Integrand size = 122, antiderivative size = 31 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=\log (5) \left (x-x^2-\frac {1}{\log \left (\left (e^2-\frac {2-x}{x}\right )^2\right )}\right ) \] Output:
ln(5)*(-1/ln((exp(2)-(2-x)/x)^2)+x-x^2)
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=\log (5) \left (x-x^2-\frac {1}{\log \left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \] Input:
Integrate[(4*Log[5] + (-2*x + 5*x^2 - 2*x^3 + E^2*(x^2 - 2*x^3))*Log[5]*Lo g[(4 - 4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))/x^2]^2)/((-2*x + x^2 + E^ 2*x^2)*Log[(4 - 4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))/x^2]^2),x]
Output:
Log[5]*(x - x^2 - Log[(-2 + x + E^2*x)^2/x^2]^(-1))
Time = 0.76 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6, 2026, 7239, 27, 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.
\(\displaystyle \int \frac {\left (-2 x^3+5 x^2+e^2 \left (x^2-2 x^3\right )-2 x\right ) \log (5) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )+4 \log (5)}{\left (e^2 x^2+x^2-2 x\right ) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (-2 x^3+5 x^2+e^2 \left (x^2-2 x^3\right )-2 x\right ) \log (5) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )+4 \log (5)}{\left (\left (1+e^2\right ) x^2-2 x\right ) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-2 x^3+5 x^2+e^2 \left (x^2-2 x^3\right )-2 x\right ) \log (5) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )+4 \log (5)}{x \left (\left (1+e^2\right ) x-2\right ) \log ^2\left (\frac {e^4 x^2+x^2+e^2 \left (2 x^2-4 x\right )-4 x+4}{x^2}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \log (5) \left (\frac {4}{\left (e^2 x+x-2\right ) x \log ^2\left (\frac {\left (e^2 x+x-2\right )^2}{x^2}\right )}-2 x+1\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (5) \int \left (-2 x+1-\frac {4}{\left (-e^2 x-x+2\right ) \log ^2\left (\frac {\left (-e^2 x-x+2\right )^2}{x^2}\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (5) \left (-x^2-\frac {1}{\log \left (\frac {\left (2-\left (1+e^2\right ) x\right )^2}{x^2}\right )}+x\right )\) |
Input:
Int[(4*Log[5] + (-2*x + 5*x^2 - 2*x^3 + E^2*(x^2 - 2*x^3))*Log[5]*Log[(4 - 4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))/x^2]^2)/((-2*x + x^2 + E^2*x^2) *Log[(4 - 4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))/x^2]^2),x]
Output:
Log[5]*(x - x^2 - Log[(2 - (1 + E^2)*x)^2/x^2]^(-1))
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(-\ln \left (5\right ) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\) | \(42\) |
default | \(-\ln \left (5\right ) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\) | \(42\) |
risch | \(-\ln \left (5\right ) x \left (-1+x \right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(47\) |
parts | \(-\ln \left (5\right ) \left (x^{2}-x \right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(52\) |
norman | \(\frac {x \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-x^{2} \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(119\) |
parallelrisch | \(-\frac {4 x^{2} \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-4 x \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )+4 \ln \left (5\right )}{4 \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(121\) |
Input:
int((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*ln(5)*ln((x^2*exp(2)^2+(2*x^2-4 *x)*exp(2)+x^2-4*x+4)/x^2)^2+4*ln(5))/(x^2*exp(2)+x^2-2*x)/ln((x^2*exp(2)^ 2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x,method=_RETURNVERBOSE)
Output:
-ln(5)*(x^2-x+1/ln(exp(2)^2-4*exp(2)/x+4/x^2+2*exp(2)-4/x+1))
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (26) = 52\).
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.48 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-\frac {{\left (x^{2} - x\right )} \log \left (5\right ) \log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right ) + \log \left (5\right )}{\log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right )} \] Input:
integrate((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*log(5)*log((x^2*exp(2)^2+ (2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2+4*log(5))/(x^2*exp(2)+x^2-2*x)/log((x ^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x, algorithm="fricas")
Output:
-((x^2 - x)*log(5)*log((x^2*e^4 + x^2 + 2*(x^2 - 2*x)*e^2 - 4*x + 4)/x^2) + log(5))/log((x^2*e^4 + x^2 + 2*(x^2 - 2*x)*e^2 - 4*x + 4)/x^2)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=- x^{2} \log {\left (5 \right )} + x \log {\left (5 \right )} - \frac {\log {\left (5 \right )}}{\log {\left (\frac {x^{2} + x^{2} e^{4} - 4 x + \left (2 x^{2} - 4 x\right ) e^{2} + 4}{x^{2}} \right )}} \] Input:
integrate((((-2*x**3+x**2)*exp(2)-2*x**3+5*x**2-2*x)*ln(5)*ln((x**2*exp(2) **2+(2*x**2-4*x)*exp(2)+x**2-4*x+4)/x**2)**2+4*ln(5))/(x**2*exp(2)+x**2-2* x)/ln((x**2*exp(2)**2+(2*x**2-4*x)*exp(2)+x**2-4*x+4)/x**2)**2,x)
Output:
-x**2*log(5) + x*log(5) - log(5)/log((x**2 + x**2*exp(4) - 4*x + (2*x**2 - 4*x)*exp(2) + 4)/x**2)
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (26) = 52\).
Time = 0.16 (sec) , antiderivative size = 217, normalized size of antiderivative = 7.00 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-{\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} e^{2} \log \left (5\right ) + {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} e^{2} \log \left (5\right ) - {\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} \log \left (5\right ) + 5 \, {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} \log \left (5\right ) - \frac {2 \, \log \left (5\right ) \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{2} + 1} - \frac {\log \left (5\right )}{2 \, {\left (\log \left (x {\left (e^{2} + 1\right )} - 2\right ) - \log \left (x\right )\right )}} \] Input:
integrate((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*log(5)*log((x^2*exp(2)^2+ (2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2+4*log(5))/(x^2*exp(2)+x^2-2*x)/log((x ^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x, algorithm="maxima")
Output:
-((x^2*(e^2 + 1) + 4*x)/(e^4 + 2*e^2 + 1) + 8*log(x*(e^2 + 1) - 2)/(e^6 + 3*e^4 + 3*e^2 + 1))*e^2*log(5) + (x/(e^2 + 1) + 2*log(x*(e^2 + 1) - 2)/(e^ 4 + 2*e^2 + 1))*e^2*log(5) - ((x^2*(e^2 + 1) + 4*x)/(e^4 + 2*e^2 + 1) + 8* log(x*(e^2 + 1) - 2)/(e^6 + 3*e^4 + 3*e^2 + 1))*log(5) + 5*(x/(e^2 + 1) + 2*log(x*(e^2 + 1) - 2)/(e^4 + 2*e^2 + 1))*log(5) - 2*log(5)*log(x*(e^2 + 1 ) - 2)/(e^2 + 1) - 1/2*log(5)/(log(x*(e^2 + 1) - 2) - log(x))
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (26) = 52\).
Time = 0.58 (sec) , antiderivative size = 125, normalized size of antiderivative = 4.03 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-\frac {x^{2} \log \left (5\right ) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - x^{2} \log \left (5\right ) \log \left (x^{2}\right ) - x \log \left (5\right ) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) + x \log \left (5\right ) \log \left (x^{2}\right ) + \log \left (5\right )}{\log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - \log \left (x^{2}\right )} \] Input:
integrate((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*log(5)*log((x^2*exp(2)^2+ (2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2+4*log(5))/(x^2*exp(2)+x^2-2*x)/log((x ^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x, algorithm="giac")
Output:
-(x^2*log(5)*log(x^2*e^4 + 2*x^2*e^2 + x^2 - 4*x*e^2 - 4*x + 4) - x^2*log( 5)*log(x^2) - x*log(5)*log(x^2*e^4 + 2*x^2*e^2 + x^2 - 4*x*e^2 - 4*x + 4) + x*log(5)*log(x^2) + log(5))/(log(x^2*e^4 + 2*x^2*e^2 + x^2 - 4*x*e^2 - 4 *x + 4) - log(x^2))
Time = 3.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=x\,\ln \left (5\right )-x^2\,\ln \left (5\right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (4\,x-2\,x^2\right )-4\,x+x^2+4}{x^2}\right )} \] Input:
int((4*log(5) - log(5)*log((x^2*exp(4) - exp(2)*(4*x - 2*x^2) - 4*x + x^2 + 4)/x^2)^2*(2*x - exp(2)*(x^2 - 2*x^3) - 5*x^2 + 2*x^3))/(log((x^2*exp(4) - exp(2)*(4*x - 2*x^2) - 4*x + x^2 + 4)/x^2)^2*(x^2*exp(2) - 2*x + x^2)), x)
Output:
x*log(5) - x^2*log(5) - log(5)/log((x^2*exp(4) - exp(2)*(4*x - 2*x^2) - 4* x + x^2 + 4)/x^2)
Time = 0.17 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.74 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=\frac {\mathrm {log}\left (5\right ) \left (-\mathrm {log}\left (\frac {e^{4} x^{2}+2 e^{2} x^{2}-4 e^{2} x +x^{2}-4 x +4}{x^{2}}\right ) x^{2}+\mathrm {log}\left (\frac {e^{4} x^{2}+2 e^{2} x^{2}-4 e^{2} x +x^{2}-4 x +4}{x^{2}}\right ) x -1\right )}{\mathrm {log}\left (\frac {e^{4} x^{2}+2 e^{2} x^{2}-4 e^{2} x +x^{2}-4 x +4}{x^{2}}\right )} \] Input:
int((((-2*x^3+x^2)*exp(2)-2*x^3+5*x^2-2*x)*log(5)*log((x^2*exp(2)^2+(2*x^2 -4*x)*exp(2)+x^2-4*x+4)/x^2)^2+4*log(5))/(x^2*exp(2)+x^2-2*x)/log((x^2*exp (2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x+4)/x^2)^2,x)
Output:
(log(5)*( - log((e**4*x**2 + 2*e**2*x**2 - 4*e**2*x + x**2 - 4*x + 4)/x**2 )*x**2 + log((e**4*x**2 + 2*e**2*x**2 - 4*e**2*x + x**2 - 4*x + 4)/x**2)*x - 1))/log((e**4*x**2 + 2*e**2*x**2 - 4*e**2*x + x**2 - 4*x + 4)/x**2)