+d \sqrt{x}}\right )}{d^4}+\frac{30 b^2 \sqrt{x} \text{Li}_4\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 a b \sqrt{x} \text{Li}_5\left (-e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (e^{c+d \sqrt{x}}\right )}{d^5}-\frac{15 b^2 \text{Li}_5\left (e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{480 a b \text{Li}_6\left (-e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 a b \text{Li}_6\left (e^{c+d \sqrt{x}}\right )}{d^6}\\ \end{align*}
Mathematica [A] time = 12.2088, size = 833, normalized size = 1.89 \[ \frac{a^2 \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \sinh ^2\left (c+d \sqrt{x}\right ) x^3}{3 \left (b+a \sinh \left (c+d \sqrt{x}\right )\right )^2}+\frac{b^2 \text{csch}\left (\frac{c}{2}\right ) \text{csch}\left (\frac{c}{2}+\frac{d \sqrt{x}}{2}\right ) \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \sinh ^2\left (c+d \sqrt{x}\right ) \sinh \left (\frac{d \sqrt{x}}{2}\right ) x^{5/2}}{d \left (b+a \sinh \left (c+d \sqrt{x}\right )\right )^2}-\frac{b^2 \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \text{sech}\left (\frac{c}{2}\right ) \text{sech}\left (\frac{c}{2}+\frac{d \sqrt{x}}{2}\right ) \sinh ^2\left (c+d \sqrt{x}\right ) \sinh \left (\frac{d \sqrt{x}}{2}\right ) x^{5/2}}{d \left (b+a \sinh \left (c+d \sqrt{x}\right )\right )^2}+\frac{2 b \left (a+b \text{csch}\left (c+d \sqrt{x}\right )\right )^2 \left (-\frac{2 b x^{5/2} d^5}{-1+e^{2 c}}+2 a x^{5/2} \log \left (1-e^{-c-d \sqrt{x}}\right ) d^5-2 a x^{5/2} \log \left (1+e^{-c-d \sqrt{x}}\right ) d^5+5 b x^2 \log \left (1-e^{-c-d \sqrt{x}}\right ) d^4+5 b x^2 \log \left (1+e^{-c-d \sqrt{x}}\right ) d^4+40 a x^{3/2} \text{PolyLog}\left (3,-e^{-c-d \sqrt{x}}\right ) d^3-40 a x^{3/2} \text{PolyLog}\left (3,e^{-c-d \sqrt{x}}\right ) d^3-60 b x \text{PolyLog}\left (3,-e^{-c-d \sqrt{x}}\right ) d^2-60 b x \text{PolyLog}\left (3,e^{-c-d \sqrt{x}}\right ) d^2+120 a x \text{PolyLog}\left (4,-e^{-c-d \sqrt{x}}\right ) d^2-120 a x \text{PolyLog}\left (4,e^{-c-d \sqrt{x}}\right ) d^2-120 b \sqrt{x} \text{PolyLog}\left (4,-e^{-c-d \sqrt{x}}\right ) d-120 b \sqrt{x} \text{PolyLog}\left (4,e^{-c-d \sqrt{x}}\right ) d+240 a \sqrt{x} \text{PolyLog}\left (5,-e^{-c-d \sqrt{x}}\right ) d-240 a \sqrt{x} \text{PolyLog}\left (5,e^{-c-d \sqrt{x}}\right ) d+10 \left (a d^4 x^2-2 b d^3 x^{3/2}\right ) \text{PolyLog}\left (2,-e^{-c-d \sqrt{x}}\right )-10 \left (a x^2 d^4+2 b x^{3/2} d^3\right ) \text{PolyLog}\left (2,e^{-c-d \sqrt{x}}\right )-120 b \text{PolyLog}\left (5,-e^{-c-d \sqrt{x}}\right )-120 b \text{PolyLog}\left (5,e^{-c-d \sqrt{x}}\right )+240 a \text{PolyLog}\left (6,-e^{-c-d \sqrt{x}}\right )-240 a \text{PolyLog}\left (6,e^{-c-d \sqrt{x}}\right )\right ) \sinh ^2\left (c+d \sqrt{x}\right )}{d^6 \left (b+a \sinh \left (c+d \sqrt{x}\right )\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.132, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b{\rm csch} \left (c+d\sqrt{x}\right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.98784, size = 670, normalized size = 1.52 \begin{align*} \frac{1}{3} \, a^{2} x^{3} - \frac{4 \, b^{2} x^{\frac{5}{2}}}{d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} - d} - \frac{4 \,{\left (d^{5} x^{\frac{5}{2}} \log \left (e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2}{\rm Li}_2\left (-e^{\left (d \sqrt{x} + c\right )}\right ) - 20 \, d^{3} x^{\frac{3}{2}}{\rm Li}_{3}(-e^{\left (d \sqrt{x} + c\right )}) + 60 \, d^{2} x{\rm Li}_{4}(-e^{\left (d \sqrt{x} + c\right )}) - 120 \, d \sqrt{x}{\rm Li}_{5}(-e^{\left (d \sqrt{x} + c\right )}) + 120 \,{\rm Li}_{6}(-e^{\left (d \sqrt{x} + c\right )})\right )} a b}{d^{6}} + \frac{4 \,{\left (d^{5} x^{\frac{5}{2}} \log \left (-e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2}{\rm Li}_2\left (e^{\left (d \sqrt{x} + c\right )}\right ) - 20 \, d^{3} x^{\frac{3}{2}}{\rm Li}_{3}(e^{\left (d \sqrt{x} + c\right )}) + 60 \, d^{2} x{\rm Li}_{4}(e^{\left (d \sqrt{x} + c\right )}) - 120 \, d \sqrt{x}{\rm Li}_{5}(e^{\left (d \sqrt{x} + c\right )}) + 120 \,{\rm Li}_{6}(e^{\left (d \sqrt{x} + c\right )})\right )} a b}{d^{6}} + \frac{10 \,{\left (d^{4} x^{2} \log \left (e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac{3}{2}}{\rm Li}_2\left (-e^{\left (d \sqrt{x} + c\right )}\right ) - 12 \, d^{2} x{\rm Li}_{3}(-e^{\left (d \sqrt{x} + c\right )}) + 24 \, d \sqrt{x}{\rm Li}_{4}(-e^{\left (d \sqrt{x} + c\right )}) - 24 \,{\rm Li}_{5}(-e^{\left (d \sqrt{x} + c\right )})\right )} b^{2}}{d^{6}} + \frac{10 \,{\left (d^{4} x^{2} \log \left (-e^{\left (d \sqrt{x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac{3}{2}}{\rm Li}_2\left (e^{\left (d \sqrt{x} + c\right )}\right ) - 12 \, d^{2} x{\rm Li}_{3}(e^{\left (d \sqrt{x} + c\right )}) + 24 \, d \sqrt{x}{\rm Li}_{4}(e^{\left (d \sqrt{x} + c\right )}) - 24 \,{\rm Li}_{5}(e^{\left (d \sqrt{x} + c\right )})\right )} b^{2}}{d^{6}} - \frac{2 \,{\left (a b d^{6} x^{3} + 3 \, b^{2} d^{5} x^{\frac{5}{2}}\right )}}{3 \, d^{6}} + \frac{2 \,{\left (a b d^{6} x^{3} - 3 \, b^{2} d^{5} x^{\frac{5}{2}}\right )}}{3 \, 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^{2} x^{2} \operatorname{csch}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b x^{2} \operatorname{csch}\left (d \sqrt{x} + c\right ) + a^{2} 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 )^{2}\, 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 )}^{2} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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