Integrand size = 18, antiderivative size = 70 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {a c}{2 x}-\frac {1}{2} a^2 c \arctan (a x)-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \]
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Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5070, 4946, 331, 209, 4940, 2438} \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {1}{2} a^2 c \arctan (a x)+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x)-\frac {c \arctan (a x)}{2 x^2}-\frac {a c}{2 x} \]
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Rule 209
Rule 331
Rule 2438
Rule 4940
Rule 4946
Rule 5070
Rubi steps \begin{align*} \text {integral}& = c \int \frac {\arctan (a x)}{x^3} \, dx+\left (a^2 c\right ) \int \frac {\arctan (a x)}{x} \, dx \\ & = -\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} (a c) \int \frac {1}{x^2 \left (1+a^2 x^2\right )} \, dx+\frac {1}{2} \left (i a^2 c\right ) \int \frac {\log (1-i a x)}{x} \, dx-\frac {1}{2} \left (i a^2 c\right ) \int \frac {\log (1+i a x)}{x} \, dx \\ & = -\frac {a c}{2 x}-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x)-\frac {1}{2} \left (a^3 c\right ) \int \frac {1}{1+a^2 x^2} \, dx \\ & = -\frac {a c}{2 x}-\frac {1}{2} a^2 c \arctan (a x)-\frac {c \arctan (a x)}{2 x^2}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {c \arctan (a x)}{2 x^2}-\frac {a c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-a^2 x^2\right )}{2 x}+\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,-i a x)-\frac {1}{2} i a^2 c \operatorname {PolyLog}(2,i a x) \]
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Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36
method | result | size |
parts | \(c \arctan \left (a x \right ) a^{2} \ln \left (x \right )-\frac {c \arctan \left (a x \right )}{2 x^{2}}-\frac {c a \left (\frac {1}{x}+a \arctan \left (a x \right )+2 a^{2} \left (-\frac {i \ln \left (x \right ) \left (\ln \left (i a x +1\right )-\ln \left (-i a x +1\right )\right )}{2 a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{2 a}\right )\right )}{2}\) | \(95\) |
derivativedivides | \(a^{2} \left (-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}+c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\) | \(96\) |
default | \(a^{2} \left (-\frac {c \arctan \left (a x \right )}{2 a^{2} x^{2}}+c \arctan \left (a x \right ) \ln \left (a x \right )-\frac {c \left (\frac {1}{a x}+\arctan \left (a x \right )-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )\right )}{2}\right )\) | \(96\) |
meijerg | \(\frac {c \,a^{2} \left (-\frac {2 i a x \operatorname {polylog}\left (2, i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}+\frac {2 i a x \operatorname {polylog}\left (2, -i \sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}\right )}{4}+\frac {c \,a^{2} \left (-\frac {2}{a x}-\frac {2 \left (a^{2} x^{2}+1\right ) \arctan \left (a x \right )}{a^{2} x^{2}}\right )}{4}\) | \(101\) |
risch | \(-\frac {i c \,a^{2} \operatorname {dilog}\left (-i a x +1\right )}{2}+\frac {i c \,a^{2} \ln \left (-i a x \right )}{4}-\frac {a c}{2 x}-\frac {i c \,a^{2} \ln \left (-i a x +1\right )}{4}-\frac {i c \ln \left (-i a x +1\right )}{4 x^{2}}+\frac {i c \,a^{2} \operatorname {dilog}\left (i a x +1\right )}{2}-\frac {i c \,a^{2} \ln \left (i a x \right )}{4}+\frac {i c \,a^{2} \ln \left (i a x +1\right )}{4}+\frac {i c \ln \left (i a x +1\right )}{4 x^{2}}\) | \(125\) |
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\[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=\int { \frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{x^{3}} \,d x } \]
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\[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=c \left (\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{3}}\, dx + \int \frac {a^{2} \operatorname {atan}{\left (a x \right )}}{x}\, dx\right ) \]
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none
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=-\frac {\pi a^{2} c x^{2} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{2} c x^{2} \arctan \left (a x\right ) \log \left (a x\right ) + 2 i \, a^{2} c x^{2} {\rm Li}_2\left (i \, a x + 1\right ) - 2 i \, a^{2} c x^{2} {\rm Li}_2\left (-i \, a x + 1\right ) + 2 \, a c x + 2 \, {\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{4 \, x^{2}} \]
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\[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=\int { \frac {{\left (a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{x^{3}} \,d x } \]
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Time = 0.66 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+a^2 c x^2\right ) \arctan (a x)}{x^3} \, dx=\left \{\begin {array}{cl} 0 & \text {\ if\ \ }a=0\\ -\frac {c\,\mathrm {atan}\left (a\,x\right )}{2\,x^2}-\frac {c\,\left (a^3\,\mathrm {atan}\left (a\,x\right )+\frac {a^2}{x}\right )}{2\,a}-\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1-a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {a^2\,c\,{\mathrm {Li}}_{\mathrm {2}}\left (1+a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }a\neq 0 \end {array}\right . \]
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