Integrand size = 111, antiderivative size = 35 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=\frac {e^{e^2} (3-x) \left (-e^{-1+x}+x\right ) \log \left (\frac {5}{x}\right )}{4-x^2} \]
Time = 3.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=e^{-1+e^2} \left (-\frac {\left (e^x (-3+x)+e (-4+3 x)\right ) \log \left (\frac {5}{x}\right )}{-4+x^2}-e \log (x)\right ) \]
Integrate[(E^E^2*(-12*x + 4*x^2 + 3*x^3 - x^4 + E^(-1 + x)*(12 - 4*x - 3*x ^2 + x^3)) + E^E^2*(12*x - 8*x^2 + 3*x^3 + E^(-1 + x)*(-8*x - 2*x^2 + 4*x^ 3 - x^4))*Log[5/x])/(16*x - 8*x^3 + x^5),x]
Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(35)=70\).
Time = 4.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.77, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2026, 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.
\(\displaystyle \int \frac {e^{e^2} \left (-x^4+3 x^3+4 x^2+e^{x-1} \left (x^3-3 x^2-4 x+12\right )-12 x\right )+e^{e^2} \left (3 x^3-8 x^2+e^{x-1} \left (-x^4+4 x^3-2 x^2-8 x\right )+12 x\right ) \log \left (\frac {5}{x}\right )}{x^5-8 x^3+16 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{e^2} \left (-x^4+3 x^3+4 x^2+e^{x-1} \left (x^3-3 x^2-4 x+12\right )-12 x\right )+e^{e^2} \left (3 x^3-8 x^2+e^{x-1} \left (-x^4+4 x^3-2 x^2-8 x\right )+12 x\right ) \log \left (\frac {5}{x}\right )}{x \left (x^4-8 x^2+16\right )}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {e^{e^2} \left (x^4-3 x^3-4 x^2-e^{x-1} \left (x^3-3 x^2-4 x+12\right )+12 x\right )-e^{e^2} \left (3 x^3-8 x^2-e^{x-1} \left (x^4-4 x^3+2 x^2+8 x\right )+12 x\right ) \log \left (\frac {5}{x}\right )}{x \left (4-x^2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {e^{e^2} \left (x^4-3 x^3-4 x^2+12 x-e^{x-1} \left (x^3-3 x^2-4 x+12\right )\right )-e^{e^2} \left (3 x^3-8 x^2+12 x-e^{x-1} \left (x^4-4 x^3+2 x^2+8 x\right )\right ) \log \left (\frac {5}{x}\right )}{x \left (4-x^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {e^{e^2} x^3}{\left (x^2-4\right )^2}-\frac {3 e^{e^2} \log \left (\frac {5}{x}\right ) x^2}{\left (x^2-4\right )^2}-\frac {3 e^{e^2} x^2}{\left (x^2-4\right )^2}+\frac {8 e^{e^2} \log \left (\frac {5}{x}\right ) x}{\left (x^2-4\right )^2}-\frac {4 e^{e^2} x}{\left (x^2-4\right )^2}-\frac {12 e^{e^2} \log \left (\frac {5}{x}\right )}{\left (x^2-4\right )^2}+\frac {12 e^{e^2}}{\left (x^2-4\right )^2}+\frac {e^{x+e^2-1} \left (\log \left (\frac {5}{x}\right ) x^4-4 \log \left (\frac {5}{x}\right ) x^3-x^3+2 \log \left (\frac {5}{x}\right ) x^2+3 x^2+8 \log \left (\frac {5}{x}\right ) x+4 x-12\right )}{\left (x^2-4\right )^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{e^2} x^2 \log \left (\frac {5}{x}\right )}{4-x^2}+\frac {3 e^{e^2} x \log \left (\frac {5}{x}\right )}{4-x^2}-\frac {e^{x+e^2-1} \log \left (\frac {5}{x}\right )}{4 (2-x)}-\frac {5 e^{x+e^2-1} \log \left (\frac {5}{x}\right )}{4 (x+2)}\) |
Int[(E^E^2*(-12*x + 4*x^2 + 3*x^3 - x^4 + E^(-1 + x)*(12 - 4*x - 3*x^2 + x ^3)) + E^E^2*(12*x - 8*x^2 + 3*x^3 + E^(-1 + x)*(-8*x - 2*x^2 + 4*x^3 - x^ 4))*Log[5/x])/(16*x - 8*x^3 + x^5),x]
-1/4*(E^(-1 + E^2 + x)*Log[5/x])/(2 - x) - (5*E^(-1 + E^2 + x)*Log[5/x])/( 4*(2 + x)) + (3*E^E^2*x*Log[5/x])/(4 - x^2) - (E^E^2*x^2*Log[5/x])/(4 - x^ 2)
3.18.22.3.1 Defintions of rubi rules used
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]
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])
Leaf count of result is larger than twice the leaf count of optimal. \(65\) vs. \(2(32)=64\).
Time = 0.96 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{{\mathrm e}^{2}} \ln \left (\frac {5}{x}\right ) x^{2}-{\mathrm e}^{-1+x} {\mathrm e}^{{\mathrm e}^{2}} \ln \left (\frac {5}{x}\right ) x -3 \,{\mathrm e}^{{\mathrm e}^{2}} \ln \left (\frac {5}{x}\right ) x +3 \,{\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{-1+x} \ln \left (\frac {5}{x}\right )}{x^{2}-4}\) | \(66\) |
risch | \(\frac {{\mathrm e}^{{\mathrm e}^{2}} \left (x \,{\mathrm e}^{-1+x}+3 x -3 \,{\mathrm e}^{-1+x}-4\right ) \ln \left (x \right )}{x^{2}-4}-\frac {{\mathrm e}^{{\mathrm e}^{2}} \left (2 x^{2} \ln \left (x \right )+2 \ln \left (5\right ) x \,{\mathrm e}^{-1+x}+6 x \ln \left (5\right )-6 \ln \left (5\right ) {\mathrm e}^{-1+x}-8 \ln \left (x \right )-8 \ln \left (5\right )\right )}{2 \left (x^{2}-4\right )}\) | \(82\) |
int((((-x^4+4*x^3-2*x^2-8*x)*exp(-1+x)+3*x^3-8*x^2+12*x)*exp(exp(2))*ln(5/ x)+((x^3-3*x^2-4*x+12)*exp(-1+x)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2)))/(x^5-8 *x^3+16*x),x,method=_RETURNVERBOSE)
(exp(exp(2))*ln(5/x)*x^2-exp(-1+x)*exp(exp(2))*ln(5/x)*x-3*exp(exp(2))*ln( 5/x)*x+3*exp(exp(2))*exp(-1+x)*ln(5/x))/(x^2-4)
Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=\frac {{\left (x^{2} - {\left (x - 3\right )} e^{\left (x - 1\right )} - 3 \, x\right )} e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right )}{x^{2} - 4} \]
integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(-1+x)+3*x^3-8*x^2+12*x)*exp(exp(2)) *log(5/x)+((x^3-3*x^2-4*x+12)*exp(-1+x)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2))) /(x^5-8*x^3+16*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (24) = 48\).
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=- e^{e^{2}} \log {\left (x \right )} + \frac {\left (- 3 x e^{e^{2}} + 4 e^{e^{2}}\right ) \log {\left (\frac {5}{x} \right )}}{x^{2} - 4} + \frac {\left (- x e^{e^{2}} \log {\left (\frac {5}{x} \right )} + 3 e^{e^{2}} \log {\left (\frac {5}{x} \right )}\right ) e^{x - 1}}{x^{2} - 4} \]
integrate((((-x**4+4*x**3-2*x**2-8*x)*exp(-1+x)+3*x**3-8*x**2+12*x)*exp(ex p(2))*ln(5/x)+((x**3-3*x**2-4*x+12)*exp(-1+x)-x**4+3*x**3+4*x**2-12*x)*exp (exp(2)))/(x**5-8*x**3+16*x),x)
-exp(exp(2))*log(x) + (-3*x*exp(exp(2)) + 4*exp(exp(2)))*log(5/x)/(x**2 - 4) + (-x*exp(exp(2))*log(5/x) + 3*exp(exp(2))*log(5/x))*exp(x - 1)/(x**2 - 4)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (28) = 56\).
Time = 0.34 (sec) , antiderivative size = 140, normalized size of antiderivative = 4.00 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=\frac {3}{8} \, {\left (\frac {4 \, x}{x^{2} - 4} - \log \left (x + 2\right ) + \log \left (x - 2\right )\right )} e^{\left (e^{2}\right )} + \frac {3}{8} \, e^{\left (e^{2}\right )} \log \left (x + 2\right ) - \frac {3}{8} \, e^{\left (e^{2}\right )} \log \left (x - 2\right ) - \frac {3 \, x {\left (2 \, \log \left (5\right ) + 1\right )} e^{\left (e^{2} + 1\right )} + 2 \, {\left (x e^{\left (e^{2}\right )} \log \left (5\right ) - 3 \, e^{\left (e^{2}\right )} \log \left (5\right ) - {\left (x e^{\left (e^{2}\right )} - 3 \, e^{\left (e^{2}\right )}\right )} \log \left (x\right )\right )} e^{x} - 8 \, e^{\left (e^{2} + 1\right )} \log \left (5\right ) + 2 \, {\left (x^{2} e^{\left (e^{2} + 1\right )} - 3 \, x e^{\left (e^{2} + 1\right )}\right )} \log \left (x\right )}{2 \, {\left (x^{2} e - 4 \, e\right )}} \]
integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(-1+x)+3*x^3-8*x^2+12*x)*exp(exp(2)) *log(5/x)+((x^3-3*x^2-4*x+12)*exp(-1+x)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2))) /(x^5-8*x^3+16*x),x, algorithm=\
3/8*(4*x/(x^2 - 4) - log(x + 2) + log(x - 2))*e^(e^2) + 3/8*e^(e^2)*log(x + 2) - 3/8*e^(e^2)*log(x - 2) - 1/2*(3*x*(2*log(5) + 1)*e^(e^2 + 1) + 2*(x *e^(e^2)*log(5) - 3*e^(e^2)*log(5) - (x*e^(e^2) - 3*e^(e^2))*log(x))*e^x - 8*e^(e^2 + 1)*log(5) + 2*(x^2*e^(e^2 + 1) - 3*x*e^(e^2 + 1))*log(x))/(x^2 *e - 4*e)
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (28) = 56\).
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.20 \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=-\frac {x^{2} e^{\left (e^{2}\right )} \log \left (x\right ) + x e^{\left (x + e^{2} - 1\right )} \log \left (\frac {5}{x}\right ) + 3 \, x e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right ) - 4 \, e^{\left (e^{2}\right )} \log \left (x\right ) - 3 \, e^{\left (x + e^{2} - 1\right )} \log \left (\frac {5}{x}\right ) - 4 \, e^{\left (e^{2}\right )} \log \left (\frac {5}{x}\right )}{x^{2} - 4} \]
integrate((((-x^4+4*x^3-2*x^2-8*x)*exp(-1+x)+3*x^3-8*x^2+12*x)*exp(exp(2)) *log(5/x)+((x^3-3*x^2-4*x+12)*exp(-1+x)-x^4+3*x^3+4*x^2-12*x)*exp(exp(2))) /(x^5-8*x^3+16*x),x, algorithm=\
-(x^2*e^(e^2)*log(x) + x*e^(x + e^2 - 1)*log(5/x) + 3*x*e^(e^2)*log(5/x) - 4*e^(e^2)*log(x) - 3*e^(x + e^2 - 1)*log(5/x) - 4*e^(e^2)*log(5/x))/(x^2 - 4)
Timed out. \[ \int \frac {e^{e^2} \left (-12 x+4 x^2+3 x^3-x^4+e^{-1+x} \left (12-4 x-3 x^2+x^3\right )\right )+e^{e^2} \left (12 x-8 x^2+3 x^3+e^{-1+x} \left (-8 x-2 x^2+4 x^3-x^4\right )\right ) \log \left (\frac {5}{x}\right )}{16 x-8 x^3+x^5} \, dx=\int -\frac {{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (12\,x+{\mathrm {e}}^{x-1}\,\left (-x^3+3\,x^2+4\,x-12\right )-4\,x^2-3\,x^3+x^4\right )-{\mathrm {e}}^{{\mathrm {e}}^2}\,\ln \left (\frac {5}{x}\right )\,\left (12\,x-{\mathrm {e}}^{x-1}\,\left (x^4-4\,x^3+2\,x^2+8\,x\right )-8\,x^2+3\,x^3\right )}{x^5-8\,x^3+16\,x} \,d x \]
int(-(exp(exp(2))*(12*x + exp(x - 1)*(4*x + 3*x^2 - x^3 - 12) - 4*x^2 - 3* x^3 + x^4) - exp(exp(2))*log(5/x)*(12*x - exp(x - 1)*(8*x + 2*x^2 - 4*x^3 + x^4) - 8*x^2 + 3*x^3))/(16*x - 8*x^3 + x^5),x)