Integrand size = 61, antiderivative size = 24 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=2+\frac {1}{e^3}-x+\frac {x}{\left (-2+e^2+2 x\right ) \log (4)} \] Output:
2+1/2*x/ln(2)/(2*x-2+exp(2))-x+1/exp(3)
Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=-\frac {-2+e^2-4 e^2 \log (4)+\left (-2+e^2+2 x\right )^2 \log (4)+e^2 \log (256)}{2 \left (-2+e^2+2 x\right ) \log (4)} \] Input:
Integrate[(-2 + E^2 + (-4 - E^4 + E^2*(4 - 4*x) + 8*x - 4*x^2)*Log[4])/((4 + E^4 - 8*x + 4*x^2 + E^2*(-4 + 4*x))*Log[4]),x]
Output:
-1/2*(-2 + E^2 - 4*E^2*Log[4] + (-2 + E^2 + 2*x)^2*Log[4] + E^2*Log[256])/ ((-2 + E^2 + 2*x)*Log[4])
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {27, 25, 2007, 2081, 1107, 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.
\(\displaystyle \int \frac {\left (-4 x^2+8 x+e^2 (4-4 x)-e^4-4\right ) \log (4)+e^2-2}{\left (4 x^2-8 x+e^2 (4 x-4)+e^4+4\right ) \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\log (4) \left (4 x^2-8 x-4 e^2 (1-x)+e^4+4\right )-e^2+2}{4 x^2-8 x-4 e^2 (1-x)+e^4+4}dx}{\log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\log (4) \left (4 x^2-8 x-4 e^2 (1-x)+e^4+4\right )-e^2+2}{4 x^2-8 x-4 e^2 (1-x)+e^4+4}dx}{\log (4)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\frac {\int \frac {\log (4) \left (4 x^2-8 x-4 e^2 (1-x)+e^4+4\right )-e^2+2}{\left (2 x+e^2-2\right )^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 2081 |
\(\displaystyle -\frac {\int \frac {4 \log (4) x^2-4 \left (2-e^2\right ) \log (4) x+\left (2-e^2\right ) \left (1+2 \log (4)-e^2 \log (4)\right )}{\left (2 x+e^2-2\right )^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 1107 |
\(\displaystyle -\frac {\int \left (\log (4)+\frac {2-e^2}{\left (2 x+e^2-2\right )^2}\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {2-e^2}{2 \left (-2 x-e^2+2\right )}+x \log (4)}{\log (4)}\) |
Input:
Int[(-2 + E^2 + (-4 - E^4 + E^2*(4 - 4*x) + 8*x - 4*x^2)*Log[4])/((4 + E^4 - 8*x + 4*x^2 + E^2*(-4 + 4*x))*Log[4]),x]
Output:
-(((2 - E^2)/(2*(2 - E^2 - 2*x)) + x*Log[4])/Log[4])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; F reeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b*e, 0] && IGtQ[p, 0] && !(EqQ[ m, 3] && NeQ[p, 1])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum [v, x]^p, x] /; FreeQ[{m, p}, x] && LinearQ[u, x] && QuadraticQ[v, x] && ! (LinearMatchQ[u, x] && QuadraticMatchQ[v, x])
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54
method | result | size |
risch | \(-x -\frac {{\mathrm e}^{2}}{4 \ln \left (2\right ) \left (2 x -2+{\mathrm e}^{2}\right )}+\frac {1}{2 \ln \left (2\right ) \left (2 x -2+{\mathrm e}^{2}\right )}\) | \(37\) |
gosper | \(\frac {2 \,{\mathrm e}^{4} \ln \left (2\right )-8 x^{2} \ln \left (2\right )-8 \,{\mathrm e}^{2} \ln \left (2\right )+8 \ln \left (2\right )-{\mathrm e}^{2}+2}{4 \ln \left (2\right ) \left (2 x -2+{\mathrm e}^{2}\right )}\) | \(47\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{4} \ln \left (2\right )-8 x^{2} \ln \left (2\right )-8 \,{\mathrm e}^{2} \ln \left (2\right )+8 \ln \left (2\right )-{\mathrm e}^{2}+2}{4 \ln \left (2\right ) \left (2 x -2+{\mathrm e}^{2}\right )}\) | \(47\) |
meijerg | \(-\frac {x}{2 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \ln \left (2\right ) \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}-\frac {{\mathrm e}^{4} x}{2 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}+\frac {2 \,{\mathrm e}^{2} x}{\left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}-\frac {2 x}{\left (\frac {{\mathrm e}^{2}}{2}-1\right ) \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}+\frac {{\mathrm e}^{2} x}{4 \left (\frac {{\mathrm e}^{2}}{2}-1\right ) \ln \left (2\right ) \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}+\left (2-{\mathrm e}^{2}\right ) \left (-\frac {2 x}{\left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}+\ln \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )\right )-\frac {\left ({\mathrm e}^{2}-2\right )^{2} \left (\frac {2 x \left (\frac {6 x}{{\mathrm e}^{2}-2}+6\right )}{3 \left ({\mathrm e}^{2}-2\right ) \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )}-2 \ln \left (1+\frac {2 x}{{\mathrm e}^{2}-2}\right )\right )}{4 \left (\frac {{\mathrm e}^{2}}{2}-1\right )}\) | \(272\) |
Input:
int(1/2*(2*(-exp(2)^2+(4-4*x)*exp(2)-4*x^2+8*x-4)*ln(2)+exp(2)-2)/(exp(2)^ 2+(-4+4*x)*exp(2)+4*x^2-8*x+4)/ln(2),x,method=_RETURNVERBOSE)
Output:
-x-1/4/ln(2)/(2*x-2+exp(2))*exp(2)+1/2/ln(2)/(2*x-2+exp(2))
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=-\frac {4 \, {\left (2 \, x^{2} + x e^{2} - 2 \, x\right )} \log \left (2\right ) + e^{2} - 2}{4 \, {\left (2 \, x + e^{2} - 2\right )} \log \left (2\right )} \] Input:
integrate(1/2*(2*(-exp(2)^2+(4-4*x)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/( exp(2)^2+(-4+4*x)*exp(2)+4*x^2-8*x+4)/log(2),x, algorithm="fricas")
Output:
-1/4*(4*(2*x^2 + x*e^2 - 2*x)*log(2) + e^2 - 2)/((2*x + e^2 - 2)*log(2))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=- x - \frac {-2 + e^{2}}{8 x \log {\left (2 \right )} - 8 \log {\left (2 \right )} + 4 e^{2} \log {\left (2 \right )}} \] Input:
integrate(1/2*(2*(-exp(2)**2+(4-4*x)*exp(2)-4*x**2+8*x-4)*ln(2)+exp(2)-2)/ (exp(2)**2+(-4+4*x)*exp(2)+4*x**2-8*x+4)/ln(2),x)
Output:
-x - (-2 + exp(2))/(8*x*log(2) - 8*log(2) + 4*exp(2)*log(2))
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=-\frac {4 \, x \log \left (2\right ) + \frac {e^{2} - 2}{2 \, x + e^{2} - 2}}{4 \, \log \left (2\right )} \] Input:
integrate(1/2*(2*(-exp(2)^2+(4-4*x)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/( exp(2)^2+(-4+4*x)*exp(2)+4*x^2-8*x+4)/log(2),x, algorithm="maxima")
Output:
-1/4*(4*x*log(2) + (e^2 - 2)/(2*x + e^2 - 2))/log(2)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=-\frac {4 \, x \log \left (2\right ) + \frac {e^{2} - 2}{2 \, x + e^{2} - 2}}{4 \, \log \left (2\right )} \] Input:
integrate(1/2*(2*(-exp(2)^2+(4-4*x)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/( exp(2)^2+(-4+4*x)*exp(2)+4*x^2-8*x+4)/log(2),x, algorithm="giac")
Output:
-1/4*(4*x*log(2) + (e^2 - 2)/(2*x + e^2 - 2))/log(2)
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=-x-\frac {\frac {{\mathrm {e}}^2}{2}-1}{2\,{\mathrm {e}}^2\,\ln \left (2\right )-4\,\ln \left (2\right )+4\,x\,\ln \left (2\right )} \] Input:
int(-(log(2)*(exp(4) - 8*x + 4*x^2 + exp(2)*(4*x - 4) + 4) - exp(2)/2 + 1) /(log(2)*(exp(4) - 8*x + 4*x^2 + exp(2)*(4*x - 4) + 4)),x)
Output:
- x - (exp(2)/2 - 1)/(2*exp(2)*log(2) - 4*log(2) + 4*x*log(2))
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {-2+e^2+\left (-4-e^4+e^2 (4-4 x)+8 x-4 x^2\right ) \log (4)}{\left (4+e^4-8 x+4 x^2+e^2 (-4+4 x)\right ) \log (4)} \, dx=\frac {x \left (-2 \,\mathrm {log}\left (2\right ) e^{2}-4 \,\mathrm {log}\left (2\right ) x +4 \,\mathrm {log}\left (2\right )+1\right )}{2 \,\mathrm {log}\left (2\right ) \left (e^{2}+2 x -2\right )} \] Input:
int(1/2*(2*(-exp(2)^2+(4-4*x)*exp(2)-4*x^2+8*x-4)*log(2)+exp(2)-2)/(exp(2) ^2+(-4+4*x)*exp(2)+4*x^2-8*x+4)/log(2),x)
Output:
(x*( - 2*log(2)*e**2 - 4*log(2)*x + 4*log(2) + 1))/(2*log(2)*(e**2 + 2*x - 2))