Optimal. Leaf size=194 \[ -\frac{6 i b x \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i b \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{12 i b \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d} \]
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Rubi [A] time = 0.170835, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {14, 5436, 4180, 2531, 6609, 2282, 6589} \[ -\frac{6 i b x \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i b \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{12 i b \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 14
Rule 5436
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x \left (a+b \text{sech}\left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a x+b x \text{sech}\left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{a x^2}{2}+b \int x \text{sech}\left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{a x^2}{2}+(2 b) \operatorname{Subst}\left (\int x^3 \text{sech}(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{(6 i b) \operatorname{Subst}\left (\int x^2 \log \left (1-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(6 i b) \operatorname{Subst}\left (\int x^2 \log \left (1+i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 i b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{(12 i b) \operatorname{Subst}\left (\int x \text{Li}_2\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{(12 i b) \operatorname{Subst}\left (\int x \text{Li}_2\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 i b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(12 i b) \operatorname{Subst}\left (\int \text{Li}_3\left (-i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{(12 i b) \operatorname{Subst}\left (\int \text{Li}_3\left (i e^{c+d x}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 i b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{(12 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^4}+\frac{(12 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{c+d \sqrt{x}}\right )}{d^4}\\ &=\frac{a x^2}{2}+\frac{4 b x^{3/2} \tan ^{-1}\left (e^{c+d \sqrt{x}}\right )}{d}-\frac{6 i b x \text{Li}_2\left (-i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{6 i b x \text{Li}_2\left (i e^{c+d \sqrt{x}}\right )}{d^2}+\frac{12 i b \sqrt{x} \text{Li}_3\left (-i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \sqrt{x} \text{Li}_3\left (i e^{c+d \sqrt{x}}\right )}{d^3}-\frac{12 i b \text{Li}_4\left (-i e^{c+d \sqrt{x}}\right )}{d^4}+\frac{12 i b \text{Li}_4\left (i e^{c+d \sqrt{x}}\right )}{d^4}\\ \end{align*}
Mathematica [A] time = 1.75242, size = 207, normalized size = 1.07 \[ \frac{a x^2}{2}+\frac{2 i b \left (-3 d^2 x \text{PolyLog}\left (2,-i e^{c+d \sqrt{x}}\right )+3 d^2 x \text{PolyLog}\left (2,i e^{c+d \sqrt{x}}\right )+6 d \sqrt{x} \text{PolyLog}\left (3,-i e^{c+d \sqrt{x}}\right )-6 d \sqrt{x} \text{PolyLog}\left (3,i e^{c+d \sqrt{x}}\right )-6 \text{PolyLog}\left (4,-i e^{c+d \sqrt{x}}\right )+6 \text{PolyLog}\left (4,i e^{c+d \sqrt{x}}\right )+d^3 x^{3/2} \log \left (1-i e^{c+d \sqrt{x}}\right )-d^3 x^{3/2} \log \left (1+i e^{c+d \sqrt{x}}\right )\right )}{d^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.069, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b{\rm sech} \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a x^{2} + 2 \, b \int \frac{x e^{\left (d \sqrt{x} + c\right )}}{e^{\left (2 \, d \sqrt{x} + 2 \, c\right )} + 1}\,{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 x \operatorname{sech}\left (d \sqrt{x} + c\right ) + a 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 x \left (a + b \operatorname{sech}{\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{sech}\left (d \sqrt{x} + c\right ) + a\right )} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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