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

Optimal. Leaf size=195 \[ \frac{d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}+\frac{\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p^2}-\frac{4 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}-\frac{x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[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])

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Rubi [A]  time = 0.380937, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {2454, 2400, 2399, 2389, 2300, 2178, 2390, 2310} \[ \frac{d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}+\frac{\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p^2}-\frac{4 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p^2}-\frac{x^6 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

Antiderivative was successfully verified.

[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 2454

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])

Rule 2400

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 2399

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 2389

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 2300

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 2178

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

Rule 2390

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 2310

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)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps

\begin{align*} \int \frac{x^8}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx &=\frac{1}{3} \operatorname{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{\operatorname{Subst}\left (\int \frac{x^2}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{p}+\frac{(2 d) \operatorname{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{\operatorname{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) \operatorname{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{\operatorname{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) \operatorname{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) \operatorname{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 ) \operatorname{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 \operatorname{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{\operatorname{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) \operatorname{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) \operatorname{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 ) \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.259922, size = 290, normalized size = 1.49 \[ \frac{\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \log \left (c \left (d+e x^3\right )^p\right ) \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )+3 d^2 \log \left (c \left (d+e x^3\right )^p\right ) \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )+3 e^2 x^6 \log \left (c \left (d+e x^3\right )^p\right ) \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )-e^2 p x^6 \left (c \left (d+e x^3\right )^p\right )^{3/p}-4 d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{\frac{1}{p}} \log \left (c \left (d+e x^3\right )^p\right ) \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )+6 d e x^3 \log \left (c \left (d+e x^3\right )^p\right ) \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )\right )}{3 e^3 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \]

Antiderivative was successfully verified.

[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])

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Maple [F]  time = 3.678, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{8}}{ \left ( \ln \left ( c \left ( e{x}^{3}+d \right ) ^{p} \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{e x^{9} + d x^{6}}{3 \,{\left (e p \log \left ({\left (e x^{3} + d\right )}^{p}\right ) + e p \log \left (c\right )\right )}} + \int \frac{3 \, e x^{8} + 2 \, d x^{5}}{e p \log \left ({\left (e x^{3} + d\right )}^{p}\right ) + e p \log \left (c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.85092, size = 494, normalized size = 2.53 \begin{align*} -\frac{4 \,{\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac{1}{p}\right )} \logintegral \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}} \logintegral \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 )} \logintegral \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.29354, size = 667, normalized size = 3.42 \begin{align*} -\frac{{\left (x^{3} e + d\right )}^{3} p}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )}} + \frac{2 \,{\left (x^{3} e + d\right )}^{2} d p}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )}} - \frac{{\left (x^{3} e + d\right )} d^{2} p}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )}} + \frac{d^{2} p{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (x^{3} e + d\right )\right ) \log \left (x^{3} e + d\right )}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )} c^{\left (\frac{1}{p}\right )}} - \frac{4 \, d p{\rm Ei}\left (\frac{2 \, \log \left (c\right )}{p} + 2 \, \log \left (x^{3} e + d\right )\right ) \log \left (x^{3} e + d\right )}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )} c^{\frac{2}{p}}} + \frac{d^{2}{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (x^{3} e + d\right )\right ) \log \left (c\right )}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )} c^{\left (\frac{1}{p}\right )}} + \frac{p{\rm Ei}\left (\frac{3 \, \log \left (c\right )}{p} + 3 \, \log \left (x^{3} e + d\right )\right ) \log \left (x^{3} e + d\right )}{{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \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 (x^{3} e + d\right )\right ) \log \left (c\right )}{3 \,{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )} c^{\frac{2}{p}}} + \frac{{\rm Ei}\left (\frac{3 \, \log \left (c\right )}{p} + 3 \, \log \left (x^{3} e + d\right )\right ) \log \left (c\right )}{{\left (p^{3} e^{3} \log \left (x^{3} e + d\right ) + p^{2} e^{3} \log \left (c\right )\right )} c^{\frac{3}{p}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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