Optimal. Leaf size=255 \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]
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Rubi [A] time = 0.324528, antiderivative size = 255, 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 = {4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{2 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x}{d} \]
Antiderivative was successfully verified.
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Rule 4204
Rule 4190
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rule 4184
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \sqrt{x} \left (a+b \sec \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sec (c+d x)+b^2 x^2 \sec ^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 \sec (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(8 a b) \operatorname{Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int x \tan (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(8 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i (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^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{\left (8 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(8 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 i b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 a b \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}+\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}\\ &=-\frac{2 i b^2 x}{d}+\frac{2}{3} a^2 x^{3/2}-\frac{8 i a b x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{4 b^2 \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{8 i a b \sqrt{x} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{8 i a b \sqrt{x} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 i b^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{8 a b \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{8 a b \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x \tan \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}
Mathematica [A] time = 0.694471, size = 247, normalized size = 0.97 \[ \frac{2 \left (12 i a b d \sqrt{x} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )-12 i a b d \sqrt{x} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )-12 a b \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )+12 a b \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )-3 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+a^2 d^3 x^{3/2}-12 i a b d^2 x \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )+3 b^2 d^2 x \tan \left (c+d \sqrt{x}\right )+6 b^2 d \sqrt{x} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )-3 i b^2 d^2 x\right )}{3 d^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.081, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \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 [B] time = 2.10303, size = 1727, normalized size = 6.77 \begin{align*} \text{result too large to display} \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} \sec \left (d \sqrt{x} + c\right )^{2} + 2 \, a b \sqrt{x} \sec \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 \sec{\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 \sec \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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