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

Optimal result

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

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))
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-3/p} \left (d^2 \left (c \left (d+e x^3\right )^p\right )^{2/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \left (2 d \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )-\left (d+e x^3\right ) \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )\right )\right )}{3 e^3 p} \] Input:

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

Output:

((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))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2904, 2846, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{3} \int \frac {x^6}{\log \left (c \left (e x^3+d\right )^p\right )}dx^3\)

\(\Big \downarrow \) 2846

\(\displaystyle \frac {1}{3} \int \left (\frac {d^2}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}-\frac {2 \left (e x^3+d\right ) d}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}+\frac {\left (e x^3+d\right )^2}{e^2 \log \left (c \left (e x^3+d\right )^p\right )}\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {d^2 \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}+\frac {\left (d+e x^3\right )^3 \left (c \left (d+e x^3\right )^p\right )^{-3/p} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}-\frac {2 d \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^3 p}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2846

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) 
]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(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 2904

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

Maple [C] (warning: unable to verify)

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

Time = 1.29 (sec) , antiderivative size = 823, normalized size of antiderivative = 5.02

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

Input:

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

Output:

-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)+c 
sgn(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)*c 
sgn(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*c 
sgn(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)

Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {c^{\frac {2}{p}} d^{2} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) - 2 \, c^{\left (\frac {1}{p}\right )} d \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right ) + \operatorname {log\_integral}\left ({\left (e^{3} x^{9} + 3 \, d e^{2} x^{6} + 3 \, d^{2} e x^{3} + d^{3}\right )} c^{\frac {3}{p}}\right )}{3 \, c^{\frac {3}{p}} e^{3} p} \] Input:

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

Output:

1/3*(c^(2/p)*d^2*log_integral((e*x^3 + d)*c^(1/p)) - 2*c^(1/p)*d*log_integ 
ral((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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.66 \[ \int \frac {x^8}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {d^{2} {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right )}{3 \, c^{\left (\frac {1}{p}\right )} e^{3} p} - \frac {2 \, d {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right )}{3 \, c^{\frac {2}{p}} e^{3} p} + \frac {{\rm Ei}\left (\frac {3 \, \log \left (c\right )}{p} + 3 \, \log \left (e x^{3} + d\right )\right )}{3 \, c^{\frac {3}{p}} e^{3} p} \] Input:

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

Output:

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

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

Output:

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

Reduce [F]

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

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

Output:

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