\(\int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} (125 e^3 \log (2)-125 e^{3+x} x \log (2))}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx\) [9814]

Optimal result

Integrand size = 78, antiderivative size = 38 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {\frac {x}{5}+\frac {e^3}{\frac {4 x}{5}-\frac {5 e^{-e^x}}{\log (2)}}}{2 x} \]

[Out]

1/2*(1/5*x+exp(3)/(4/5*x-5/ln(2)/exp(exp(x))))/x

Rubi [F]

\[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx \]

[In]

Int[(-40*E^(3 + 2*E^x)*x*Log[2]^2 + E^E^x*(125*E^3*Log[2] - 125*E^(3 + x)*x*Log[2]))/(1250*x^2 - 400*E^E^x*x^3
*Log[2] + 32*E^(2*E^x)*x^4*Log[2]^2),x]

[Out]

(-125*Log[2]*Defer[Int][E^(3 + E^x)/(x^2*(-25 + E^E^x*x*Log[16])^2), x])/2 - (125*Log[2]*Defer[Int][E^(3 + E^x
+ x)/(x*(-25 + E^E^x*x*Log[16])^2), x])/2 - (20*Log[2]^2*Defer[Int][E^(3 + E^x)/(x^2*(-25 + E^E^x*x*Log[16]))
, x])/Log[16]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{3+e^x} \log (2) \left (25-25 e^x x-8 e^{e^x} x \log (2)\right )}{2 x^2 \left (25-e^{e^x} x \log (16)\right )^2} \, dx \\ & = \frac {1}{2} (5 \log (2)) \int \frac {e^{3+e^x} \left (25-25 e^x x-8 e^{e^x} x \log (2)\right )}{x^2 \left (25-e^{e^x} x \log (16)\right )^2} \, dx \\ & = \frac {1}{2} (5 \log (2)) \int \left (-\frac {25 e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2}-\frac {e^{3+e^x} \left (-25+8 e^{e^x} x \log (2)\right )}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2}\right ) \, dx \\ & = -\left (\frac {1}{2} (5 \log (2)) \int \frac {e^{3+e^x} \left (-25+8 e^{e^x} x \log (2)\right )}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx \\ & = -\left (\frac {1}{2} (5 \log (2)) \int \left (\frac {25 e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2}+\frac {8 e^{3+e^x} \log (2)}{x^2 \log (16) \left (-25+e^{e^x} x \log (16)\right )}\right ) \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx \\ & = -\left (\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )^2} \, dx\right )-\frac {1}{2} (125 \log (2)) \int \frac {e^{3+e^x+x}}{x \left (-25+e^{e^x} x \log (16)\right )^2} \, dx-\frac {\left (20 \log ^2(2)\right ) \int \frac {e^{3+e^x}}{x^2 \left (-25+e^{e^x} x \log (16)\right )} \, dx}{\log (16)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=-\frac {5 e^{3+e^x} \log (2)}{2 \left (25 x-e^{e^x} x^2 \log (16)\right )} \]

[In]

Integrate[(-40*E^(3 + 2*E^x)*x*Log[2]^2 + E^E^x*(125*E^3*Log[2] - 125*E^(3 + x)*x*Log[2]))/(1250*x^2 - 400*E^E
^x*x^3*Log[2] + 32*E^(2*E^x)*x^4*Log[2]^2),x]

[Out]

(-5*E^(3 + E^x)*Log[2])/(2*(25*x - E^E^x*x^2*Log[16]))

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66

method	result	size
norman	\(\frac {5 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{3} \ln \left (2\right )}{2 x \left (4 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x}} x -25\right )}\)	\(25\)
parallelrisch	\(\frac {5 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{3} \ln \left (2\right )}{2 x \left (4 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x}} x -25\right )}\)	\(25\)
risch	\(\frac {5 \,{\mathrm e}^{3}}{8 x^{2}}+\frac {125 \,{\mathrm e}^{3}}{8 x^{2} \left (4 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{x}} x -25\right )}\)	\(28\)

[In]

int((-40*x*exp(3)*ln(2)^2*exp(exp(x))^2+(-125*x*exp(3)*ln(2)*exp(x)+125*exp(3)*ln(2))*exp(exp(x)))/(32*x^4*ln(
2)^2*exp(exp(x))^2-400*x^3*ln(2)*exp(exp(x))+1250*x^2),x,method=_RETURNVERBOSE)

[Out]

5/2*exp(exp(x))*exp(3)*ln(2)/x/(4*ln(2)*exp(exp(x))*x-25)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \]

[In]

integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x)+125*exp(3)*log(2))*exp(exp(x)))/(3
2*x^4*log(2)^2*exp(exp(x))^2-400*x^3*log(2)*exp(exp(x))+1250*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {125 e^{3}}{32 x^{3} e^{e^{x}} \log {\left (2 \right )} - 200 x^{2}} + \frac {5 e^{3}}{8 x^{2}} \]

[In]

integrate((-40*x*exp(3)*ln(2)**2*exp(exp(x))**2+(-125*x*exp(3)*ln(2)*exp(x)+125*exp(3)*ln(2))*exp(exp(x)))/(32
*x**4*ln(2)**2*exp(exp(x))**2-400*x**3*ln(2)*exp(exp(x))+1250*x**2),x)

[Out]

125*exp(3)/(32*x**3*exp(exp(x))*log(2) - 200*x**2) + 5*exp(3)/(8*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \]

[In]

integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x)+125*exp(3)*log(2))*exp(exp(x)))/(3
2*x^4*log(2)^2*exp(exp(x))^2-400*x^3*log(2)*exp(exp(x))+1250*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \]

[In]

integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x)+125*exp(3)*log(2))*exp(exp(x)))/(3
2*x^4*log(2)^2*exp(exp(x))^2-400*x^3*log(2)*exp(exp(x))+1250*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.94 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5\,{\mathrm {e}}^3}{8\,x^2}+\frac {125\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^{x+3}\right )}{32\,x\,\left (x^2\,\ln \left (2\right )+x^3\,{\mathrm {e}}^x\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}-\frac {25}{4\,x\,\ln \left (2\right )}\right )} \]

[In]

int((exp(exp(x))*(125*exp(3)*log(2) - 125*x*exp(3)*exp(x)*log(2)) - 40*x*exp(3)*exp(2*exp(x))*log(2)^2)/(1250*
x^2 + 32*x^4*exp(2*exp(x))*log(2)^2 - 400*x^3*exp(exp(x))*log(2)),x)

[Out]

(5*exp(3))/(8*x^2) + (125*(exp(3) + x*exp(x + 3)))/(32*x*(x^2*log(2) + x^3*exp(x)*log(2))*(exp(exp(x)) - 25/(4
*x*log(2))))