Integrand size = 19, antiderivative size = 157 \[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {2 c n x^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}-\frac {2 c n x^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}+\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m} \]
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Time = 0.15 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2605, 844, 66} \[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {2 c n x^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c n x^{m+2} \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{(m+1) (m+2) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {x^{m+1} \log \left (d \left (a+b x+c x^2\right )^n\right )}{m+1} \]
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Rule 66
Rule 844
Rule 2605
Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {n \int \frac {x^{1+m} (b+2 c x)}{a+b x+c x^2} \, dx}{1+m} \\ & = \frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {n \int \left (\frac {2 c x^{1+m}}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {2 c x^{1+m}}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{1+m} \\ & = \frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m}-\frac {(2 c n) \int \frac {x^{1+m}}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{1+m}-\frac {(2 c n) \int \frac {x^{1+m}}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{1+m} \\ & = -\frac {2 c n x^{2+m} \, _2F_1\left (1,2+m;3+m;-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}-\frac {2 c n x^{2+m} \, _2F_1\left (1,2+m;3+m;-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) (1+m) (2+m)}+\frac {x^{1+m} \log \left (d \left (a+b x+c x^2\right )^n\right )}{1+m} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.87 \[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {x^{1+m} \left (\left (b+\sqrt {b^2-4 a c}\right ) n x \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (b-\sqrt {b^2-4 a c}\right ) n x \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )-2 a (2+m) \log \left (d (a+x (b+c x))^n\right )\right )}{2 a \left (2+3 m+m^2\right )} \]
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\[\int x^{m} \ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )d x\]
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\[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right ) \,d x } \]
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Timed out. \[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Timed out} \]
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\[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right ) \,d x } \]
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\[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\int { x^{m} \log \left ({\left (c x^{2} + b x + a\right )}^{n} d\right ) \,d x } \]
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Timed out. \[ \int x^m \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\int x^m\,\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right ) \,d x \]
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