Integrand size = 18, antiderivative size = 141 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}+\frac {2 \left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e^2 p^2}-\frac {x^3 \left (d+e x^3\right )}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]
-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)
Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-2/p} \left (e p x^3 \left (c \left (d+e x^3\right )^p\right )^{2/p}+d \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )-2 \left (d+e x^3\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e^2 p^2 \log \left (c \left (d+e x^3\right )^p\right )} \]
-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])
Time = 0.53 (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 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^2 p^2}+\frac {2 \left (\frac {\left (d+e x^3\right )^2 \left (c \left (d+e x^3\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^2 p}-\frac {d \left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{e^2 p}\right )}{p}-\frac {x^3 \left (d+e x^3\right )}{e p \log \left (c \left (d+e x^3\right )^p\right )}\right )\) |
((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
3.2.49.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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]
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])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.96 (sec) , antiderivative size = 1487, normalized size of antiderivative = 10.55
-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...
Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {{\left (d p \log \left (e x^{3} + d\right ) + d \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (e^{2} p x^{6} + d e p x^{3}\right )} c^{\frac {2}{p}} - 2 \, {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{6} + 2 \, d e x^{3} + d^{2}\right )} c^{\frac {2}{p}}\right )}{3 \, {\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} \]
-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))
\[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{5}}{\log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}\, dx \]
\[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int { \frac {x^{5}}{\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}} \,d x } \]
-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)
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (137) = 274\).
Time = 0.31 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.22 \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {1}{3} \, d {\left (\frac {{\left (e x^{3} + d\right )} p}{e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e^{2} p^{3} \log \left (e x^{3} + d\right ) + e^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {\frac {{\left (e x^{3} + d\right )}^{2} p}{e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )} - \frac {2 \, p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} - \frac {2 \, {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{{\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}}}{3 \, e} \]
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
Timed out. \[ \int \frac {x^5}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^5}{{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2} \,d x \]