3.98.76 \(\int \frac {e^{\frac {e^x}{x^2+x^2 \log (\frac {1}{x})}} (e^x (90-90 x)+e^x (180-90 x) \log (\frac {1}{x}))+e^{\frac {2 e^x}{x^2+x^2 \log (\frac {1}{x})}} (e^x (-50+50 x)+e^x (-100+50 x) \log (\frac {1}{x}))}{9 x^3+18 x^3 \log (\frac {1}{x})+9 x^3 \log ^2(\frac {1}{x})} \, dx\)

Optimal. Leaf size=27 \[ \left (3-\frac {5}{3} e^{\frac {e^x}{x \left (x+x \log \left (\frac {1}{x}\right )\right )}}\right )^2 \]

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Rubi [A]  time = 3.96, antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 4, number of rules used = 3, integrand size = 119, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6688, 12, 6708} \begin {gather*} \frac {25}{9} e^{\frac {2 e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}-10 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 90*x)*Log[x^(-1)]) + E^((2*E^x)/(x^2 + x^2*
Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x*(-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Log[x^(-
1)]^2),x]

[Out]

-10*E^(E^x/(x^2*(1 + Log[x^(-1)]))) + (25*E^((2*E^x)/(x^2*(1 + Log[x^(-1)]))))/9

Rule 12

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

Rule 6688

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

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{9 x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {10}{9} \int \frac {e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {2}{9} \operatorname {Subst}\left (\int (9+x) \, dx,x,-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right )\\ &=-10 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {25}{9} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 3.32, size = 46, normalized size = 1.70 \begin {gather*} \frac {10}{9} \left (-9 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {5}{2} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^x/(x^2 + x^2*Log[x^(-1)]))*(E^x*(90 - 90*x) + E^x*(180 - 90*x)*Log[x^(-1)]) + E^((2*E^x)/(x^2
+ x^2*Log[x^(-1)]))*(E^x*(-50 + 50*x) + E^x*(-100 + 50*x)*Log[x^(-1)]))/(9*x^3 + 18*x^3*Log[x^(-1)] + 9*x^3*Lo
g[x^(-1)]^2),x]

[Out]

(10*(-9*E^(E^x/(x^2*(1 + Log[x^(-1)]))) + (5*E^((2*E^x)/(x^2*(1 + Log[x^(-1)]))))/2))/9

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fricas [A]  time = 0.78, size = 42, normalized size = 1.56 \begin {gather*} \frac {25}{9} \, e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 10 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2))^2+((-90*x+180)*exp(x)*
log(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algori
thm="fricas")

[Out]

25/9*e^(2*e^x/(x^2*log(1/x) + x^2)) - 10*e^(e^x/(x^2*log(1/x) + x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 \, {\left (5 \, {\left ({\left (x - 2\right )} e^{x} \log \left (\frac {1}{x}\right ) + {\left (x - 1\right )} e^{x}\right )} e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 9 \, {\left ({\left (x - 2\right )} e^{x} \log \left (\frac {1}{x}\right ) + {\left (x - 1\right )} e^{x}\right )} e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )}\right )}}{9 \, {\left (x^{3} \log \left (\frac {1}{x}\right )^{2} + 2 \, x^{3} \log \left (\frac {1}{x}\right ) + x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2))^2+((-90*x+180)*exp(x)*
log(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algori
thm="giac")

[Out]

integrate(10/9*(5*((x - 2)*e^x*log(1/x) + (x - 1)*e^x)*e^(2*e^x/(x^2*log(1/x) + x^2)) - 9*((x - 2)*e^x*log(1/x
) + (x - 1)*e^x)*e^(e^x/(x^2*log(1/x) + x^2)))/(x^3*log(1/x)^2 + 2*x^3*log(1/x) + x^3), x)

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maple [A]  time = 0.14, size = 34, normalized size = 1.26

method	result	size
risch	\(\frac {25 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}}{x^{2} \left (\ln \relax (x )-1\right )}}}{9}-10 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{x^{2} \left (\ln \relax (x )-1\right )}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*ln(1/x)+x^2))^2+((-90*x+180)*exp(x)*ln(1/x)+
(-90*x+90)*exp(x))*exp(exp(x)/(x^2*ln(1/x)+x^2)))/(9*x^3*ln(1/x)^2+18*x^3*ln(1/x)+9*x^3),x,method=_RETURNVERBO
SE)

[Out]

25/9*exp(-2*exp(x)/x^2/(ln(x)-1))-10*exp(-exp(x)/x^2/(ln(x)-1))

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maxima [A]  time = 0.46, size = 43, normalized size = 1.59 \begin {gather*} -\frac {5}{9} \, {\left (18 \, e^{\left (\frac {e^{x}}{x^{2} \log \relax (x) - x^{2}}\right )} - 5\right )} e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \relax (x) - x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x-100)*exp(x)*log(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2))^2+((-90*x+180)*exp(x)*
log(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x^2*log(1/x)+x^2)))/(9*x^3*log(1/x)^2+18*x^3*log(1/x)+9*x^3),x, algori
thm="maxima")

[Out]

-5/9*(18*e^(e^x/(x^2*log(x) - x^2)) - 5)*e^(-2*e^x/(x^2*log(x) - x^2))

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mupad [B]  time = 6.06, size = 42, normalized size = 1.56 \begin {gather*} \frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}\,\left (5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}-18\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*exp(x))/(x^2*log(1/x) + x^2))*(exp(x)*(50*x - 50) + log(1/x)*exp(x)*(50*x - 100)) - exp(exp(x)/(x^
2*log(1/x) + x^2))*(exp(x)*(90*x - 90) + log(1/x)*exp(x)*(90*x - 180)))/(18*x^3*log(1/x) + 9*x^3 + 9*x^3*log(1
/x)^2),x)

[Out]

(5*exp(exp(x)/(x^2*log(1/x) + x^2))*(5*exp(exp(x)/(x^2*log(1/x) + x^2)) - 18))/9

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sympy [A]  time = 0.88, size = 39, normalized size = 1.44 \begin {gather*} \frac {25 e^{\frac {2 e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}}}{9} - 10 e^{\frac {e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x-100)*exp(x)*ln(1/x)+(50*x-50)*exp(x))*exp(exp(x)/(x**2*ln(1/x)+x**2))**2+((-90*x+180)*exp(x)
*ln(1/x)+(-90*x+90)*exp(x))*exp(exp(x)/(x**2*ln(1/x)+x**2)))/(9*x**3*ln(1/x)**2+18*x**3*ln(1/x)+9*x**3),x)

[Out]

25*exp(2*exp(x)/(x**2*log(1/x) + x**2))/9 - 10*exp(exp(x)/(x**2*log(1/x) + x**2))

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