\(\int \frac {\log (c (d+e x^2)^p)}{x^3 (f+g x^2)} \, dx\) [342]

Optimal result

Integrand size = 25, antiderivative size = 176 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )} \, dx=\frac {e p \log (x)}{d f}-\frac {e p \log \left (d+e x^2\right )}{2 d f}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 f x^2}-\frac {g \log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 f^2}+\frac {g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )}{2 f^2}+\frac {g p \operatorname {PolyLog}\left (2,-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{2 f^2}-\frac {g p \operatorname {PolyLog}\left (2,1+\frac {e x^2}{d}\right )}{2 f^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )} \, dx=\frac {e p \log (x)}{d f}-\frac {e p \log \left (d+e x^2\right )}{2 d f}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 f x^2}-\frac {g \left (\log \left (-\frac {e x^2}{d}\right ) \log \left (c \left (d+e x^2\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {d+e x^2}{d}\right )\right )}{2 f^2}+\frac {g \log \left (c \left (d+e x^2\right )^p\right ) \log \left (\frac {e \left (f+g x^2\right )}{e f-d g}\right )+g p \operatorname {PolyLog}\left (2,-\frac {g \left (d+e x^2\right )}{e f-d g}\right )}{2 f^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95, 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.

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

\(\Big \downarrow \) 2925

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((e*p*Log[x^2])/(d*f) - (e*p*Log[d + e*x^2])/(d*f) - Log[c*(d + e*x^2)^p]/ 
(f*x^2) - (g*Log[-((e*x^2)/d)]*Log[c*(d + e*x^2)^p])/f^2 + (g*Log[c*(d + e 
*x^2)^p]*Log[(e*(f + g*x^2))/(e*f - d*g)])/f^2 + (g*p*PolyLog[2, -((g*(d + 
 e*x^2))/(e*f - d*g))])/f^2 - (g*p*PolyLog[2, 1 + (e*x^2)/d])/f^2)/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 = 4.08 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.68

method	result	size
parts	\(-\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2 f \,x^{2}}-\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) g \ln \left (x \right )}{f^{2}}+\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) g \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-e p \left (\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 )}{2 f^{2} e}+\frac {\ln \left (e \,x^{2}+d \right )}{2 f d}-\frac {\ln \left (x \right )}{f d}-\frac {2 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^{2}}\right )\)	\(471\)
risch	\(\text {Expression too large to display}\)	\(617\)

Input:

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

Output:

-1/2*ln(c*(e*x^2+d)^p)/f/x^2-ln(c*(e*x^2+d)^p)*g/f^2*ln(x)+1/2*ln(c*(e*x^2 
+d)^p)*g/f^2*ln(g*x^2+f)-e*p*(1/2*g/f^2/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)/ 
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)))-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*_al 
pha*e*g-d*g+e*f,index=2)),_alpha=RootOf(_Z^2*e+d))+1/2/f/d*ln(e*x^2+d)-1/f 
/d*ln(x)-2*g/f^2*(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))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (d+e x^2\right )^p\right )}{x^3 \left (f+g x^2\right )} \, dx=\frac {1}{2} \, e p {\left (\frac {{\left (2 \, \log \left (\frac {e x^{2}}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x^{2}}{d}\right )\right )} g}{e f^{2}} - \frac {{\left (\log \left (g x^{2} + f\right ) \log \left (-\frac {e g x^{2} + e f}{e f - d g} + 1\right ) + {\rm Li}_2\left (\frac {e g x^{2} + e f}{e f - d g}\right )\right )} g}{e f^{2}} - \frac {\log \left (e x^{2} + d\right )}{d f} + \frac {2 \, \log \left (x\right )}{d f}\right )} + \frac {1}{2} \, {\left (\frac {g \log \left (g x^{2} + f\right )}{f^{2}} - \frac {g \log \left (x^{2}\right )}{f^{2}} - \frac {1}{f x^{2}}\right )} \log \left ({\left (e x^{2} + d\right )}^{p} c\right ) \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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