Optimal. Leaf size=126 \[ \frac{i a b \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{2 d^2}+\frac{a^2 x^4}{4}-\frac{a b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d}+\frac{1}{2} i a b x^4+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}-\frac{b^2 x^4}{4} \]
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Rubi [A] time = 0.239084, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3747, 3722, 3719, 2190, 2279, 2391, 3720, 3475, 30} \[ \frac{a^2 x^4}{4}+\frac{i a b \text{Li}_2\left (-e^{2 i \left (d x^2+c\right )}\right )}{2 d^2}-\frac{a b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d}+\frac{1}{2} i a b x^4+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}-\frac{b^2 x^4}{4} \]
Antiderivative was successfully verified.
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Rule 3747
Rule 3722
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rule 3720
Rule 3475
Rule 30
Rubi steps
\begin{align*} \int x^3 \left (a+b \tan \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b \tan (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x+2 a b x \tan (c+d x)+b^2 x \tan ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}+(a b) \operatorname{Subst}\left (\int x \tan (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x \tan ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}+\frac{1}{2} i a b x^4+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}-(2 i a b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,x^2\right )-\frac{1}{2} b^2 \operatorname{Subst}\left (\int x \, dx,x,x^2\right )-\frac{b^2 \operatorname{Subst}\left (\int \tan (c+d x) \, dx,x,x^2\right )}{2 d}\\ &=\frac{a^2 x^4}{4}+\frac{1}{2} i a b x^4-\frac{b^2 x^4}{4}-\frac{a b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}+\frac{(a b) \operatorname{Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,x^2\right )}{d}\\ &=\frac{a^2 x^4}{4}+\frac{1}{2} i a b x^4-\frac{b^2 x^4}{4}-\frac{a b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}-\frac{(i a b) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d x^2\right )}\right )}{2 d^2}\\ &=\frac{a^2 x^4}{4}+\frac{1}{2} i a b x^4-\frac{b^2 x^4}{4}-\frac{a b x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{i a b \text{Li}_2\left (-e^{2 i \left (c+d x^2\right )}\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}\\ \end{align*}
Mathematica [B] time = 6.51332, size = 295, normalized size = 2.34 \[ -\frac{a b \csc (c) \sec (c) \left (d^2 x^4 e^{-i \tan ^{-1}(\cot (c))}-\frac{\cot (c) \left (i \text{PolyLog}\left (2,e^{2 i \left (d x^2-\tan ^{-1}(\cot (c))\right )}\right )+i d x^2 \left (-2 \tan ^{-1}(\cot (c))-\pi \right )-2 \left (d x^2-\tan ^{-1}(\cot (c))\right ) \log \left (1-e^{2 i \left (d x^2-\tan ^{-1}(\cot (c))\right )}\right )-2 \tan ^{-1}(\cot (c)) \log \left (\sin \left (d x^2-\tan ^{-1}(\cot (c))\right )\right )-\pi \log \left (1+e^{-2 i d x^2}\right )+\pi \log \left (\cos \left (d x^2\right )\right )\right )}{\sqrt{\cot ^2(c)+1}}\right )}{2 d^2 \sqrt{\csc ^2(c) \left (\sin ^2(c)+\cos ^2(c)\right )}}+\frac{1}{4} x^4 \sec (c) \left (a^2 \cos (c)+2 a b \sin (c)-b^2 \cos (c)\right )+\frac{b^2 \sec (c) \left (d x^2 \sin (c)+\cos (c) \log \left (\cos (c) \cos \left (d x^2\right )-\sin (c) \sin \left (d x^2\right )\right )\right )}{2 d^2 \left (\sin ^2(c)+\cos ^2(c)\right )}+\frac{b^2 x^2 \sec (c) \sin \left (d x^2\right ) \sec \left (c+d x^2\right )}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.245, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\tan \left ( d{x}^{2}+c \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.44262, size = 539, normalized size = 4.28 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{{\left (2 \, a b + i \, b^{2}\right )} d^{2} x^{4} -{\left (4 \, a b d x^{2} - 2 \, b^{2} + 2 \,{\left (2 \, a b d x^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right ) +{\left (4 i \, a b d x^{2} - 2 i \, b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} \arctan \left (\sin \left (2 \, d x^{2} + 2 \, c\right ), \cos \left (2 \, d x^{2} + 2 \, c\right ) + 1\right ) +{\left ({\left (2 \, a b + i \, b^{2}\right )} d^{2} x^{4} - 4 \, b^{2} d x^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right ) +{\left (2 \, a b \cos \left (2 \, d x^{2} + 2 \, c\right ) + 2 i \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + 2 \, a b\right )}{\rm Li}_2\left (-e^{\left (2 i \, d x^{2} + 2 i \, c\right )}\right ) -{\left (-2 i \, a b d x^{2} + i \, b^{2} +{\left (-2 i \, a b d x^{2} + i \, b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right ) +{\left (2 \, a b d x^{2} - b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} \log \left (\cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{2} + 2 \, c\right ) + 1\right ) -{\left ({\left (-2 i \, a b + b^{2}\right )} d^{2} x^{4} + 4 i \, b^{2} d x^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{-4 i \, d^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) + 4 \, d^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) - 4 i \, d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.68201, size = 470, normalized size = 3.73 \begin{align*} \frac{{\left (a^{2} - b^{2}\right )} d^{2} x^{4} + 2 \, b^{2} d x^{2} \tan \left (d x^{2} + c\right ) - i \, a b{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1} + 1\right ) + i \, a b{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1} + 1\right ) -{\left (2 \, a b d x^{2} - b^{2}\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) -{\left (2 \, a b d x^{2} - b^{2}\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (d x^{2} + c\right ) - 1\right )}}{\tan \left (d x^{2} + c\right )^{2} + 1}\right )}{4 \, d^{2}} \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^{3} \left (a + b \tan{\left (c + d x^{2} \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 x^{2} + c\right ) + a\right )}^{2} x^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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