\(\int \frac {e^{e^x} (-32-16 x-10 x^2-5 x^3+(-16 x+20 x^2+5 x^3+e^x (32 x+16 x^2+10 x^3+5 x^4)) \log (x))}{(16 x+16 x^2+4 x^3) \log ^2(x)} \, dx\) [6577]

Optimal result

Integrand size = 81, antiderivative size = 24 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {e^{e^x} \left (4+\frac {5 x^2}{4}\right )}{(2+x) \log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^E^x*(-32 - 16*x - 10*x^2 - 5*x^3 + (-16*x + 20*x^2 + 5*x^3 + E^x*(32*x + 16*x^2 + 10*x^3 + 5*x^4))*Log[
x]))/((16*x + 16*x^2 + 4*x^3)*Log[x]^2),x]

[Out]

(-5*Defer[Int][E^E^x/Log[x]^2, x])/4 - 2*Defer[Int][E^E^x/(x*Log[x]^2), x] + (9*Defer[Int][E^E^x/((2 + x)*Log[
x]^2), x])/2 + (5*Defer[Int][E^E^x/Log[x], x])/4 - (5*Defer[Int][E^(E^x + x)/Log[x], x])/2 + (5*Defer[Int][(E^
(E^x + x)*x)/Log[x], x])/4 - 9*Defer[Int][E^E^x/((2 + x)^2*Log[x]), x] + 9*Defer[Int][E^(E^x + x)/((2 + x)*Log
[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{x \left (16+16 x+4 x^2\right ) \log ^2(x)} \, dx \\ & = \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{4 x (2+x)^2 \log ^2(x)} \, dx \\ & = \frac {1}{4} \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{x (2+x)^2 \log ^2(x)} \, dx \\ & = \frac {1}{4} \int \left (-\frac {16 e^{e^x}}{(2+x)^2 \log ^2(x)}-\frac {32 e^{e^x}}{x (2+x)^2 \log ^2(x)}-\frac {10 e^{e^x} x}{(2+x)^2 \log ^2(x)}-\frac {5 e^{e^x} x^2}{(2+x)^2 \log ^2(x)}-\frac {16 e^{e^x}}{(2+x)^2 \log (x)}+\frac {20 e^{e^x} x}{(2+x)^2 \log (x)}+\frac {5 e^{e^x} x^2}{(2+x)^2 \log (x)}+\frac {e^{e^x+x} \left (16+5 x^2\right )}{(2+x) \log (x)}\right ) \, dx \\ & = \frac {1}{4} \int \frac {e^{e^x+x} \left (16+5 x^2\right )}{(2+x) \log (x)} \, dx-\frac {5}{4} \int \frac {e^{e^x} x^2}{(2+x)^2 \log ^2(x)} \, dx+\frac {5}{4} \int \frac {e^{e^x} x^2}{(2+x)^2 \log (x)} \, dx-\frac {5}{2} \int \frac {e^{e^x} x}{(2+x)^2 \log ^2(x)} \, dx-4 \int \frac {e^{e^x}}{(2+x)^2 \log ^2(x)} \, dx-4 \int \frac {e^{e^x}}{(2+x)^2 \log (x)} \, dx+5 \int \frac {e^{e^x} x}{(2+x)^2 \log (x)} \, dx-8 \int \frac {e^{e^x}}{x (2+x)^2 \log ^2(x)} \, dx \\ & = \frac {1}{4} \int \left (-\frac {10 e^{e^x+x}}{\log (x)}+\frac {5 e^{e^x+x} x}{\log (x)}+\frac {36 e^{e^x+x}}{(2+x) \log (x)}\right ) \, dx-\frac {5}{4} \int \left (\frac {e^{e^x}}{\log ^2(x)}+\frac {4 e^{e^x}}{(2+x)^2 \log ^2(x)}-\frac {4 e^{e^x}}{(2+x) \log ^2(x)}\right ) \, dx+\frac {5}{4} \int \left (\frac {e^{e^x}}{\log (x)}+\frac {4 e^{e^x}}{(2+x)^2 \log (x)}-\frac {4 e^{e^x}}{(2+x) \log (x)}\right ) \, dx-\frac {5}{2} \int \left (-\frac {2 e^{e^x}}{(2+x)^2 \log ^2(x)}+\frac {e^{e^x}}{(2+x) \log ^2(x)}\right ) \, dx-4 \int \frac {e^{e^x}}{(2+x)^2 \log ^2(x)} \, dx-4 \int \frac {e^{e^x}}{(2+x)^2 \log (x)} \, dx+5 \int \left (-\frac {2 e^{e^x}}{(2+x)^2 \log (x)}+\frac {e^{e^x}}{(2+x) \log (x)}\right ) \, dx-8 \int \left (\frac {e^{e^x}}{4 x \log ^2(x)}-\frac {e^{e^x}}{2 (2+x)^2 \log ^2(x)}-\frac {e^{e^x}}{4 (2+x) \log ^2(x)}\right ) \, dx \\ & = -\left (\frac {5}{4} \int \frac {e^{e^x}}{\log ^2(x)} \, dx\right )+\frac {5}{4} \int \frac {e^{e^x}}{\log (x)} \, dx+\frac {5}{4} \int \frac {e^{e^x+x} x}{\log (x)} \, dx-2 \int \frac {e^{e^x}}{x \log ^2(x)} \, dx+2 \int \frac {e^{e^x}}{(2+x) \log ^2(x)} \, dx-\frac {5}{2} \int \frac {e^{e^x}}{(2+x) \log ^2(x)} \, dx-\frac {5}{2} \int \frac {e^{e^x+x}}{\log (x)} \, dx-4 \int \frac {e^{e^x}}{(2+x)^2 \log (x)} \, dx+5 \int \frac {e^{e^x}}{(2+x) \log ^2(x)} \, dx+5 \int \frac {e^{e^x}}{(2+x)^2 \log (x)} \, dx+9 \int \frac {e^{e^x+x}}{(2+x) \log (x)} \, dx-10 \int \frac {e^{e^x}}{(2+x)^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {e^{e^x} \left (16+5 x^2\right )}{4 (2+x) \log (x)} \]

[In]

Integrate[(E^E^x*(-32 - 16*x - 10*x^2 - 5*x^3 + (-16*x + 20*x^2 + 5*x^3 + E^x*(32*x + 16*x^2 + 10*x^3 + 5*x^4)
)*Log[x]))/((16*x + 16*x^2 + 4*x^3)*Log[x]^2),x]

[Out]

(E^E^x*(16 + 5*x^2))/(4*(2 + x)*Log[x])

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\left (5 x^{2}+16\right ) {\mathrm e}^{{\mathrm e}^{x}}}{4 \left (2+x \right ) \ln \left (x \right )}\)	\(22\)
parallelrisch	\(-\frac {-10 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-32 \,{\mathrm e}^{{\mathrm e}^{x}}}{8 \left (2+x \right ) \ln \left (x \right )}\)	\(26\)

[In]

int((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*ln(x)-5*x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x
^2+16*x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*(5*x^2+16)/(2+x)/ln(x)*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} + 16\right )} e^{\left (e^{x}\right )}}{4 \, {\left (x + 2\right )} \log \left (x\right )} \]

[In]

integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5*x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x
^3+16*x^2+16*x)/log(x)^2,x, algorithm="fricas")

[Out]

1/4*(5*x^2 + 16)*e^(e^x)/((x + 2)*log(x))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {\left (5 x^{2} + 16\right ) e^{e^{x}}}{4 x \log {\left (x \right )} + 8 \log {\left (x \right )}} \]

[In]

integrate((((5*x**4+10*x**3+16*x**2+32*x)*exp(x)+5*x**3+20*x**2-16*x)*ln(x)-5*x**3-10*x**2-16*x-32)*exp(exp(x)
)/(4*x**3+16*x**2+16*x)/ln(x)**2,x)

[Out]

(5*x**2 + 16)*exp(exp(x))/(4*x*log(x) + 8*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} + 16\right )} e^{\left (e^{x}\right )}}{4 \, {\left (x + 2\right )} \log \left (x\right )} \]

[In]

integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5*x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x
^3+16*x^2+16*x)/log(x)^2,x, algorithm="maxima")

[Out]

1/4*(5*x^2 + 16)*e^(e^x)/((x + 2)*log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {5 \, x^{2} e^{\left (x + e^{x}\right )} + 16 \, e^{\left (x + e^{x}\right )}}{4 \, {\left (x e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )\right )}} \]

[In]

integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5*x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x
^3+16*x^2+16*x)/log(x)^2,x, algorithm="giac")

[Out]

1/4*(5*x^2*e^(x + e^x) + 16*e^(x + e^x))/(x*e^x*log(x) + 2*e^x*log(x))

Mupad [B] (verification not implemented)

Time = 12.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (5\,x^2+16\right )}{4\,\ln \left (x\right )\,\left (x+2\right )} \]

[In]

int(-(exp(exp(x))*(16*x - log(x)*(exp(x)*(32*x + 16*x^2 + 10*x^3 + 5*x^4) - 16*x + 20*x^2 + 5*x^3) + 10*x^2 +
5*x^3 + 32))/(log(x)^2*(16*x + 16*x^2 + 4*x^3)),x)

[Out]

(exp(exp(x))*(5*x^2 + 16))/(4*log(x)*(x + 2))