Optimal. Leaf size=274 \[ \frac{6 i a b x \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 a b \sqrt{x} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{3 i a b \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{6 i b^2 \sqrt{x} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{3 b^2 \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{a^2 x^2}{2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+i a b x^2+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{3/2}}{d}-\frac{1}{2} b^2 x^2 \]
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Rubi [A] time = 0.46905, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {3747, 3722, 3719, 2190, 2531, 6609, 2282, 6589, 3720, 30} \[ \frac{a^2 x^2}{2}+\frac{6 i a b x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 a b \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{3 i a b \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+i a b x^2-\frac{6 i b^2 \sqrt{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{3 b^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{3/2}}{d}-\frac{1}{2} b^2 x^2 \]
Antiderivative was successfully verified.
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Rule 3747
Rule 3722
Rule 3719
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 3720
Rule 30
Rubi steps
\begin{align*} \int x \left (a+b \tan \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^3 (a+b \tan (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^3+2 a b x^3 \tan (c+d x)+b^2 x^3 \tan ^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^2}{2}+(4 a b) \operatorname{Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^3 \tan ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^2}{2}+i a b x^2+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}-(8 i a b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^3}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )-\left (2 b^2\right ) \operatorname{Subst}\left (\int x^3 \, dx,x,\sqrt{x}\right )-\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int x^2 \tan (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+i a b x^2-\frac{b^2 x^2}{2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(12 a b) \operatorname{Subst}\left (\int x^2 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (12 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^2}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+i a b x^2-\frac{b^2 x^2}{2}+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{6 i a b x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(12 i a b) \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int x \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 i b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+i a b x^2-\frac{b^2 x^2}{2}+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{6 i b^2 \sqrt{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{6 i a b x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 a b \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(6 a b) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=-\frac{2 i b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+i a b x^2-\frac{b^2 x^2}{2}+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{6 i b^2 \sqrt{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{6 i a b x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{6 a b \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(3 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}\\ &=-\frac{2 i b^2 x^{3/2}}{d}+\frac{a^2 x^2}{2}+i a b x^2-\frac{b^2 x^2}{2}+\frac{6 b^2 x \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{6 i b^2 \sqrt{x} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{6 i a b x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{3 b^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{6 a b \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{3 i a b \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{2 b^2 x^{3/2} \tan \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}
Mathematica [A] time = 1.98856, size = 365, normalized size = 1.33 \[ \frac{b \left (-6 i \left (1+e^{2 i c}\right ) d \sqrt{x} \left (a d \sqrt{x}-b\right ) \text{PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )+3 \left (1+e^{2 i c}\right ) \left (b-2 a d \sqrt{x}\right ) \text{PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )+3 i a e^{2 i c} \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )+3 i a \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )-4 a e^{2 i c} d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )-4 a d^3 x^{3/2} \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )-2 i a d^4 x^2+6 b e^{2 i c} d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )+6 b d^2 x \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )+4 i b d^3 x^{3/2}\right )}{\left (1+e^{2 i c}\right ) d^4}+\frac{1}{2} x^2 \left (a^2+2 a b \tan (c)-b^2\right )+\frac{2 b^2 x^{3/2} \sec (c) \sin \left (d \sqrt{x}\right ) \sec \left (c+d \sqrt{x}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.254, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\tan \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 [B] time = 2.16195, size = 1733, normalized size = 6.32 \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} x \tan \left (d \sqrt{x} + c\right )^{2} + 2 \, a b x \tan \left (d \sqrt{x} + c\right ) + a^{2} 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 \tan{\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 \tan \left (d \sqrt{x} + c\right ) + a\right )}^{2} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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