Integrand size = 26, antiderivative size = 154 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=b c^2 e \arctan (c x)+a c^2 e \log (x)-\frac {1}{2} a c^2 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x) \]
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Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4946, 331, 209, 5141, 815, 649, 266, 4940, 2438} \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=-\frac {(a+b \arctan (c x)) \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x^2}-\frac {1}{2} a c^2 e \log \left (c^2 x^2+1\right )+a c^2 e \log (x)-\frac {1}{2} b c^2 \arctan (c x) \left (e \log \left (c^2 x^2+1\right )+d\right )+b c^2 e \arctan (c x)-\frac {b c \left (e \log \left (c^2 x^2+1\right )+d\right )}{2 x}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x) \]
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Rule 209
Rule 266
Rule 331
Rule 649
Rule 815
Rule 2438
Rule 4940
Rule 4946
Rule 5141
Rubi steps \begin{align*} \text {integral}& = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}-\left (2 c^2 e\right ) \int \left (\frac {-a-b c x}{2 x \left (1+c^2 x^2\right )}-\frac {b \arctan (c x)}{2 x}\right ) \, dx \\ & = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}-\left (c^2 e\right ) \int \frac {-a-b c x}{x \left (1+c^2 x^2\right )} \, dx+\left (b c^2 e\right ) \int \frac {\arctan (c x)}{x} \, dx \\ & = -\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}-\left (c^2 e\right ) \int \left (-\frac {a}{x}+\frac {c (-b+a c x)}{1+c^2 x^2}\right ) \, dx+\frac {1}{2} \left (i b c^2 e\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b c^2 e\right ) \int \frac {\log (1+i c x)}{x} \, dx \\ & = a c^2 e \log (x)-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x)-\left (c^3 e\right ) \int \frac {-b+a c x}{1+c^2 x^2} \, dx \\ & = a c^2 e \log (x)-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x)+\left (b c^3 e\right ) \int \frac {1}{1+c^2 x^2} \, dx-\left (a c^4 e\right ) \int \frac {x}{1+c^2 x^2} \, dx \\ & = b c^2 e \arctan (c x)+a c^2 e \log (x)-\frac {1}{2} a c^2 e \log \left (1+c^2 x^2\right )-\frac {b c \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x}-\frac {1}{2} b c^2 \arctan (c x) \left (d+e \log \left (1+c^2 x^2\right )\right )-\frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^2 e \operatorname {PolyLog}(2,i c x) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=-\frac {a d+b c d x+b d \arctan (c x)+b c^2 d x^2 \arctan (c x)-2 b c^2 e x^2 \arctan (c x)-2 a c^2 e x^2 \log (x)+a e \log \left (1+c^2 x^2\right )+b c e x \log \left (1+c^2 x^2\right )+a c^2 e x^2 \log \left (1+c^2 x^2\right )+b e \arctan (c x) \log \left (1+c^2 x^2\right )+b c^2 e x^2 \arctan (c x) \log \left (1+c^2 x^2\right )-i b c^2 e x^2 \operatorname {PolyLog}(2,-i c x)+i b c^2 e x^2 \operatorname {PolyLog}(2,i c x)}{2 x^2} \]
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\[\int \frac {\left (a +b \arctan \left (c x \right )\right ) \left (d +e \ln \left (c^{2} x^{2}+1\right )\right )}{x^{3}}d x\]
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\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \]
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\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e \log {\left (c^{2} x^{2} + 1 \right )}\right )}{x^{3}}\, dx \]
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\[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )} {\left (e \log \left (c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \]
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Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+b \arctan (c x)) \left (d+e \log \left (1+c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (c^2\,x^2+1\right )\right )}{x^3} \,d x \]
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