\(\int \frac {-20 x^2+e^{5 x} (e^x (30-150 x)+5 x-125 x^2)+e^x (5 x-25 x^2)+(-5 x^2-5 e^x x^2+e^{5 x} (e^x (5-30 x)-25 x^2)) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} (4 e^{2 x}+8 e^x x+4 x^2)+e^{5 x} (8 e^{2 x} x+16 e^x x^2+8 x^3)} \, dx\) [1234]

Optimal result

Integrand size = 161, antiderivative size = 29 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 x \left (1+\frac {1}{4} (1+\log (x))\right )}{\left (e^x+x\right ) \left (e^{5 x}+x\right )} \]

[Out]

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

Rubi [F(-1)]

Timed out. \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\text {\$Aborted} \]

[In]

Int[(-20*x^2 + E^(5*x)*(E^x*(30 - 150*x) + 5*x - 125*x^2) + E^x*(5*x - 25*x^2) + (-5*x^2 - 5*E^x*x^2 + E^(5*x)
*(E^x*(5 - 30*x) - 25*x^2))*Log[x])/(4*E^(2*x)*x^2 + 8*E^x*x^3 + 4*x^4 + E^(10*x)*(4*E^(2*x) + 8*E^x*x + 4*x^2
) + E^(5*x)*(8*E^(2*x)*x + 16*E^x*x^2 + 8*x^3)),x]

[Out]

$Aborted

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 x (5+\log (x))}{4 \left (e^{6 x}+e^x x+e^{5 x} x+x^2\right )} \]

[In]

Integrate[(-20*x^2 + E^(5*x)*(E^x*(30 - 150*x) + 5*x - 125*x^2) + E^x*(5*x - 25*x^2) + (-5*x^2 - 5*E^x*x^2 + E
^(5*x)*(E^x*(5 - 30*x) - 25*x^2))*Log[x])/(4*E^(2*x)*x^2 + 8*E^x*x^3 + 4*x^4 + E^(10*x)*(4*E^(2*x) + 8*E^x*x +
4*x^2) + E^(5*x)*(8*E^(2*x)*x + 16*E^x*x^2 + 8*x^3)),x]

[Out]

(5*x*(5 + Log[x]))/(4*(E^(6*x) + E^x*x + E^(5*x)*x + x^2))

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(-\frac {-5 x \ln \left (x \right )-25 x}{4 \left ({\mathrm e}^{x} {\mathrm e}^{5 x}+{\mathrm e}^{x} x +x \,{\mathrm e}^{5 x}+x^{2}\right )}\)	\(35\)
risch	\(\frac {5 x \ln \left (x \right )}{4 \left (x \,{\mathrm e}^{5 x}+{\mathrm e}^{6 x}+x^{2}+{\mathrm e}^{x} x \right )}+\frac {25 x}{4 \left (x \,{\mathrm e}^{5 x}+{\mathrm e}^{6 x}+x^{2}+{\mathrm e}^{x} x \right )}\)	\(50\)

[In]

int(((((-30*x+5)*exp(x)-25*x^2)*exp(5*x)-5*exp(x)*x^2-5*x^2)*ln(x)+((-150*x+30)*exp(x)-125*x^2+5*x)*exp(5*x)+(
-25*x^2+5*x)*exp(x)-20*x^2)/((4*exp(x)^2+8*exp(x)*x+4*x^2)*exp(5*x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3)*exp(5
*x)+4*exp(x)^2*x^2+8*exp(x)*x^3+4*x^4),x,method=_RETURNVERBOSE)

[Out]

-1/4*(-5*x*ln(x)-25*x)/(exp(x)*exp(5*x)+exp(x)*x+x*exp(5*x)+x^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 \, {\left (x \log \left (x\right ) + 5 \, x\right )}}{4 \, {\left (x^{2} + x e^{\left (5 \, x\right )} + x e^{x} + e^{\left (6 \, x\right )}\right )}} \]

[In]

integrate(((((-30*x+5)*exp(x)-25*x^2)*exp(5*x)-5*exp(x)*x^2-5*x^2)*log(x)+((-150*x+30)*exp(x)-125*x^2+5*x)*exp
(5*x)+(-25*x^2+5*x)*exp(x)-20*x^2)/((4*exp(x)^2+8*exp(x)*x+4*x^2)*exp(5*x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3
)*exp(5*x)+4*exp(x)^2*x^2+8*exp(x)*x^3+4*x^4),x, algorithm="fricas")

[Out]

5/4*(x*log(x) + 5*x)/(x^2 + x*e^(5*x) + x*e^x + e^(6*x))

Sympy [A] (verification not implemented)

Time = 1.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 x \log {\left (x \right )} + 25 x}{4 x^{2} + 4 x e^{5 x} + 4 x e^{x} + 4 e^{6 x}} \]

[In]

integrate(((((-30*x+5)*exp(x)-25*x**2)*exp(5*x)-5*exp(x)*x**2-5*x**2)*ln(x)+((-150*x+30)*exp(x)-125*x**2+5*x)*
exp(5*x)+(-25*x**2+5*x)*exp(x)-20*x**2)/((4*exp(x)**2+8*exp(x)*x+4*x**2)*exp(5*x)**2+(8*x*exp(x)**2+16*exp(x)*
x**2+8*x**3)*exp(5*x)+4*exp(x)**2*x**2+8*exp(x)*x**3+4*x**4),x)

[Out]

(5*x*log(x) + 25*x)/(4*x**2 + 4*x*exp(5*x) + 4*x*exp(x) + 4*exp(6*x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 \, {\left (x \log \left (x\right ) + 5 \, x\right )}}{4 \, {\left (x^{2} + x e^{\left (5 \, x\right )} + x e^{x} + e^{\left (6 \, x\right )}\right )}} \]

[In]

integrate(((((-30*x+5)*exp(x)-25*x^2)*exp(5*x)-5*exp(x)*x^2-5*x^2)*log(x)+((-150*x+30)*exp(x)-125*x^2+5*x)*exp
(5*x)+(-25*x^2+5*x)*exp(x)-20*x^2)/((4*exp(x)^2+8*exp(x)*x+4*x^2)*exp(5*x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3
)*exp(5*x)+4*exp(x)^2*x^2+8*exp(x)*x^3+4*x^4),x, algorithm="maxima")

[Out]

5/4*(x*log(x) + 5*x)/(x^2 + x*e^(5*x) + x*e^x + e^(6*x))

Giac [A] (verification not implemented)

none

Time = 30.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5 \, {\left (x \log \left (x\right ) + 5 \, x\right )}}{4 \, {\left (x^{2} + x e^{\left (5 \, x\right )} + x e^{x} + e^{\left (6 \, x\right )}\right )}} \]

[In]

integrate(((((-30*x+5)*exp(x)-25*x^2)*exp(5*x)-5*exp(x)*x^2-5*x^2)*log(x)+((-150*x+30)*exp(x)-125*x^2+5*x)*exp
(5*x)+(-25*x^2+5*x)*exp(x)-20*x^2)/((4*exp(x)^2+8*exp(x)*x+4*x^2)*exp(5*x)^2+(8*x*exp(x)^2+16*exp(x)*x^2+8*x^3
)*exp(5*x)+4*exp(x)^2*x^2+8*exp(x)*x^3+4*x^4),x, algorithm="giac")

[Out]

5/4*(x*log(x) + 5*x)/(x^2 + x*e^(5*x) + x*e^x + e^(6*x))

Mupad [B] (verification not implemented)

Time = 8.95 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {-20 x^2+e^{5 x} \left (e^x (30-150 x)+5 x-125 x^2\right )+e^x \left (5 x-25 x^2\right )+\left (-5 x^2-5 e^x x^2+e^{5 x} \left (e^x (5-30 x)-25 x^2\right )\right ) \log (x)}{4 e^{2 x} x^2+8 e^x x^3+4 x^4+e^{10 x} \left (4 e^{2 x}+8 e^x x+4 x^2\right )+e^{5 x} \left (8 e^{2 x} x+16 e^x x^2+8 x^3\right )} \, dx=\frac {5\,x\,\left (\ln \left (x\right )+5\right )}{4\,\left ({\mathrm {e}}^{6\,x}+x\,{\mathrm {e}}^{5\,x}+x\,{\mathrm {e}}^x+x^2\right )} \]

[In]

int(-(exp(5*x)*(exp(x)*(150*x - 30) - 5*x + 125*x^2) + log(x)*(5*x^2*exp(x) + exp(5*x)*(exp(x)*(30*x - 5) + 25
*x^2) + 5*x^2) - exp(x)*(5*x - 25*x^2) + 20*x^2)/(8*x^3*exp(x) + exp(5*x)*(8*x*exp(2*x) + 16*x^2*exp(x) + 8*x^
3) + 4*x^2*exp(2*x) + exp(10*x)*(4*exp(2*x) + 8*x*exp(x) + 4*x^2) + 4*x^4),x)

[Out]

(5*x*(log(x) + 5))/(4*(exp(6*x) + x*exp(5*x) + x*exp(x) + x^2))