\(\int \frac {\log (c (d+e x^2)^p)}{x^3 (f+g x^2)^2} \, dx\) [352]

Optimal result

Integrand size = 25, antiderivative size = 251 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\frac {e p \log (x)}{d f^2}-\frac {e p \log \left (d+e x^2\right )}{2 d f^2}+\frac {e g p \log \left (d+e x^2\right )}{2 f^2 (e f-d g)}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 x^2}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{2 f^2 \left (f+g x^2\right )}-\frac {g \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^3}-\frac {e g p \log \left (f+g x^2\right )}{2 f^2 (e f-d g)}+\frac {g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{f^3}+\frac {g p \operatorname {PolyLog}\left (2,-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{f^3}-\frac {g p \operatorname {PolyLog}\left (2,1+\frac {e x^2}{d}\right )}{f^3} \] Output:

e*p*ln(x)/d/f^2-1/2*e*p*ln(e*x^2+d)/d/f^2+1/2*e*g*p*ln(e*x^2+d)/f^2/(-d*g+ 
e*f)-1/2*ln(c*(e*x^2+d)^p)/f^2/x^2-1/2*g*ln(c*(e*x^2+d)^p)/f^2/(g*x^2+f)-g 
*ln(-e*x^2/d)*ln(c*(e*x^2+d)^p)/f^3-1/2*e*g*p*ln(g*x^2+f)/f^2/(-d*g+e*f)+g 
*ln(c*(e*x^2+d)^p)*ln(e*(g*x^2+f)/(-d*g+e*f))/f^3+g*p*polylog(2,-g*(e*x^2+ 
d)/(-d*g+e*f))/f^3-g*p*polylog(2,1+e*x^2/d)/f^3
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.89 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\frac {\frac {2 e f p \log (x)}{d}-\frac {e f p \log \left (d+e x^2\right )}{d}+\frac {e f g p \log \left (d+e x^2\right )}{e f-d g}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{x^2}-\frac {f g \log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}-2 g \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+\frac {e f g p \log \left (f+g x^2\right )}{-e f+d g}+2 g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )+2 g p \operatorname {PolyLog}\left (2,\frac {g \left (d+e x^2\right )}{-e f+d g}\right )-2 g p \operatorname {PolyLog}\left (2,1+\frac {e x^2}{d}\right )}{2 f^3} \] Input:

Integrate[Log[c*(d + e*x^2)^p]/(x^3*(f + g*x^2)^2),x]
 

Output:

((2*e*f*p*Log[x])/d - (e*f*p*Log[d + e*x^2])/d + (e*f*g*p*Log[d + e*x^2])/ 
(e*f - d*g) - (f*Log[c*(d + e*x^2)^p])/x^2 - (f*g*Log[c*(d + e*x^2)^p])/(f 
 + g*x^2) - 2*g*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p] + (e*f*g*p*Log[f + 
g*x^2])/(-(e*f) + d*g) + 2*g*Log[c*(d + e*x^2)^p]*Log[(e*(f + g*x^2))/(e*f 
 - d*g)] + 2*g*p*PolyLog[2, (g*(d + e*x^2))/(-(e*f) + d*g)] - 2*g*p*PolyLo 
g[2, 1 + (e*x^2)/d])/(2*f^3)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2925, 2863, 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.

\(\displaystyle \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx\)

\(\Big \downarrow \) 2925

\(\displaystyle \frac {1}{2} \int \frac {\log \left (c \left (e x^2+d\right )^p\right )}{x^4 \left (g x^2+f\right )^2}dx^2\)

\(\Big \downarrow \) 2863

\(\displaystyle \frac {1}{2} \int \left (\frac {2 \log \left (c \left (e x^2+d\right )^p\right ) g^2}{f^3 \left (g x^2+f\right )}+\frac {\log \left (c \left (e x^2+d\right )^p\right ) g^2}{f^2 \left (g x^2+f\right )^2}-\frac {2 \log \left (c \left (e x^2+d\right )^p\right ) g}{f^3 x^2}+\frac {\log \left (c \left (e x^2+d\right )^p\right )}{f^2 x^4}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {2 g \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{f^3}+\frac {2 g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{f^3}-\frac {g \log \left (c \left (d+e x^2\right )^p\right )}{f^2 \left (f+g x^2\right )}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f^2 x^2}+\frac {2 g p \operatorname {PolyLog}\left (2,-\frac {g \left (e x^2+d\right )}{e f-d g}\right )}{f^3}-\frac {2 g p \operatorname {PolyLog}\left (2,\frac {e x^2}{d}+1\right )}{f^3}+\frac {e g p \log \left (d+e x^2\right )}{f^2 (e f-d g)}-\frac {e g p \log \left (f+g x^2\right )}{f^2 (e f-d g)}+\frac {e p \log \left (x^2\right )}{d f^2}-\frac {e p \log \left (d+e x^2\right )}{d f^2}\right )\)

Input:

Int[Log[c*(d + e*x^2)^p]/(x^3*(f + g*x^2)^2),x]
 

Output:

((e*p*Log[x^2])/(d*f^2) - (e*p*Log[d + e*x^2])/(d*f^2) + (e*g*p*Log[d + e* 
x^2])/(f^2*(e*f - d*g)) - Log[c*(d + e*x^2)^p]/(f^2*x^2) - (g*Log[c*(d + e 
*x^2)^p])/(f^2*(f + g*x^2)) - (2*g*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p]) 
/f^3 - (e*g*p*Log[f + g*x^2])/(f^2*(e*f - d*g)) + (2*g*Log[c*(d + e*x^2)^p 
]*Log[(e*(f + g*x^2))/(e*f - d*g)])/f^3 + (2*g*p*PolyLog[2, -((g*(d + e*x^ 
2))/(e*f - d*g))])/f^3 - (2*g*p*PolyLog[2, 1 + (e*x^2)/d])/f^3)/2
 

Defintions of rubi rules used

rule 2009

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

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

rule 2925

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Si 
mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], 
x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer 
Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 
] || IGtQ[q, 0])
 

Maple [C] (warning: unable to verify)

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

Time = 9.01 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.13

method	result	size
parts	\(-\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2 f^{2} x^{2}}-\frac {2 \ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) g \ln \left (x \right )}{f^{3}}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) g \ln \left (g \,x^{2}+f \right )}{f^{3}}-\frac {g \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2 f^{2} \left (g \,x^{2}+f \right )}-e p \left (-\frac {4 g \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )+\ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )\right )}{2 e}+\frac {\operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )+\operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e}\right )}{f^{3}}+\frac {g \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (e \,\textit {\_Z}^{2}+d \right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (g \,x^{2}+f \right )-\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 )-\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 )\right )\right )}{f^{3} e}-\frac {-\frac {\left (2 d g -e f \right ) \ln \left (e \,x^{2}+d \right )}{2 d \left (d g -e f \right )}+\frac {\ln \left (x \right )}{d}+\frac {g \ln \left (g \,x^{2}+f \right )}{2 d g -2 e f}}{f^{2}}\right )\)	\(535\)
risch	\(\text {Expression too large to display}\)	\(732\)

Input:

int(ln(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2*ln(c*(e*x^2+d)^p)/f^2/x^2-2*ln(c*(e*x^2+d)^p)/f^3*g*ln(x)+ln(c*(e*x^2 
+d)^p)*g/f^3*ln(g*x^2+f)-1/2*g*ln(c*(e*x^2+d)^p)/f^2/(g*x^2+f)-e*p*(-4*g/f 
^3*(1/2*ln(x)*(ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+ln((e*x+(-d*e)^(1/2))/ 
(-d*e)^(1/2)))/e+1/2*(dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+dilog((e*x+( 
-d*e)^(1/2))/(-d*e)^(1/2)))/e)+g/f^3/e*sum(ln(x-_alpha)*ln(g*x^2+f)-ln(x-_ 
alpha)*(ln((RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1)-x+_alpha)/Roo 
tOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1))+ln((RootOf(_Z^2*e*g+2*_Z*_a 
lpha*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)))-dilog((RootOf(_Z^2*e*g+2*_Z*_alpha*e*g-d*g+e*f,index=1)-x+_al 
pha)/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)),_alpha=RootOf(_Z^2*e+d))-1/f^2*(-1/2*(2*d*g-e*f)/d/ 
(d*g-e*f)*ln(e*x^2+d)+1/d*ln(x)+1/2*g/(d*g-e*f)*ln(g*x^2+f)))
 

Fricas [F]

\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2} x^{3}} \,d x } \] Input:

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="fricas")
 

Output:

integral(log((e*x^2 + d)^p*c)/(g^2*x^7 + 2*f*g*x^5 + f^2*x^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(ln(c*(e*x**2+d)**p)/x**3/(g*x**2+f)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.18 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=-\frac {1}{2} \, {\left (f {\left (\frac {e \log \left (e x^{2} + d\right )}{d e f^{3} - d^{2} f^{2} g} - \frac {g \log \left (g x^{2} + f\right )}{e f^{4} - d f^{3} g} - \frac {\log \left (x^{2}\right )}{d f^{3}}\right )} - 2 \, g {\left (\frac {\log \left (e x^{2} + d\right )}{e f^{3} - d f^{2} g} - \frac {\log \left (g x^{2} + f\right )}{e f^{3} - d f^{2} g}\right )} - \frac {2 \, {\left (2 \, \log \left (\frac {e x^{2}}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x^{2}}{d}\right )\right )} g}{e f^{3}} + \frac {2 \, {\left (\log \left (g x^{2} + f\right ) \log \left (-\frac {e g x^{2} + e f}{e f - d g} + 1\right ) + {\rm Li}_2\left (\frac {e g x^{2} + e f}{e f - d g}\right )\right )} g}{e f^{3}}\right )} e p - \frac {1}{2} \, {\left (\frac {2 \, g x^{2} + f}{f^{2} g x^{4} + f^{3} x^{2}} - \frac {2 \, g \log \left (g x^{2} + f\right )}{f^{3}} + \frac {2 \, g \log \left (x^{2}\right )}{f^{3}}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \] Input:

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="maxima")
 

Output:

-1/2*(f*(e*log(e*x^2 + d)/(d*e*f^3 - d^2*f^2*g) - g*log(g*x^2 + f)/(e*f^4 
- d*f^3*g) - log(x^2)/(d*f^3)) - 2*g*(log(e*x^2 + d)/(e*f^3 - d*f^2*g) - l 
og(g*x^2 + f)/(e*f^3 - d*f^2*g)) - 2*(2*log(e*x^2/d + 1)*log(x) + dilog(-e 
*x^2/d))*g/(e*f^3) + 2*(log(g*x^2 + f)*log(-(e*g*x^2 + e*f)/(e*f - d*g) + 
1) + dilog((e*g*x^2 + e*f)/(e*f - d*g)))*g/(e*f^3))*e*p - 1/2*((2*g*x^2 + 
f)/(f^2*g*x^4 + f^3*x^2) - 2*g*log(g*x^2 + f)/f^3 + 2*g*log(x^2)/f^3)*log( 
(e*x^2 + d)^p*c)
 

Giac [F]

\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int { \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2} x^{3}} \,d x } \] Input:

integrate(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x, algorithm="giac")
 

Output:

integrate(log((e*x^2 + d)^p*c)/((g*x^2 + f)^2*x^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int \frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{x^3\,{\left (g\,x^2+f\right )}^2} \,d x \] Input:

int(log(c*(d + e*x^2)^p)/(x^3*(f + g*x^2)^2),x)
 

Output:

int(log(c*(d + e*x^2)^p)/(x^3*(f + g*x^2)^2), x)
 

Reduce [F]

\[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )^2} \, dx=\int \frac {\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right )}{g^{2} x^{7}+2 f g \,x^{5}+f^{2} x^{3}}d x \] Input:

int(log(c*(e*x^2+d)^p)/x^3/(g*x^2+f)^2,x)
                                                                                    
                                                                                    
 

Output:

int(log((d + e*x**2)**p*c)/(f**2*x**3 + 2*f*g*x**5 + g**2*x**7),x)