\(\int \frac {\log (g (a+b x+c x^2)^n)}{d+e x+f x^2} \, dx\) [95]

Optimal result

Integrand size = 28, antiderivative size = 782 \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=-\frac {n \log \left (-\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c e-b f+\sqrt {b^2-4 a c} f-c \sqrt {e^2-4 d f}}\right ) \log \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{\sqrt {e^2-4 d f}}-\frac {n \log \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}\right ) \log \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{\sqrt {e^2-4 d f}}+\frac {n \log \left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right ) \log \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{\sqrt {e^2-4 d f}}+\frac {n \log \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right ) \log \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{\sqrt {e^2-4 d f}}+\frac {\log \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f}}-\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \] Output:

-n*ln(-f*(b-(-4*a*c+b^2)^(1/2)+2*c*x)/(c*e-b*f+f*(-4*a*c+b^2)^(1/2)-c*(-4* 
d*f+e^2)^(1/2)))*ln(e-(-4*d*f+e^2)^(1/2)+2*f*x)/(-4*d*f+e^2)^(1/2)-n*ln(f* 
(b+(-4*a*c+b^2)^(1/2)+2*c*x)/((b+(-4*a*c+b^2)^(1/2))*f-c*(e-(-4*d*f+e^2)^( 
1/2))))*ln(e-(-4*d*f+e^2)^(1/2)+2*f*x)/(-4*d*f+e^2)^(1/2)+n*ln(f*(b-(-4*a* 
c+b^2)^(1/2)+2*c*x)/((b-(-4*a*c+b^2)^(1/2))*f-c*(e+(-4*d*f+e^2)^(1/2))))*l 
n(2*f*x+(-4*d*f+e^2)^(1/2)+e)/(-4*d*f+e^2)^(1/2)+n*ln(f*(b+(-4*a*c+b^2)^(1 
/2)+2*c*x)/((b+(-4*a*c+b^2)^(1/2))*f-c*(e+(-4*d*f+e^2)^(1/2))))*ln(2*f*x+( 
-4*d*f+e^2)^(1/2)+e)/(-4*d*f+e^2)^(1/2)+ln(e-(-4*d*f+e^2)^(1/2)+2*f*x)*ln( 
g*(c*x^2+b*x+a)^n)/(-4*d*f+e^2)^(1/2)-ln(2*f*x+(-4*d*f+e^2)^(1/2)+e)*ln(g* 
(c*x^2+b*x+a)^n)/(-4*d*f+e^2)^(1/2)-n*polylog(2,-c*(e-(-4*d*f+e^2)^(1/2)+2 
*f*x)/((b-(-4*a*c+b^2)^(1/2))*f-c*(e-(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2)^(1 
/2)-n*polylog(2,-c*(e-(-4*d*f+e^2)^(1/2)+2*f*x)/((b+(-4*a*c+b^2)^(1/2))*f- 
c*(e-(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2)^(1/2)+n*polylog(2,-c*(2*f*x+(-4*d* 
f+e^2)^(1/2)+e)/((b-(-4*a*c+b^2)^(1/2))*f-c*(e+(-4*d*f+e^2)^(1/2))))/(-4*d 
*f+e^2)^(1/2)+n*polylog(2,-c*(2*f*x+(-4*d*f+e^2)^(1/2)+e)/((b+(-4*a*c+b^2) 
^(1/2))*f-c*(e+(-4*d*f+e^2)^(1/2))))/(-4*d*f+e^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 663, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\frac {-n \log \left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{-c e+b f-\sqrt {b^2-4 a c} f+c \sqrt {e^2-4 d f}}\right ) \log \left (e-\sqrt {e^2-4 d f}+2 f x\right )-n \log \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f+c \left (-e+\sqrt {e^2-4 d f}\right )}\right ) \log \left (e-\sqrt {e^2-4 d f}+2 f x\right )+n \log \left (\frac {f \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{\left (-b+\sqrt {b^2-4 a c}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )}\right ) \log \left (e+\sqrt {e^2-4 d f}+2 f x\right )+n \log \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right ) \log \left (e+\sqrt {e^2-4 d f}+2 f x\right )+\log \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log \left (g (a+x (b+c x))^n\right )-\log \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log \left (g (a+x (b+c x))^n\right )-n \operatorname {PolyLog}\left (2,\frac {c \left (-e+\sqrt {e^2-4 d f}-2 f x\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f+c \left (-e+\sqrt {e^2-4 d f}\right )}\right )-n \operatorname {PolyLog}\left (2,\frac {c \left (-e+\sqrt {e^2-4 d f}-2 f x\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f+c \left (-e+\sqrt {e^2-4 d f}\right )}\right )+n \operatorname {PolyLog}\left (2,\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{\left (-b+\sqrt {b^2-4 a c}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )}\right )+n \operatorname {PolyLog}\left (2,\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{-\left (\left (b+\sqrt {b^2-4 a c}\right ) f\right )+c \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \] Input:

Integrate[Log[g*(a + b*x + c*x^2)^n]/(d + e*x + f*x^2),x]
 

Output:

(-(n*Log[(f*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(-(c*e) + b*f - Sqrt[b^2 - 4* 
a*c]*f + c*Sqrt[e^2 - 4*d*f])]*Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x]) - n*Log 
[(f*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((b + Sqrt[b^2 - 4*a*c])*f + c*(-e + 
Sqrt[e^2 - 4*d*f]))]*Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x] + n*Log[(f*(-b + S 
qrt[b^2 - 4*a*c] - 2*c*x))/((-b + Sqrt[b^2 - 4*a*c])*f + c*(e + Sqrt[e^2 - 
 4*d*f]))]*Log[e + Sqrt[e^2 - 4*d*f] + 2*f*x] + n*Log[(f*(b + Sqrt[b^2 - 4 
*a*c] + 2*c*x))/((b + Sqrt[b^2 - 4*a*c])*f - c*(e + Sqrt[e^2 - 4*d*f]))]*L 
og[e + Sqrt[e^2 - 4*d*f] + 2*f*x] + Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x]*Log 
[g*(a + x*(b + c*x))^n] - Log[e + Sqrt[e^2 - 4*d*f] + 2*f*x]*Log[g*(a + x* 
(b + c*x))^n] - n*PolyLog[2, (c*(-e + Sqrt[e^2 - 4*d*f] - 2*f*x))/((b - Sq 
rt[b^2 - 4*a*c])*f + c*(-e + Sqrt[e^2 - 4*d*f]))] - n*PolyLog[2, (c*(-e + 
Sqrt[e^2 - 4*d*f] - 2*f*x))/((b + Sqrt[b^2 - 4*a*c])*f + c*(-e + Sqrt[e^2 
- 4*d*f]))] + n*PolyLog[2, (c*(e + Sqrt[e^2 - 4*d*f] + 2*f*x))/((-b + Sqrt 
[b^2 - 4*a*c])*f + c*(e + Sqrt[e^2 - 4*d*f]))] + n*PolyLog[2, (c*(e + Sqrt 
[e^2 - 4*d*f] + 2*f*x))/(-((b + Sqrt[b^2 - 4*a*c])*f) + c*(e + Sqrt[e^2 - 
4*d*f]))])/Sqrt[e^2 - 4*d*f]
 

Rubi [A] (verified)

Time = 1.91 (sec) , antiderivative size = 782, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3008, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx\)

\(\Big \downarrow \) 3008

\(\displaystyle \int \left (\frac {2 f \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f} \left (-\sqrt {e^2-4 d f}+e+2 f x\right )}-\frac {2 f \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f} \left (\sqrt {e^2-4 d f}+e+2 f x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+2 f x-\sqrt {e^2-4 d f}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+2 f x-\sqrt {e^2-4 d f}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+2 f x+\sqrt {e^2-4 d f}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,-\frac {c \left (e+2 f x+\sqrt {e^2-4 d f}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \log \left (-\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (-\frac {f \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{f \sqrt {b^2-4 a c}-b f-c \sqrt {e^2-4 d f}+c e}\right )}{\sqrt {e^2-4 d f}}-\frac {n \log \left (-\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (\frac {f \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{f \left (\sqrt {b^2-4 a c}+b\right )-c \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \log \left (\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (\frac {f \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{f \left (b-\sqrt {b^2-4 a c}\right )-c \left (\sqrt {e^2-4 d f}+e\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \log \left (\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (\frac {f \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{f \left (\sqrt {b^2-4 a c}+b\right )-c \left (\sqrt {e^2-4 d f}+e\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {\log \left (-\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (\sqrt {e^2-4 d f}+e+2 f x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{\sqrt {e^2-4 d f}}\)

Input:

Int[Log[g*(a + b*x + c*x^2)^n]/(d + e*x + f*x^2),x]
 

Output:

-((n*Log[-((f*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*e - b*f + Sqrt[b^2 - 4*a 
*c]*f - c*Sqrt[e^2 - 4*d*f]))]*Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x])/Sqrt[e^ 
2 - 4*d*f]) - (n*Log[(f*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((b + Sqrt[b^2 - 
4*a*c])*f - c*(e - Sqrt[e^2 - 4*d*f]))]*Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x] 
)/Sqrt[e^2 - 4*d*f] + (n*Log[(f*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/((b - Sqr 
t[b^2 - 4*a*c])*f - c*(e + Sqrt[e^2 - 4*d*f]))]*Log[e + Sqrt[e^2 - 4*d*f] 
+ 2*f*x])/Sqrt[e^2 - 4*d*f] + (n*Log[(f*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/( 
(b + Sqrt[b^2 - 4*a*c])*f - c*(e + Sqrt[e^2 - 4*d*f]))]*Log[e + Sqrt[e^2 - 
 4*d*f] + 2*f*x])/Sqrt[e^2 - 4*d*f] + (Log[e - Sqrt[e^2 - 4*d*f] + 2*f*x]* 
Log[g*(a + b*x + c*x^2)^n])/Sqrt[e^2 - 4*d*f] - (Log[e + Sqrt[e^2 - 4*d*f] 
 + 2*f*x]*Log[g*(a + b*x + c*x^2)^n])/Sqrt[e^2 - 4*d*f] - (n*PolyLog[2, -( 
(c*(e - Sqrt[e^2 - 4*d*f] + 2*f*x))/((b - Sqrt[b^2 - 4*a*c])*f - c*(e - Sq 
rt[e^2 - 4*d*f])))])/Sqrt[e^2 - 4*d*f] - (n*PolyLog[2, -((c*(e - Sqrt[e^2 
- 4*d*f] + 2*f*x))/((b + Sqrt[b^2 - 4*a*c])*f - c*(e - Sqrt[e^2 - 4*d*f])) 
)])/Sqrt[e^2 - 4*d*f] + (n*PolyLog[2, -((c*(e + Sqrt[e^2 - 4*d*f] + 2*f*x) 
)/((b - Sqrt[b^2 - 4*a*c])*f - c*(e + Sqrt[e^2 - 4*d*f])))])/Sqrt[e^2 - 4* 
d*f] + (n*PolyLog[2, -((c*(e + Sqrt[e^2 - 4*d*f] + 2*f*x))/((b + Sqrt[b^2 
- 4*a*c])*f - c*(e + Sqrt[e^2 - 4*d*f])))])/Sqrt[e^2 - 4*d*f]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3008

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With 
[{u = ExpandIntegrand[(a + b*Log[c*RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u 
]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalFuncti 
onQ[RGx, x] && IGtQ[n, 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.21 (sec) , antiderivative size = 637, normalized size of antiderivative = 0.81

method	result	size
risch	\(\text {Expression too large to display}\)	\(637\)

Input:

int(ln(g*(c*x^2+b*x+a)^n)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

Output:

2*(ln((c*x^2+b*x+a)^n)-n*ln(c*x^2+b*x+a))/(4*d*f-e^2)^(1/2)*arctan((2*f*x+ 
e)/(4*d*f-e^2)^(1/2))+n*sum(1/(2*_alpha*f+e)*(ln(x-_alpha)*ln(c*x^2+b*x+a) 
-ln(x-_alpha)*ln((RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_alpha*f-_alpha* 
c*e+f*a-c*d,index=1)-x+_alpha)/RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_al 
pha*f-_alpha*c*e+f*a-c*d,index=1))-ln(x-_alpha)*ln((RootOf(_Z^2*c*f+(2*_al 
pha*c*f+b*f)*_Z+b*_alpha*f-_alpha*c*e+f*a-c*d,index=2)-x+_alpha)/RootOf(_Z 
^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_alpha*f-_alpha*c*e+f*a-c*d,index=2))-dilog 
((RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_alpha*f-_alpha*c*e+f*a-c*d,inde 
x=1)-x+_alpha)/RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_alpha*f-_alpha*c*e 
+f*a-c*d,index=1))-dilog((RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)*_Z+b*_alpha*f 
-_alpha*c*e+f*a-c*d,index=2)-x+_alpha)/RootOf(_Z^2*c*f+(2*_alpha*c*f+b*f)* 
_Z+b*_alpha*f-_alpha*c*e+f*a-c*d,index=2))),_alpha=RootOf(_Z^2*f+_Z*e+d))+ 
2*(-1/2*I*Pi*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*g)*csgn(I*g*(c*x^2+b*x+a)^n)+1 
/2*I*Pi*csgn(I*(c*x^2+b*x+a)^n)*csgn(I*g*(c*x^2+b*x+a)^n)^2+1/2*I*Pi*csgn( 
I*g)*csgn(I*g*(c*x^2+b*x+a)^n)^2-1/2*I*Pi*csgn(I*g*(c*x^2+b*x+a)^n)^3+ln(g 
))/(4*d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2))
 

Fricas [F]

\[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} g\right )}{f x^{2} + e x + d} \,d x } \] Input:

integrate(log(g*(c*x^2+b*x+a)^n)/(f*x^2+e*x+d),x, algorithm="fricas")
 

Output:

integral(log((c*x^2 + b*x + a)^n*g)/(f*x^2 + e*x + d), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\text {Timed out} \] Input:

integrate(ln(g*(c*x**2+b*x+a)**n)/(f*x**2+e*x+d),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(log(g*(c*x^2+b*x+a)^n)/(f*x^2+e*x+d),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for 
 more deta
 

Giac [F]

\[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} g\right )}{f x^{2} + e x + d} \,d x } \] Input:

integrate(log(g*(c*x^2+b*x+a)^n)/(f*x^2+e*x+d),x, algorithm="giac")
 

Output:

integrate(log((c*x^2 + b*x + a)^n*g)/(f*x^2 + e*x + d), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\int \frac {\ln \left (g\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{f\,x^2+e\,x+d} \,d x \] Input:

int(log(g*(a + b*x + c*x^2)^n)/(d + e*x + f*x^2),x)
 

Output:

int(log(g*(a + b*x + c*x^2)^n)/(d + e*x + f*x^2), x)
 

Reduce [F]

\[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x+f x^2} \, dx=\int \frac {\mathrm {log}\left (\left (c \,x^{2}+b x +a \right )^{n} g \right )}{f \,x^{2}+e x +d}d x \] Input:

int(log(g*(c*x^2+b*x+a)^n)/(f*x^2+e*x+d),x)
 

Output:

int(log((a + b*x + c*x**2)**n*g)/(d + e*x + f*x**2),x)