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

Optimal. Leaf size=164 \[ \frac{d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}+\frac{\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}-\frac{2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p} \]

[Out]

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

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Rubi [A]  time = 0.237425, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {2454, 2399, 2389, 2300, 2178, 2390, 2310} \[ \frac{d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}+\frac{\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}-\frac{2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p} \]

Antiderivative was successfully verified.

[In]

Int[x^8/Log[c*(d + e*x^3)^p],x]

[Out]

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

Mathematica [A]  time = 0.231072, size = 146, normalized size = 0.89 \[ \frac{\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \text{Ei}\left (\frac{\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \left (2 d \left (c \left (d+e x^3\right )^p\right )^{\frac{1}{p}} \text{Ei}\left (\frac{2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \text{Ei}\left (\frac{3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )\right )\right )}{3 e^3 p} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/Log[c*(d + e*x^3)^p],x]

[Out]

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

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Maple [F]  time = 0.621, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{8}}{\ln \left ( c \left ( e{x}^{3}+d \right ) ^{p} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{\log \left ({\left (e x^{3} + d\right )}^{p} 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.90462, size = 271, normalized size = 1.65 \begin{align*} \frac{c^{\frac{2}{p}} d^{2} \logintegral \left ({\left (e x^{3} + d\right )} c^{\left (\frac{1}{p}\right )}\right ) - 2 \, c^{\left (\frac{1}{p}\right )} d \logintegral \left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac{2}{p}}\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 \, c^{\frac{3}{p}} e^{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),x, algorithm="fricas")

[Out]

1/3*(c^(2/p)*d^2*log_integral((e*x^3 + d)*c^(1/p)) - 2*c^(1/p)*d*log_integral((e^2*x^6 + 2*d*e*x^3 + d^2)*c^(2
/p)) + log_integral((e^3*x^9 + 3*d*e^2*x^6 + 3*d^2*e*x^3 + d^3)*c^(3/p)))/(c^(3/p)*e^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),x)

[Out]

Timed out

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Giac [A]  time = 1.32653, size = 146, normalized size = 0.89 \begin{align*} \frac{d^{2}{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (x^{3} e + d\right )\right ) e^{\left (-3\right )}}{3 \, c^{\left (\frac{1}{p}\right )} p} - \frac{2 \, d{\rm Ei}\left (\frac{2 \, \log \left (c\right )}{p} + 2 \, \log \left (x^{3} e + d\right )\right ) e^{\left (-3\right )}}{3 \, c^{\frac{2}{p}} p} + \frac{{\rm Ei}\left (\frac{3 \, \log \left (c\right )}{p} + 3 \, \log \left (x^{3} e + d\right )\right ) e^{\left (-3\right )}}{3 \, c^{\frac{3}{p}} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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