3.32.68 \(\int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x))}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+(-10 x^2-50 e^{e^3} x^2) \log ^2(x)+25 x \log ^4(x)} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {-5+x}{x+e^{-e^3} \left (\frac {x}{5}-\log ^2(x)\right )}} \]

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Rubi [F]  time = 15.20, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^E^3*(25 - 5*x))/(-x - 5*E^E^3*x + 5*Log[x]^2))*(25*E^E^3*x + 125*E^(2*E^3)*x + E^E^3*(-250 + 50*x)*
Log[x] - 25*E^E^3*x*Log[x]^2))/(x^3 + 10*E^E^3*x^3 + 25*E^(2*E^3)*x^3 + (-10*x^2 - 50*E^E^3*x^2)*Log[x]^2 + 25
*x*Log[x]^4),x]

[Out]

25*(1 + 5*E^E^3)*Defer[Int][E^(E^3 + (5*E^E^3*(-5 + x))/(x + 5*E^E^3*x - 5*Log[x]^2))/((1 + 5*E^E^3)*x - 5*Log
[x]^2)^2, x] - 5*(1 + 5*E^E^3)*Defer[Int][(E^(E^3 + (5*E^E^3*(-5 + x))/(x + 5*E^E^3*x - 5*Log[x]^2))*x)/((1 +
5*E^E^3)*x - 5*Log[x]^2)^2, x] + 50*Defer[Int][(E^(E^3 + (5*E^E^3*(-5 + x))/(x + 5*E^E^3*x - 5*Log[x]^2))*Log[
x])/((1 + 5*E^E^3)*x - 5*Log[x]^2)^2, x] - 250*Defer[Int][(E^(E^3 + (5*E^E^3*(-5 + x))/(x + 5*E^E^3*x - 5*Log[
x]^2))*Log[x])/(x*((1 + 5*E^E^3)*x - 5*Log[x]^2)^2), x] + 5*Defer[Int][E^(E^3 + (5*E^E^3*(-5 + x))/(x + 5*E^E^
3*x - 5*Log[x]^2))/((1 + 5*E^E^3)*x - 5*Log[x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{25 e^{2 e^3} x^3+\left (1+10 e^{e^3}\right ) x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{\left (1+10 e^{e^3}+25 e^{2 e^3}\right ) x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {25 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x+2 (-5+x) \log (x)-x \log ^2(x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=25 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x+2 (-5+x) \log (x)-x \log ^2(x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=25 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) (5-x) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{5 x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{5 \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) (5-x) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+5 \int \left (\frac {5 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-\left (\left (1+5 e^{e^3}\right ) x\right )+10 \log (x)\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-\left (\left (1+5 e^{e^3}\right ) x\right )+10 \log (x)\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+25 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+5 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-1-5 e^{e^3}\right ) x}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {10 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx+25 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (1+5 e^{e^3}\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}-\frac {10 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+50 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx-250 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx-\left (5 \left (1+5 e^{e^3}\right )\right ) \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) x}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+\left (25 \left (1+5 e^{e^3}\right )\right ) \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 32, normalized size = 1.07 \begin {gather*} e^{-\frac {5 e^{e^3} (-5+x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^E^3*(25 - 5*x))/(-x - 5*E^E^3*x + 5*Log[x]^2))*(25*E^E^3*x + 125*E^(2*E^3)*x + E^E^3*(-250 +
50*x)*Log[x] - 25*E^E^3*x*Log[x]^2))/(x^3 + 10*E^E^3*x^3 + 25*E^(2*E^3)*x^3 + (-10*x^2 - 50*E^E^3*x^2)*Log[x]^
2 + 25*x*Log[x]^4),x]

[Out]

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

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fricas [A]  time = 0.52, size = 25, normalized size = 0.83 \begin {gather*} e^{\left (\frac {5 \, {\left (x - 5\right )} e^{\left (e^{3}\right )}}{5 \, x e^{\left (e^{3}\right )} - 5 \, \log \relax (x)^{2} + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x*exp(exp(3))^2+25*x*exp(exp(3)))*exp(
(-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x*exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*
x^3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {25 \, {\left (x e^{\left (e^{3}\right )} \log \relax (x)^{2} - 2 \, {\left (x - 5\right )} e^{\left (e^{3}\right )} \log \relax (x) - 5 \, x e^{\left (2 \, e^{3}\right )} - x e^{\left (e^{3}\right )}\right )} e^{\left (\frac {5 \, {\left (x - 5\right )} e^{\left (e^{3}\right )}}{5 \, x e^{\left (e^{3}\right )} - 5 \, \log \relax (x)^{2} + x}\right )}}{25 \, x \log \relax (x)^{4} + 25 \, x^{3} e^{\left (2 \, e^{3}\right )} + 10 \, x^{3} e^{\left (e^{3}\right )} + x^{3} - 10 \, {\left (5 \, x^{2} e^{\left (e^{3}\right )} + x^{2}\right )} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x*exp(exp(3))^2+25*x*exp(exp(3)))*exp(
(-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x*exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*
x^3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="giac")

[Out]

integrate(-25*(x*e^(e^3)*log(x)^2 - 2*(x - 5)*e^(e^3)*log(x) - 5*x*e^(2*e^3) - x*e^(e^3))*e^(5*(x - 5)*e^(e^3)
/(5*x*e^(e^3) - 5*log(x)^2 + x))/(25*x*log(x)^4 + 25*x^3*e^(2*e^3) + 10*x^3*e^(e^3) + x^3 - 10*(5*x^2*e^(e^3)
+ x^2)*log(x)^2), x)

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maple [A]  time = 0.05, size = 28, normalized size = 0.93

method	result	size
risch	\({\mathrm e}^{-\frac {5 \left (x -5\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5 \ln \relax (x )^{2}-5 x \,{\mathrm e}^{{\mathrm e}^{3}}-x}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-25*x*exp(exp(3))*ln(x)^2+(50*x-250)*exp(exp(3))*ln(x)+125*x*exp(exp(3))^2+25*x*exp(exp(3)))*exp((-5*x+25
)*exp(exp(3))/(5*ln(x)^2-5*x*exp(exp(3))-x))/(25*x*ln(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*ln(x)^2+25*x^3*exp(exp
(3))^2+10*x^3*exp(exp(3))+x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.30, size = 81, normalized size = 2.70 \begin {gather*} e^{\left (-\frac {25 \, e^{\left (e^{3}\right )} \log \relax (x)^{2}}{5 \, {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} \log \relax (x)^{2} - x {\left (25 \, e^{\left (2 \, e^{3}\right )} + 10 \, e^{\left (e^{3}\right )} + 1\right )}} - \frac {25 \, e^{\left (e^{3}\right )}}{x {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} - 5 \, \log \relax (x)^{2}} + \frac {5 \, e^{\left (e^{3}\right )}}{5 \, e^{\left (e^{3}\right )} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x*exp(exp(3))^2+25*x*exp(exp(3)))*exp(
(-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x*exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*
x^3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="maxima")

[Out]

e^(-25*e^(e^3)*log(x)^2/(5*(5*e^(e^3) + 1)*log(x)^2 - x*(25*e^(2*e^3) + 10*e^(e^3) + 1)) - 25*e^(e^3)/(x*(5*e^
(e^3) + 1) - 5*log(x)^2) + 5*e^(e^3)/(5*e^(e^3) + 1))

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mupad [B]  time = 2.72, size = 31, normalized size = 1.03 \begin {gather*} {\mathrm {e}}^{-\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^3}-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}{-5\,{\ln \relax (x)}^2+x+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(exp(3))*(5*x - 25))/(x - 5*log(x)^2 + 5*x*exp(exp(3))))*(125*x*exp(2*exp(3)) + 25*x*exp(exp(3))
+ exp(exp(3))*log(x)*(50*x - 250) - 25*x*exp(exp(3))*log(x)^2))/(10*x^3*exp(exp(3)) + 25*x*log(x)^4 - log(x)^2
*(50*x^2*exp(exp(3)) + 10*x^2) + 25*x^3*exp(2*exp(3)) + x^3),x)

[Out]

exp(-(25*exp(exp(3)) - 5*x*exp(exp(3)))/(x - 5*log(x)^2 + 5*x*exp(exp(3))))

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sympy [A]  time = 0.74, size = 27, normalized size = 0.90 \begin {gather*} e^{\frac {\left (25 - 5 x\right ) e^{e^{3}}}{- 5 x e^{e^{3}} - x + 5 \log {\relax (x )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-25*x*exp(exp(3))*ln(x)**2+(50*x-250)*exp(exp(3))*ln(x)+125*x*exp(exp(3))**2+25*x*exp(exp(3)))*exp(
(-5*x+25)*exp(exp(3))/(5*ln(x)**2-5*x*exp(exp(3))-x))/(25*x*ln(x)**4+(-50*x**2*exp(exp(3))-10*x**2)*ln(x)**2+2
5*x**3*exp(exp(3))**2+10*x**3*exp(exp(3))+x**3),x)

[Out]

exp((25 - 5*x)*exp(exp(3))/(-5*x*exp(exp(3)) - x + 5*log(x)**2))

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