Optimal. Leaf size=256 \[ \frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}+\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{a^2 c x^2+c}} \]
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Rubi [A] time = 0.153852, antiderivative size = 256, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4890, 4888, 4181, 2531, 2282, 6589} \[ \frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}+\frac{2 \sqrt{a^2 x^2+1} \text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{a^2 c x^2+c}}-\frac{2 i \sqrt{a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{a^2 c x^2+c}} \]
Antiderivative was successfully verified.
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Rule 4890
Rule 4888
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(a x)^2}{\sqrt{c+a^2 c x^2}} \, dx &=\frac{\sqrt{1+a^2 x^2} \int \frac{\tan ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{\sqrt{c+a^2 c x^2}}\\ &=\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{c+a^2 c x^2}}-\frac{\left (2 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt{c+a^2 c x^2}}+\frac{\left (2 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{\left (2 i \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt{c+a^2 c x^2}}+\frac{\left (2 i \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{\left (2 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}+\frac{\left (2 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}\\ &=-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt{c+a^2 c x^2}}+\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{2 i \sqrt{1+a^2 x^2} \tan ^{-1}(a x) \text{Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}-\frac{2 \sqrt{1+a^2 x^2} \text{Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}+\frac{2 \sqrt{1+a^2 x^2} \text{Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt{c+a^2 c x^2}}\\ \end{align*}
Mathematica [A] time = 0.115355, size = 140, normalized size = 0.55 \[ \frac{2 \sqrt{c \left (a^2 x^2+1\right )} \left (i \tan ^{-1}(a x) \text{PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )-i \tan ^{-1}(a x) \text{PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )-\text{PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )+\text{PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )-i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{a c \sqrt{a^2 x^2+1}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.45, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \arctan \left ( ax \right ) \right ) ^{2}{\frac{1}{\sqrt{{a}^{2}c{x}^{2}+c}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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 (\frac{\arctan \left (a x\right )^{2}}{\sqrt{a^{2} c x^{2} + c}}, 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 \frac{\operatorname{atan}^{2}{\left (a x \right )}}{\sqrt{c \left (a^{2} x^{2} + 1\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{\arctan \left (a x\right )^{2}}{\sqrt{a^{2} c x^{2} + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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