Integrand size = 117, antiderivative size = 24 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \] Output:
x*(ln(5)+exp(2))+ln(-ln(x/(2+x))+x)^8
Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (e^2+\log (5)\right )+\log ^8\left (x-\log \left (\frac {x}{2+x}\right )\right ) \] Input:
Integrate[(E^2*(-2*x^2 - x^3) + (-2*x^2 - x^3)*Log[5] + (E^2*(2*x + x^2) + (2*x + x^2)*Log[5])*Log[x/(2 + x)] + (16 - 16*x - 8*x^2)*Log[x - Log[x/(2 + x)]]^7)/(-2*x^2 - x^3 + (2*x + x^2)*Log[x/(2 + x)]),x]
Output:
x*(E^2 + Log[5]) + Log[x - Log[x/(2 + x)]]^8
Time = 0.94 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6, 7292, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (-8 x^2-16 x+16\right ) \log ^7\left (x-\log \left (\frac {x}{x+2}\right )\right )+\left (e^2 \left (x^2+2 x\right )+\left (x^2+2 x\right ) \log (5)\right ) \log \left (\frac {x}{x+2}\right )+e^2 \left (-x^3-2 x^2\right )+\left (-x^3-2 x^2\right ) \log (5)}{-x^3-2 x^2+\left (x^2+2 x\right ) \log \left (\frac {x}{x+2}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (-8 x^2-16 x+16\right ) \log ^7\left (x-\log \left (\frac {x}{x+2}\right )\right )+\left (e^2 \left (x^2+2 x\right )+\left (x^2+2 x\right ) \log (5)\right ) \log \left (\frac {x}{x+2}\right )+\left (-x^3-2 x^2\right ) \left (e^2+\log (5)\right )}{-x^3-2 x^2+\left (x^2+2 x\right ) \log \left (\frac {x}{x+2}\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (-8 x^2-16 x+16\right ) \log ^7\left (x-\log \left (\frac {x}{x+2}\right )\right )\right )-\left (e^2 \left (x^2+2 x\right )+\left (x^2+2 x\right ) \log (5)\right ) \log \left (\frac {x}{x+2}\right )-\left (-x^3-2 x^2\right ) \left (e^2+\log (5)\right )}{x (x+2) \left (x-\log \left (\frac {x}{x+2}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 \left (x^2+2 x-2\right ) \log ^7\left (x-\log \left (\frac {x}{x+2}\right )\right )}{x (x+2) \left (x-\log \left (\frac {x}{x+2}\right )\right )}+e^2 \left (1+\frac {\log (5)}{e^2}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log ^8\left (x-\log \left (\frac {x}{x+2}\right )\right )+x \left (e^2+\log (5)\right )\) |
Input:
Int[(E^2*(-2*x^2 - x^3) + (-2*x^2 - x^3)*Log[5] + (E^2*(2*x + x^2) + (2*x + x^2)*Log[5])*Log[x/(2 + x)] + (16 - 16*x - 8*x^2)*Log[x - Log[x/(2 + x)] ]^7)/(-2*x^2 - x^3 + (2*x + x^2)*Log[x/(2 + x)]),x]
Output:
x*(E^2 + Log[5]) + Log[x - Log[x/(2 + x)]]^8
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Leaf count of result is larger than twice the leaf count of optimal. \(59\) vs. \(2(23)=46\).
Time = 2.88 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50
method | result | size |
default | \(\ln \left (-\frac {\frac {\ln \left (\frac {x}{2+x}\right ) x}{2+x}+\frac {2 x}{2+x}-\ln \left (\frac {x}{2+x}\right )}{\frac {x}{2+x}-1}\right )^{8}+{\mathrm e}^{2} x +x \ln \left (5\right )\) | \(60\) |
Input:
int(((-8*x^2-16*x+16)*ln(-ln(x/(2+x))+x)^7+((x^2+2*x)*ln(5)+(x^2+2*x)*exp( 2))*ln(x/(2+x))+(-x^3-2*x^2)*ln(5)+(-x^3-2*x^2)*exp(2))/((x^2+2*x)*ln(x/(2 +x))-x^3-2*x^2),x,method=_RETURNVERBOSE)
Output:
ln(-(ln(x/(2+x))*x/(2+x)+2*x/(2+x)-ln(x/(2+x)))/(x/(2+x)-1))^8+exp(2)*x+x* ln(5)
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\log \left (x - \log \left (\frac {x}{x + 2}\right )\right )^{8} + x e^{2} + x \log \left (5\right ) \] Input:
integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+ 2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2 *x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="fricas")
Output:
log(x - log(x/(x + 2)))^8 + x*e^2 + x*log(5)
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=x \left (\log {\left (5 \right )} + e^{2}\right ) + \log {\left (x - \log {\left (\frac {x}{x + 2} \right )} \right )}^{8} \] Input:
integrate(((-8*x**2-16*x+16)*ln(-ln(x/(2+x))+x)**7+((x**2+2*x)*ln(5)+(x**2 +2*x)*exp(2))*ln(x/(2+x))+(-x**3-2*x**2)*ln(5)+(-x**3-2*x**2)*exp(2))/((x* *2+2*x)*ln(x/(2+x))-x**3-2*x**2),x)
Output:
x*(log(5) + exp(2)) + log(x - log(x/(x + 2)))**8
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\log \left (x + \log \left (x + 2\right ) - \log \left (x\right )\right )^{8} + x {\left (e^{2} + \log \left (5\right )\right )} \] Input:
integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+ 2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2 *x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="maxima")
Output:
log(x + log(x + 2) - log(x))^8 + x*(e^2 + log(5))
Timed out. \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\text {Timed out} \] Input:
integrate(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+ 2*x)*exp(2))*log(x/(2+x))+(-x^3-2*x^2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2 *x)*log(x/(2+x))-x^3-2*x^2),x, algorithm="giac")
Output:
Timed out
Time = 3.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx={\ln \left (x-\ln \left (\frac {x}{x+2}\right )\right )}^8+x\,\left ({\mathrm {e}}^2+\ln \left (5\right )\right ) \] Input:
int((exp(2)*(2*x^2 + x^3) + log(5)*(2*x^2 + x^3) + log(x - log(x/(x + 2))) ^7*(16*x + 8*x^2 - 16) - log(x/(x + 2))*(exp(2)*(2*x + x^2) + log(5)*(2*x + x^2)))/(2*x^2 + x^3 - log(x/(x + 2))*(2*x + x^2)),x)
Output:
x*(exp(2) + log(5)) + log(x - log(x/(x + 2)))^8
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^2 \left (-2 x^2-x^3\right )+\left (-2 x^2-x^3\right ) \log (5)+\left (e^2 \left (2 x+x^2\right )+\left (2 x+x^2\right ) \log (5)\right ) \log \left (\frac {x}{2+x}\right )+\left (16-16 x-8 x^2\right ) \log ^7\left (x-\log \left (\frac {x}{2+x}\right )\right )}{-2 x^2-x^3+\left (2 x+x^2\right ) \log \left (\frac {x}{2+x}\right )} \, dx=\mathrm {log}\left (-\mathrm {log}\left (\frac {x}{x +2}\right )+x \right )^{8}+\mathrm {log}\left (5\right ) x +e^{2} x \] Input:
int(((-8*x^2-16*x+16)*log(-log(x/(2+x))+x)^7+((x^2+2*x)*log(5)+(x^2+2*x)*e xp(2))*log(x/(2+x))+(-x^3-2*x^2)*log(5)+(-x^3-2*x^2)*exp(2))/((x^2+2*x)*lo g(x/(2+x))-x^3-2*x^2),x)
Output:
log( - log(x/(x + 2)) + x)**8 + log(5)*x + e**2*x