\(\int \frac {x^4 \log (c (d+e x^2)^p)}{(f+g x^2)^2} \, dx\) [353]

Optimal result

Integrand size = 25, antiderivative size = 802 \[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {2 p x}{g^2}+\frac {2 \sqrt {d} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e} g^2}+\frac {\sqrt {d} \sqrt {e} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{g^2 (e f-d g)}-\frac {e (-f)^{3/2} p \log \left (\sqrt {-f}-\sqrt {g} x\right )}{2 g^{5/2} (e f-d g)}-\frac {3 \sqrt {f} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{g^{5/2}}+\frac {3 \sqrt {f} 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 )}{2 g^{5/2}}+\frac {3 \sqrt {f} 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 )}{2 g^{5/2}}+\frac {e (-f)^{3/2} p \log \left (\sqrt {-f}+\sqrt {g} x\right )}{2 g^{5/2} (e f-d g)}+\frac {x \log \left (c \left (d+e x^2\right )^p\right )}{g^2}-\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 g^{5/2} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {f \log \left (c \left (d+e x^2\right )^p\right )}{4 g^{5/2} \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {3 \sqrt {f} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (d+e x^2\right )^p\right )}{2 g^{5/2}}+\frac {3 i \sqrt {f} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 g^{5/2}}-\frac {3 i \sqrt {f} 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 )}{4 g^{5/2}}-\frac {3 i \sqrt {f} 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 )}{4 g^{5/2}} \] Output:

-2*p*x/g^2+2*d^(1/2)*p*arctan(e^(1/2)*x/d^(1/2))/e^(1/2)/g^2+d^(1/2)*e^(1/ 
2)*f*p*arctan(e^(1/2)*x/d^(1/2))/g^2/(-d*g+e*f)-1/2*e*(-f)^(3/2)*p*ln((-f) 
^(1/2)-g^(1/2)*x)/g^(5/2)/(-d*g+e*f)-3*f^(1/2)*p*arctan(g^(1/2)*x/f^(1/2)) 
*ln(2*f^(1/2)/(f^(1/2)-I*g^(1/2)*x))/g^(5/2)+3/2*f^(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))/g^(5/2)+3/2*f^(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))/g^(5/2)+1/2*e*(-f)^(3/2) 
*p*ln((-f)^(1/2)+g^(1/2)*x)/g^(5/2)/(-d*g+e*f)+x*ln(c*(e*x^2+d)^p)/g^2-1/4 
*f*ln(c*(e*x^2+d)^p)/g^(5/2)/((-f)^(1/2)-g^(1/2)*x)+1/4*f*ln(c*(e*x^2+d)^p 
)/g^(5/2)/((-f)^(1/2)+g^(1/2)*x)-3/2*f^(1/2)*arctan(g^(1/2)*x/f^(1/2))*ln( 
c*(e*x^2+d)^p)/g^(5/2)+3/2*I*f^(1/2)*p*polylog(2,1-2*f^(1/2)/(f^(1/2)-I*g^ 
(1/2)*x))/g^(5/2)-3/4*I*f^(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)) 
/g^(5/2)-3/4*I*f^(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))/g^(5/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.07 (sec) , antiderivative size = 915, normalized size of antiderivative = 1.14 \[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx =\text {Too large to display} \] Input:

Integrate[(x^4*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]
 

Output:

(-8*Sqrt[g]*p*x + (8*Sqrt[d]*Sqrt[g]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e 
] + (2*Sqrt[-d]*Sqrt[e]*f*Sqrt[g]*p*Log[Sqrt[-d] - Sqrt[e]*x])/(-(e*f) + d 
*g) + (2*Sqrt[-d]*Sqrt[e]*f*Sqrt[g]*p*Log[Sqrt[-d] + Sqrt[e]*x])/(e*f - d* 
g) + (2*e*Sqrt[-f]*f*p*Log[Sqrt[-f] - Sqrt[g]*x])/(e*f - d*g) + (2*e*(-f)^ 
(3/2)*p*Log[Sqrt[-f] + Sqrt[g]*x])/(e*f - d*g) + (3*I)*Sqrt[f]*p*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]] + (3*I)*Sqrt[f]*p*Log[(Sqrt[g]*(Sqrt[-d] + Sqrt[e 
]*x))/((-I)*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sqrt[g])]*Log[1 - (I*Sqrt[g]*x)/Sqr 
t[f]] - (3*I)*Sqrt[f]*p*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]] - (3*I)*Sqrt[ 
f]*p*Log[(Sqrt[g]*(Sqrt[-d] + Sqrt[e]*x))/(I*Sqrt[e]*Sqrt[f] + Sqrt[-d]*Sq 
rt[g])]*Log[1 + (I*Sqrt[g]*x)/Sqrt[f]] + 4*Sqrt[g]*x*Log[c*(d + e*x^2)^p] 
- (f*Log[c*(d + e*x^2)^p])/(Sqrt[-f] - Sqrt[g]*x) + (f*Log[c*(d + e*x^2)^p 
])/(Sqrt[-f] + Sqrt[g]*x) - 6*Sqrt[f]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[c*(d 
 + e*x^2)^p] + (3*I)*Sqrt[f]*p*PolyLog[2, (Sqrt[e]*(Sqrt[f] - I*Sqrt[g]*x) 
)/(Sqrt[e]*Sqrt[f] - I*Sqrt[-d]*Sqrt[g])] + (3*I)*Sqrt[f]*p*PolyLog[2, (Sq 
rt[e]*(Sqrt[f] - I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g])] - ( 
3*I)*Sqrt[f]*p*PolyLog[2, (Sqrt[e]*(Sqrt[f] + I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[ 
f] - I*Sqrt[-d]*Sqrt[g])] - (3*I)*Sqrt[f]*p*PolyLog[2, (Sqrt[e]*(Sqrt[f] + 
 I*Sqrt[g]*x))/(Sqrt[e]*Sqrt[f] + I*Sqrt[-d]*Sqrt[g])])/(4*g^(5/2))
 

Rubi [A] (verified)

Time = 2.66 (sec) , antiderivative size = 802, 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.

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

\(\Big \downarrow \) 2926

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e p \log \left (\sqrt {-f}-\sqrt {g} x\right ) (-f)^{3/2}}{2 g^{5/2} (e f-d g)}+\frac {e p \log \left (\sqrt {g} x+\sqrt {-f}\right ) (-f)^{3/2}}{2 g^{5/2} (e f-d g)}-\frac {2 p x}{g^2}+\frac {\sqrt {d} \sqrt {e} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{g^2 (e f-d g)}+\frac {2 \sqrt {d} p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e} g^2}-\frac {3 \sqrt {f} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{g^{5/2}}+\frac {3 \sqrt {f} 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 )}{2 g^{5/2}}+\frac {3 \sqrt {f} p \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (\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 g^{5/2}}+\frac {x \log \left (c \left (e x^2+d\right )^p\right )}{g^2}-\frac {3 \sqrt {f} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right ) \log \left (c \left (e x^2+d\right )^p\right )}{2 g^{5/2}}-\frac {f \log \left (c \left (e x^2+d\right )^p\right )}{4 g^{5/2} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {f \log \left (c \left (e x^2+d\right )^p\right )}{4 g^{5/2} \left (\sqrt {g} x+\sqrt {-f}\right )}+\frac {3 i \sqrt {f} p \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {f}}{\sqrt {f}-i \sqrt {g} x}\right )}{2 g^{5/2}}-\frac {3 i \sqrt {f} 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 )}{4 g^{5/2}}-\frac {3 i \sqrt {f} 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 )}{4 g^{5/2}}\)

Input:

Int[(x^4*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]
 

Output:

(-2*p*x)/g^2 + (2*Sqrt[d]*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[e]*g^2) + ( 
Sqrt[d]*Sqrt[e]*f*p*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(g^2*(e*f - d*g)) - (e*(- 
f)^(3/2)*p*Log[Sqrt[-f] - Sqrt[g]*x])/(2*g^(5/2)*(e*f - d*g)) - (3*Sqrt[f] 
*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/g 
^(5/2) + (3*Sqrt[f]*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))])/(2*g^(5/2)) + (3*Sqrt[f]*p*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Lo 
g[(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))])/(2*g^(5/2)) + (e*(-f)^(3/2)*p*Log[Sq 
rt[-f] + Sqrt[g]*x])/(2*g^(5/2)*(e*f - d*g)) + (x*Log[c*(d + e*x^2)^p])/g^ 
2 - (f*Log[c*(d + e*x^2)^p])/(4*g^(5/2)*(Sqrt[-f] - Sqrt[g]*x)) + (f*Log[c 
*(d + e*x^2)^p])/(4*g^(5/2)*(Sqrt[-f] + Sqrt[g]*x)) - (3*Sqrt[f]*ArcTan[(S 
qrt[g]*x)/Sqrt[f]]*Log[c*(d + e*x^2)^p])/(2*g^(5/2)) + (((3*I)/2)*Sqrt[f]* 
p*PolyLog[2, 1 - (2*Sqrt[f])/(Sqrt[f] - I*Sqrt[g]*x)])/g^(5/2) - (((3*I)/4 
)*Sqrt[f]*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))])/g^(5/2) - ( 
((3*I)/4)*Sqrt[f]*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))])/g^( 
5/2)
 

Defintions of rubi rules used

rule 2009

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

rule 2926

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]
 

Maple [F]

\[\int \frac {x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{\left (g \,x^{2}+f \right )^{2}}d x\]

Input:

int(x^4*ln(c*(e*x^2+d)^p)/(g*x^2+f)^2,x)
 

Output:

int(x^4*ln(c*(e*x^2+d)^p)/(g*x^2+f)^2,x)
 

Fricas [F]

\[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {x^{4} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}} \,d x } \] Input:

integrate(x^4*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="fricas")
 

Output:

integral(x^4*log((e*x^2 + d)^p*c)/(g^2*x^4 + 2*f*g*x^2 + f^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**4*ln(c*(e*x**2+d)**p)/(g*x**2+f)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,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
 

Giac [F]

\[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\int { \frac {x^{4} \log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{{\left (g x^{2} + f\right )}^{2}} \,d x } \] Input:

integrate(x^4*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="giac")
 

Output:

integrate(x^4*log((e*x^2 + d)^p*c)/(g*x^2 + f)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\int \frac {x^4\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{{\left (g\,x^2+f\right )}^2} \,d x \] Input:

int((x^4*log(c*(d + e*x^2)^p))/(f + g*x^2)^2,x)
 

Output:

int((x^4*log(c*(d + e*x^2)^p))/(f + g*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^4 \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\text {too large to display} \] Input:

int(x^4*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x)
 

Output:

(2*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**3*f*g**4*p + 2*sqrt(e) 
*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**3*g**5*p*x**2 - 10*sqrt(e)*sqrt( 
d)*atan((e*x)/(sqrt(e)*sqrt(d)))*d**2*e*f**2*g**3*p - 10*sqrt(e)*sqrt(d)*a 
tan((e*x)/(sqrt(e)*sqrt(d)))*d**2*e*f*g**4*p*x**2 + 14*sqrt(e)*sqrt(d)*ata 
n((e*x)/(sqrt(e)*sqrt(d)))*d*e**2*f**3*g**2*p + 14*sqrt(e)*sqrt(d)*atan((e 
*x)/(sqrt(e)*sqrt(d)))*d*e**2*f**2*g**3*p*x**2 + 4*sqrt(g)*sqrt(f)*atan((g 
*x)/(sqrt(g)*sqrt(f)))*d**2*e**2*f**2*g**2*p + 4*sqrt(g)*sqrt(f)*atan((g*x 
)/(sqrt(g)*sqrt(f)))*d**2*e**2*f*g**3*p*x**2 - 8*sqrt(g)*sqrt(f)*atan((g*x 
)/(sqrt(g)*sqrt(f)))*d*e**3*f**3*g*p - 8*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt( 
g)*sqrt(f)))*d*e**3*f**2*g**2*p*x**2 - 2*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt( 
g)*sqrt(f)))*e**4*f**4*p - 2*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f))) 
*e**4*f**3*g*p*x**2 + 3*int(log((d + e*x**2)**p*c)/(d**2*f**2*g**2 + 2*d** 
2*f*g**3*x**2 + d**2*g**4*x**4 - 2*d*e*f**3*g - 4*d*e*f**2*g**2*x**2 - 2*d 
*e*f*g**3*x**4 + e**2*f**4 + 2*e**2*f**3*g*x**2 + e**2*f**2*g**2*x**4),x)* 
d**4*e**2*f**4*g**5 + 3*int(log((d + e*x**2)**p*c)/(d**2*f**2*g**2 + 2*d** 
2*f*g**3*x**2 + d**2*g**4*x**4 - 2*d*e*f**3*g - 4*d*e*f**2*g**2*x**2 - 2*d 
*e*f*g**3*x**4 + e**2*f**4 + 2*e**2*f**3*g*x**2 + e**2*f**2*g**2*x**4),x)* 
d**4*e**2*f**3*g**6*x**2 - 9*int(log((d + e*x**2)**p*c)/(d**2*f**2*g**2 + 
2*d**2*f*g**3*x**2 + d**2*g**4*x**4 - 2*d*e*f**3*g - 4*d*e*f**2*g**2*x**2 
- 2*d*e*f*g**3*x**4 + e**2*f**4 + 2*e**2*f**3*g*x**2 + e**2*f**2*g**2*x...