Integrand size = 18, antiderivative size = 51 \[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e p} \] Output:
1/3*(e*x^3+d)*Ei(ln(c*(e*x^3+d)^p)/p)/e/p/((c*(e*x^3+d)^p)^(1/p))
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e p} \] Input:
Integrate[x^2/Log[c*(d + e*x^3)^p],x]
Output:
((d + e*x^3)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p])/(3*e*p*(c*(d + e*x^3)^ p)^p^(-1))
Time = 0.45 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2904, 2836, 2737, 2609}
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.
\(\displaystyle \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{3} \int \frac {1}{\log \left (c \left (e x^3+d\right )^p\right )}dx^3\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {\int \frac {1}{\log \left (c \left (e x^3+d\right )^p\right )}d\left (e x^3+d\right )}{3 e}\) |
\(\Big \downarrow \) 2737 |
\(\displaystyle \frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \int \frac {\left (c \left (e x^3+d\right )^p\right )^{\frac {1}{p}}}{x^3}d\log \left (c \left (e x^3+d\right )^p\right )}{3 e p}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e p}\) |
Input:
Int[x^2/Log[c*(d + e*x^3)^p],x]
Output:
((d + e*x^3)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p])/(3*e*p*(c*(d + e*x^3)^ p)^p^(-1))
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[((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 = 2.91 (sec) , antiderivative size = 272, normalized size of antiderivative = 5.33
method | result | size |
risch | \(-\frac {\left (e \,x^{3}+d \right ) {\left (\left (e \,x^{3}+d \right )^{p}\right )}^{-\frac {1}{p}} c^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right )\right )}{2 p}} \operatorname {expIntegral}_{1}\left (-\ln \left (e \,x^{3}+d \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (e \,x^{3}+d \right )^{p}\right )-2 p \ln \left (e \,x^{3}+d \right )}{2 p}\right )}{3 e p}\) | \(272\) |
Input:
int(x^2/ln(c*(e*x^3+d)^p),x,method=_RETURNVERBOSE)
Output:
-1/3/e/p*(e*x^3+d)*((e*x^3+d)^p)^(-1/p)*c^(-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)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right )}{3 \, c^{\left (\frac {1}{p}\right )} e p} \] Input:
integrate(x^2/log(c*(e*x^3+d)^p),x, algorithm="fricas")
Output:
1/3*log_integral((e*x^3 + d)*c^(1/p))/(c^(1/p)*e*p)
\[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^{2}}{\log {\left (c \left (d + e x^{3}\right )^{p} \right )}}\, dx \] Input:
integrate(x**2/ln(c*(e*x**3+d)**p),x)
Output:
Integral(x**2/log(c*(d + e*x**3)**p), x)
\[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\int { \frac {x^{2}}{\log \left ({\left (e x^{3} + d\right )}^{p} c\right )} \,d x } \] Input:
integrate(x^2/log(c*(e*x^3+d)^p),x, algorithm="maxima")
Output:
integrate(x^2/log((e*x^3 + d)^p*c), x)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right )}{3 \, c^{\left (\frac {1}{p}\right )} e p} \] Input:
integrate(x^2/log(c*(e*x^3+d)^p),x, algorithm="giac")
Output:
1/3*Ei(log(c)/p + log(e*x^3 + d))/(c^(1/p)*e*p)
Timed out. \[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\int \frac {x^2}{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )} \,d x \] Input:
int(x^2/log(c*(d + e*x^3)^p),x)
Output:
int(x^2/log(c*(d + e*x^3)^p), x)
\[ \int \frac {x^2}{\log \left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {3 \left (\int \frac {x^{5}}{\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) d +\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) e \,x^{3}}d x \right ) e^{2} p +\mathrm {log}\left (\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right )\right ) d}{3 e p} \] Input:
int(x^2/log(c*(e*x^3+d)^p),x)
Output:
(3*int(x**5/(log((d + e*x**3)**p*c)*d + log((d + e*x**3)**p*c)*e*x**3),x)* e**2*p + log(log((d + e*x**3)**p*c))*d)/(3*e*p)