Integrand size = 59, antiderivative size = 20 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=2-2 x+\log (x)+x \log \left (e^5 x (-4+x+\log (4))\right ) \] Output:
x*ln(x*exp(5)*(x-4+2*ln(2)))-2*x+2+ln(x)
Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=3 x+\log (x)+x \log (x (-4+x+\log (4))) \] Input:
Integrate[(-4 + 5*x + (1 - x)*Log[4] + (-4*x + x^2 + x*Log[4])*Log[E^5*(-4 *x + x^2) + E^5*x*Log[4]])/(-4*x + x^2 + x*Log[4]),x]
Output:
3*x + Log[x] + x*Log[x*(-4 + x + Log[4])]
Time = 0.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 2026, 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.
\(\displaystyle \int \frac {\left (x^2-4 x+x \log (4)\right ) \log \left (e^5 \left (x^2-4 x\right )+e^5 x \log (4)\right )+5 x+(1-x) \log (4)-4}{x^2-4 x+x \log (4)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (x^2-4 x+x \log (4)\right ) \log \left (e^5 \left (x^2-4 x\right )+e^5 x \log (4)\right )+5 x+(1-x) \log (4)-4}{x^2+x (\log (4)-4)}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^2-4 x+x \log (4)\right ) \log \left (e^5 \left (x^2-4 x\right )+e^5 x \log (4)\right )+5 x+(1-x) \log (4)-4}{x (x-4+\log (4))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-(x (5-\log (4)))+4-\log (4)}{x (-x+4-\log (4))}+\log (x (x-4+\log (4)))+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 x+x \log (-x (-x+4-\log (4)))+\log (x)\) |
Input:
Int[(-4 + 5*x + (1 - x)*Log[4] + (-4*x + x^2 + x*Log[4])*Log[E^5*(-4*x + x ^2) + E^5*x*Log[4]])/(-4*x + x^2 + x*Log[4]),x]
Output:
3*x + Log[x] + x*Log[-(x*(4 - x - Log[4]))]
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]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.76 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \(-16+x \ln \left (x \,{\mathrm e}^{5} \left (x -4+2 \ln \left (2\right )\right )\right )+\ln \left (x \right )+8 \ln \left (2\right )-2 x\) | \(26\) |
norman | \(x \ln \left (2 x \,{\mathrm e}^{5} \ln \left (2\right )+\left (x^{2}-4 x \right ) {\mathrm e}^{5}\right )-2 x +\ln \left (x \right )\) | \(28\) |
risch | \(x \ln \left (2 x \,{\mathrm e}^{5} \ln \left (2\right )+\left (x^{2}-4 x \right ) {\mathrm e}^{5}\right )-2 x +\ln \left (x \right )\) | \(28\) |
default | \(\ln \left (x \right )-\left (2 \ln \left (2\right )-4\right ) \ln \left (x -4+2 \ln \left (2\right )\right )+\ln \left (2 x \,{\mathrm e}^{5} \ln \left (2\right )+x^{2} {\mathrm e}^{5}-4 x \,{\mathrm e}^{5}\right ) x -2 x -2 \left (2-\ln \left (2\right )\right ) \ln \left (x -4+2 \ln \left (2\right )\right )\) | \(61\) |
parts | \(\ln \left (x \right )-\left (2 \ln \left (2\right )-4\right ) \ln \left (x -4+2 \ln \left (2\right )\right )+\ln \left (2 x \,{\mathrm e}^{5} \ln \left (2\right )+x^{2} {\mathrm e}^{5}-4 x \,{\mathrm e}^{5}\right ) x -2 x -2 \left (2-\ln \left (2\right )\right ) \ln \left (x -4+2 \ln \left (2\right )\right )\) | \(61\) |
Input:
int(((2*x*ln(2)+x^2-4*x)*ln(2*x*exp(5)*ln(2)+(x^2-4*x)*exp(5))+2*(1-x)*ln( 2)+5*x-4)/(2*x*ln(2)+x^2-4*x),x,method=_RETURNVERBOSE)
Output:
-16+x*ln(x*exp(5)*(x-4+2*ln(2)))+ln(x)+8*ln(2)-2*x
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=x \log \left (2 \, x e^{5} \log \left (2\right ) + {\left (x^{2} - 4 \, x\right )} e^{5}\right ) - 2 \, x + \log \left (x\right ) \] Input:
integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2* (1-x)*log(2)+5*x-4)/(2*x*log(2)+x^2-4*x),x, algorithm="fricas")
Output:
x*log(2*x*e^5*log(2) + (x^2 - 4*x)*e^5) - 2*x + log(x)
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=x \log {\left (2 x e^{5} \log {\left (2 \right )} + \left (x^{2} - 4 x\right ) e^{5} \right )} - 2 x + \log {\left (x \right )} \] Input:
integrate(((2*x*ln(2)+x**2-4*x)*ln(2*x*exp(5)*ln(2)+(x**2-4*x)*exp(5))+2*( 1-x)*ln(2)+5*x-4)/(2*x*ln(2)+x**2-4*x),x)
Output:
x*log(2*x*exp(5)*log(2) + (x**2 - 4*x)*exp(5)) - 2*x + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (21) = 42\).
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 5.10 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=-{\left (\frac {\log \left (x + 2 \, \log \left (2\right ) - 4\right )}{\log \left (2\right ) - 2} - \frac {\log \left (x\right )}{\log \left (2\right ) - 2}\right )} \log \left (2\right ) + {\left (x + 2 \, \log \left (2\right ) - 4\right )} \log \left (x + 2 \, \log \left (2\right ) - 4\right ) - 2 \, \log \left (2\right ) \log \left (x + 2 \, \log \left (2\right ) - 4\right ) + x \log \left (x\right ) + 3 \, x + \frac {2 \, \log \left (x + 2 \, \log \left (2\right ) - 4\right )}{\log \left (2\right ) - 2} - \frac {2 \, \log \left (x\right )}{\log \left (2\right ) - 2} + 5 \, \log \left (x + 2 \, \log \left (2\right ) - 4\right ) \] Input:
integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2* (1-x)*log(2)+5*x-4)/(2*x*log(2)+x^2-4*x),x, algorithm="maxima")
Output:
-(log(x + 2*log(2) - 4)/(log(2) - 2) - log(x)/(log(2) - 2))*log(2) + (x + 2*log(2) - 4)*log(x + 2*log(2) - 4) - 2*log(2)*log(x + 2*log(2) - 4) + x*l og(x) + 3*x + 2*log(x + 2*log(2) - 4)/(log(2) - 2) - 2*log(x)/(log(2) - 2) + 5*log(x + 2*log(2) - 4)
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=x \log \left (x^{2} + 2 \, x \log \left (2\right ) - 4 \, x\right ) + 3 \, x + \log \left (x\right ) \] Input:
integrate(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2* (1-x)*log(2)+5*x-4)/(2*x*log(2)+x^2-4*x),x, algorithm="giac")
Output:
x*log(x^2 + 2*x*log(2) - 4*x) + 3*x + log(x)
Time = 3.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=\ln \left (x\right )-2\,x+x\,\ln \left (2\,x\,{\mathrm {e}}^5\,\ln \left (2\right )-{\mathrm {e}}^5\,\left (4\,x-x^2\right )\right ) \] Input:
int((5*x + log(2*x*exp(5)*log(2) - exp(5)*(4*x - x^2))*(2*x*log(2) - 4*x + x^2) - 2*log(2)*(x - 1) - 4)/(2*x*log(2) - 4*x + x^2),x)
Output:
log(x) - 2*x + x*log(2*x*exp(5)*log(2) - exp(5)*(4*x - x^2))
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 5.65 \[ \int \frac {-4+5 x+(1-x) \log (4)+\left (-4 x+x^2+x \log (4)\right ) \log \left (e^5 \left (-4 x+x^2\right )+e^5 x \log (4)\right )}{-4 x+x^2+x \log (4)} \, dx=2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right ) e^{5} x +e^{5} x^{2}-4 e^{5} x \right ) \mathrm {log}\left (2\right )+\mathrm {log}\left (2 \,\mathrm {log}\left (2\right ) e^{5} x +e^{5} x^{2}-4 e^{5} x \right ) x -4 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right ) e^{5} x +e^{5} x^{2}-4 e^{5} x \right )-2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )+x -4\right ) \mathrm {log}\left (2\right )+4 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )+x -4\right )-2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )+5 \,\mathrm {log}\left (x \right )-2 x \] Input:
int(((2*x*log(2)+x^2-4*x)*log(2*x*exp(5)*log(2)+(x^2-4*x)*exp(5))+2*(1-x)* log(2)+5*x-4)/(2*x*log(2)+x^2-4*x),x)
Output:
2*log(2*log(2)*e**5*x + e**5*x**2 - 4*e**5*x)*log(2) + log(2*log(2)*e**5*x + e**5*x**2 - 4*e**5*x)*x - 4*log(2*log(2)*e**5*x + e**5*x**2 - 4*e**5*x) - 2*log(2*log(2) + x - 4)*log(2) + 4*log(2*log(2) + x - 4) - 2*log(x)*log (2) + 5*log(x) - 2*x