\(\int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x (-20+30 x+10 x^2)+(20-10 e^x-30 x-10 x^2) \log (x)+5 \log ^2(x)} \, dx\) [885]

Optimal result

Integrand size = 88, antiderivative size = 21 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 x}{5 \left (-2+e^x+x (3+x)-\log (x)\right )} \]

[Out]

18/5*x/(exp(x)+(3+x)*x-ln(x)-2)

Rubi [F]

\[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx \]

[In]

Int[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 60*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*(-20 +
30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]

[Out]

(18*Defer[Int][(-2 + E^x + 3*x + x^2 - Log[x])^(-2), x])/5 - 18*Defer[Int][x/(-2 + E^x + 3*x + x^2 - Log[x])^2
, x] + (18*Defer[Int][x^2/(-2 + E^x + 3*x + x^2 - Log[x])^2, x])/5 + (18*Defer[Int][x^3/(-2 + E^x + 3*x + x^2
- Log[x])^2, x])/5 + (18*Defer[Int][(-2 + E^x + 3*x + x^2 - Log[x])^(-1), x])/5 - (18*Defer[Int][x/(-2 + E^x +
3*x + x^2 - Log[x]), x])/5 - (18*Defer[Int][(x*Log[x])/(-2 + E^x + 3*x + x^2 - Log[x])^2, x])/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {18 \left (-1-e^x (-1+x)-x^2-\log (x)\right )}{5 \left (2-e^x-3 x-x^2+\log (x)\right )^2} \, dx \\ & = \frac {18}{5} \int \frac {-1-e^x (-1+x)-x^2-\log (x)}{\left (2-e^x-3 x-x^2+\log (x)\right )^2} \, dx \\ & = \frac {18}{5} \int \left (-\frac {-1+x}{-2+e^x+3 x+x^2-\log (x)}+\frac {1-5 x+x^2+x^3-x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}\right ) \, dx \\ & = -\left (\frac {18}{5} \int \frac {-1+x}{-2+e^x+3 x+x^2-\log (x)} \, dx\right )+\frac {18}{5} \int \frac {1-5 x+x^2+x^3-x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx \\ & = -\left (\frac {18}{5} \int \left (-\frac {1}{-2+e^x+3 x+x^2-\log (x)}+\frac {x}{-2+e^x+3 x+x^2-\log (x)}\right ) \, dx\right )+\frac {18}{5} \int \left (\frac {1}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}-\frac {5 x}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}+\frac {x^2}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}+\frac {x^3}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}-\frac {x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2}\right ) \, dx \\ & = \frac {18}{5} \int \frac {1}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {x^2}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {x^3}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx+\frac {18}{5} \int \frac {1}{-2+e^x+3 x+x^2-\log (x)} \, dx-\frac {18}{5} \int \frac {x}{-2+e^x+3 x+x^2-\log (x)} \, dx-\frac {18}{5} \int \frac {x \log (x)}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx-18 \int \frac {x}{\left (-2+e^x+3 x+x^2-\log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=-\frac {18 x}{5 \left (2-e^x-3 x-x^2+\log (x)\right )} \]

[In]

Integrate[(-18 + E^x*(18 - 18*x) - 18*x^2 - 18*Log[x])/(20 + 5*E^(2*x) - 60*x + 25*x^2 + 30*x^3 + 5*x^4 + E^x*
(-20 + 30*x + 10*x^2) + (20 - 10*E^x - 30*x - 10*x^2)*Log[x] + 5*Log[x]^2),x]

[Out]

(-18*x)/(5*(2 - E^x - 3*x - x^2 + Log[x]))

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {18 x}{5 \left (x^{2}+3 x +{\mathrm e}^{x}-\ln \left (x \right )-2\right )}\)	\(20\)
parallelrisch	\(\frac {18 x}{5 \left (x^{2}+3 x +{\mathrm e}^{x}-\ln \left (x \right )-2\right )}\)	\(20\)

[In]

int((-18*ln(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*ln(x)^2+(-10*exp(x)-10*x^2-30*x+20)*ln(x)+5*exp(x)^2+(10*x^2+30
*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x,method=_RETURNVERBOSE)

[Out]

18/5*x/(x^2+3*x+exp(x)-ln(x)-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \]

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="fricas")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 x}{5 x^{2} + 15 x + 5 e^{x} - 5 \log {\left (x \right )} - 10} \]

[In]

integrate((-18*ln(x)+(-18*x+18)*exp(x)-18*x**2-18)/(5*ln(x)**2+(-10*exp(x)-10*x**2-30*x+20)*ln(x)+5*exp(x)**2+
(10*x**2+30*x-20)*exp(x)+5*x**4+30*x**3+25*x**2-60*x+20),x)

[Out]

18*x/(5*x**2 + 15*x + 5*exp(x) - 5*log(x) - 10)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \]

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="maxima")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18 \, x}{5 \, {\left (x^{2} + 3 \, x + e^{x} - \log \left (x\right ) - 2\right )}} \]

[In]

integrate((-18*log(x)+(-18*x+18)*exp(x)-18*x^2-18)/(5*log(x)^2+(-10*exp(x)-10*x^2-30*x+20)*log(x)+5*exp(x)^2+(
10*x^2+30*x-20)*exp(x)+5*x^4+30*x^3+25*x^2-60*x+20),x, algorithm="giac")

[Out]

18/5*x/(x^2 + 3*x + e^x - log(x) - 2)

Mupad [B] (verification not implemented)

Time = 9.88 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-18+e^x (18-18 x)-18 x^2-18 \log (x)}{20+5 e^{2 x}-60 x+25 x^2+30 x^3+5 x^4+e^x \left (-20+30 x+10 x^2\right )+\left (20-10 e^x-30 x-10 x^2\right ) \log (x)+5 \log ^2(x)} \, dx=\frac {18\,x}{5\,\left (3\,x+{\mathrm {e}}^x-\ln \left (x\right )+x^2-2\right )} \]

[In]

int(-(18*log(x) + exp(x)*(18*x - 18) + 18*x^2 + 18)/(5*exp(2*x) - 60*x + 5*log(x)^2 + exp(x)*(30*x + 10*x^2 -
20) + 25*x^2 + 30*x^3 + 5*x^4 - log(x)*(30*x + 10*exp(x) + 10*x^2 - 20) + 20),x)

[Out]

(18*x)/(5*(3*x + exp(x) - log(x) + x^2 - 2))