\(\int \frac {x^8}{\log ^2(c (d+e x^3)^p)} \, dx\) [148]

Optimal result

Integrand size = 18, antiderivative size = 195 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^3 p^2}-\frac {4 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^3 p^2}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{e^3 p^2}-\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \] Output:

1/3*d^2*(e*x^3+d)*Ei(ln(c*(e*x^3+d)^p)/p)/e^3/p^2/((c*(e*x^3+d)^p)^(1/p))- 
4/3*d*(e*x^3+d)^2*Ei(2*ln(c*(e*x^3+d)^p)/p)/e^3/p^2/((c*(e*x^3+d)^p)^(2/p) 
)+(e*x^3+d)^3*Ei(3*ln(c*(e*x^3+d)^p)/p)/e^3/p^2/((c*(e*x^3+d)^p)^(3/p))-1/ 
3*x^6*(e*x^3+d)/e/p/ln(c*(e*x^3+d)^p)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.49 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (-e^2 p x^6 \left (c \left (d+e x^3\right )^p\right )^{3/p}+d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )-4 d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+3 d^2 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+6 d e x^3 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+3 e^2 x^6 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^3 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \] Input:

Integrate[x^8/Log[c*(d + e*x^3)^p]^2,x]
 

Output:

((d + e*x^3)*(-(e^2*p*x^6*(c*(d + e*x^3)^p)^(3/p)) + d^2*(c*(d + e*x^3)^p) 
^(2/p)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p]*Log[c*(d + e*x^3)^p] - 4*d*(d 
 + e*x^3)*(c*(d + e*x^3)^p)^p^(-1)*ExpIntegralEi[(2*Log[c*(d + e*x^3)^p])/ 
p]*Log[c*(d + e*x^3)^p] + 3*d^2*ExpIntegralEi[(3*Log[c*(d + e*x^3)^p])/p]* 
Log[c*(d + e*x^3)^p] + 6*d*e*x^3*ExpIntegralEi[(3*Log[c*(d + e*x^3)^p])/p] 
*Log[c*(d + e*x^3)^p] + 3*e^2*x^6*ExpIntegralEi[(3*Log[c*(d + e*x^3)^p])/p 
]*Log[c*(d + e*x^3)^p]))/(3*e^3*p^2*(c*(d + e*x^3)^p)^(3/p)*Log[c*(d + e*x 
^3)^p])
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.58, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2904, 2847, 2846, 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.

\(\displaystyle \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx\)

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{3} \int \frac {x^6}{\log ^2\left (c \left (e x^3+d\right )^p\right )}dx^3\)

\(\Big \downarrow \) 2847

\(\displaystyle \frac {1}{3} \left (\frac {2 d \int \frac {x^3}{\log \left (c \left (e x^3+d\right )^p\right )}dx^3}{e p}+\frac {3 \int \frac {x^6}{\log \left (c \left (e x^3+d\right )^p\right )}dx^3}{p}-\frac {x^6 \left (d+e x^3\right )}{e p \log \left (c \left (d+e x^3\right )^p\right )}\right )\)

\(\Big \downarrow \) 2846

\(\displaystyle \frac {1}{3} \left (\frac {3 \int \left (\frac {d^2}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}-\frac {2 \left (e x^3+d\right ) d}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}+\frac {\left (e x^3+d\right )^2}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}\right )dx^3}{p}+\frac {2 d \int \left (\frac {e x^3+d}{e \log \left (c \left (e x^3+d\right )^p\right )}-\frac {d}{e \log \left (c \left (e x^3+d\right )^p\right )}\right )dx^3}{e p}-\frac {x^6 \left (d+e x^3\right )}{e p \log \left (c \left (d+e x^3\right )^p\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {3 \left (\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}-\frac {2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}\right )}{p}+\frac {2 d \left (\frac {\left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^2 p}-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^2 p}\right )}{e p}-\frac {x^6 \left (d+e x^3\right )}{e p \log \left (c \left (d+e x^3\right )^p\right )}\right )\)

Input:

Int[x^8/Log[c*(d + e*x^3)^p]^2,x]
 

Output:

((2*d*(-((d*(d + e*x^3)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p])/(e^2*p*(c*( 
d + e*x^3)^p)^p^(-1))) + ((d + e*x^3)^2*ExpIntegralEi[(2*Log[c*(d + e*x^3) 
^p])/p])/(e^2*p*(c*(d + e*x^3)^p)^(2/p))))/(e*p) + (3*((d^2*(d + e*x^3)*Ex 
pIntegralEi[Log[c*(d + e*x^3)^p]/p])/(e^3*p*(c*(d + e*x^3)^p)^p^(-1)) - (2 
*d*(d + e*x^3)^2*ExpIntegralEi[(2*Log[c*(d + e*x^3)^p])/p])/(e^3*p*(c*(d + 
 e*x^3)^p)^(2/p)) + ((d + e*x^3)^3*ExpIntegralEi[(3*Log[c*(d + e*x^3)^p])/ 
p])/(e^3*p*(c*(d + e*x^3)^p)^(3/p))))/p - (x^6*(d + e*x^3))/(e*p*Log[c*(d 
+ e*x^3)^p]))/3
 

Defintions of rubi rules used

rule 2009

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

rule 2846

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) 
]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q/(a + b*Log[c*(d + e* 
x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] & 
& IGtQ[q, 0]
 

rule 2847

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)*(f + g*x)^q*((a + b*Log[c*(d + e 
*x)^n])^(p + 1)/(b*e*n*(p + 1))), x] + (-Simp[(q + 1)/(b*n*(p + 1))   Int[( 
f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Simp[q*((e*f - d*g) 
/(b*e*n*(p + 1)))   Int[(f + g*x)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1 
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && Lt 
Q[p, -1] && GtQ[q, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, 
 x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & 
&  !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
 

Maple [C] (warning: unable to verify)

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

Time = 1.42 (sec) , antiderivative size = 2564, normalized size of antiderivative = 13.15

method	result	size
risch	\(\text {Expression too large to display}\)	\(2564\)

Input:

int(x^8/ln(c*(e*x^3+d)^p)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/3/p/e*x^6*(e*x^3+d)/(I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-I 
*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-I*Pi*csgn(I*c*(e*x 
^3+d)^p)^3+I*Pi*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c)+2*ln((e*x^3+d)^p 
))-1/3/p^2/e^2*d^2*((e*x^3+d)^p)^(-1/p)*c^(-1/p)*exp(1/2*I*Pi*csgn(I*c*(e* 
x^3+d)^p)*(-csgn(I*c*(e*x^3+d)^p)+csgn(I*c))*(-csgn(I*c*(e*x^3+d)^p)+csgn( 
I*(e*x^3+d)^p))/p)*Ei(1,-ln(e*x^3+d)-1/2*(I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I* 
c*(e*x^3+d)^p)^2-I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)- 
I*Pi*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c 
)+2*ln((e*x^3+d)^p)-2*p*ln(e*x^3+d))/p)*x^3-1/3/p^2/e^3*d^3*((e*x^3+d)^p)^ 
(-1/p)*c^(-1/p)*exp(1/2*I*Pi*csgn(I*c*(e*x^3+d)^p)*(-csgn(I*c*(e*x^3+d)^p) 
+csgn(I*c))*(-csgn(I*c*(e*x^3+d)^p)+csgn(I*(e*x^3+d)^p))/p)*Ei(1,-ln(e*x^3 
+d)-1/2*(I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-I*Pi*csgn(I*(e*x 
^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-I*Pi*csgn(I*c*(e*x^3+d)^p)^3+I*Pi 
*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c)+2*ln((e*x^3+d)^p)-2*p*ln(e*x^3+ 
d))/p)-1/p^2*((e*x^3+d)^p)^(-3/p)*c^(-3/p)*exp(3/2*I*Pi*csgn(I*c*(e*x^3+d) 
^p)*(-csgn(I*c*(e*x^3+d)^p)+csgn(I*c))*(-csgn(I*c*(e*x^3+d)^p)+csgn(I*(e*x 
^3+d)^p))/p)*Ei(1,-3*ln(e*x^3+d)-3/2*(I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e 
*x^3+d)^p)^2-I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-I*Pi 
*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c)+2* 
ln((e*x^3+d)^p)-2*p*ln(e*x^3+d))/p)*x^9-3/p^2/e*((e*x^3+d)^p)^(-3/p)*c^...
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.08 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {4 \, {\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right ) - {\left (d^{2} p \log \left (e x^{3} + d\right ) + d^{2} \log \left (c\right )\right )} c^{\frac {2}{p}} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (e^{3} p x^{9} + d e^{2} p x^{6}\right )} c^{\frac {3}{p}} - 3 \, {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e^{3} x^{9} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{3} + d^{3}\right )} c^{\frac {3}{p}}\right )}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {3}{p}}} \] Input:

integrate(x^8/log(c*(e*x^3+d)^p)^2,x, algorithm="fricas")
 

Output:

-1/3*(4*(d*p*log(e*x^3 + d) + d*log(c))*c^(1/p)*log_integral((e^2*x^6 + 2* 
d*e*x^3 + d^2)*c^(2/p)) - (d^2*p*log(e*x^3 + d) + d^2*log(c))*c^(2/p)*log_ 
integral((e*x^3 + d)*c^(1/p)) + (e^3*p*x^9 + d*e^2*p*x^6)*c^(3/p) - 3*(p*l 
og(e*x^3 + d) + log(c))*log_integral((e^3*x^9 + 3*d*e^2*x^6 + 3*d^2*e*x^3 
+ d^3)*c^(3/p)))/((e^3*p^3*log(e*x^3 + d) + e^3*p^2*log(c))*c^(3/p))
 

Sympy [F]

\[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{8}}{\log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}\, dx \] Input:

integrate(x**8/ln(c*(e*x**3+d)**p)**2,x)
 

Output:

Integral(x**8/log(c*(d + e*x**3)**p)**2, x)
 

Maxima [F]

\[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int { \frac {x^{8}}{\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}} \,d x } \] Input:

integrate(x^8/log(c*(e*x^3+d)^p)^2,x, algorithm="maxima")
 

Output:

-1/3*(e*x^9 + d*x^6)/(e*p*log((e*x^3 + d)^p) + e*p*log(c)) + integrate((3* 
e*x^8 + 2*d*x^5)/(e*p*log((e*x^3 + d)^p) + e*p*log(c)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (193) = 386\).

Time = 0.15 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.50 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {1}{3} \, d^{2} {\left (\frac {{\left (e x^{3} + d\right )} p}{e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {{\left (e x^{3} + d\right )}^{3} p}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )}} + \frac {2 \, {\left (e x^{3} + d\right )}^{2} d p}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )}} - \frac {4 \, d p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} + \frac {p {\rm Ei}\left (\frac {3 \, \log \left (c\right )}{p} + 3 \, \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {3}{p}}} - \frac {4 \, d {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} + \frac {{\rm Ei}\left (\frac {3 \, \log \left (c\right )}{p} + 3 \, \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {3}{p}}} \] Input:

integrate(x^8/log(c*(e*x^3+d)^p)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^8}{{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2} \,d x \] Input:

int(x^8/log(c*(d + e*x^3)^p)^2,x)
 

Output:

int(x^8/log(c*(d + e*x^3)^p)^2, x)
 

Reduce [F]

\[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{8}}{{\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right )}^{2}}d x \] Input:

int(x^8/log(c*(e*x^3+d)^p)^2,x)
 

Output:

int(x**8/log((d + e*x**3)**p*c)**2,x)