Integrand size = 19, antiderivative size = 129 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=-n \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )+\log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )-n \operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2604, 2404, 2354, 2438} \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=-n \operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )-n \log (x) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )-n \log (x) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )+\log (x) \log \left (d \left (a+b x+c x^2\right )^n\right ) \]
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Rule 2354
Rule 2404
Rule 2438
Rule 2604
Rubi steps \begin{align*} \text {integral}& = \log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )-n \int \frac {(b+2 c x) \log (x)}{a+b x+c x^2} \, dx \\ & = \log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )-n \int \left (\frac {2 c \log (x)}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {2 c \log (x)}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx \\ & = \log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )-(2 c n) \int \frac {\log (x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx-(2 c n) \int \frac {\log (x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx \\ & = -n \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )+\log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )+n \int \frac {\log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{x} \, dx+n \int \frac {\log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{x} \, dx \\ & = -n \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )+\log (x) \log \left (d \left (a+b x+c x^2\right )^n\right )-n \text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-n \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right ) \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\log (x) \log \left (d (a+x (b+c x))^n\right )-n \left (\log (x) \log \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )+\log (x) \log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )+\operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )+\operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.29
method | result | size |
parts | \(\ln \left (x \right ) \ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )-n \left (\ln \left (x \right ) \ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )+\ln \left (x \right ) \ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )+\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )+\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )\right )\) | \(166\) |
risch | \(\ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right ) \ln \left (x \right )-\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) \ln \left (x \right ) n -\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) \ln \left (x \right ) n -\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) n -\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) n +\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right ) \operatorname {csgn}\left (i d \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2} \operatorname {csgn}\left (i d \right )}{2}+\ln \left (d \right )\right ) \ln \left (x \right )\) | \(308\) |
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\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{x} \,d x } \]
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\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\int \frac {\log {\left (d \left (a + b x + c x^{2}\right )^{n} \right )}}{x}\, dx \]
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\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{x} \,d x } \]
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\[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \, dx=\int \frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{x} \,d x \]
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