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

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 21, 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} \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p^2}-\frac {4 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}-\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

Int[x^8/Log[c*(d + e*x^3)^p]^2,x]

[Out]

(d^2*(d + e*x^3)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p])/(3*e^3*p^2*(c*(d + e*x^3)^p)^p^(-1)) - (4*d*(d + e*x^3
)^2*ExpIntegralEi[(2*Log[c*(d + e*x^3)^p])/p])/(3*e^3*p^2*(c*(d + e*x^3)^p)^(2/p)) + ((d + e*x^3)^3*ExpIntegra
lEi[(3*Log[c*(d + e*x^3)^p])/p])/(e^3*p^2*(c*(d + e*x^3)^p)^(3/p)) - (x^6*(d + e*x^3))/(3*e*p*Log[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^2}{\log ^2\left (c (d+e x)^p\right )} \, dx,x,x^3\right ) \\ & = -\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\text {Subst}\left (\int \frac {x^2}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{p}+\frac {(2 d) \text {Subst}\left (\int \frac {x}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e p} \\ & = -\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\text {Subst}\left (\int \left (\frac {d^2}{e^2 \log \left (c (d+e x)^p\right )}-\frac {2 d (d+e x)}{e^2 \log \left (c (d+e x)^p\right )}+\frac {(d+e x)^2}{e^2 \log \left (c (d+e x)^p\right )}\right ) \, dx,x,x^3\right )}{p}+\frac {(2 d) \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 e p} \\ & = -\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\text {Subst}\left (\int \frac {(d+e x)^2}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{e^2 p}+\frac {(2 d) \text {Subst}\left (\int \frac {d+e x}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e^2 p}-\frac {(2 d) \text {Subst}\left (\int \frac {d+e x}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{e^2 p}-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {1}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e^2 p}+\frac {d^2 \text {Subst}\left (\int \frac {1}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{e^2 p} \\ & = -\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\text {Subst}\left (\int \frac {x^2}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{e^3 p}+\frac {(2 d) \text {Subst}\left (\int \frac {x}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^3 p}-\frac {(2 d) \text {Subst}\left (\int \frac {x}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{e^3 p}-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^3 p}+\frac {d^2 \text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{e^3 p} \\ & = -\frac {x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\left (\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {3 x}{p}}}{x} \, dx,x,\log \left (c \left (d+e x^3\right )^p\right )\right )}{e^3 p^2}+\frac {\left (2 d \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^3 p^2}-\frac {\left (2 d \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 )}{e^3 p^2}-\frac {\left (2 d^2 \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^3 p^2}+\frac {\left (d^2 \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 )}{e^3 p^2} \\ & = \frac {d^2 \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^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} \text {Ei}\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} \text {Ei}\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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.49 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (-e^2 p x^6 \left (c \left (d+e x^3\right )^p\right )^{3/p}+d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )-4 d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+3 d^2 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+6 d e x^3 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )+3 e^2 x^6 \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^3 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

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

[Out]

((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*E
xpIntegralEi[(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])

Maple [C] (warning: unable to verify)

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

Time = 1.14 (sec) , antiderivative size = 2564, normalized size of antiderivative = 13.15

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

[In]

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

[Out]

-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)*csg
n(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*P
i*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*csg
n(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*csg
n(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)*cs
gn(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^(-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)*d*x^6-3/p^2/e^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)*d^2*x^3-1/p^2/e^3*((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)*d^3+4/3/p^2/e*d*((e*x^3+d)^p)^(-2/p)*c^(-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+8/3/p^2/
e^2*d^2*((e*x^3+d)^p)^(-2/p)*c^(-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^3+4/3/p^2/e^3*d^3*((e*x^3+d)^p)^(-2/p)*c^(-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)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.08 \[ \int \frac {x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {4 \, {\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right ) - {\left (d^{2} p \log \left (e x^{3} + d\right ) + d^{2} \log \left (c\right )\right )} c^{\frac {2}{p}} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (e^{3} p x^{9} + d e^{2} p x^{6}\right )} c^{\frac {3}{p}} - 3 \, {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e^{3} x^{9} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{3} + d^{3}\right )} c^{\frac {3}{p}}\right )}{3 \, {\left (e^{3} p^{3} \log \left (e x^{3} + d\right ) + e^{3} p^{2} \log \left (c\right )\right )} c^{\frac {3}{p}}} \]

[In]

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

[Out]

-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
*log(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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-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 = 0.33 (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}}} \]

[In]

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

[Out]

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

[In]

int(x^8/log(c*(d + e*x^3)^p)^2,x)

[Out]

int(x^8/log(c*(d + e*x^3)^p)^2, x)