Optimal. Leaf size=497 \[ -\frac{20 i a b x^2 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 i a b x^{3/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 i a b x \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 i a b \sqrt{x} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 i a b \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )}{d^6}-\frac{20 b^2 x^{3/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{30 b^2 \sqrt{x} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{15 b^2 \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}+\frac{8 a b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b^2 x^2 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{5/2}}{d} \]
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Rubi [A] time = 0.602376, antiderivative size = 497, 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 = {5436, 4190, 4180, 2531, 6609, 2282, 6589, 4184, 3718, 2190} \[ -\frac{20 i a b x^2 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 i a b x^{3/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 i a b x \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 i a b \sqrt{x} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 i a b \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )}{d^6}-\frac{20 b^2 x^{3/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{30 b^2 \sqrt{x} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{15 b^2 \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}+\frac{8 a b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{10 b^2 x^2 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x^{5/2}}{d} \]
Antiderivative was successfully verified.
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Rule 5436
Rule 4190
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4184
Rule 3718
Rule 2190
Rubi steps
\begin{align*} \int x^2 \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^5 (a+b \text{sech}(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{sech}(c+d x)+b^2 x^5 \text{sech}^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{sech}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^5 \text{sech}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}+\frac{8 a b x^{5/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(20 i a b) \operatorname{Subst}\left (\int x^4 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(20 i a b) \operatorname{Subst}\left (\int x^4 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (10 b^2\right ) \operatorname{Subst}\left (\int x^4 \tanh (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} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(80 i a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(80 i a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (i 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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{80 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(240 i a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{(240 i a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (i 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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i 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 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{240 i a b x \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(480 i a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(480 i a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (i 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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i 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 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i 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 i a b x \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{480 i a b \sqrt{x} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(480 i a b) \operatorname{Subst}\left (\int \text{Li}_5\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(480 i a b) \operatorname{Subst}\left (\int \text{Li}_5\left (i 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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i 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 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i 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 i a b x \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i 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 i a b \sqrt{x} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(480 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^6}+\frac{(480 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(i 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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i 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 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i 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 i a b x \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i 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 i a b \sqrt{x} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{Li}_5\left (i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 i a b \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}+\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} \tan ^{-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{20 i a b x^2 \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (i 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 i a b x^{3/2} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{80 i a b x^{3/2} \text{Li}_3\left (i 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 i a b x \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i 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 i a b \sqrt{x} \text{Li}_5\left (-i e^{c+d \sqrt{x}}\right )}{d^5}-\frac{480 i a b \sqrt{x} \text{Li}_5\left (i 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 i a b \text{Li}_6\left (-i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{480 i a b \text{Li}_6\left (i e^{c+d \sqrt{x}}\right )}{d^6}+\frac{2 b^2 x^{5/2} \tanh \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}
Mathematica [A] time = 7.81533, size = 573, normalized size = 1.15 \[ \frac{\cosh \left (c+d \sqrt{x}\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \left (\frac{3 b \cosh \left (c+d \sqrt{x}\right ) \left (\frac{4 b e^{2 c} d^5 x^{5/2}}{e^{2 c}+1}+i \left (-20 a d^4 x^2 \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+20 a d^4 x^2 \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+80 a d^3 x^{3/2} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-80 a d^3 x^{3/2} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )-240 a d^2 x \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )+240 a d^2 x \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )+480 a d \sqrt{x} \text{PolyLog}\left (5,-i e^{c+d \sqrt{x}}\right )-480 a d \sqrt{x} \text{PolyLog}\left (5,i e^{c+d \sqrt{x}}\right )-480 a \text{PolyLog}\left (6,-i e^{c+d \sqrt{x}}\right )+480 a \text{PolyLog}\left (6,i e^{c+d \sqrt{x}}\right )+20 i b d^3 x^{3/2} \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )-30 i b d^2 x \text{PolyLog}\left (3,-e^{2 \left (c+d \sqrt{x}\right )}\right )+30 i b d \sqrt{x} \text{PolyLog}\left (4,-e^{2 \left (c+d \sqrt{x}\right )}\right )-15 i b \text{PolyLog}\left (5,-e^{2 \left (c+d \sqrt{x}\right )}\right )+4 a d^5 x^{5/2} \log \left (1-i e^{c+d \sqrt{x}}\right )-4 a d^5 x^{5/2} \log \left (1+i e^{c+d \sqrt{x}}\right )+10 i b d^4 x^2 \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )\right )\right )}{d^6}+a^2 x^3 \cosh \left (c+d \sqrt{x}\right )+\frac{6 b^2 x^{5/2} \text{sech}(c) \sinh \left (d \sqrt{x}\right )}{d}\right )}{3 \left (a \cosh \left (c+d \sqrt{x}\right )+b\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b{\rm sech} \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} d x^{3} e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + a^{2} d x^{3} - 12 \, b^{2} x^{\frac{5}{2}}}{3 \,{\left (d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d\right )}} + \int \frac{2 \,{\left (2 \, a b d x^{2} e^{\left (d \sqrt{x} + c\right )} + 5 \, b^{2} x^{\frac{3}{2}}\right )}}{d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d}\,{d x} \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{sech}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b x^{2} \operatorname{sech}\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{sech}{\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{sech}\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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