Optimal. Leaf size=165 \[ \frac {n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {n (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac {n (2 c d-b e) \log (d+e x)}{e \left (a e^2-b d e+c d^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \]
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Rubi [A] time = 0.25, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2525, 800, 634, 618, 206, 628} \[ \frac {n \sqrt {b^2-4 a c} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a e^2-b d e+c d^2}+\frac {n (2 c d-b e) \log \left (a+b x+c x^2\right )}{2 e \left (a e^2-b d e+c d^2\right )}-\frac {n (2 c d-b e) \log (d+e x)}{e \left (a e^2-b d e+c d^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)} \]
Antiderivative was successfully verified.
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Rule 206
Rule 618
Rule 628
Rule 634
Rule 800
Rule 2525
Rubi steps
\begin {align*} \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{(d+e x)^2} \, dx &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}+\frac {n \int \frac {b+2 c x}{(d+e x) \left (a+b x+c x^2\right )} \, dx}{e}\\ &=-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}+\frac {n \int \left (\frac {e (-2 c d+b e)}{\left (c d^2-b d e+a e^2\right ) (d+e x)}+\frac {b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}\right ) \, dx}{e}\\ &=-\frac {(2 c d-b e) n \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}+\frac {n \int \frac {b c d-b^2 e+2 a c e+c (2 c d-b e) x}{a+b x+c x^2} \, dx}{e \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {(2 c d-b e) n \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}-\frac {\left (\left (b^2-4 a c\right ) n\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 \left (c d^2-b d e+a e^2\right )}+\frac {((2 c d-b e) n) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 e \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {(2 c d-b e) n \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}+\frac {(2 c d-b e) n \log \left (a+b x+c x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}+\frac {\left (\left (b^2-4 a c\right ) n\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c d^2-b d e+a e^2}\\ &=\frac {\sqrt {b^2-4 a c} n \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c d^2-b d e+a e^2}-\frac {(2 c d-b e) n \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}+\frac {(2 c d-b e) n \log \left (a+b x+c x^2\right )}{2 e \left (c d^2-b d e+a e^2\right )}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{e (d+e x)}\\ \end {align*}
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Mathematica [A] time = 0.32, size = 166, normalized size = 1.01 \[ -\frac {n \sqrt {4 a c-b^2} \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{e (b d-a e)-c d^2}+\frac {n (b e-2 c d) \log (d+e x)}{e \left (e (a e-b d)+c d^2\right )}-\frac {n (b e-2 c d) \log (a+x (b+c x))}{2 e \left (e (a e-b d)+c d^2\right )}-\frac {\log \left (d (a+x (b+c x))^n\right )}{e (d+e x)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 429, normalized size = 2.60 \[ \left [\frac {{\left (e^{2} n x + d e n\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left ({\left (2 \, c d e - b e^{2}\right )} n x + {\left (b d e - 2 \, a e^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (2 \, c d e - b e^{2}\right )} n x + {\left (2 \, c d^{2} - b d e\right )} n\right )} \log \left (e x + d\right ) - 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \log \relax (d)}{2 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3} + {\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x\right )}}, \frac {2 \, {\left (e^{2} n x + d e n\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (2 \, c d e - b e^{2}\right )} n x + {\left (b d e - 2 \, a e^{2}\right )} n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (2 \, c d e - b e^{2}\right )} n x + {\left (2 \, c d^{2} - b d e\right )} n\right )} \log \left (e x + d\right ) - 2 \, {\left (c d^{2} - b d e + a e^{2}\right )} \log \relax (d)}{2 \, {\left (c d^{3} e - b d^{2} e^{2} + a d e^{3} + {\left (c d^{2} e^{2} - b d e^{3} + a e^{4}\right )} x\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 284, normalized size = 1.72 \[ \frac {{\left (2 \, c d n - b n e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c d^{2} e - b d e^{2} + a e^{3}\right )}} - \frac {{\left (b^{2} n - 4 \, a c n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c d^{2} - b d e + a e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c d n x e \log \left (x e + d\right ) + c d^{2} n \log \left (c x^{2} + b x + a\right ) - b d n e \log \left (c x^{2} + b x + a\right ) + 2 \, c d^{2} n \log \left (x e + d\right ) - b n x e^{2} \log \left (x e + d\right ) - b d n e \log \left (x e + d\right ) + a n e^{2} \log \left (c x^{2} + b x + a\right ) + c d^{2} \log \relax (d) - b d e \log \relax (d) + a e^{2} \log \relax (d)}{c d^{2} x e^{2} + c d^{3} e - b d x e^{3} - b d^{2} e^{2} + a x e^{4} + a d e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.72, size = 6540, normalized size = 39.64 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.34, size = 590, normalized size = 3.58 \[ \frac {\ln \left (d+e\,x\right )\,\left (b\,e\,n-2\,c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{e\,\left (d+e\,x\right )}-\frac {\ln \left (\frac {2\,b\,c^2\,n^2}{e}+\frac {4\,c^3\,n^2\,x}{e}-\frac {n\,\left (b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (c^2\,n\,x\,\left (b\,e-2\,c\,d\right )-c\,n\,\left (-e\,b^2+c\,d\,b+2\,a\,c\,e\right )+\frac {c\,e\,n\,\left (b\,e-2\,c\,d+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )\,\left (e\,\left (\frac {b\,n}{2}+\frac {n\,\sqrt {b^2-4\,a\,c}}{2}\right )-c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3}-\frac {\ln \left (\frac {2\,b\,c^2\,n^2}{e}+\frac {4\,c^3\,n^2\,x}{e}-\frac {n\,\left (2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (c\,n\,\left (-e\,b^2+c\,d\,b+2\,a\,c\,e\right )-c^2\,n\,x\,\left (b\,e-2\,c\,d\right )+\frac {c\,e\,n\,\left (2\,c\,d-b\,e+e\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,d\,e+2\,x\,b^2\,e^2+b\,c\,d^2-2\,x\,b\,c\,d\,e+a\,b\,e^2+2\,x\,c^2\,d^2-8\,a\,c\,d\,e-6\,a\,x\,c\,e^2\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )}{2\,\left (c\,d^2\,e-b\,d\,e^2+a\,e^3\right )}\right )\,\left (e\,\left (\frac {b\,n}{2}-\frac {n\,\sqrt {b^2-4\,a\,c}}{2}\right )-c\,d\,n\right )}{c\,d^2\,e-b\,d\,e^2+a\,e^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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