Integrand size = 14, antiderivative size = 39 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=a \log (x)+\frac {1}{4} i b \operatorname {PolyLog}\left (2,-i c x^2\right )-\frac {1}{4} i b \operatorname {PolyLog}\left (2,i c x^2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4944, 4940, 2438} \[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=a \log (x)+\frac {1}{4} i b \operatorname {PolyLog}\left (2,-i c x^2\right )-\frac {1}{4} i b \operatorname {PolyLog}\left (2,i c x^2\right ) \]
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Rule 2438
Rule 4940
Rule 4944
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {a+b \arctan (c x)}{x} \, dx,x,x^2\right ) \\ & = a \log (x)+\frac {1}{4} (i b) \text {Subst}\left (\int \frac {\log (1-i c x)}{x} \, dx,x,x^2\right )-\frac {1}{4} (i b) \text {Subst}\left (\int \frac {\log (1+i c x)}{x} \, dx,x,x^2\right ) \\ & = a \log (x)+\frac {1}{4} i b \operatorname {PolyLog}\left (2,-i c x^2\right )-\frac {1}{4} i b \operatorname {PolyLog}\left (2,i c x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=a \log (x)+\frac {1}{4} i b \operatorname {PolyLog}\left (2,-i c x^2\right )-\frac {1}{4} i b \operatorname {PolyLog}\left (2,i c x^2\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.75 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.62
method | result | size |
default | \(a \ln \left (x \right )+b \ln \left (x \right ) \arctan \left (c \,x^{2}\right )-\frac {b \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}}\right )}{2 c}\) | \(63\) |
parts | \(a \ln \left (x \right )+b \ln \left (x \right ) \arctan \left (c \,x^{2}\right )-\frac {b \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}}\right )}{2 c}\) | \(63\) |
risch | \(\frac {i \ln \left (-i c \,x^{2}+1\right ) \ln \left (x \right ) b}{2}-\frac {i \ln \left (x \right ) \ln \left (1-i x \sqrt {-i c}\right ) b}{2}-\frac {i \ln \left (x \right ) \ln \left (1+i x \sqrt {-i c}\right ) b}{2}-\frac {i \operatorname {dilog}\left (1-i x \sqrt {-i c}\right ) b}{2}-\frac {i \operatorname {dilog}\left (1+i x \sqrt {-i c}\right ) b}{2}+a \ln \left (x \right )-\frac {i \ln \left (i c \,x^{2}+1\right ) \ln \left (x \right ) b}{2}+\frac {i \ln \left (x \right ) \ln \left (1+i x \sqrt {i c}\right ) b}{2}+\frac {i \ln \left (x \right ) \ln \left (1-i x \sqrt {i c}\right ) b}{2}+\frac {i \operatorname {dilog}\left (1+i x \sqrt {i c}\right ) b}{2}+\frac {i \operatorname {dilog}\left (1-i x \sqrt {i c}\right ) b}{2}\) | \(182\) |
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\[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=\int { \frac {b \arctan \left (c x^{2}\right ) + a}{x} \,d x } \]
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\[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=\int \frac {a + b \operatorname {atan}{\left (c x^{2} \right )}}{x}\, dx \]
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\[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=\int { \frac {b \arctan \left (c x^{2}\right ) + a}{x} \,d x } \]
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\[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=\int { \frac {b \arctan \left (c x^{2}\right ) + a}{x} \,d x } \]
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Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \arctan \left (c x^2\right )}{x} \, dx=a\,\ln \left (x\right )-\frac {b\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x^2\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1{}\mathrm {i}\,c\,x^2+1\right )\right )\,1{}\mathrm {i}}{4} \]
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