\(\int \frac {144+e^{e^{2 x}} (-4+(4-8 e^{2 x} x) \log (x))}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx\) [6370]

Optimal result

Integrand size = 67, antiderivative size = 22 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {x}{9-x+\frac {1}{4} e^{e^{2 x}} \log (x)} \]

[Out]

x/(9-x+1/4*ln(x)*exp(exp(x)^2))

Rubi [F]

\[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx \]

[In]

Int[(144 + E^E^(2*x)*(-4 + (4 - 8*E^(2*x)*x)*Log[x]))/(1296 - 288*x + 16*x^2 + E^E^(2*x)*(72 - 8*x)*Log[x] + E
^(2*E^(2*x))*Log[x]^2),x]

[Out]

16*Defer[Int][x/(-36 + 4*x - E^E^(2*x)*Log[x])^2, x] - 16*Defer[Int][x/(Log[x]*(-36 + 4*x - E^E^(2*x)*Log[x])^
2), x] - 4*Defer[Int][(-36 + 4*x - E^E^(2*x)*Log[x])^(-1), x] + 144*Defer[Int][1/(Log[x]*(36 - 4*x + E^E^(2*x)
*Log[x])^2), x] - 8*Defer[Int][(E^(E^(2*x) + 2*x)*x*Log[x])/(36 - 4*x + E^E^(2*x)*Log[x])^2, x] - 4*Defer[Int]
[1/(Log[x]*(36 - 4*x + E^E^(2*x)*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx \\ & = \int \left (\frac {4 \left (36-e^{e^{2 x}}+e^{e^{2 x}} \log (x)\right )}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}-\frac {8 e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx \\ & = 4 \int \frac {36-e^{e^{2 x}}+e^{e^{2 x}} \log (x)}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx \\ & = 4 \int \left (\frac {-1+\log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )}+\frac {4 (9-x+x \log (x))}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx \\ & = 4 \int \frac {-1+\log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \frac {9-x+x \log (x)}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx \\ & = 4 \int \left (-\frac {1}{-36+4 x-e^{e^{2 x}} \log (x)}-\frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )}\right ) \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \left (\frac {x}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}-\frac {x}{\log (x) \left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2}+\frac {9}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2}\right ) \, dx \\ & = -\left (4 \int \frac {1}{-36+4 x-e^{e^{2 x}} \log (x)} \, dx\right )-4 \int \frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )} \, dx-8 \int \frac {e^{e^{2 x}+2 x} x \log (x)}{\left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx+16 \int \frac {x}{\left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx-16 \int \frac {x}{\log (x) \left (-36+4 x-e^{e^{2 x}} \log (x)\right )^2} \, dx+144 \int \frac {1}{\log (x) \left (36-4 x+e^{e^{2 x}} \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {4 x}{36-4 x+e^{e^{2 x}} \log (x)} \]

[In]

Integrate[(144 + E^E^(2*x)*(-4 + (4 - 8*E^(2*x)*x)*Log[x]))/(1296 - 288*x + 16*x^2 + E^E^(2*x)*(72 - 8*x)*Log[
x] + E^(2*E^(2*x))*Log[x]^2),x]

[Out]

(4*x)/(36 - 4*x + E^E^(2*x)*Log[x])

Maple [A] (verified)

Time = 10.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {4 x}{-{\mathrm e}^{{\mathrm e}^{2 x}} \ln \left (x \right )+4 x -36}\)	\(20\)
parallelrisch	\(-\frac {4 x}{-{\mathrm e}^{{\mathrm e}^{2 x}} \ln \left (x \right )+4 x -36}\)	\(20\)

[In]

int((((-8*x*exp(x)^2+4)*ln(x)-4)*exp(exp(x)^2)+144)/(ln(x)^2*exp(exp(x)^2)^2+(-8*x+72)*ln(x)*exp(exp(x)^2)+16*
x^2-288*x+1296),x,method=_RETURNVERBOSE)

[Out]

-4*x/(-exp(exp(2*x))*ln(x)+4*x-36)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \left (x\right ) - 4 \, x + 36} \]

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="fricas")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {4 x}{- 4 x + e^{e^{2 x}} \log {\left (x \right )} + 36} \]

[In]

integrate((((-8*x*exp(x)**2+4)*ln(x)-4)*exp(exp(x)**2)+144)/(ln(x)**2*exp(exp(x)**2)**2+(-8*x+72)*ln(x)*exp(ex
p(x)**2)+16*x**2-288*x+1296),x)

[Out]

4*x/(-4*x + exp(exp(2*x))*log(x) + 36)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \left (x\right ) - 4 \, x + 36} \]

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="maxima")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\frac {4 \, x}{e^{\left (e^{\left (2 \, x\right )}\right )} \log \left (x\right ) - 4 \, x + 36} \]

[In]

integrate((((-8*x*exp(x)^2+4)*log(x)-4)*exp(exp(x)^2)+144)/(log(x)^2*exp(exp(x)^2)^2+(-8*x+72)*log(x)*exp(exp(
x)^2)+16*x^2-288*x+1296),x, algorithm="giac")

[Out]

4*x/(e^(e^(2*x))*log(x) - 4*x + 36)

Mupad [F(-1)]

Timed out. \[ \int \frac {144+e^{e^{2 x}} \left (-4+\left (4-8 e^{2 x} x\right ) \log (x)\right )}{1296-288 x+16 x^2+e^{e^{2 x}} (72-8 x) \log (x)+e^{2 e^{2 x}} \log ^2(x)} \, dx=\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\left (\ln \left (x\right )\,\left (8\,x\,{\mathrm {e}}^{2\,x}-4\right )+4\right )-144}{{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\ln \left (x\right )}^2-288\,x+16\,x^2-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\ln \left (x\right )\,\left (8\,x-72\right )+1296} \,d x \]

[In]

int(-(exp(exp(2*x))*(log(x)*(8*x*exp(2*x) - 4) + 4) - 144)/(exp(2*exp(2*x))*log(x)^2 - 288*x + 16*x^2 - exp(ex
p(2*x))*log(x)*(8*x - 72) + 1296),x)

[Out]

int(-(exp(exp(2*x))*(log(x)*(8*x*exp(2*x) - 4) + 4) - 144)/(exp(2*exp(2*x))*log(x)^2 - 288*x + 16*x^2 - exp(ex
p(2*x))*log(x)*(8*x - 72) + 1296), x)