Integrand size = 25, antiderivative size = 581 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\frac {2 \sqrt {e} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 \sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{3/2}}+\frac {\sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{3/2}}+\frac {\sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{f^{3/2}}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f x}-\frac {\sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{3/2}}+\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{3/2}}-\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,1+\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{3/2}}-\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{3/2}} \] Output:
2*e^(1/2)*p*arctan(e^(1/2)*x/d^(1/2))/d^(1/2)/f-2*g^(1/2)*p*arctan(g^(1/2) *x/f^(1/2))*ln(2*f^(1/2)/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)+g^(1/2)*p*arctan(g ^(1/2)*x/f^(1/2))*ln(-2*f^(1/2)*g^(1/2)*((-d)^(1/2)-e^(1/2)*x)/(I*e^(1/2)* f^(1/2)-(-d)^(1/2)*g^(1/2))/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)+g^(1/2)*p*arcta n(g^(1/2)*x/f^(1/2))*ln(2*f^(1/2)*g^(1/2)*((-d)^(1/2)+e^(1/2)*x)/(I*e^(1/2 )*f^(1/2)+(-d)^(1/2)*g^(1/2))/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)-ln(c*(e*x^2+d )^p)/f/x-g^(1/2)*arctan(g^(1/2)*x/f^(1/2))*ln(c*(e*x^2+d)^p)/f^(3/2)+I*g^( 1/2)*p*polylog(2,1-2*f^(1/2)/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)-1/2*I*g^(1/2)* p*polylog(2,1+2*f^(1/2)*g^(1/2)*((-d)^(1/2)-e^(1/2)*x)/(I*e^(1/2)*f^(1/2)- (-d)^(1/2)*g^(1/2))/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)-1/2*I*g^(1/2)*p*polylog (2,1-2*f^(1/2)*g^(1/2)*((-d)^(1/2)+e^(1/2)*x)/(I*e^(1/2)*f^(1/2)+(-d)^(1/2 )*g^(1/2))/(f^(1/2)-I*g^(1/2)*x))/f^(3/2)
Time = 0.42 (sec) , antiderivative size = 613, normalized size of antiderivative = 1.06 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\frac {\frac {2 \sqrt {e} \sqrt {f} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {\sqrt {f} \log \left (c \left (d+e x^2\right )^p\right )}{x}-\sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{2} i \sqrt {g} p \left (\log \left (\frac {\sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}}\right ) \log \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right )+\log \left (\frac {\sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{-i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}}\right ) \log \left (1-\frac {i \sqrt {g} x}{\sqrt {f}}\right )-\log \left (\frac {\sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{-i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}}\right ) \log \left (1+\frac {i \sqrt {g} x}{\sqrt {f}}\right )-\log \left (\frac {\sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}}\right ) \log \left (1+\frac {i \sqrt {g} x}{\sqrt {f}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt {f}-i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}-i \sqrt {-d} \sqrt {g}}\right )+\operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt {f}-i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}+i \sqrt {-d} \sqrt {g}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt {f}+i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}-i \sqrt {-d} \sqrt {g}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {e} \left (\sqrt {f}+i \sqrt {g} x\right )}{\sqrt {e} \sqrt {f}+i \sqrt {-d} \sqrt {g}}\right )\right )}{f^{3/2}} \] Input:
Integrate[Log[c*(d + e*x^2)^p]/(x^2*(f + g*x^2)),x]
Output:
((2*Sqrt[e]*Sqrt[f]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] - (Sqrt[f]*Log[ c*(d + e*x^2)^p])/x - Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[c*(d + e*x^2 )^p] + (I/2)*Sqrt[g]*p*(Log[(Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/(I*Sqrt[e]*Sq rt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 - (I*Sqrt[g]*x)/Sqrt[f]] + Log[(Sqrt[g]*( Sqrt[-d] + Sqrt[e]*x))/((-I)*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 - (I*Sqrt[g]*x)/Sqrt[f]] - Log[(Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/((-I)*Sqrt[e ]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 + (I*Sqrt[g]*x)/Sqrt[f]] - Log[(Sqrt[ g]*(Sqrt[-d] + Sqrt[e]*x))/(I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 + (I*Sqrt[g]*x)/Sqrt[f]] + PolyLog[2, (Sqrt[e]*(Sqrt[f] - I*Sqrt[g]*x))/(Sq rt[e]*Sqrt[f] - I*Sqrt[-d]*Sqrt[g])] + PolyLog[2, (Sqrt[e]*(Sqrt[f] - I*Sq rt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g])] - PolyLog[2, (Sqrt[e]*(S qrt[f] + I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] - I*Sqrt[-d]*Sqrt[g])] - PolyLog[2 , (Sqrt[e]*(Sqrt[f] + I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g]) ]))/f^(3/2)
Time = 1.39 (sec) , antiderivative size = 581, 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 = {2926, 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.
\(\displaystyle \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx\) |
\(\Big \downarrow \) 2926 |
\(\displaystyle \int \left (\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f x^2}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f \left (f+g x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^{3/2}}+\frac {\sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (-\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{3/2}}+\frac {\sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (\sqrt {-d} \sqrt {g}+i \sqrt {e} \sqrt {f}\right )}\right )}{f^{3/2}}+\frac {2 \sqrt {e} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {2 \sqrt {g} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{3/2}}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f x}-\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (i \sqrt {e} \sqrt {f}-\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}+1\right )}{2 f^{3/2}}-\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f} \sqrt {g} \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (i \sqrt {e} \sqrt {f}+\sqrt {-d} \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}\right )}{2 f^{3/2}}+\frac {i \sqrt {g} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{f^{3/2}}\) |
Input:
Int[Log[c*(d + e*x^2)^p]/(x^2*(f + g*x^2)),x]
Output:
(2*Sqrt[e]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*f) - (2*Sqrt[g]*p*ArcTa n[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/f^(3/2) + (Sqrt[g]*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(-2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/((I*Sqrt[e]*Sqrt[f] - Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g] *x))])/f^(3/2) + (Sqrt[g]*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f]*Sqr t[g]*(Sqrt[-d] + Sqrt[e]*x))/((I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])*(Sqrt [f] - I*Sqrt[g]*x))])/f^(3/2) - Log[c*(d + e*x^2)^p]/(f*x) - (Sqrt[g]*ArcT an[(Sqrt[g]*x)/Sqrt[f]]*Log[c*(d + e*x^2)^p])/f^(3/2) + (I*Sqrt[g]*p*PolyL og[2, 1 - (2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/f^(3/2) - ((I/2)*Sqrt[g]*p *PolyLog[2, 1 + (2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] - Sqrt[e]*x))/((I*Sqrt[e]*Sqr t[f] - Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f^(3/2) - ((I/2)*Sqrt[ g]*p*PolyLog[2, 1 - (2*Sqrt[f]*Sqrt[g]*(Sqrt[-d] + Sqrt[e]*x))/((I*Sqrt[e] *Sqrt[f] + Sqrt[-d]*Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/f^(3/2)
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b *Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e , f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & & IntegerQ[s]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 3.10 (sec) , antiderivative size = 526, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\left (\ln \left (\left (e \,x^{2}+d \right )^{p}\right )-p \ln \left (e \,x^{2}+d \right )\right ) \left (-\frac {1}{f x}-\frac {g \arctan \left (\frac {g x}{\sqrt {g f}}\right )}{f \sqrt {g f}}\right )-\frac {p \ln \left (e \,x^{2}+d \right )}{f x}+\frac {2 p e \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{f \sqrt {d e}}+p \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (g \,\textit {\_Z}^{2}+f \right )}{\sum }\left (-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (e \,x^{2}+d \right )-2 e \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \left (\ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =1\right )-x +\underline {\hspace {1.25 ex}}\alpha }{\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =1\right )}\right )+\ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =2\right )-x +\underline {\hspace {1.25 ex}}\alpha }{\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =2\right )}\right )\right )}{2 e}+\frac {\operatorname {dilog}\left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =1\right )-x +\underline {\hspace {1.25 ex}}\alpha }{\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =1\right )}\right )+\operatorname {dilog}\left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =2\right )-x +\underline {\hspace {1.25 ex}}\alpha }{\operatorname {RootOf}\left (\textit {\_Z}^{2} e g +2 \underline {\hspace {1.25 ex}}\alpha \textit {\_Z} g e +d g -e f , \operatorname {index} =2\right )}\right )}{2 e}\right )}{2 f \underline {\hspace {1.25 ex}}\alpha }\right )\right )+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {1}{f x}-\frac {g \arctan \left (\frac {g x}{\sqrt {g f}}\right )}{f \sqrt {g f}}\right )\) | \(526\) |
Input:
int(ln(c*(e*x^2+d)^p)/x^2/(g*x^2+f),x,method=_RETURNVERBOSE)
Output:
(ln((e*x^2+d)^p)-p*ln(e*x^2+d))*(-1/f/x-g/f/(g*f)^(1/2)*arctan(g*x/(g*f)^( 1/2)))-p/f/x*ln(e*x^2+d)+2*p/f*e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+p*Sum (-1/2*(ln(x-_alpha)*ln(e*x^2+d)-2*e*(1/2*ln(x-_alpha)*(ln((RootOf(_Z^2*e*g +2*_Z*_alpha*e*g+d*g-e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e* g+d*g-e*f,index=1))+ln((RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2)-x +_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2)))/e+1/2*(dilog(( RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=1)-x+_alpha)/RootOf(_Z^2*e*g +2*_Z*_alpha*e*g+d*g-e*f,index=1))+dilog((RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+ d*g-e*f,index=2)-x+_alpha)/RootOf(_Z^2*e*g+2*_Z*_alpha*e*g+d*g-e*f,index=2 )))/e))/f/_alpha,_alpha=RootOf(_Z^2*g+f))+(1/2*I*Pi*csgn(I*(e*x^2+d)^p)*cs gn(I*c*(e*x^2+d)^p)^2-1/2*I*Pi*csgn(I*(e*x^2+d)^p)*csgn(I*c*(e*x^2+d)^p)*c sgn(I*c)-1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^3+1/2*I*Pi*csgn(I*c*(e*x^2+d)^p)^2 *csgn(I*c)+ln(c))*(-1/f/x-g/f/(g*f)^(1/2)*arctan(g*x/(g*f)^(1/2)))
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )} x^{2}} \,d x } \] Input:
integrate(log(c*(e*x^2+d)^p)/x^2/(g*x^2+f),x, algorithm="fricas")
Output:
integral(log((e*x^2 + d)^p*c)/(g*x^4 + f*x^2), x)
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\int \frac {\log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{x^{2} \left (f + g x^{2}\right )}\, dx \] Input:
integrate(ln(c*(e*x**2+d)**p)/x**2/(g*x**2+f),x)
Output:
Integral(log(c*(d + e*x**2)**p)/(x**2*(f + g*x**2)), x)
Exception generated. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(c*(e*x^2+d)^p)/x^2/(g*x^2+f),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(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )} x^{2}} \,d x } \] Input:
integrate(log(c*(e*x^2+d)^p)/x^2/(g*x^2+f),x, algorithm="giac")
Output:
integrate(log((e*x^2 + d)^p*c)/((g*x^2 + f)*x^2), x)
Timed out. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{x^2\,\left (g\,x^2+f\right )} \,d x \] Input:
int(log(c*(d + e*x^2)^p)/(x^2*(f + g*x^2)),x)
Output:
int(log(c*(d + e*x^2)^p)/(x^2*(f + g*x^2)), x)
\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^2 \left (f+g x^2\right )} \, dx=\int \frac {\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right )}{g \,x^{4}+f \,x^{2}}d x \] Input:
int(log(c*(e*x^2+d)^p)/x^2/(g*x^2+f),x)
Output:
int(log((d + e*x**2)**p*c)/(f*x**2 + g*x**4),x)