\(\int \frac {e^{-e^5} (e^{e^5} x+(8 x+e^{e^5} (1-3 x^2)) \log (x)+2 e^{e^5} x \log (x) \log (\log (x)))}{\log (x)} \, dx\) [5266]

Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=x^2 \left (4 e^{-e^5}+\frac {1}{x}-x+\log (\log (x))\right ) \]

[Out]

x^2*(4/exp(exp(5))+ln(ln(x))+1/x-x)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.113, Rules used = {12, 6874, 6820, 2346, 2209, 2602} \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=-x^3+4 e^{-e^5} x^2+x^2 \log (\log (x))+x \]

[In]

Int[(E^E^5*x + (8*x + E^E^5*(1 - 3*x^2))*Log[x] + 2*E^E^5*x*Log[x]*Log[Log[x]])/(E^E^5*Log[x]),x]

[Out]

x + (4*x^2)/E^E^5 - x^3 + x^2*Log[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 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 2346

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 2602

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

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = e^{-e^5} \int \frac {e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))}{\log (x)} \, dx \\ & = e^{-e^5} \int \left (-\frac {-e^{e^5} x-e^{e^5} \log (x)-8 x \log (x)+3 e^{e^5} x^2 \log (x)}{\log (x)}+2 e^{e^5} x \log (\log (x))\right ) \, dx \\ & = 2 \int x \log (\log (x)) \, dx-e^{-e^5} \int \frac {-e^{e^5} x-e^{e^5} \log (x)-8 x \log (x)+3 e^{e^5} x^2 \log (x)}{\log (x)} \, dx \\ & = x^2 \log (\log (x))-e^{-e^5} \int \left (-8 x+e^{e^5} \left (-1+3 x^2\right )-\frac {e^{e^5} x}{\log (x)}\right ) \, dx-\int \frac {x}{\log (x)} \, dx \\ & = 4 e^{-e^5} x^2+x^2 \log (\log (x))-\int \left (-1+3 x^2\right ) \, dx+\int \frac {x}{\log (x)} \, dx-\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = x+4 e^{-e^5} x^2-x^3-\text {Ei}(2 \log (x))+x^2 \log (\log (x))+\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = x+4 e^{-e^5} x^2-x^3+x^2 \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=x+4 e^{-e^5} x^2-x^3+x^2 \log (\log (x)) \]

[In]

Integrate[(E^E^5*x + (8*x + E^E^5*(1 - 3*x^2))*Log[x] + 2*E^E^5*x*Log[x]*Log[Log[x]])/(E^E^5*Log[x]),x]

[Out]

x + (4*x^2)/E^E^5 - x^3 + x^2*Log[Log[x]]

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
norman	\(x +x^{2} \ln \left (\ln \left (x \right )\right )-x^{3}+4 \,{\mathrm e}^{-{\mathrm e}^{5}} x^{2}\)	\(25\)
risch	\(x +x^{2} \ln \left (\ln \left (x \right )\right )-x^{3}+4 \,{\mathrm e}^{-{\mathrm e}^{5}} x^{2}\)	\(25\)
parts	\(x^{2} \ln \left (\ln \left (x \right )\right )-{\mathrm e}^{-{\mathrm e}^{5}} \left ({\mathrm e}^{{\mathrm e}^{5}} x^{3}-x \,{\mathrm e}^{{\mathrm e}^{5}}-4 x^{2}\right )\)	\(35\)
parallelrisch	\({\mathrm e}^{-{\mathrm e}^{5}} \left (-{\mathrm e}^{{\mathrm e}^{5}} x^{3}+{\mathrm e}^{{\mathrm e}^{5}} x^{2} \ln \left (\ln \left (x \right )\right )+x \,{\mathrm e}^{{\mathrm e}^{5}}+4 x^{2}\right )\)	\(36\)
default	\({\mathrm e}^{-{\mathrm e}^{5}} \left (-{\mathrm e}^{{\mathrm e}^{5}} \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+4 x^{2}-{\mathrm e}^{{\mathrm e}^{5}} x^{3}+2 \,{\mathrm e}^{{\mathrm e}^{5}} \left (\frac {x^{2} \ln \left (\ln \left (x \right )\right )}{2}+\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )}{2}\right )+x \,{\mathrm e}^{{\mathrm e}^{5}}\right )\)	\(59\)

[In]

int((2*x*exp(exp(5))*ln(x)*ln(ln(x))+((-3*x^2+1)*exp(exp(5))+8*x)*ln(x)+x*exp(exp(5)))/exp(exp(5))/ln(x),x,met
hod=_RETURNVERBOSE)

[Out]

x+x^2*ln(ln(x))-x^3+4/exp(exp(5))*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx={\left (x^{2} e^{\left (e^{5}\right )} \log \left (\log \left (x\right )\right ) + 4 \, x^{2} - {\left (x^{3} - x\right )} e^{\left (e^{5}\right )}\right )} e^{\left (-e^{5}\right )} \]

[In]

integrate((2*x*exp(exp(5))*log(x)*log(log(x))+((-3*x^2+1)*exp(exp(5))+8*x)*log(x)+x*exp(exp(5)))/exp(exp(5))/l
og(x),x, algorithm="fricas")

[Out]

(x^2*e^(e^5)*log(log(x)) + 4*x^2 - (x^3 - x)*e^(e^5))*e^(-e^5)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=- x^{3} + x^{2} \log {\left (\log {\left (x \right )} \right )} + \frac {4 x^{2}}{e^{e^{5}}} + x \]

[In]

integrate((2*x*exp(exp(5))*ln(x)*ln(ln(x))+((-3*x**2+1)*exp(exp(5))+8*x)*ln(x)+x*exp(exp(5)))/exp(exp(5))/ln(x
),x)

[Out]

-x**3 + x**2*log(log(x)) + 4*x**2*exp(-exp(5)) + x

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.43 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=-{\left (x^{3} e^{\left (e^{5}\right )} - 4 \, x^{2} - {\left (x^{2} \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (2 \, \log \left (x\right )\right )\right )} e^{\left (e^{5}\right )} - x e^{\left (e^{5}\right )} - {\rm Ei}\left (2 \, \log \left (x\right )\right ) e^{\left (e^{5}\right )}\right )} e^{\left (-e^{5}\right )} \]

[In]

integrate((2*x*exp(exp(5))*log(x)*log(log(x))+((-3*x^2+1)*exp(exp(5))+8*x)*log(x)+x*exp(exp(5)))/exp(exp(5))/l
og(x),x, algorithm="maxima")

[Out]

-(x^3*e^(e^5) - 4*x^2 - (x^2*log(log(x)) - Ei(2*log(x)))*e^(e^5) - x*e^(e^5) - Ei(2*log(x))*e^(e^5))*e^(-e^5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=-{\left (x^{3} e^{\left (e^{5}\right )} - x^{2} e^{\left (e^{5}\right )} \log \left (\log \left (x\right )\right ) - 4 \, x^{2} - x e^{\left (e^{5}\right )}\right )} e^{\left (-e^{5}\right )} \]

[In]

integrate((2*x*exp(exp(5))*log(x)*log(log(x))+((-3*x^2+1)*exp(exp(5))+8*x)*log(x)+x*exp(exp(5)))/exp(exp(5))/l
og(x),x, algorithm="giac")

[Out]

-(x^3*e^(e^5) - x^2*e^(e^5)*log(log(x)) - 4*x^2 - x*e^(e^5))*e^(-e^5)

Mupad [B] (verification not implemented)

Time = 14.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-e^5} \left (e^{e^5} x+\left (8 x+e^{e^5} \left (1-3 x^2\right )\right ) \log (x)+2 e^{e^5} x \log (x) \log (\log (x))\right )}{\log (x)} \, dx=x+4\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^5}+x^2\,\ln \left (\ln \left (x\right )\right )-x^3 \]

[In]

int((exp(-exp(5))*(log(x)*(8*x - exp(exp(5))*(3*x^2 - 1)) + x*exp(exp(5)) + 2*x*log(log(x))*exp(exp(5))*log(x)
))/log(x),x)

[Out]

x + 4*x^2*exp(-exp(5)) + x^2*log(log(x)) - x^3