Optimal. Leaf size=229 \[ -\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 i a b \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{2 b^2 \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{4 b^2 \sqrt{x} \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x}{d} \]
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Rubi [A] time = 0.321019, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5436, 4190, 4180, 2531, 2282, 6589, 4184, 3718, 2190, 2279, 2391} \[ -\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 i a b \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{2 b^2 \text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{4 b^2 \sqrt{x} \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )}{d^2}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{2 b^2 x}{d} \]
Antiderivative was successfully verified.
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Rule 5436
Rule 4190
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rule 4184
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \sqrt{x} \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^2 (a+b \text{sech}(c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \text{sech}(c+d x)+b^2 x^2 \text{sech}^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}+(4 a b) \operatorname{Subst}\left (\int x^2 \text{sech}(c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^2 \text{sech}^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}-\frac{(8 i a b) \operatorname{Subst}\left (\int x \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(8 i a b) \operatorname{Subst}\left (\int x \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int x \tanh (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(8 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(8 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 (c+d x)} x}{1+e^{2 (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{(8 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(8 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^3}+\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 i a b \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}\\ &=\frac{2 b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}+\frac{8 a b x \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}-\frac{2 b^2 \text{Li}_2\left (-e^{2 \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 i a b \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{8 i a b \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}+\frac{2 b^2 x \tanh \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}
Mathematica [A] time = 5.64705, size = 309, normalized size = 1.35 \[ \frac{2 \cosh \left (c+d \sqrt{x}\right ) \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right )^2 \left (\frac{3 b \cosh \left (c+d \sqrt{x}\right ) \left (2 i a \left (-2 d \sqrt{x} \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+2 d \sqrt{x} \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+2 \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-2 \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )+d^2 x \log \left (1-i e^{c+d \sqrt{x}}\right )-d^2 x \log \left (1+i e^{c+d \sqrt{x}}\right )\right )-b \left (\text{PolyLog}\left (2,-e^{2 \left (c+d \sqrt{x}\right )}\right )+2 d \sqrt{x} \log \left (e^{2 \left (c+d \sqrt{x}\right )}+1\right )\right )+\frac{2 b e^{2 c} d^2 x}{e^{2 c}+1}\right )}{d^3}+a^2 x^{3/2} \cosh \left (c+d \sqrt{x}\right )+\frac{3 b^2 x \text{sech}(c) \sinh \left (d \sqrt{x}\right )}{d}\right )}{3 \left (a \cosh \left (c+d \sqrt{x}\right )+b\right )^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.065, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm sech} \left (c+d\sqrt{x}\right ) \right ) ^{2}\sqrt{x}\, 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{2 \,{\left (a^{2} d x^{\frac{3}{2}} e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + a^{2} d x^{\frac{3}{2}} - 6 \, b^{2} x\right )}}{3 \,{\left (d e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d\right )}} + \int \frac{4 \,{\left (a b d x^{\frac{3}{2}} e^{\left (d \sqrt{x} + c\right )} + b^{2} x\right )}}{d x e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + d x}\,{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} \sqrt{x} \operatorname{sech}\left (d \sqrt{x} + c\right )^{2} + 2 \, a b \sqrt{x} \operatorname{sech}\left (d \sqrt{x} + c\right ) + a^{2} \sqrt{x}, 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 \sqrt{x} \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} \sqrt{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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