Integrand size = 25, antiderivative size = 762 \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=-\frac {n \log \left (\frac {\sqrt {e} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}+\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}-\sqrt {e} x\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \log \left (\frac {\sqrt {e} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}-\sqrt {e} x\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \log \left (-\frac {\sqrt {e} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}-\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \log \left (-\frac {\sqrt {e} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}-\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 c \sqrt {-d}+\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 c \sqrt {-d}+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{2 c \sqrt {-d}-\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{2 c \sqrt {-d}-\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]
1/2*ln(g*(c*x^2+b*x+a)^n)*ln((-d)^(1/2)-x*e^(1/2))/(-d)^(1/2)/e^(1/2)-1/2* ln(g*(c*x^2+b*x+a)^n)*ln((-d)^(1/2)+x*e^(1/2))/(-d)^(1/2)/e^(1/2)+1/2*n*ln ((-d)^(1/2)+x*e^(1/2))*ln(-(b+2*c*x-(-4*a*c+b^2)^(1/2))*e^(1/2)/(2*c*(-d)^ (1/2)-(b-(-4*a*c+b^2)^(1/2))*e^(1/2)))/(-d)^(1/2)/e^(1/2)-1/2*n*ln((-d)^(1 /2)-x*e^(1/2))*ln((b+2*c*x-(-4*a*c+b^2)^(1/2))*e^(1/2)/(2*c*(-d)^(1/2)+(b- (-4*a*c+b^2)^(1/2))*e^(1/2)))/(-d)^(1/2)/e^(1/2)+1/2*n*ln((-d)^(1/2)+x*e^( 1/2))*ln(-(b+2*c*x+(-4*a*c+b^2)^(1/2))*e^(1/2)/(2*c*(-d)^(1/2)-(b+(-4*a*c+ b^2)^(1/2))*e^(1/2)))/(-d)^(1/2)/e^(1/2)-1/2*n*ln((-d)^(1/2)-x*e^(1/2))*ln ((b+2*c*x+(-4*a*c+b^2)^(1/2))*e^(1/2)/(2*c*(-d)^(1/2)+(b+(-4*a*c+b^2)^(1/2 ))*e^(1/2)))/(-d)^(1/2)/e^(1/2)+1/2*n*polylog(2,2*c*((-d)^(1/2)+x*e^(1/2)) /(2*c*(-d)^(1/2)-(b-(-4*a*c+b^2)^(1/2))*e^(1/2)))/(-d)^(1/2)/e^(1/2)-1/2*n *polylog(2,2*c*((-d)^(1/2)-x*e^(1/2))/(2*c*(-d)^(1/2)+(b-(-4*a*c+b^2)^(1/2 ))*e^(1/2)))/(-d)^(1/2)/e^(1/2)+1/2*n*polylog(2,2*c*((-d)^(1/2)+x*e^(1/2)) /(2*c*(-d)^(1/2)-(b+(-4*a*c+b^2)^(1/2))*e^(1/2)))/(-d)^(1/2)/e^(1/2)-1/2*n *polylog(2,2*c*((-d)^(1/2)-x*e^(1/2))/(2*c*(-d)^(1/2)+(b+(-4*a*c+b^2)^(1/2 ))*e^(1/2)))/(-d)^(1/2)/e^(1/2)
Time = 0.64 (sec) , antiderivative size = 626, normalized size of antiderivative = 0.82 \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\frac {-n \log \left (\frac {\sqrt {e} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}+\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}-\sqrt {e} x\right )-n \log \left (\frac {\sqrt {e} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{2 c \sqrt {-d}+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}-\sqrt {e} x\right )+n \log \left (\frac {\sqrt {e} \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c \sqrt {-d}+\left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )+n \log \left (\frac {\sqrt {e} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c \sqrt {-d}+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )+\log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (g (a+x (b+c x))^n\right )-\log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (g (a+x (b+c x))^n\right )-n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 c \sqrt {-d}+\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )-n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 c \sqrt {-d}+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )+n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{2 c \sqrt {-d}+\left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )+n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{2 c \sqrt {-d}-\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]
(-(n*Log[(Sqrt[e]*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] + (b - Sq rt[b^2 - 4*a*c])*Sqrt[e])]*Log[Sqrt[-d] - Sqrt[e]*x]) - n*Log[(Sqrt[e]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt [e])]*Log[Sqrt[-d] - Sqrt[e]*x] + n*Log[(Sqrt[e]*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(2*c*Sqrt[-d] + (-b + Sqrt[b^2 - 4*a*c])*Sqrt[e])]*Log[Sqrt[-d] + Sqrt[e]*x] + n*Log[(Sqrt[e]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*Sqrt[- d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[e])]*Log[Sqrt[-d] + Sqrt[e]*x] + Log[Sqr t[-d] - Sqrt[e]*x]*Log[g*(a + x*(b + c*x))^n] - Log[Sqrt[-d] + Sqrt[e]*x]* Log[g*(a + x*(b + c*x))^n] - n*PolyLog[2, (2*c*(Sqrt[-d] - Sqrt[e]*x))/(2* c*Sqrt[-d] + (b - Sqrt[b^2 - 4*a*c])*Sqrt[e])] - n*PolyLog[2, (2*c*(Sqrt[- d] - Sqrt[e]*x))/(2*c*Sqrt[-d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[e])] + n*Pol yLog[2, (2*c*(Sqrt[-d] + Sqrt[e]*x))/(2*c*Sqrt[-d] + (-b + Sqrt[b^2 - 4*a* c])*Sqrt[e])] + n*PolyLog[2, (2*c*(Sqrt[-d] + Sqrt[e]*x))/(2*c*Sqrt[-d] - (b + Sqrt[b^2 - 4*a*c])*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])
Time = 1.47 (sec) , antiderivative size = 762, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.
| \(\displaystyle \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx\) |
| \(\Big \downarrow \) 3008 |
| \(\displaystyle \int \left (\frac {\sqrt {-d} \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 d \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {\sqrt {-d} \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 d \left (\sqrt {-d}+\sqrt {e} x\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 \sqrt {-d} c+\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{2 \sqrt {-d} c+\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{2 c \sqrt {-d}-\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{2 c \sqrt {-d}-\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (\frac {\sqrt {e} \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{\sqrt {e} \left (b-\sqrt {b^2-4 a c}\right )+2 c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (\frac {\sqrt {e} \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{\sqrt {e} \left (\sqrt {b^2-4 a c}+b\right )+2 c \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (-\frac {\sqrt {e} \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c \sqrt {-d}-\sqrt {e} \left (b-\sqrt {b^2-4 a c}\right )}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (-\frac {\sqrt {e} \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{2 c \sqrt {-d}-\sqrt {e} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (g \left (a+b x+c x^2\right )^n\right )}{2 \sqrt {-d} \sqrt {e}}\) |
-1/2*(n*Log[(Sqrt[e]*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] + (b - Sqrt[b^2 - 4*a*c])*Sqrt[e])]*Log[Sqrt[-d] - Sqrt[e]*x])/(Sqrt[-d]*Sqrt[e] ) - (n*Log[(Sqrt[e]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[e])]*Log[Sqrt[-d] - Sqrt[e]*x])/(2*Sqrt[-d]*Sqrt[e ]) + (n*Log[-((Sqrt[e]*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] - (b - Sqrt[b^2 - 4*a*c])*Sqrt[e]))]*Log[Sqrt[-d] + Sqrt[e]*x])/(2*Sqrt[-d]*Sq rt[e]) + (n*Log[-((Sqrt[e]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(2*c*Sqrt[-d] - (b + Sqrt[b^2 - 4*a*c])*Sqrt[e]))]*Log[Sqrt[-d] + Sqrt[e]*x])/(2*Sqrt[-d ]*Sqrt[e]) + (Log[Sqrt[-d] - Sqrt[e]*x]*Log[g*(a + b*x + c*x^2)^n])/(2*Sqr t[-d]*Sqrt[e]) - (Log[Sqrt[-d] + Sqrt[e]*x]*Log[g*(a + b*x + c*x^2)^n])/(2 *Sqrt[-d]*Sqrt[e]) - (n*PolyLog[2, (2*c*(Sqrt[-d] - Sqrt[e]*x))/(2*c*Sqrt[ -d] + (b - Sqrt[b^2 - 4*a*c])*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e]) - (n*PolyLog [2, (2*c*(Sqrt[-d] - Sqrt[e]*x))/(2*c*Sqrt[-d] + (b + Sqrt[b^2 - 4*a*c])*S qrt[e])])/(2*Sqrt[-d]*Sqrt[e]) + (n*PolyLog[2, (2*c*(Sqrt[-d] + Sqrt[e]*x) )/(2*c*Sqrt[-d] - (b - Sqrt[b^2 - 4*a*c])*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e]) + (n*PolyLog[2, (2*c*(Sqrt[-d] + Sqrt[e]*x))/(2*c*Sqrt[-d] - (b + Sqrt[b^2 - 4*a*c])*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])
3.1.94.3.1 Defintions of rubi rules used
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.88 (sec) , antiderivative size = 555, normalized size of antiderivative = 0.73
(ln((c*x^2+b*x+a)^n)-n*ln(c*x^2+b*x+a))/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2) )+1/2*n/e*sum(1/_alpha*(ln(x-_alpha)*ln(c*x^2+b*x+a)-ln(x-_alpha)*ln((Root Of(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index=1)-x+_alpha)/Ro otOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index=1))-ln(x-_alp ha)*ln((RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index=2)- x+_alpha)/RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index=2 ))-dilog((RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index=1 )-x+_alpha)/RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index =1))-dilog((RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,index =2)-x+_alpha)/RootOf(_Z^2*c*e+(2*_alpha*c*e+b*e)*_Z+b*_alpha*e+a*e-c*d,ind ex=2))),_alpha=RootOf(_Z^2*e+d))+(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*(c*x^2+b*x+a)^n)*csgn(I*g*(c*x^2+b*x+ a)^n)*csgn(I*g)-1/2*I*Pi*csgn(I*g*(c*x^2+b*x+a)^n)^3+1/2*I*Pi*csgn(I*g*(c* x^2+b*x+a)^n)^2*csgn(I*g)+ln(g))/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))
\[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} g\right )}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\int { \frac {\log \left ({\left (c x^{2} + b x + a\right )}^{n} g\right )}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {\log \left (g \left (a+b x+c x^2\right )^n\right )}{d+e x^2} \, dx=\int \frac {\ln \left (g\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{e\,x^2+d} \,d x \]