Optimal. Leaf size=311 \[ \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e}-\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e} \]
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Rubi [A] time = 0.24, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4980, 4856, 2402, 2315, 2447} \[ -\frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{4 e}-\frac {i b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{4 e}+\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e}-\frac {\log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 2447
Rule 4856
Rule 4980
Rubi steps
\begin {align*} \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{d+e x^2} \, dx &=\int \left (-\frac {a+b \tan ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {a+b \tan ^{-1}(c x)}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}\right ) \, dx\\ &=-\frac {\int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx}{2 \sqrt {e}}+\frac {\int \frac {a+b \tan ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx}{2 \sqrt {e}}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+2 \frac {(b c) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{2 e}-\frac {(b c) \int \frac {\log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{2 e}-\frac {(b c) \int \frac {\log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{2 e}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}+2 \frac {(i b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )}{2 e}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{2 e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}-\frac {i b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 441, normalized size = 1.42 \[ \frac {a \log \left (d+e x^2\right )}{2 e}+\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} (1-i c x)}{i c \sqrt {-d}-\sqrt {e}}\right )}{4 e}+\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (1-i c x)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 e}-\frac {i b \text {Li}_2\left (-\frac {\sqrt {e} (i c x+1)}{i c \sqrt {-d}-\sqrt {e}}\right )}{4 e}-\frac {i b \text {Li}_2\left (\frac {\sqrt {e} (i c x+1)}{i \sqrt {-d} c+\sqrt {e}}\right )}{4 e}-\frac {i b \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e}+\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}-\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e}+\frac {i b \log (1-i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}-i \sqrt {e}}\right )}{4 e}-\frac {i b \log (1+i c x) \log \left (\frac {c \left (\sqrt {-d}+\sqrt {e} x\right )}{c \sqrt {-d}+i \sqrt {e}}\right )}{4 e} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \arctan \left (c x\right ) + a x}{e x^{2} + d}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.36, size = 646, normalized size = 2.08 \[ \frac {a \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{2 e}+\frac {b \ln \left (c^{2} e \,x^{2}+c^{2} d \right ) \arctan \left (c x \right )}{2 e}-\frac {i b \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e}-\frac {i b \ln \left (c x -i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e}+\frac {i b \ln \left (c x -i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 e}-\frac {i b \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e}-\frac {i b \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x +i}{\RootOf \left (e \,\textit {\_Z}^{2}+2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e}+\frac {i b \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e}+\frac {i b \ln \left (c x +i\right ) \ln \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e}-\frac {i b \ln \left (c x +i\right ) \ln \left (c^{2} e \,x^{2}+c^{2} d \right )}{4 e}+\frac {i b \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =1\right )}\right )}{4 e}+\frac {i b \dilog \left (\frac {\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )-c x -i}{\RootOf \left (e \,\textit {\_Z}^{2}-2 i \textit {\_Z} e +c^{2} d -e , \mathit {index} =2\right )}\right )}{4 e} \]
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 \[ 2 \, b \int \frac {x \arctan \left (c x\right )}{2 \, {\left (e x^{2} + d\right )}}\,{d x} + \frac {a \log \left (e x^{2} + d\right )}{2 \, e} \]
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 \[ \int \frac {x\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{e\,x^2+d} \,d x \]
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 \[ \int \frac {x \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{d + e x^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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