Integrand size = 103, antiderivative size = 36 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=-1+x \left (-x+\frac {5-x+\frac {x}{e^6-x^2}}{x}\right )+\frac {x}{\log (x)} \] Output:
x*((5-x+x/(exp(6)-x^2))/x-x)+x/ln(x)-1
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=-x-x^2-\frac {x}{-e^6+x^2}+\frac {x}{\log (x)} \] Input:
Integrate[(-E^12 + 2*E^6*x^2 - x^4 + (E^12 - 2*E^6*x^2 + x^4)*Log[x] + (E^ 12*(-1 - 2*x) + x^2 - x^4 - 2*x^5 + E^6*(1 + 2*x^2 + 4*x^3))*Log[x]^2)/((E ^12 - 2*E^6*x^2 + x^4)*Log[x]^2),x]
Output:
-x - x^2 - x/(-E^6 + x^2) + x/Log[x]
Time = 0.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {1380, 25, 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.
\(\displaystyle \int \frac {-x^4+2 e^6 x^2+\left (x^4-2 e^6 x^2+e^{12}\right ) \log (x)+\left (-2 x^5-x^4+x^2+e^6 \left (4 x^3+2 x^2+1\right )+e^{12} (-2 x-1)\right ) \log ^2(x)-e^{12}}{\left (x^4-2 e^6 x^2+e^{12}\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {x^4-2 e^6 x^2-\left (x^4-2 e^6 x^2+e^{12}\right ) \log (x)-\left (-2 x^5-x^4+x^2+e^6 \left (4 x^3+2 x^2+1\right )-e^{12} (2 x+1)\right ) \log ^2(x)+e^{12}}{\left (e^6-x^2\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {x^4-2 e^6 x^2-\left (-2 x^5-x^4+x^2-e^{12} (2 x+1)+e^6 \left (4 x^3+2 x^2+1\right )\right ) \log ^2(x)-\left (x^4-2 e^6 x^2+e^{12}\right ) \log (x)+e^{12}}{\left (e^6-x^2\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {2 x^5+x^4-4 e^6 x^3-\left (1+2 e^6\right ) x^2+2 e^{12} x-e^6 \left (1-e^6\right )}{\left (e^6-x^2\right )^2}-\frac {1}{\log (x)}+\frac {1}{\log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x}{e^6-x^2}-\frac {1}{4} (2 x+1)^2+\frac {x}{\log (x)}\) |
Input:
Int[(-E^12 + 2*E^6*x^2 - x^4 + (E^12 - 2*E^6*x^2 + x^4)*Log[x] + (E^12*(-1 - 2*x) + x^2 - x^4 - 2*x^5 + E^6*(1 + 2*x^2 + 4*x^3))*Log[x]^2)/((E^12 - 2*E^6*x^2 + x^4)*Log[x]^2),x]
Output:
-1/4*(1 + 2*x)^2 + x/(E^6 - x^2) + x/Log[x]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 8.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {x}{\ln \left (x \right )}-x^{2}-x +\frac {x}{{\mathrm e}^{6}-x^{2}}\) | \(28\) |
parts | \(\frac {x}{\ln \left (x \right )}-x^{2}-x +\frac {x}{{\mathrm e}^{6}-x^{2}}\) | \(28\) |
risch | \(-\frac {x \left (-x^{3}+x \,{\mathrm e}^{6}-x^{2}+{\mathrm e}^{6}-1\right )}{{\mathrm e}^{6}-x^{2}}+\frac {x}{\ln \left (x \right )}\) | \(39\) |
norman | \(\frac {x \,{\mathrm e}^{6}+x^{3} \ln \left (x \right )+x^{4} \ln \left (x \right )-{\mathrm e}^{12} \ln \left (x \right )+\left (-{\mathrm e}^{6}+1\right ) x \ln \left (x \right )-x^{3}}{\left ({\mathrm e}^{6}-x^{2}\right ) \ln \left (x \right )}\) | \(56\) |
parallelrisch | \(-\frac {-x^{4} \ln \left (x \right )-x^{3} \ln \left (x \right )+{\mathrm e}^{12} \ln \left (x \right )+{\mathrm e}^{6} \ln \left (x \right ) x +x^{3}-x \,{\mathrm e}^{6}-x \ln \left (x \right )}{\ln \left (x \right ) \left ({\mathrm e}^{6}-x^{2}\right )}\) | \(58\) |
Input:
int((((-1-2*x)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*ln(x)^2+(exp (6)^2-2*x^2*exp(6)+x^4)*ln(x)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2*e xp(6)+x^4)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
x/ln(x)-x^2-x+x/(exp(6)-x^2)
Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x^{3} - x e^{6} - {\left (x^{4} + x^{3} - {\left (x^{2} + x\right )} e^{6} + x\right )} \log \left (x\right )}{{\left (x^{2} - e^{6}\right )} \log \left (x\right )} \] Input:
integrate((((-1-2*x)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x) ^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2 -2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="fricas")
Output:
(x^3 - x*e^6 - (x^4 + x^3 - (x^2 + x)*e^6 + x)*log(x))/((x^2 - e^6)*log(x) )
Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.47 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=- x^{2} - x + \frac {x}{\log {\left (x \right )}} - \frac {x}{x^{2} - e^{6}} \] Input:
integrate((((-2*x-1)*exp(6)**2+(4*x**3+2*x**2+1)*exp(6)-2*x**5-x**4+x**2)* ln(x)**2+(exp(6)**2-2*x**2*exp(6)+x**4)*ln(x)-exp(6)**2+2*x**2*exp(6)-x**4 )/(exp(6)**2-2*x**2*exp(6)+x**4)/ln(x)**2,x)
Output:
-x**2 - x + x/log(x) - x/(x**2 - exp(6))
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.36 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x^{3} - x e^{6} - {\left (x^{4} + x^{3} - x^{2} e^{6} - x {\left (e^{6} - 1\right )}\right )} \log \left (x\right )}{{\left (x^{2} - e^{6}\right )} \log \left (x\right )} \] Input:
integrate((((-1-2*x)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x) ^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2 -2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="maxima")
Output:
(x^3 - x*e^6 - (x^4 + x^3 - x^2*e^6 - x*(e^6 - 1))*log(x))/((x^2 - e^6)*lo g(x))
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.67 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=-\frac {x^{4} \log \left (x\right ) + x^{3} \log \left (x\right ) - x^{2} e^{6} \log \left (x\right ) - x^{3} - x e^{6} \log \left (x\right ) + x e^{6} + 2 \, x \log \left (x\right )}{x^{2} \log \left (x\right ) - e^{6} \log \left (x\right )} \] Input:
integrate((((-1-2*x)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x) ^2+(exp(6)^2-2*x^2*exp(6)+x^4)*log(x)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2 -2*x^2*exp(6)+x^4)/log(x)^2,x, algorithm="giac")
Output:
-(x^4*log(x) + x^3*log(x) - x^2*e^6*log(x) - x^3 - x*e^6*log(x) + x*e^6 + 2*x*log(x))/(x^2*log(x) - e^6*log(x))
Time = 4.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x}{\ln \left (x\right )}-x+\frac {x}{{\mathrm {e}}^6-x^2}-x^2 \] Input:
int(-(exp(12) - log(x)*(exp(12) - 2*x^2*exp(6) + x^4) - 2*x^2*exp(6) + log (x)^2*(x^4 - x^2 - exp(6)*(2*x^2 + 4*x^3 + 1) + 2*x^5 + exp(12)*(2*x + 1)) + x^4)/(log(x)^2*(exp(12) - 2*x^2*exp(6) + x^4)),x)
Output:
x/log(x) - x + x/(exp(6) - x^2) - x^2
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53 \[ \int \frac {-e^{12}+2 e^6 x^2-x^4+\left (e^{12}-2 e^6 x^2+x^4\right ) \log (x)+\left (e^{12} (-1-2 x)+x^2-x^4-2 x^5+e^6 \left (1+2 x^2+4 x^3\right )\right ) \log ^2(x)}{\left (e^{12}-2 e^6 x^2+x^4\right ) \log ^2(x)} \, dx=\frac {x \left (-\mathrm {log}\left (x \right ) e^{6} x -\mathrm {log}\left (x \right ) e^{6}+\mathrm {log}\left (x \right ) x^{3}+\mathrm {log}\left (x \right ) x^{2}+\mathrm {log}\left (x \right )+e^{6}-x^{2}\right )}{\mathrm {log}\left (x \right ) \left (e^{6}-x^{2}\right )} \] Input:
int((((-2*x-1)*exp(6)^2+(4*x^3+2*x^2+1)*exp(6)-2*x^5-x^4+x^2)*log(x)^2+(ex p(6)^2-2*x^2*exp(6)+x^4)*log(x)-exp(6)^2+2*x^2*exp(6)-x^4)/(exp(6)^2-2*x^2 *exp(6)+x^4)/log(x)^2,x)
Output:
(x*( - log(x)*e**6*x - log(x)*e**6 + log(x)*x**3 + log(x)*x**2 + log(x) + e**6 - x**2))/(log(x)*(e**6 - x**2))