3.51.23 \(\int \frac {e^{-e^2} (-3 x^2+e^{e^2} (4+x^2)+8 e^{e^2} \log (x)+e^{e^2} (-x^3+e^x x^3) \log ^2(x))}{x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ -2+e^x-x+\frac {-1+3 e^{-e^2}-\frac {4}{x^2}}{\log (x)} \]

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Rubi [A]  time = 1.02, antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 15, number of rules used = 9, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.141, Rules used = {12, 6742, 2194, 2353, 2306, 2309, 2178, 2302, 30} \begin {gather*} -\frac {4}{x^2 \log (x)}-x+e^x-\frac {1-3 e^{-e^2}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x^2 + E^E^2*(4 + x^2) + 8*E^E^2*Log[x] + E^E^2*(-x^3 + E^x*x^3)*Log[x]^2)/(E^E^2*x^3*Log[x]^2),x]

[Out]

E^x - x - (1 - 3/E^E^2)/Log[x] - 4/(x^2*Log[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-e^2} \int \frac {-3 x^2+e^{e^2} \left (4+x^2\right )+8 e^{e^2} \log (x)+e^{e^2} \left (-x^3+e^x x^3\right ) \log ^2(x)}{x^3 \log ^2(x)} \, dx\\ &=e^{-e^2} \int \left (e^{e^2+x}+\frac {4 e^{e^2}-3 \left (1-\frac {e^{e^2}}{3}\right ) x^2+8 e^{e^2} \log (x)-e^{e^2} x^3 \log ^2(x)}{x^3 \log ^2(x)}\right ) \, dx\\ &=e^{-e^2} \int e^{e^2+x} \, dx+e^{-e^2} \int \frac {4 e^{e^2}-3 \left (1-\frac {e^{e^2}}{3}\right ) x^2+8 e^{e^2} \log (x)-e^{e^2} x^3 \log ^2(x)}{x^3 \log ^2(x)} \, dx\\ &=e^x+e^{-e^2} \int \left (-e^{e^2}+\frac {4 e^{e^2}-\left (3-e^{e^2}\right ) x^2}{x^3 \log ^2(x)}+\frac {8 e^{e^2}}{x^3 \log (x)}\right ) \, dx\\ &=e^x-x+8 \int \frac {1}{x^3 \log (x)} \, dx+e^{-e^2} \int \frac {4 e^{e^2}+\left (-3+e^{e^2}\right ) x^2}{x^3 \log ^2(x)} \, dx\\ &=e^x-x+8 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{x} \, dx,x,\log (x)\right )+e^{-e^2} \int \left (\frac {4 e^{e^2}}{x^3 \log ^2(x)}+\frac {-3+e^{e^2}}{x \log ^2(x)}\right ) \, dx\\ &=e^x-x+8 \text {Ei}(-2 \log (x))+4 \int \frac {1}{x^3 \log ^2(x)} \, dx+\left (1-3 e^{-e^2}\right ) \int \frac {1}{x \log ^2(x)} \, dx\\ &=e^x-x+8 \text {Ei}(-2 \log (x))-\frac {4}{x^2 \log (x)}-8 \int \frac {1}{x^3 \log (x)} \, dx+\left (1-3 e^{-e^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=e^x-x+8 \text {Ei}(-2 \log (x))-\frac {1-3 e^{-e^2}}{\log (x)}-\frac {4}{x^2 \log (x)}-8 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{x} \, dx,x,\log (x)\right )\\ &=e^x-x-\frac {1-3 e^{-e^2}}{\log (x)}-\frac {4}{x^2 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 28, normalized size = 0.97 \begin {gather*} e^x-x+\frac {-1+3 e^{-e^2}-\frac {4}{x^2}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^2 + E^E^2*(4 + x^2) + 8*E^E^2*Log[x] + E^E^2*(-x^3 + E^x*x^3)*Log[x]^2)/(E^E^2*x^3*Log[x]^2),x
]

[Out]

E^x - x + (-1 + 3/E^E^2 - 4/x^2)/Log[x]

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fricas [A]  time = 0.56, size = 46, normalized size = 1.59 \begin {gather*} -\frac {{\left ({\left (x^{3} - x^{2} e^{x}\right )} e^{\left (e^{2}\right )} \log \relax (x) - 3 \, x^{2} + {\left (x^{2} + 4\right )} e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}}{x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^3-x^3)*exp(exp(2))*log(x)^2+8*exp(exp(2))*log(x)+(x^2+4)*exp(exp(2))-3*x^2)/x^3/exp(exp(2
))/log(x)^2,x, algorithm="fricas")

[Out]

-((x^3 - x^2*e^x)*e^(e^2)*log(x) - 3*x^2 + (x^2 + 4)*e^(e^2))*e^(-e^2)/(x^2*log(x))

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giac [A]  time = 0.20, size = 53, normalized size = 1.83 \begin {gather*} -\frac {{\left (x^{3} e^{\left (e^{2}\right )} \log \relax (x) - x^{2} e^{\left (x + e^{2}\right )} \log \relax (x) + x^{2} e^{\left (e^{2}\right )} - 3 \, x^{2} + 4 \, e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}}{x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^3-x^3)*exp(exp(2))*log(x)^2+8*exp(exp(2))*log(x)+(x^2+4)*exp(exp(2))-3*x^2)/x^3/exp(exp(2
))/log(x)^2,x, algorithm="giac")

[Out]

-(x^3*e^(e^2)*log(x) - x^2*e^(x + e^2)*log(x) + x^2*e^(e^2) - 3*x^2 + 4*e^(e^2))*e^(-e^2)/(x^2*log(x))

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maple [A]  time = 0.04, size = 39, normalized size = 1.34

method	result	size
risch	\(-x +{\mathrm e}^{x}-\frac {{\mathrm e}^{-{\mathrm e}^{2}} \left (x^{2} {\mathrm e}^{{\mathrm e}^{2}}-3 x^{2}+4 \,{\mathrm e}^{{\mathrm e}^{2}}\right )}{x^{2} \ln \relax (x )}\)	\(39\)
default	\({\mathrm e}^{-{\mathrm e}^{2}} \left ({\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{x}-\frac {{\mathrm e}^{{\mathrm e}^{2}}}{\ln \relax (x )}+\frac {3}{\ln \relax (x )}-\frac {4 \,{\mathrm e}^{{\mathrm e}^{2}}}{x^{2} \ln \relax (x )}-x \,{\mathrm e}^{{\mathrm e}^{2}}\right )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x^3-x^3)*exp(exp(2))*ln(x)^2+8*exp(exp(2))*ln(x)+(x^2+4)*exp(exp(2))-3*x^2)/x^3/exp(exp(2))/ln(x)
^2,x,method=_RETURNVERBOSE)

[Out]

-x+exp(x)-exp(-exp(2))/x^2*(x^2*exp(exp(2))-3*x^2+4*exp(exp(2)))/ln(x)

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maxima [C]  time = 0.38, size = 55, normalized size = 1.90 \begin {gather*} -{\left (x e^{\left (e^{2}\right )} - 8 \, {\rm Ei}\left (-2 \, \log \relax (x)\right ) e^{\left (e^{2}\right )} + 8 \, e^{\left (e^{2}\right )} \Gamma \left (-1, 2 \, \log \relax (x)\right ) + \frac {e^{\left (e^{2}\right )}}{\log \relax (x)} - \frac {3}{\log \relax (x)} - e^{\left (x + e^{2}\right )}\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^3-x^3)*exp(exp(2))*log(x)^2+8*exp(exp(2))*log(x)+(x^2+4)*exp(exp(2))-3*x^2)/x^3/exp(exp(2
))/log(x)^2,x, algorithm="maxima")

[Out]

-(x*e^(e^2) - 8*Ei(-2*log(x))*e^(e^2) + 8*e^(e^2)*gamma(-1, 2*log(x)) + e^(e^2)/log(x) - 3/log(x) - e^(x + e^2
))*e^(-e^2)

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mupad [B]  time = 3.25, size = 32, normalized size = 1.10 \begin {gather*} {\mathrm {e}}^x-x-\frac {1}{\ln \relax (x)}-\frac {4}{x^2\,\ln \relax (x)}+\frac {3\,{\mathrm {e}}^{-{\mathrm {e}}^2}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(2))*(8*exp(exp(2))*log(x) + exp(exp(2))*(x^2 + 4) - 3*x^2 + exp(exp(2))*log(x)^2*(x^3*exp(x) - x
^3)))/(x^3*log(x)^2),x)

[Out]

exp(x) - x - 1/log(x) - 4/(x^2*log(x)) + (3*exp(-exp(2)))/log(x)

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sympy [A]  time = 0.28, size = 36, normalized size = 1.24 \begin {gather*} - x + e^{x} + \frac {- x^{2} e^{e^{2}} + 3 x^{2} - 4 e^{e^{2}}}{x^{2} e^{e^{2}} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x**3-x**3)*exp(exp(2))*ln(x)**2+8*exp(exp(2))*ln(x)+(x**2+4)*exp(exp(2))-3*x**2)/x**3/exp(e
xp(2))/ln(x)**2,x)

[Out]

-x + exp(x) + (-x**2*exp(exp(2)) + 3*x**2 - 4*exp(exp(2)))*exp(-exp(2))/(x**2*log(x))

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