Integrand size = 106, antiderivative size = 18 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {1}{-2-\frac {4 x}{2-e^2}+\log (x)} \] Output:
1/(ln(x)-2-4/(2-exp(2))*x)
Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {-2+e^2}{4-2 e^2+4 x+\left (-2+e^2\right ) \log (x)} \] Input:
Integrate[(-4 - E^4 + E^2*(4 - 4*x) + 8*x)/(16*x + 4*E^4*x + 32*x^2 + 16*x ^3 + E^2*(-16*x - 16*x^2) + (-16*x - 4*E^4*x - 16*x^2 + E^2*(16*x + 8*x^2) )*Log[x] + (4*x - 4*E^2*x + E^4*x)*Log[x]^2),x]
Output:
(-2 + E^2)/(4 - 2*E^2 + 4*x + (-2 + E^2)*Log[x])
Time = 0.47 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.94, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6, 7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {e^2 (4-4 x)+8 x-e^4-4}{16 x^3+32 x^2+e^2 \left (-16 x^2-16 x\right )+\left (-16 x^2+e^2 \left (8 x^2+16 x\right )-4 e^4 x-16 x\right ) \log (x)+4 e^4 x+16 x+\left (e^4 x-4 e^2 x+4 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^2 (4-4 x)+8 x-e^4-4}{16 x^3+32 x^2+e^2 \left (-16 x^2-16 x\right )+\left (-16 x^2+e^2 \left (8 x^2+16 x\right )-4 e^4 x-16 x\right ) \log (x)+\left (16+4 e^4\right ) x+\left (e^4 x-4 e^2 x+4 x\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (2-e^2\right ) \left (4 x+e^2-2\right )}{x \left (4 x+\left (e^2-2\right ) \log (x)+4 \left (1-\frac {e^2}{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (2-e^2\right ) \int -\frac {-4 x-e^2+2}{x \left (4 x-\left (2-e^2\right ) \log (x)+2 \left (2-e^2\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\left (\left (2-e^2\right ) \int \frac {-4 x-e^2+2}{x \left (4 x-\left (2-e^2\right ) \log (x)+2 \left (2-e^2\right )\right )^2}dx\right )\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {2-e^2}{4 x-\left (2-e^2\right ) \log (x)+2 \left (2-e^2\right )}\) |
Input:
Int[(-4 - E^4 + E^2*(4 - 4*x) + 8*x)/(16*x + 4*E^4*x + 32*x^2 + 16*x^3 + E ^2*(-16*x - 16*x^2) + (-16*x - 4*E^4*x - 16*x^2 + E^2*(16*x + 8*x^2))*Log[ x] + (4*x - 4*E^2*x + E^4*x)*Log[x]^2),x]
Output:
-((2 - E^2)/(2*(2 - E^2) + 4*x - (2 - E^2)*Log[x]))
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[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.92 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44
method | result | size |
norman | \(\frac {{\mathrm e}^{2}-2}{{\mathrm e}^{2} \ln \left (x \right )-2 \ln \left (x \right )-2 \,{\mathrm e}^{2}+4 x +4}\) | \(26\) |
default | \(-\frac {2-{\mathrm e}^{2}}{{\mathrm e}^{2} \ln \left (x \right )-2 \ln \left (x \right )-2 \,{\mathrm e}^{2}+4 x +4}\) | \(29\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{2}-8}{4 \,{\mathrm e}^{2} \ln \left (x \right )-8 \ln \left (x \right )-8 \,{\mathrm e}^{2}+16 x +16}\) | \(29\) |
risch | \(\frac {{\mathrm e}^{2}}{{\mathrm e}^{2} \ln \left (x \right )-2 \ln \left (x \right )-2 \,{\mathrm e}^{2}+4 x +4}-\frac {2}{{\mathrm e}^{2} \ln \left (x \right )-2 \ln \left (x \right )-2 \,{\mathrm e}^{2}+4 x +4}\) | \(47\) |
Input:
int((-exp(2)^2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*ln(x)^2+ (-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-16*x^2-16*x)*ln(x)+4*x*exp(2)^2+(-16*x^ 2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x,method=_RETURNVERBOSE)
Output:
(exp(2)-2)/(exp(2)*ln(x)-2*ln(x)-2*exp(2)+4*x+4)
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {e^{2} - 2}{{\left (e^{2} - 2\right )} \log \left (x\right ) + 4 \, x - 2 \, e^{2} + 4} \] Input:
integrate((-exp(2)^2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*lo g(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2 +(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="fricas")
Output:
(e^2 - 2)/((e^2 - 2)*log(x) + 4*x - 2*e^2 + 4)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {-2 + e^{2}}{4 x + \left (-2 + e^{2}\right ) \log {\left (x \right )} - 2 e^{2} + 4} \] Input:
integrate((-exp(2)**2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)**2-4*exp(2)*x+4*x)* ln(x)**2+(-4*x*exp(2)**2+(8*x**2+16*x)*exp(2)-16*x**2-16*x)*ln(x)+4*x*exp( 2)**2+(-16*x**2-16*x)*exp(2)+16*x**3+32*x**2+16*x),x)
Output:
(-2 + exp(2))/(4*x + (-2 + exp(2))*log(x) - 2*exp(2) + 4)
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {e^{2} - 2}{{\left (e^{2} - 2\right )} \log \left (x\right ) + 4 \, x - 2 \, e^{2} + 4} \] Input:
integrate((-exp(2)^2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*lo g(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2 +(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="maxima")
Output:
(e^2 - 2)/((e^2 - 2)*log(x) + 4*x - 2*e^2 + 4)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {e^{2} - 2}{e^{2} \log \left (x\right ) + 4 \, x - 2 \, e^{2} - 2 \, \log \left (x\right ) + 4} \] Input:
integrate((-exp(2)^2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*lo g(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2 +(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="giac")
Output:
(e^2 - 2)/(e^2*log(x) + 4*x - 2*e^2 - 2*log(x) + 4)
Time = 3.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^2-2}{4\,x-2\,{\mathrm {e}}^2-2\,\ln \left (x\right )+{\mathrm {e}}^2\,\ln \left (x\right )+4} \] Input:
int(-(exp(4) - 8*x + exp(2)*(4*x - 4) + 4)/(16*x - log(x)*(16*x - exp(2)*( 16*x + 8*x^2) + 4*x*exp(4) + 16*x^2) - exp(2)*(16*x + 16*x^2) + 4*x*exp(4) + log(x)^2*(4*x - 4*x*exp(2) + x*exp(4)) + 32*x^2 + 16*x^3),x)
Output:
(exp(2) - 2)/(4*x - 2*exp(2) - 2*log(x) + exp(2)*log(x) + 4)
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56 \[ \int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx=\frac {e^{2}-2}{\mathrm {log}\left (x \right ) e^{2}-2 \,\mathrm {log}\left (x \right )-2 e^{2}+4 x +4} \] Input:
int((-exp(2)^2+(4-4*x)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*log(x)^2 +(-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2+(-16* x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x)
Output:
(e**2 - 2)/(log(x)*e**2 - 2*log(x) - 2*e**2 + 4*x + 4)