Optimal. Leaf size=373 \[ -\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {c \sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}-\frac {c \sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {a^2 c x^2+c}}-\frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^2}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^2} \]
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Rubi [A] time = 0.36, antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {4930, 4880, 4890, 4888, 4181, 2531, 2282, 6589, 217, 206} \[ -\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}-\frac {c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {a^2 c x^2+c}}+\frac {\left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}+\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {a^2 c x^2+c}}-\frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{2 a}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^2}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^2} \]
Antiderivative was successfully verified.
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Rule 206
Rule 217
Rule 2282
Rule 2531
Rule 4181
Rule 4880
Rule 4888
Rule 4890
Rule 4930
Rule 6589
Rubi steps
\begin {align*} \int x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3 \, dx &=\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2 \, dx}{a}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {c \int \frac {\tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx}{2 a}-\frac {c \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{a}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {c \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{a}-\frac {\left (c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{2 a \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^2}-\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 a^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^2}+\frac {\left (c \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^2 \sqrt {c+a^2 c x^2}}-\frac {\left (c \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^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^2}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (i c \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^2 \sqrt {c+a^2 c x^2}}-\frac {\left (i c \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^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^2}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (c \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^2 \sqrt {c+a^2 c x^2}}-\frac {\left (c \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^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a^2}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{2 a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a^2 \sqrt {c+a^2 c x^2}}+\frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^3}{3 a^2 c}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a^2}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}+\frac {c \sqrt {1+a^2 x^2} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}-\frac {c \sqrt {1+a^2 x^2} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}
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Mathematica [A] time = 0.59, size = 206, normalized size = 0.55 \[ \frac {\sqrt {a^2 c x^2+c} \left (\left (a^2 x^2+1\right ) \tan ^{-1}(a x) \left (4 \tan ^{-1}(a x)^2-3 \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )+6 \cos \left (2 \tan ^{-1}(a x)\right )+6\right )+\frac {12 \left (-\tanh ^{-1}\left (\frac {a x}{\sqrt {a^2 x^2+1}}\right )-i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+\text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )-\text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )+i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{\sqrt {a^2 x^2+1}}\right )}{12 a^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a^{2} c x^{2} + c} x \arctan \left (a x\right )^{3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.33, size = 370, normalized size = 0.99 \[ \frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right ) \left (2 \arctan \left (a x \right )^{2} x^{2} a^{2}-3 \arctan \left (a x \right ) x a +2 \arctan \left (a x \right )^{2}+6\right )}{6 a^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (-i \arctan \left (a x \right )^{3}+3 \arctan \left (a x \right )^{2} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{6 a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (-i \arctan \left (a x \right )^{3}+3 \arctan \left (a x \right )^{2} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{6 a^{2} \sqrt {a^{2} x^{2}+1}}+\frac {2 i \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2} x^{2}+1}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a^{2} c x^{2} + c} x \arctan \left (a x\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,{\mathrm {atan}\left (a\,x\right )}^3\,\sqrt {c\,a^2\,x^2+c} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}^{3}{\left (a x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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