\(\int \frac {x^5}{\log ^2(c (d+e x^3)^p)} \, dx\) [149]

Optimal result

Integrand size = 18, antiderivative size = 141 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}+\frac {2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2504, 2447, 2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

Int[x^5/Log[c*(d + e*x^3)^p]^2,x]

[Out]

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2437

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

Rule 2446

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(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 2447

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] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[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] && LtQ[p, -1] && GtQ[q, 0]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {x}{\log ^2\left (c (d+e x)^p\right )} \, dx,x,x^3\right ) \\ & = -\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {2 \text {Subst}\left (\int \frac {x}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 p}+\frac {d \text {Subst}\left (\int \frac {1}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e p} \\ & = -\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {2 \text {Subst}\left (\int \left (-\frac {d}{e \log \left (c (d+e x)^p\right )}+\frac {d+e x}{e \log \left (c (d+e x)^p\right )}\right ) \, dx,x,x^3\right )}{3 p}+\frac {d \text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^2 p} \\ & = -\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {2 \text {Subst}\left (\int \frac {d+e x}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e p}-\frac {(2 d) \text {Subst}\left (\int \frac {1}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e p}+\frac {\left (d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{p}}}{x} \, dx,x,\log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2} \\ & = \frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {2 \text {Subst}\left (\int \frac {x}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^2 p}-\frac {(2 d) \text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^2 p} \\ & = \frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\left (2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {2 x}{p}}}{x} \, dx,x,\log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2}-\frac {\left (2 d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{p}}}{x} \, dx,x,\log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2} \\ & = -\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}+\frac {2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \text {Ei}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-2/p} \left (e p x^3 \left (c \left (d+e x^3\right )^p\right )^{2/p}+d \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )-2 \left (d+e x^3\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

Integrate[x^5/Log[c*(d + e*x^3)^p]^2,x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.96 (sec) , antiderivative size = 1487, normalized size of antiderivative = 10.55

method	result	size
risch	\(\text {Expression too large to display}\)	\(1487\)

[In]

int(x^5/ln(c*(e*x^3+d)^p)^2,x,method=_RETURNVERBOSE)

[Out]

-2/3/p/e*x^3*(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))-2/3/p^2*c^(-2/p)*((e*x^3+d)^p)^(-2/p)*exp(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,-2*ln(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)-2*p*ln(e*x^3+d))/p)*x^6-4/3/p^2/e*c^(-2/p)*((e*x^3+d)^p)^(-2/p
)*exp(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,-2*ln(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)-2*p*ln(e*x^3+d))/p)*d*x^3-2/3/p^2/e^2*c^(-2/p)*((e*x^3+d)^p)^(-2/p)*exp(I*Pi*csgn(I*c*(e*x^3+d)^p)*(-cs
gn(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,-2*ln(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)-2*p*ln(e*x^3+d))/p)*d^2+1/3/p^
2/e*d*c^(-1/p)*((e*x^3+d)^p)^(-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))*(-cs
gn(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^2*d^2*c^(-1/p)*((e*x^3+d)^p)^
(-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)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {{\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (e^{2} p x^{6} + d e p x^{3}\right )} c^{\frac {2}{p}} - 2 \, {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right )}{3 \, {\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} \]

[In]

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="fricas")

[Out]

-1/3*((d*p*log(e*x^3 + d) + d*log(c))*c^(1/p)*log_integral((e*x^3 + d)*c^(1/p)) + (e^2*p*x^6 + d*e*p*x^3)*c^(2
/p) - 2*(p*log(e*x^3 + d) + log(c))*log_integral((e^2*x^6 + 2*d*e*x^3 + d^2)*c^(2/p)))/((e^2*p^3*log(e*x^3 + d
) + e^2*p^2*log(c))*c^(2/p))

Sympy [F]

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

[In]

integrate(x**5/ln(c*(e*x**3+d)**p)**2,x)

[Out]

Integral(x**5/log(c*(d + e*x**3)**p)**2, x)

Maxima [F]

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

[In]

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="maxima")

[Out]

-1/3*(e*x^6 + d*x^3)/(e*p*log((e*x^3 + d)^p) + e*p*log(c)) + integrate((2*e*x^5 + d*x^2)/(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. 313 vs. \(2 (137) = 274\).

Time = 0.31 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.22 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {1}{3} \, d {\left (\frac {{\left (e x^{3} + d\right )} p}{e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {\frac {{\left (e x^{3} + d\right )}^{2} p}{e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )} - \frac {2 \, p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} - \frac {2 \, {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}}}{3 \, e} \]

[In]

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^5}{{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2} \,d x \]

[In]

int(x^5/log(c*(d + e*x^3)^p)^2,x)

[Out]

int(x^5/log(c*(d + e*x^3)^p)^2, x)