Integrand size = 19, antiderivative size = 77 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=-\frac {b c d}{2 x}-\frac {1}{2} b c^2 d \arctan (c x)-\frac {d (a+b \arctan (c x))}{2 x^2}+a e \log (x)+\frac {1}{2} i b e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e \operatorname {PolyLog}(2,i c x) \]
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Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5100, 4946, 331, 209, 4940, 2438} \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=-\frac {d (a+b \arctan (c x))}{2 x^2}+a e \log (x)-\frac {1}{2} b c^2 d \arctan (c x)-\frac {b c d}{2 x}+\frac {1}{2} i b e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e \operatorname {PolyLog}(2,i c x) \]
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Rule 209
Rule 331
Rule 2438
Rule 4940
Rule 4946
Rule 5100
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d (a+b \arctan (c x))}{x^3}+\frac {e (a+b \arctan (c x))}{x}\right ) \, dx \\ & = d \int \frac {a+b \arctan (c x)}{x^3} \, dx+e \int \frac {a+b \arctan (c x)}{x} \, dx \\ & = -\frac {d (a+b \arctan (c x))}{2 x^2}+a e \log (x)+\frac {1}{2} (b c d) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} (i b e) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} (i b e) \int \frac {\log (1+i c x)}{x} \, dx \\ & = -\frac {b c d}{2 x}-\frac {d (a+b \arctan (c x))}{2 x^2}+a e \log (x)+\frac {1}{2} i b e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e \operatorname {PolyLog}(2,i c x)-\frac {1}{2} \left (b c^3 d\right ) \int \frac {1}{1+c^2 x^2} \, dx \\ & = -\frac {b c d}{2 x}-\frac {1}{2} b c^2 d \arctan (c x)-\frac {d (a+b \arctan (c x))}{2 x^2}+a e \log (x)+\frac {1}{2} i b e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e \operatorname {PolyLog}(2,i c x) \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.12 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=-\frac {a d}{2 x^2}-\frac {b d \arctan (c x)}{2 x^2}-\frac {b c d \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )}{2 x}+a e \log (x)+\frac {1}{2} i b e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b e \operatorname {PolyLog}(2,i c x) \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (63 ) = 126\).
Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.68
method | result | size |
derivativedivides | \(c^{2} \left (\frac {a e \ln \left (c x \right )}{c^{2}}-\frac {a d}{2 c^{2} x^{2}}+\frac {b \left (\arctan \left (c x \right ) e \ln \left (c x \right )-\frac {\arctan \left (c x \right ) d}{2 x^{2}}+\frac {i e \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i e \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i e \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i e \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {d \,c^{2} \left (-\frac {1}{c x}-\arctan \left (c x \right )\right )}{2}\right )}{c^{2}}\right )\) | \(129\) |
default | \(c^{2} \left (\frac {a e \ln \left (c x \right )}{c^{2}}-\frac {a d}{2 c^{2} x^{2}}+\frac {b \left (\arctan \left (c x \right ) e \ln \left (c x \right )-\frac {\arctan \left (c x \right ) d}{2 x^{2}}+\frac {i e \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i e \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i e \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i e \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {d \,c^{2} \left (-\frac {1}{c x}-\arctan \left (c x \right )\right )}{2}\right )}{c^{2}}\right )\) | \(129\) |
parts | \(a e \ln \left (x \right )-\frac {a d}{2 x^{2}}+b \,c^{2} \left (\frac {\arctan \left (c x \right ) \ln \left (c x \right ) e}{c^{2}}-\frac {\arctan \left (c x \right ) d}{2 c^{2} x^{2}}-\frac {-i e \ln \left (c x \right ) \ln \left (i c x +1\right )+i e \ln \left (c x \right ) \ln \left (-i c x +1\right )-i e \operatorname {dilog}\left (i c x +1\right )+i e \operatorname {dilog}\left (-i c x +1\right )-d \,c^{2} \left (-\frac {1}{c x}-\arctan \left (c x \right )\right )}{2 c^{2}}\right )\) | \(129\) |
risch | \(-\frac {i b e \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c^{2} b d \ln \left (-i c x \right )}{4}-\frac {b c d}{2 x}-\frac {i c^{2} b d \ln \left (c^{2} x^{2}+1\right )}{8}-\frac {b \,c^{2} d \arctan \left (c x \right )}{4}-\frac {i b d \ln \left (-i c x +1\right )}{4 x^{2}}+a e \ln \left (-i c x \right )-\frac {a d}{2 x^{2}}+\frac {i b e \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i b \,c^{2} d \ln \left (i c x \right )}{4}+\frac {i b \,c^{2} d \ln \left (i c x +1\right )}{4}+\frac {i b d \ln \left (i c x +1\right )}{4 x^{2}}\) | \(157\) |
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\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
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\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{3}}\, dx \]
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\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
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\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]
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Time = 0.85 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))}{x^3} \, dx=\left \{\begin {array}{cl} a\,e\,\ln \left (x\right )-\frac {a\,d}{2\,x^2} & \text {\ if\ \ }c=0\\ a\,e\,\ln \left (x\right )-\frac {a\,d}{2\,x^2}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}-\frac {b\,d\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-\frac {b\,e\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
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