\(\int \frac {x^5}{\log ^2(c (d+e x^3)^p)} \, dx\) [149]

Optimal result

Integrand size = 18, antiderivative size = 141 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}+\frac {2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-2/p} \left (e p x^3 \left (c \left (d+e x^3\right )^p\right )^{2/p}+d \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )-2 \left (d+e x^3\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \] Input:

Integrate[x^5/Log[c*(d + e*x^3)^p]^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.37, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2904, 2847, 2836, 2737, 2609, 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.

\(\displaystyle \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx\)

\(\Big \downarrow \) 2904

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

\(\Big \downarrow \) 2847

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

\(\Big \downarrow \) 2836

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

\(\Big \downarrow \) 2737

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

\(\Big \downarrow \) 2609

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

\(\Big \downarrow \) 2846

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^5/Log[c*(d + e*x^3)^p]^2,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2609

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2737

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[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 2836

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : 
> Simp[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 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 2847

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] + (-Simp[(q + 1)/(b*n*(p + 1))   Int[( 
f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Simp[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] && Lt 
Q[p, -1] && GtQ[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 = 1.23 (sec) , antiderivative size = 1487, normalized size of antiderivative = 10.55

method	result	size
risch	\(\text {Expression too large to display}\)	\(1487\)

Input:

int(x^5/ln(c*(e*x^3+d)^p)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/3/p/e*x^3*(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 
))-2/3/p^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)*x^6-4/3/p^2/e*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*c 
sgn(I*c)+2*ln(c)+2*ln((e*x^3+d)^p)-2*p*ln(e*x^3+d))/p)*d*x^3-2/3/p^2/e^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)*d^2+1/3/p^2/e*d*c^(-1/p)*((e*x^3+d)^p)^(-1/p)*exp(1/2*I*P...
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {{\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (e^{2} p x^{6} + d e p x^{3}\right )} c^{\frac {2}{p}} - 2 \, {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right )}{3 \, {\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} \] Input:

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="fricas")
 

Output:

-1/3*((d*p*log(e*x^3 + d) + d*log(c))*c^(1/p)*log_integral((e*x^3 + d)*c^( 
1/p)) + (e^2*p*x^6 + d*e*p*x^3)*c^(2/p) - 2*(p*log(e*x^3 + d) + log(c))*lo 
g_integral((e^2*x^6 + 2*d*e*x^3 + d^2)*c^(2/p)))/((e^2*p^3*log(e*x^3 + d) 
+ e^2*p^2*log(c))*c^(2/p))
 

Sympy [F]

\[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{5}}{\log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}\, dx \] Input:

integrate(x**5/ln(c*(e*x**3+d)**p)**2,x)
 

Output:

Integral(x**5/log(c*(d + e*x**3)**p)**2, x)
 

Maxima [F]

\[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int { \frac {x^{5}}{\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}} \,d x } \] Input:

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="maxima")
 

Output:

-1/3*(e*x^6 + d*x^3)/(e*p*log((e*x^3 + d)^p) + e*p*log(c)) + integrate((2* 
e*x^5 + d*x^2)/(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. 313 vs. \(2 (137) = 274\).

Time = 0.13 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.22 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {1}{3} \, d {\left (\frac {{\left (e x^{3} + d\right )} p}{e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {\frac {{\left (e x^{3} + d\right )}^{2} p}{e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )} - \frac {2 \, p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} - \frac {2 \, {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}}}{3 \, e} \] Input:

integrate(x^5/log(c*(e*x^3+d)^p)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^5}{{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2} \,d x \] Input:

int(x^5/log(c*(d + e*x^3)^p)^2,x)
                                                                                    
                                                                                    
 

Output:

int(x^5/log(c*(d + e*x^3)^p)^2, x)
 

Reduce [F]

\[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{5}}{{\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right )}^{2}}d x \] Input:

int(x^5/log(c*(e*x^3+d)^p)^2,x)
 

Output:

int(x**5/log((d + e*x**3)**p*c)**2,x)