\(\int \frac {2+e^{16-x} (1+x)+\log (x)-\log (x^2)}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+(2+2 e^{16-x}-2 x) \log (x)+\log ^2(x)+(-2-2 e^{16-x}+2 x-2 \log (x)) \log (x^2)+\log ^2(x^2)} \, dx\) [7925]

Optimal result

Integrand size = 102, antiderivative size = 24 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x}{1+e^{16-x}-x+\log (x)-\log \left (x^2\right )} \]

[Out]

x/(exp(16-x)+1-x-ln(x^2)+ln(x))

Rubi [F]

\[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx \]

[In]

Int[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 + (2 + 2
*E^(16 - x) - 2*x)*Log[x] + Log[x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),x]

[Out]

Defer[Int][(E^x*Log[x])/((1 - x + Log[x] - Log[x^2])*(E^16 + E^x - E^x*x + E^x*Log[x] - E^x*Log[x^2])), x] + D
efer[Int][E^(16 + x)/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*Log[x^2])^2), x] +
Defer[Int][(E^(16 + x)*x^2)/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*Log[x^2])^2)
, x] - Defer[Int][(E^(16 + x)*x*Log[x])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log[x] + E^x*
Log[x^2])^2), x] + Defer[Int][(E^(16 + x)*x*Log[x^2])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x
*Log[x] + E^x*Log[x^2])^2), x] + 2*Defer[Int][E^x/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*Log
[x] + E^x*Log[x^2])), x] - Defer[Int][(E^x*Log[x^2])/((-1 + x - Log[x] + Log[x^2])*(-E^16 - E^x + E^x*x - E^x*
Log[x] + E^x*Log[x^2])), x]

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

Mathematica [A] (verified)

Time = 2.51 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {e^x x}{e^{16}+e^x-e^x x+e^x \log (x)-e^x \log \left (x^2\right )} \]

[In]

Integrate[(2 + E^(16 - x)*(1 + x) + Log[x] - Log[x^2])/(1 + E^(32 - 2*x) + E^(16 - x)*(2 - 2*x) - 2*x + x^2 +
(2 + 2*E^(16 - x) - 2*x)*Log[x] + Log[x]^2 + (-2 - 2*E^(16 - x) + 2*x - 2*Log[x])*Log[x^2] + Log[x^2]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(-\frac {x}{x -{\mathrm e}^{16-x}-\ln \left (x \right )+\ln \left (x^{2}\right )-1}\)	\(25\)
risch	\(-\frac {2 x}{-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x -2 \,{\mathrm e}^{16-x}+2 \ln \left (x \right )-2}\)	\(72\)

[In]

int((-ln(x^2)+ln(x)+(1+x)*exp(16-x)+2)/(ln(x^2)^2+(-2*ln(x)-2*exp(16-x)+2*x-2)*ln(x^2)+ln(x)^2+(2*exp(16-x)-2*
x+2)*ln(x)+exp(16-x)^2+(2-2*x)*exp(16-x)+x^2-2*x+1),x,method=_RETURNVERBOSE)

[Out]

-x/(x-exp(16-x)-ln(x)+ln(x^2)-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x}{x - e^{\left (-x + 16\right )} + \log \left (x\right ) - 1} \]

[In]

integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2-2*x)*exp(16-x)+x^2-2*x+1),x, algorithm="fricas")

[Out]

-x/(x - e^(-x + 16) + log(x) - 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x}{- x + e^{16 - x} - \log {\left (x \right )} + 1} \]

[In]

integrate((-ln(x**2)+ln(x)+(1+x)*exp(16-x)+2)/(ln(x**2)**2+(-2*ln(x)-2*exp(16-x)+2*x-2)*ln(x**2)+ln(x)**2+(2*e
xp(16-x)-2*x+2)*ln(x)+exp(16-x)**2+(2-2*x)*exp(16-x)+x**2-2*x+1),x)

[Out]

x/(-x + exp(16 - x) - log(x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x e^{x}}{{\left (x + \log \left (x\right ) - 1\right )} e^{x} - e^{16}} \]

[In]

integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2-2*x)*exp(16-x)+x^2-2*x+1),x, algorithm="maxima")

[Out]

-x*e^x/((x + log(x) - 1)*e^x - e^16)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=-\frac {x}{x - e^{\left (-x + 16\right )} + \log \left (x\right ) - 1} \]

[In]

integrate((-log(x^2)+log(x)+(1+x)*exp(16-x)+2)/(log(x^2)^2+(-2*log(x)-2*exp(16-x)+2*x-2)*log(x^2)+log(x)^2+(2*
exp(16-x)-2*x+2)*log(x)+exp(16-x)^2+(2-2*x)*exp(16-x)+x^2-2*x+1),x, algorithm="giac")

[Out]

-x/(x - e^(-x + 16) + log(x) - 1)

Mupad [B] (verification not implemented)

Time = 13.79 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {2+e^{16-x} (1+x)+\log (x)-\log \left (x^2\right )}{1+e^{32-2 x}+e^{16-x} (2-2 x)-2 x+x^2+\left (2+2 e^{16-x}-2 x\right ) \log (x)+\log ^2(x)+\left (-2-2 e^{16-x}+2 x-2 \log (x)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx=\frac {x\,{\mathrm {e}}^{x-16}}{{\mathrm {e}}^{x-16}-x\,{\mathrm {e}}^{x-16}+{\mathrm {e}}^{x-16}\,\ln \left (x\right )-\ln \left (x^2\right )\,{\mathrm {e}}^{x-16}+1} \]

[In]

int((log(x) - log(x^2) + exp(16 - x)*(x + 1) + 2)/(exp(32 - 2*x) - 2*x - log(x^2)*(2*exp(16 - x) - 2*x + 2*log
(x) + 2) - exp(16 - x)*(2*x - 2) + log(x)^2 + log(x)*(2*exp(16 - x) - 2*x + 2) + log(x^2)^2 + x^2 + 1),x)

[Out]

(x*exp(x - 16))/(exp(x - 16) - x*exp(x - 16) + exp(x - 16)*log(x) - log(x^2)*exp(x - 16) + 1)