Optimal. Leaf size=122 \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{4 d^2 \left (a^2+b^2\right )}+\frac{b x^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}+\frac{x^4}{4 (a+i b)} \]
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Rubi [A] time = 0.207091, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {3747, 3732, 2190, 2279, 2391} \[ -\frac{i b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (d x^2+c\right )}}{(a+i b)^2}\right )}{4 d^2 \left (a^2+b^2\right )}+\frac{b x^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}+\frac{x^4}{4 (a+i b)} \]
Antiderivative was successfully verified.
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Rule 3747
Rule 3732
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{x^3}{a+b \tan \left (c+d x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{a+b \tan (c+d x)} \, dx,x,x^2\right )\\ &=\frac{x^4}{4 (a+i b)}+(i b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,x^2\right )\\ &=\frac{x^4}{4 (a+i b)}+\frac{b x^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}-\frac{b \operatorname{Subst}\left (\int \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,x^2\right )}{2 \left (a^2+b^2\right ) d}\\ &=\frac{x^4}{4 (a+i b)}+\frac{b x^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d x^2\right )}\right )}{4 \left (a^2+b^2\right ) d^2}\\ &=\frac{x^4}{4 (a+i b)}+\frac{b x^2 \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d}-\frac{i b \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i \left (c+d x^2\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^2}\\ \end{align*}
Mathematica [A] time = 1.51816, size = 110, normalized size = 0.9 \[ \frac{i b \text{PolyLog}\left (2,\frac{(-a-i b) e^{-2 i \left (c+d x^2\right )}}{a-i b}\right )+d x^2 \left (2 b \log \left (1+\frac{(a+i b) e^{-2 i \left (c+d x^2\right )}}{a-i b}\right )+d x^2 (a+i b)\right )}{4 d^2 \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.216, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{a+b\tan \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.42896, size = 360, normalized size = 2.95 \begin{align*} \frac{{\left (a - i \, b\right )} d^{2} x^{4} - 2 i \, b d x^{2} \arctan \left (\frac{2 \, a b \cos \left (2 \, d x^{2} + 2 \, c\right ) -{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}, \frac{2 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) + a^{2} + b^{2} +{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) + b d x^{2} \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + 4 \, a b \sin \left (2 \, d x^{2} + 2 \, c\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + a^{2} + b^{2} + 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, d x^{2} + 2 \, c\right )}{a^{2} + b^{2}}\right ) - i \, b{\rm Li}_2\left (\frac{{\left (i \, a + b\right )} e^{\left (2 i \, d x^{2} + 2 i \, c\right )}}{-i \, a + b}\right )}{4 \,{\left (a^{2} + b^{2}\right )} d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.92591, size = 1257, normalized size = 10.3 \begin{align*} \frac{2 \, a d^{2} x^{4} - 2 \, b c \log \left (\frac{{\left (i \, a b + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} - a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - 2 \, b c \log \left (\frac{{\left (i \, a b - b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + i \, a b +{\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{\tan \left (d x^{2} + c\right )^{2} + 1}\right ) - i \, b{\rm Li}_2\left (-\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + i \, b{\rm Li}_2\left (-\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \,{\left (b d x^{2} + b c\right )} \log \left (\frac{{\left (2 i \, a b + 2 \, b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + 2 \, a^{2} - 2 i \, a b +{\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right ) + 2 \,{\left (b d x^{2} + b c\right )} \log \left (\frac{{\left (-2 i \, a b + 2 \, b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + 2 \, a^{2} + 2 i \, a b +{\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (d x^{2} + c\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (d x^{2} + c\right )^{2} + a^{2} + b^{2}}\right )}{8 \,{\left (a^{2} + b^{2}\right )} 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 \frac{x^{3}}{a + b \tan{\left (c + d x^{2} \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 \frac{x^{3}}{b \tan \left (d x^{2} + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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