3.2.40 \(\int \frac {x^2}{\log (c (d+e x^3)^p)} \, dx\) [140]

Optimal. Leaf size=51 \[ \frac {\left (d+e x^3\right ) \left (c \left (d+e x^3\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (d+e x^3\right )^p\right )}{p}\right )}{3 e p} \]

[Out]

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))

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Rubi [A]
time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2504, 2436, 2337, 2209} \begin {gather*} \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 (e x^3+d\right )^p\right )}{p}\right )}{3 e p} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Log[c*(d + e*x^3)^p],x]

[Out]

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

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

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Mathematica [A]
time = 0.04, size = 51, normalized size = 1.00 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Log[c*(d + e*x^3)^p],x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.58, size = 272, normalized size = 5.33

method	result	size
risch	\(-\frac {\left (e \,x^{3}+d \right ) c^{-\frac {1}{p}} \left (\left (e \,x^{3}+d \right )^{p}\right )^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \left (-\mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\mathrm {csgn}\left (i c \right )\right ) \left (-\mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )+\mathrm {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right )\right )}{2 p}} \expIntegral \left (1, -\ln \left (e \,x^{3}+d \right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \right )-i \pi \mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )^{3}+i \pi \mathrm {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (e \,x^{3}+d \right )^{p}\right )-2 p \ln \left (e \,x^{3}+d \right )}{2 p}\right )}{3 e p}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/e/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)+cs
gn(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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 29, normalized size = 0.57 \begin {gather*} \frac {e^{\left (-1\right )} \operatorname {log\_integral}\left ({\left (x^{3} e + d\right )} c^{\left (\frac {1}{p}\right )}\right )}{3 \, c^{\left (\frac {1}{p}\right )} p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\log {\left (c \left (d + e x^{3}\right )^{p} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.75, size = 31, normalized size = 0.61 \begin {gather*} \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (x^{3} e + d\right )\right ) e^{\left (-1\right )}}{3 \, c^{\left (\frac {1}{p}\right )} p} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/log(c*(d + e*x^3)^p),x)

[Out]

int(x^2/log(c*(d + e*x^3)^p), x)

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