\(\int \frac {x^8}{\log (c (d+e x^3)^p)} \, dx\) [138]

Optimal result

Integrand size = 18, antiderivative size = 164 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^3 p}-\frac {2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^3 p}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^3 p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2504, 2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p}-\frac {2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e^3 p} \]

[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 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 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 \left (c (d+e x)^p\right )} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \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 ) \\ & = \frac {\text {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) \text {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 \text {Subst}\left (\int \frac {1}{\log \left (c (d+e x)^p\right )} \, dx,x,x^3\right )}{3 e^2} \\ & = \frac {\text {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) \text {Subst}\left (\int \frac {x}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e^3}+\frac {d^2 \text {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 ) \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 )}{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 ) \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}+\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 )}{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] (verified)

Time = 0.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \left (2 d \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )\right )\right )}{3 e^3 p} \]

[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))

Maple [C] (warning: unable to verify)

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

Time = 1.03 (sec) , antiderivative size = 823, normalized size of antiderivative = 5.02

method	result	size
risch	\(\text {Expression too large to display}\)	\(823\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {c^{\frac {2}{p}} d^{2} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) - 2 \, c^{\left (\frac {1}{p}\right )} d \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right ) + \operatorname {log\_integral}\left ({\left (e^{3} x^{9} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{3} + d^{3}\right )} c^{\frac {3}{p}}\right )}{3 \, c^{\frac {3}{p}} e^{3} p} \]

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^{2} {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right )}{3 \, c^{\left (\frac {1}{p}\right )} e^{3} p} - \frac {2 \, d {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right )}{3 \, c^{\frac {2}{p}} e^{3} p} + \frac {{\rm Ei}\left (\frac {3 \, \log \left (c\right )}{p} + 3 \, \log \left (e x^{3} + d\right )\right )}{3 \, c^{\frac {3}{p}} e^{3} p} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^8/log(c*(d + e*x^3)^p),x)

[Out]

int(x^8/log(c*(d + e*x^3)^p), x)