\(\int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} (-50 x-40 x^2)+e^{e^{\frac {2}{e^{2 x}-x}}} (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} (10 x-20 e^{2 x} x))+(5 e^{4 x}-10 e^{2 x} x+5 x^2) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx\) [1464]

Optimal result

Integrand size = 148, antiderivative size = 30 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 \left (3+x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right )\right ) \]

[Out]

15+5*x*(exp(exp(2/(exp(x)^2-x)))+2*x+ln(x)+4)

Rubi [F]

\[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx \]

[In]

Int[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2) + E^E^(2/(E^(2*x) - x))*(5*E^(4*x) - 10*
E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x) - x))*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[x])/(E
^(4*x) - 2*E^(2*x)*x + x^2),x]

[Out]

20*x + 10*x^2 + 5*x*Log[x] + 5*Defer[Int][E^E^(2/(E^(2*x) - x)), x] + 10*Defer[Int][(E^(E^(2/(E^(2*x) - x)) +
2/(E^(2*x) - x))*x)/(E^(2*x) - x)^2, x] - 20*Defer[Int][(E^(E^(2/(E^(2*x) - x)) + 2/(E^(2*x) - x))*x)/(E^(2*x)
- x), x] - 20*Defer[Int][(E^(E^(2/(E^(2*x) - x)) + 2/(E^(2*x) - x))*x^2)/(E^(2*x) - x)^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{\left (e^{2 x}-x\right )^2} \, dx \\ & = \int \left (-\frac {20 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x}-\frac {10 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2}+5 \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right )\right ) \, dx \\ & = 5 \int \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right ) \, dx-10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx \\ & = 25 x+10 x^2+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+5 \int \log (x) \, dx-10 \int \left (-\frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2}+\frac {2 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2}\right ) \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx \\ & = 20 x+10 x^2+5 x \log (x)+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=5 x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right ) \]

[In]

Integrate[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2) + E^E^(2/(E^(2*x) - x))*(5*E^(4*x)
- 10*E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x) - x))*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[
x])/(E^(4*x) - 2*E^(2*x)*x + x^2),x]

[Out]

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

Maple [A] (verified)

Time = 52.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07

method	result	size
risch	\(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{-\frac {2}{x -{\mathrm e}^{2 x}}}}+20 x\)	\(32\)
parallelrisch	\(10 x^{2}+5 x \ln \left (x \right )+5 x \,{\mathrm e}^{{\mathrm e}^{\frac {2}{{\mathrm e}^{2 x}-x}}}+20 x\)	\(32\)

[In]

int((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))+(5*ex
p(x)^4-10*x*exp(x)^2+5*x^2)*ln(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x*exp(
x)^2+x^2),x,method=_RETURNVERBOSE)

[Out]

10*x^2+5*x*ln(x)+5*x*exp(exp(-2/(x-exp(2*x))))+20*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="fricas")

[Out]

10*x^2 + 5*x*e^(e^(-2/(x - e^(2*x)))) + 5*x*log(x) + 20*x

Sympy [F(-1)]

Timed out. \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\text {Timed out} \]

[In]

integrate((((-20*x*exp(x)**2+10*x)*exp(2/(exp(x)**2-x))+5*exp(x)**4-10*x*exp(x)**2+5*x**2)*exp(exp(2/(exp(x)**
2-x)))+(5*exp(x)**4-10*x*exp(x)**2+5*x**2)*ln(x)+(20*x+25)*exp(x)**4+(-40*x**2-50*x)*exp(x)**2+20*x**3+25*x**2
)/(exp(x)**4-2*x*exp(x)**2+x**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \left (x\right ) + 20 \, x \]

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="maxima")

[Out]

10*x^2 + 5*x*e^(e^(-2/(x - e^(2*x)))) + 5*x*log(x) + 20*x

Giac [F]

\[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=\int { \frac {5 \, {\left (4 \, x^{3} + 5 \, x^{2} + {\left (4 \, x + 5\right )} e^{\left (4 \, x\right )} - 2 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x e^{\left (2 \, x\right )} - x\right )} e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )} + e^{\left (4 \, x\right )}\right )} e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + {\left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}\right )} \log \left (x\right )\right )}}{x^{2} - 2 \, x e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )}} \,d x } \]

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx=20\,x+5\,x\,\ln \left (x\right )+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2}{x-{\mathrm {e}}^{2\,x}}}}+10\,x^2 \]

[In]

int((log(x)*(5*exp(4*x) - 10*x*exp(2*x) + 5*x^2) - exp(2*x)*(50*x + 40*x^2) + exp(4*x)*(20*x + 25) + 25*x^2 +
20*x^3 + exp(exp(-2/(x - exp(2*x))))*(5*exp(4*x) - 10*x*exp(2*x) + exp(-2/(x - exp(2*x)))*(10*x - 20*x*exp(2*x
)) + 5*x^2))/(exp(4*x) - 2*x*exp(2*x) + x^2),x)

[Out]

20*x + 5*x*log(x) + 5*x*exp(exp(-2/(x - exp(2*x)))) + 10*x^2