\(\int \frac {x^2}{\log ^2(c (d+e x^3)^p)} \, dx\) [150]

Optimal result

Integrand size = 18, antiderivative size = 83 \[ \int \frac {x^2}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e p^2}-\frac {d+e x^3}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2504, 2436, 2334, 2337, 2209} \[ \int \frac {x^2}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (e x^3+d\right )^p\right )}{p}\right )}{3 e p^2}-\frac {d+e x^3}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

Int[x^2/Log[c*(d + e*x^3)^p]^2,x]

[Out]

((d + e*x^3)*ExpIntegralEi[Log[c*(d + e*x^3)^p]/p])/(3*e*p^2*(c*(d + e*x^3)^p)^p^(-1)) - (d + e*x^3)/(3*e*p*Lo
g[c*(d + e*x^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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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 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 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 {1}{\log ^2\left (c (d+e x)^p\right )} \, dx,x,x^3\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{\log ^2\left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e} \\ & = -\frac {d+e x^3}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,d+e x^3\right )}{3 e p} \\ & = -\frac {d+e x^3}{3 e p \log \left (c \left (d+e x^3\right )^p\right )}+\frac {\left (\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 p^2} \\ & = \frac {\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 p^2}-\frac {d+e x^3}{3 e p \log \left (c \left (d+e x^3\right )^p\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \left (p \left (c \left (d+e x^3\right )^p\right )^{\frac {1}{p}}-\operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right ) \log \left (c \left (d+e x^3\right )^p\right )\right )}{3 e p^2 \log \left (c \left (d+e x^3\right )^p\right )} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 1.84 (sec) , antiderivative size = 421, normalized size of antiderivative = 5.07

method	result	size
risch	\(-\frac {2 \left (e \,x^{3}+d \right )}{3 \left (i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (e \,x^{3}+d \right )^{p}\right )\right ) p e}-\frac {\left (e \,x^{3}+d \right ) c^{-\frac {1}{p}} {\left (\left (e \,x^{3}+d \right )^{p}\right )}^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right )\right )}{2 p}} \operatorname {Ei}_{1}\left (-\ln \left (e \,x^{3}+d \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (e \,x^{3}+d \right )^{p}\right )-2 p \ln \left (e \,x^{3}+d \right )}{2 p}\right )}{3 p^{2} e}\)	\(421\)

[In]

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

[Out]

-2/3/(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))/p/e*(e*x^3+d)
-1/3/p^2/e*(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)*csg
n(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*P
i*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 = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {{\left (e p x^{3} + d p\right )} c^{\left (\frac {1}{p}\right )} - {\left (p \log \left (e x^{3} + d\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (e x^{3} + d\right )} c^{\left (\frac {1}{p}\right )}\right )}{3 \, {\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.70 \[ \int \frac {x^2}{\log ^2\left (c \left (d+e x^3\right )^p\right )} \, dx=-\frac {{\left (e x^{3} + d\right )} p}{3 \, {\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )}} + \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (e x^{3} + d\right )}{3 \, {\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} + \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (e x^{3} + d\right )\right ) \log \left (c\right )}{3 \, {\left (e p^{3} \log \left (e x^{3} + d\right ) + e p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^2/log(c*(d + e*x^3)^p)^2,x)

[Out]

int(x^2/log(c*(d + e*x^3)^p)^2, x)