\(\int \frac {e^x (-60+20 x)+e^x (200 x-60 x^2) \log (x)+e^x (60 x-100 x^2+20 x^3) \log ^2(x)}{(9 x-6 x^2+x^3) \log ^2(x)+(-18 x^2+12 x^3-2 x^4) \log ^3(x)+(9 x^3-6 x^4+x^5) \log ^4(x)} \, dx\) [506]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 108, antiderivative size = 22 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]

[Out]

4*exp(x)/ln(x)/(-3+x)/(1/5*x*ln(x)-1/5)

Rubi [F]

\[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx \]

[In]

Int[(E^x*(-60 + 20*x) + E^x*(200*x - 60*x^2)*Log[x] + E^x*(60*x - 100*x^2 + 20*x^3)*Log[x]^2)/((9*x - 6*x^2 +
x^3)*Log[x]^2 + (-18*x^2 + 12*x^3 - 2*x^4)*Log[x]^3 + (9*x^3 - 6*x^4 + x^5)*Log[x]^4),x]

[Out]

(20*Defer[Int][E^x/((-3 + x)*Log[x]^2), x])/3 - (20*Defer[Int][E^x/(x*Log[x]^2), x])/3 + 20*Defer[Int][E^x/((-
3 + x)^2*Log[x]), x] - 20*Defer[Int][E^x/((-3 + x)*Log[x]), x] - 20*Defer[Int][E^x/(-1 + x*Log[x])^2, x] - 80*
Defer[Int][E^x/((-3 + x)*(-1 + x*Log[x])^2), x] + 20*Defer[Int][E^x/(-1 + x*Log[x]), x] - 60*Defer[Int][E^x/((
-3 + x)^2*(-1 + x*Log[x])), x] + 40*Defer[Int][E^x/((-3 + x)*(-1 + x*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {20 e^x \left (-3+x+(10-3 x) x \log (x)+x \left (3-5 x+x^2\right ) \log ^2(x)\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2} \, dx \\ & = 20 \int \frac {e^x \left (-3+x+(10-3 x) x \log (x)+x \left (3-5 x+x^2\right ) \log ^2(x)\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2} \, dx \\ & = 20 \int \left (\frac {e^x}{(-3+x) x \log ^2(x)}+\frac {e^x (4-x)}{(-3+x)^2 \log (x)}+\frac {e^x (-1-x)}{(-3+x) (-1+x \log (x))^2}+\frac {e^x (-4+x) x}{(-3+x)^2 (-1+x \log (x))}\right ) \, dx \\ & = 20 \int \frac {e^x}{(-3+x) x \log ^2(x)} \, dx+20 \int \frac {e^x (4-x)}{(-3+x)^2 \log (x)} \, dx+20 \int \frac {e^x (-1-x)}{(-3+x) (-1+x \log (x))^2} \, dx+20 \int \frac {e^x (-4+x) x}{(-3+x)^2 (-1+x \log (x))} \, dx \\ & = 20 \int \left (\frac {e^x}{3 (-3+x) \log ^2(x)}-\frac {e^x}{3 x \log ^2(x)}\right ) \, dx+20 \int \left (\frac {e^x}{(-3+x)^2 \log (x)}-\frac {e^x}{(-3+x) \log (x)}\right ) \, dx+20 \int \left (-\frac {e^x}{(-1+x \log (x))^2}-\frac {4 e^x}{(-3+x) (-1+x \log (x))^2}\right ) \, dx+20 \int \left (\frac {e^x}{-1+x \log (x)}-\frac {3 e^x}{(-3+x)^2 (-1+x \log (x))}+\frac {2 e^x}{(-3+x) (-1+x \log (x))}\right ) \, dx \\ & = \frac {20}{3} \int \frac {e^x}{(-3+x) \log ^2(x)} \, dx-\frac {20}{3} \int \frac {e^x}{x \log ^2(x)} \, dx+20 \int \frac {e^x}{(-3+x)^2 \log (x)} \, dx-20 \int \frac {e^x}{(-3+x) \log (x)} \, dx-20 \int \frac {e^x}{(-1+x \log (x))^2} \, dx+20 \int \frac {e^x}{-1+x \log (x)} \, dx+40 \int \frac {e^x}{(-3+x) (-1+x \log (x))} \, dx-60 \int \frac {e^x}{(-3+x)^2 (-1+x \log (x))} \, dx-80 \int \frac {e^x}{(-3+x) (-1+x \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]

[In]

Integrate[(E^x*(-60 + 20*x) + E^x*(200*x - 60*x^2)*Log[x] + E^x*(60*x - 100*x^2 + 20*x^3)*Log[x]^2)/((9*x - 6*
x^2 + x^3)*Log[x]^2 + (-18*x^2 + 12*x^3 - 2*x^4)*Log[x]^3 + (9*x^3 - 6*x^4 + x^5)*Log[x]^4),x]

[Out]

(20*E^x)/((-3 + x)*Log[x]*(-1 + x*Log[x]))

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method result size
risch \(\frac {20 \,{\mathrm e}^{x}}{\left (-3+x \right ) \ln \left (x \right ) \left (x \ln \left (x \right )-1\right )}\) \(22\)
parallelrisch \(\frac {20 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (x^{2} \ln \left (x \right )-3 x \ln \left (x \right )-x +3\right )}\) \(27\)

[In]

int(((20*x^3-100*x^2+60*x)*exp(x)*ln(x)^2+(-60*x^2+200*x)*exp(x)*ln(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9*x^3)*ln
(x)^4+(-2*x^4+12*x^3-18*x^2)*ln(x)^3+(x^3-6*x^2+9*x)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

20*exp(x)/(-3+x)/ln(x)/(x*ln(x)-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]

[In]

integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*log(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9
*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)*log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm="fricas")

[Out]

20*e^x/((x^2 - 3*x)*log(x)^2 - (x - 3)*log(x))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^{x}}{x^{2} \log {\left (x \right )}^{2} - 3 x \log {\left (x \right )}^{2} - x \log {\left (x \right )} + 3 \log {\left (x \right )}} \]

[In]

integrate(((20*x**3-100*x**2+60*x)*exp(x)*ln(x)**2+(-60*x**2+200*x)*exp(x)*ln(x)+(20*x-60)*exp(x))/((x**5-6*x*
*4+9*x**3)*ln(x)**4+(-2*x**4+12*x**3-18*x**2)*ln(x)**3+(x**3-6*x**2+9*x)*ln(x)**2),x)

[Out]

20*exp(x)/(x**2*log(x)**2 - 3*x*log(x)**2 - x*log(x) + 3*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]

[In]

integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*log(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9
*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)*log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm="maxima")

[Out]

20*e^x/((x^2 - 3*x)*log(x)^2 - (x - 3)*log(x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{x^{2} \log \left (x\right )^{2} - 3 \, x \log \left (x\right )^{2} - x \log \left (x\right ) + 3 \, \log \left (x\right )} \]

[In]

integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*log(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9
*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)*log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm="giac")

[Out]

20*e^x/(x^2*log(x)^2 - 3*x*log(x)^2 - x*log(x) + 3*log(x))

Mupad [B] (verification not implemented)

Time = 7.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=-\frac {20\,{\mathrm {e}}^x}{\left (\ln \left (x\right )-x\,{\ln \left (x\right )}^2\right )\,\left (x-3\right )} \]

[In]

int((exp(x)*(20*x - 60) + exp(x)*log(x)^2*(60*x - 100*x^2 + 20*x^3) + exp(x)*log(x)*(200*x - 60*x^2))/(log(x)^
4*(9*x^3 - 6*x^4 + x^5) - log(x)^3*(18*x^2 - 12*x^3 + 2*x^4) + log(x)^2*(9*x - 6*x^2 + x^3)),x)

[Out]

-(20*exp(x))/((log(x) - x*log(x)^2)*(x - 3))