\(\int \frac {-5 e^x x^2 \log (5)+(-24-5 x^2) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+(-4 x+15 x^2+5 x^3) \log (5)+20 x \log (5) \log (x)} \, dx\) [3220]

Optimal result

Integrand size = 68, antiderivative size = 36 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=\log \left (\frac {3}{4 \left (3+e^x+x-\frac {\frac {1}{5} \left (4-\frac {x}{\log (5)}\right )-4 \log (x)}{x}\right )}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=\int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx \]

[In]

Int[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x^2 + 5*E^x*x^2*Log[5] + (-4*x + 15*x^2 + 5
*x^3)*Log[5] + 20*x*Log[5]*Log[x]),x]

[Out]

-x + 4*Log[5]*Defer[Int][(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*Log[5]*Log[x])^(-1
), x] + 24*Log[5]*Defer[Int][1/(x*(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*Log[5]*Lo
g[x])), x] - (1 + 10*Log[5])*Defer[Int][x/(4*Log[5] - 5*E^x*x*Log[5] - 5*x^2*Log[5] - x*(1 + 15*Log[5]) - 20*L
og[5]*Log[x]), x] + 5*Log[5]*Defer[Int][x^2/(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1 + 15*Log[5]) + 2
0*Log[5]*Log[x]), x] + 20*Log[5]*Defer[Int][Log[x]/(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1 + 15*Log[
5]) + 20*Log[5]*Log[x]), x] + 20*Log[5]*Defer[Int][Log[x]/(x*(-4*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + x*(1
+ 15*Log[5]) + 20*Log[5]*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\log (5) \left (-24-5 x^2-5 e^x x^2+20 \log (x)\right )}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx \\ & = \log (5) \int \frac {-24-5 x^2-5 e^x x^2+20 \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx \\ & = \log (5) \int \left (-\frac {1}{\log (5)}+\frac {24 \log (5)+4 x \log (5)-5 x^3 \log (5)-x^2 (1+10 \log (5))-20 \log (5) \log (x)-20 x \log (5) \log (x)}{x \log (5) \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )}\right ) \, dx \\ & = -x+\int \frac {24 \log (5)+4 x \log (5)-5 x^3 \log (5)-x^2 (1+10 \log (5))-20 \log (5) \log (x)-20 x \log (5) \log (x)}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )} \, dx \\ & = -x+\int \left (\frac {x (-1-10 \log (5))}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)}+\frac {4 \log (5)}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)}+\frac {24 \log (5)}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )}+\frac {5 x^2 \log (5)}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)}+\frac {20 \log (5) \log (x)}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)}+\frac {20 \log (5) \log (x)}{x \left (-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)\right )}\right ) \, dx \\ & = -x+(-1-10 \log (5)) \int \frac {x}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)} \, dx+(4 \log (5)) \int \frac {1}{4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)} \, dx+(5 \log (5)) \int \frac {x^2}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)} \, dx+(20 \log (5)) \int \frac {\log (x)}{-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)} \, dx+(20 \log (5)) \int \frac {\log (x)}{x \left (-4 \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+x (1+15 \log (5))+20 \log (5) \log (x)\right )} \, dx+(24 \log (5)) \int \frac {1}{x \left (4 \log (5)-5 e^x x \log (5)-5 x^2 \log (5)-x (1+15 \log (5))-20 \log (5) \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.42 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log (5) \left (-\frac {\log (x)}{\log (5)}+\frac {\log \left (x-4 \log (5)+15 x \log (5)+5 e^x x \log (5)+5 x^2 \log (5)+20 \log (5) \log (x)\right )}{\log (5)}\right ) \]

[In]

Integrate[(-5*E^x*x^2*Log[5] + (-24 - 5*x^2)*Log[5] + 20*Log[5]*Log[x])/(x^2 + 5*E^x*x^2*Log[5] + (-4*x + 15*x
^2 + 5*x^3)*Log[5] + 20*x*Log[5]*Log[x]),x]

[Out]

-(Log[5]*(-(Log[x]/Log[5]) + Log[x - 4*Log[5] + 15*x*Log[5] + 5*E^x*x*Log[5] + 5*x^2*Log[5] + 20*Log[5]*Log[x]
]/Log[5]))

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06

method	result	size
norman	\(\ln \left (x \right )-\ln \left (5 x \,{\mathrm e}^{x} \ln \left (5\right )+5 x^{2} \ln \left (5\right )+20 \ln \left (5\right ) \ln \left (x \right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x \right )\)	\(38\)
risch	\(\ln \left (x \right )-\ln \left (\ln \left (x \right )+\frac {5 x^{2} \ln \left (5\right )+5 x \,{\mathrm e}^{x} \ln \left (5\right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x}{20 \ln \left (5\right )}\right )\)	\(41\)
parallelrisch	\(-\ln \left (\frac {5 x \,{\mathrm e}^{x} \ln \left (5\right )+5 x^{2} \ln \left (5\right )+20 \ln \left (5\right ) \ln \left (x \right )+15 x \ln \left (5\right )-4 \ln \left (5\right )+x}{5 \ln \left (5\right )}\right )+\ln \left (x \right )\)	\(44\)

[In]

int((20*ln(5)*ln(x)-5*x^2*ln(5)*exp(x)+(-5*x^2-24)*ln(5))/(20*x*ln(5)*ln(x)+5*x^2*ln(5)*exp(x)+(5*x^3+15*x^2-4
*x)*ln(5)+x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(5*x*exp(x)*ln(5)+5*x^2*ln(5)+20*ln(5)*ln(x)+15*x*ln(5)-4*ln(5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (5 \, x e^{x} \log \left (5\right ) + {\left (5 \, x^{2} + 15 \, x - 4\right )} \log \left (5\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) + x\right ) + \log \left (x\right ) \]

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="fricas")

[Out]

-log(5*x*e^x*log(5) + (5*x^2 + 15*x - 4)*log(5) + 20*log(5)*log(x) + x) + log(x)

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=- \log {\left (e^{x} + \frac {5 x^{2} \log {\left (5 \right )} + x + 15 x \log {\left (5 \right )} + 20 \log {\left (5 \right )} \log {\left (x \right )} - 4 \log {\left (5 \right )}}{5 x \log {\left (5 \right )}} \right )} \]

[In]

integrate((20*ln(5)*ln(x)-5*x**2*ln(5)*exp(x)+(-5*x**2-24)*ln(5))/(20*x*ln(5)*ln(x)+5*x**2*ln(5)*exp(x)+(5*x**
3+15*x**2-4*x)*ln(5)+x**2),x)

[Out]

-log(exp(x) + (5*x**2*log(5) + x + 15*x*log(5) + 20*log(5)*log(x) - 4*log(5))/(5*x*log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (\frac {5 \, x^{2} \log \left (5\right ) + 5 \, x e^{x} \log \left (5\right ) + x {\left (15 \, \log \left (5\right ) + 1\right )} + 20 \, \log \left (5\right ) \log \left (x\right ) - 4 \, \log \left (5\right )}{5 \, x \log \left (5\right )}\right ) \]

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="maxima")

[Out]

-log(1/5*(5*x^2*log(5) + 5*x*e^x*log(5) + x*(15*log(5) + 1) + 20*log(5)*log(x) - 4*log(5))/(x*log(5)))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=-\log \left (5 \, x^{2} \log \left (5\right ) + 5 \, x e^{x} \log \left (5\right ) + 15 \, x \log \left (5\right ) + 20 \, \log \left (5\right ) \log \left (x\right ) + x - 4 \, \log \left (5\right )\right ) + \log \left (x\right ) \]

[In]

integrate((20*log(5)*log(x)-5*x^2*log(5)*exp(x)+(-5*x^2-24)*log(5))/(20*x*log(5)*log(x)+5*x^2*log(5)*exp(x)+(5
*x^3+15*x^2-4*x)*log(5)+x^2),x, algorithm="giac")

[Out]

-log(5*x^2*log(5) + 5*x*e^x*log(5) + 15*x*log(5) + 20*log(5)*log(x) + x - 4*log(5)) + log(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-5 e^x x^2 \log (5)+\left (-24-5 x^2\right ) \log (5)+20 \log (5) \log (x)}{x^2+5 e^x x^2 \log (5)+\left (-4 x+15 x^2+5 x^3\right ) \log (5)+20 x \log (5) \log (x)} \, dx=\int -\frac {\ln \left (5\right )\,\left (5\,x^2+24\right )-20\,\ln \left (5\right )\,\ln \left (x\right )+5\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )}{\ln \left (5\right )\,\left (5\,x^3+15\,x^2-4\,x\right )+x^2+20\,x\,\ln \left (5\right )\,\ln \left (x\right )+5\,x^2\,{\mathrm {e}}^x\,\ln \left (5\right )} \,d x \]

[In]

int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5)*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*
x*log(5)*log(x) + 5*x^2*exp(x)*log(5)),x)

[Out]

int(-(log(5)*(5*x^2 + 24) - 20*log(5)*log(x) + 5*x^2*exp(x)*log(5))/(log(5)*(15*x^2 - 4*x + 5*x^3) + x^2 + 20*
x*log(5)*log(x) + 5*x^2*exp(x)*log(5)), x)