\(\int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} (12 e^6+3 e x^2)-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx\) [489]

Optimal result

Integrand size = 77, antiderivative size = 24 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=5+x-4 \log \left (-4-3 e^{1+\frac {e^5}{x}}+x+\log (5)\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=\int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx \]

[In]

Int[(8*x^2 - x^3 + E^(E^5/x)*(12*E^6 + 3*E*x^2) - x^2*Log[5])/(4*x^2 + 3*E^(1 + E^5/x)*x^2 - x^3 - x^2*Log[5])
,x]

[Out]

(-4*E^5)/x + x + 4*Defer[Int][(3*E^(1 + E^5/x) - x + 4*(1 - Log[5]/4))^(-1), x] - 4*E^5*(4 - Log[5])*Defer[Int
][1/(x^2*(3*E^(1 + E^5/x) - x + 4*(1 - Log[5]/4))), x] + 4*E^5*Defer[Int][1/(x*(3*E^(1 + E^5/x) - x + 4*(1 - L
og[5]/4))), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{3 e^{1+\frac {e^5}{x}} x^2-x^3+x^2 (4-\log (5))} \, dx \\ & = \int \frac {-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )+x^2 (8-\log (5))}{3 e^{1+\frac {e^5}{x}} x^2-x^3+x^2 (4-\log (5))} \, dx \\ & = \int \left (\frac {4 e^5+x^2}{x^2}+\frac {4 \left (e^5 x+x^2-e^5 (4-\log (5))\right )}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}\right ) \, dx \\ & = 4 \int \frac {e^5 x+x^2-e^5 (4-\log (5))}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx+\int \frac {4 e^5+x^2}{x^2} \, dx \\ & = 4 \int \left (\frac {1}{3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )}+\frac {e^5}{x \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}+\frac {e^5 (-4+\log (5))}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )}\right ) \, dx+\int \left (1+\frac {4 e^5}{x^2}\right ) \, dx \\ & = -\frac {4 e^5}{x}+x+4 \int \frac {1}{3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )} \, dx+\left (4 e^5\right ) \int \frac {1}{x \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx-\left (4 e^5 (4-\log (5))\right ) \int \frac {1}{x^2 \left (3 e^{1+\frac {e^5}{x}}-x+4 \left (1-\frac {\log (5)}{4}\right )\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=x-4 \log \left (4+3 e^{1+\frac {e^5}{x}}-x-\log (5)\right ) \]

[In]

Integrate[(8*x^2 - x^3 + E^(E^5/x)*(12*E^6 + 3*E*x^2) - x^2*Log[5])/(4*x^2 + 3*E^(1 + E^5/x)*x^2 - x^3 - x^2*L
og[5]),x]

[Out]

x - 4*Log[4 + 3*E^(1 + E^5/x) - x - Log[5]]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
norman	\(x -4 \ln \left (x -4+\ln \left (5\right )-3 \,{\mathrm e}^{\frac {{\mathrm e}^{5}}{x}} {\mathrm e}\right )\)	\(22\)
parallelrisch	\(x -4 \ln \left (x -4+\ln \left (5\right )-3 \,{\mathrm e}^{\frac {{\mathrm e}^{5}}{x}} {\mathrm e}\right )\)	\(22\)
risch	\(x -4 \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{5}}{x}}-\frac {\left (\ln \left (5\right )-4+x \right ) {\mathrm e}^{-1}}{3}\right )\)	\(23\)

[In]

int(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*ln(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2*ln(5)-
x^3+4*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=x - 4 \, \log \left (-x + 3 \, e^{\left (\frac {x + e^{5}}{x}\right )} - \log \left (5\right ) + 4\right ) \]

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="fricas")

[Out]

x - 4*log(-x + 3*e^((x + e^5)/x) - log(5) + 4)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=x - 4 \log {\left (\frac {- x - \log {\left (5 \right )} + 4}{3 e} + e^{\frac {e^{5}}{x}} \right )} \]

[In]

integrate(((12*exp(1)*exp(5)+3*x**2*exp(1))*exp(exp(5)/x)-x**2*ln(5)-x**3+8*x**2)/(3*x**2*exp(1)*exp(exp(5)/x)
-x**2*ln(5)-x**3+4*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=x - 4 \, \log \left (-\frac {1}{3} \, {\left (x - 3 \, e^{\left (\frac {e^{5}}{x} + 1\right )} + \log \left (5\right ) - 4\right )} e^{\left (-1\right )}\right ) \]

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="maxima")

[Out]

x - 4*log(-1/3*(x - 3*e^(e^5/x + 1) + log(5) - 4)*e^(-1))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (22) = 44\).

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.79 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=-x {\left (\frac {4 \, e^{10} \log \left (-\frac {e^{5} \log \left (5\right )}{x} + \frac {4 \, e^{5}}{x} + \frac {3 \, e^{\left (\frac {e^{5}}{x} + 6\right )}}{x} - e^{5}\right )}{x} - \frac {4 \, e^{10} \log \left (\frac {e^{5}}{x}\right )}{x} - e^{10}\right )} e^{\left (-10\right )} \]

[In]

integrate(((12*exp(1)*exp(5)+3*x^2*exp(1))*exp(exp(5)/x)-x^2*log(5)-x^3+8*x^2)/(3*x^2*exp(1)*exp(exp(5)/x)-x^2
*log(5)-x^3+4*x^2),x, algorithm="giac")

[Out]

-x*(4*e^10*log(-e^5*log(5)/x + 4*e^5/x + 3*e^(e^5/x + 6)/x - e^5)/x - 4*e^10*log(e^5/x)/x - e^10)*e^(-10)

Mupad [B] (verification not implemented)

Time = 7.83 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {8 x^2-x^3+e^{\frac {e^5}{x}} \left (12 e^6+3 e x^2\right )-x^2 \log (5)}{4 x^2+3 e^{1+\frac {e^5}{x}} x^2-x^3-x^2 \log (5)} \, dx=x+4\,\ln \left (\frac {1}{x}\right )-4\,\ln \left (\frac {x+\ln \left (5\right )-3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{x}}\,\mathrm {e}-4}{x}\right ) \]

[In]

int(-(exp(exp(5)/x)*(12*exp(6) + 3*x^2*exp(1)) - x^2*log(5) + 8*x^2 - x^3)/(x^2*log(5) - 4*x^2 + x^3 - 3*x^2*e
xp(exp(5)/x)*exp(1)),x)

[Out]

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