\(\int \frac {e^x (84+6 e^5)+e^x (-70-70 x+14 x^2+e^5 (-5-5 x+x^2)) \log (x)}{1+(-10+2 x) \log (x)+(25-10 x+x^2) \log ^2(x)} \, dx\) [5101]

Optimal result

Integrand size = 66, antiderivative size = 28 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (2 x-\left (16+e^5\right ) x\right )}{-1+(5-x) \log (x)} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx \]

[In]

Int[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x^2))*Log[x])/(1 + (-10 + 2*x)*Log[x] + (25
- 10*x + x^2)*Log[x]^2),x]

[Out]

6*(14 + E^5)*Defer[Int][E^x/(1 - 5*Log[x] + x*Log[x])^2, x] + 5*(14 + E^5)*Defer[Int][E^x/((-5 + x)*(1 - 5*Log
[x] + x*Log[x])^2), x] - (14 + E^5)*Defer[Int][(E^x*x)/(1 - 5*Log[x] + x*Log[x])^2, x] - 5*(14 + E^5)*Defer[In
t][E^x/((-5 + x)*(1 - 5*Log[x] + x*Log[x])), x] + (14 + E^5)*Defer[Int][(E^x*x)/(1 - 5*Log[x] + x*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (14+e^5\right ) \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx \\ & = \left (14+e^5\right ) \int \frac {e^x \left (6+\left (-5-5 x+x^2\right ) \log (x)\right )}{(1+(-5+x) \log (x))^2} \, dx \\ & = \left (14+e^5\right ) \int \left (\frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2}+\frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))}\right ) \, dx \\ & = \left (14+e^5\right ) \int \frac {e^x \left (-25+11 x-x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x \left (-5-5 x+x^2\right )}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx \\ & = \left (14+e^5\right ) \int \left (\frac {6 e^x}{(1-5 \log (x)+x \log (x))^2}+\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2}-\frac {e^x x}{(1-5 \log (x)+x \log (x))^2}\right ) \, dx+\left (14+e^5\right ) \int \left (-\frac {5 e^x}{(-5+x) (1-5 \log (x)+x \log (x))}+\frac {e^x x}{1-5 \log (x)+x \log (x)}\right ) \, dx \\ & = \left (-14-e^5\right ) \int \frac {e^x x}{(1-5 \log (x)+x \log (x))^2} \, dx+\left (14+e^5\right ) \int \frac {e^x x}{1-5 \log (x)+x \log (x)} \, dx+\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))^2} \, dx-\left (5 \left (14+e^5\right )\right ) \int \frac {e^x}{(-5+x) (1-5 \log (x)+x \log (x))} \, dx+\left (6 \left (14+e^5\right )\right ) \int \frac {e^x}{(1-5 \log (x)+x \log (x))^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (14+e^5\right ) x}{1-5 \log (x)+x \log (x)} \]

[In]

Integrate[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x^2))*Log[x])/(1 + (-10 + 2*x)*Log[x]
+ (25 - 10*x + x^2)*Log[x]^2),x]

[Out]

(E^x*(14 + E^5)*x)/(1 - 5*Log[x] + x*Log[x])

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {\left (14+{\mathrm e}^{5}\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\)	\(21\)
risch	\(\frac {\left (14+{\mathrm e}^{5}\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\)	\(21\)
parallelrisch	\(\frac {x \,{\mathrm e}^{5} {\mathrm e}^{x}+14 \,{\mathrm e}^{x} x}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\)	\(26\)

[In]

int((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*ln(x)^2+(2*x-10)*ln
(x)+1),x,method=_RETURNVERBOSE)

[Out]

(14+exp(5))*x*exp(x)/(x*ln(x)-5*ln(x)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\left (x e^{5} + 14 \, x\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \]

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="fricas")

[Out]

(x*e^5 + 14*x)*e^x/((x - 5)*log(x) + 1)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (14 x + x e^{5}\right ) e^{x}}{x \log {\left (x \right )} - 5 \log {\left (x \right )} + 1} \]

[In]

integrate((((x**2-5*x-5)*exp(5)+14*x**2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84)*exp(x))/((x**2-10*x+25)*ln(x)**2+(
2*x-10)*ln(x)+1),x)

[Out]

(14*x + x*exp(5))*exp(x)/(x*log(x) - 5*log(x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x {\left (e^{5} + 14\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \]

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="maxima")

[Out]

x*(e^5 + 14)*e^x/((x - 5)*log(x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e^{\left (x + 5\right )} + 14 \, x e^{x}}{x \log \left (x\right ) - 5 \, \log \left (x\right ) + 1} \]

[In]

integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x))/((x^2-10*x+25)*log(x)^2+(2*
x-10)*log(x)+1),x, algorithm="giac")

[Out]

(x*e^(x + 5) + 14*x*e^x)/(x*log(x) - 5*log(x) + 1)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (6\,{\mathrm {e}}^5+84\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (70\,x+{\mathrm {e}}^5\,\left (-x^2+5\,x+5\right )-14\,x^2+70\right )}{\left (x^2-10\,x+25\right )\,{\ln \left (x\right )}^2+\left (2\,x-10\right )\,\ln \left (x\right )+1} \,d x \]

[In]

int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10)
+ log(x)^2*(x^2 - 10*x + 25) + 1),x)

[Out]

int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10)
+ log(x)^2*(x^2 - 10*x + 25) + 1), x)