Integrand size = 100, antiderivative size = 23 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=x+\frac {e^{-147+x}+x}{x \left (1+e^5+x^2\right )} \]
Time = 3.97 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=\frac {e^x+e^{152} x^2+e^{147} x \left (1+x+x^3\right )}{e^{147} x \left (1+e^5+x^2\right )} \]
Integrate[(x^2 + E^10*x^2 - 2*x^3 + 2*x^4 + x^6 + E^(-147 + x)*(-1 + E^5*( -1 + x) + x - 3*x^2 + x^3) + E^5*(2*x^2 + 2*x^4))/(x^2 + E^10*x^2 + 2*x^4 + x^6 + E^5*(2*x^2 + 2*x^4)),x]
Leaf count is larger than twice the leaf count of optimal. \(233\) vs. \(2(23)=46\).
Time = 1.02 (sec) , antiderivative size = 233, normalized size of antiderivative = 10.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 6, 2026, 1380, 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.
\(\displaystyle \int \frac {x^6+2 x^4-2 x^3+e^{10} x^2+x^2+e^5 \left (2 x^4+2 x^2\right )+e^{x-147} \left (x^3-3 x^2+x+e^5 (x-1)-1\right )}{x^6+2 x^4+e^{10} x^2+x^2+e^5 \left (2 x^4+2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^6+2 x^4-2 x^3+e^{10} x^2+x^2+e^5 \left (2 x^4+2 x^2\right )+e^{x-147} \left (x^3-3 x^2+x+e^5 (x-1)-1\right )}{x^6+2 x^4+\left (1+e^{10}\right ) x^2+e^5 \left (2 x^4+2 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^6+2 x^4-2 x^3+\left (1+e^{10}\right ) x^2+e^5 \left (2 x^4+2 x^2\right )+e^{x-147} \left (x^3-3 x^2+x+e^5 (x-1)-1\right )}{x^6+2 x^4+\left (1+e^{10}\right ) x^2+e^5 \left (2 x^4+2 x^2\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^6+2 x^4-2 x^3+\left (1+e^{10}\right ) x^2+e^5 \left (2 x^4+2 x^2\right )+e^{x-147} \left (x^3-3 x^2+x+e^5 (x-1)-1\right )}{x^2 \left (x^4+2 \left (1+e^5\right ) x^2+\left (1+e^5\right )^2\right )}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {x^6+2 x^4-2 x^3+\left (1+e^{10}\right ) x^2+2 e^5 \left (x^4+x^2\right )-e^{x-147} \left (-x^3+3 x^2-x+e^5 (1-x)+1\right )}{x^2 \left (x^2+e^5+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 x^2}{\left (x^2+e^5+1\right )^2}-\frac {2 x}{\left (x^2+e^5+1\right )^2}+\frac {2 e^5 \left (x^2+1\right )}{\left (x^2+e^5+1\right )^2}+\frac {1+e^{10}}{\left (x^2+e^5+1\right )^2}+\frac {x^4}{\left (x^2+e^5+1\right )^2}+\frac {e^{x-147} \left (x^3-3 x^2+\left (1+e^5\right ) x-e^5-1\right )}{\left (x^2+e^5+1\right )^2 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (1+e^{10}\right ) \arctan \left (\frac {x}{\sqrt {1+e^5}}\right )}{2 \left (1+e^5\right )^{3/2}}+\frac {e^5 \left (2+e^5\right ) \arctan \left (\frac {x}{\sqrt {1+e^5}}\right )}{\left (1+e^5\right )^{3/2}}-\frac {3}{2} \sqrt {1+e^5} \arctan \left (\frac {x}{\sqrt {1+e^5}}\right )+\frac {\arctan \left (\frac {x}{\sqrt {1+e^5}}\right )}{\sqrt {1+e^5}}+\frac {\left (1+e^{10}\right ) x}{2 \left (1+e^5\right ) \left (x^2+e^5+1\right )}-\frac {e^{10} x}{\left (1+e^5\right ) \left (x^2+e^5+1\right )}-\frac {x}{x^2+e^5+1}+\frac {1}{x^2+e^5+1}-\frac {x^3}{2 \left (x^2+e^5+1\right )}+\frac {e^{x-147} \left (x^3+\left (1+e^5\right ) x\right )}{\left (x^2+e^5+1\right )^2 x^2}+\frac {3 x}{2}\) |
Int[(x^2 + E^10*x^2 - 2*x^3 + 2*x^4 + x^6 + E^(-147 + x)*(-1 + E^5*(-1 + x ) + x - 3*x^2 + x^3) + E^5*(2*x^2 + 2*x^4))/(x^2 + E^10*x^2 + 2*x^4 + x^6 + E^5*(2*x^2 + 2*x^4)),x]
(3*x)/2 + (1 + E^5 + x^2)^(-1) - x/(1 + E^5 + x^2) - (E^10*x)/((1 + E^5)*( 1 + E^5 + x^2)) + ((1 + E^10)*x)/(2*(1 + E^5)*(1 + E^5 + x^2)) - x^3/(2*(1 + E^5 + x^2)) + (E^(-147 + x)*((1 + E^5)*x + x^3))/(x^2*(1 + E^5 + x^2)^2 ) + ArcTan[x/Sqrt[1 + E^5]]/Sqrt[1 + E^5] - (3*Sqrt[1 + E^5]*ArcTan[x/Sqrt [1 + E^5]])/2 + (E^5*(2 + E^5)*ArcTan[x/Sqrt[1 + E^5]])/(1 + E^5)^(3/2) + ((1 + E^10)*ArcTan[x/Sqrt[1 + E^5]])/(2*(1 + E^5)^(3/2))
3.28.90.3.1 Defintions of rubi rules used
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[(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])
Time = 1.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26
method | result | size |
risch | \(x +\frac {1}{1+x^{2}+{\mathrm e}^{5}}+\frac {{\mathrm e}^{x -147}}{x \left (1+x^{2}+{\mathrm e}^{5}\right )}\) | \(29\) |
norman | \(\frac {x^{4}+x +\left ({\mathrm e}^{5}+1\right ) x^{2}+{\mathrm e}^{x -147}}{x \left (1+x^{2}+{\mathrm e}^{5}\right )}\) | \(31\) |
parallelrisch | \(\frac {x^{4}+x^{2} {\mathrm e}^{5}+x^{2}+{\mathrm e}^{x -147}+x}{x \left (1+x^{2}+{\mathrm e}^{5}\right )}\) | \(32\) |
default | \(\text {Expression too large to display}\) | \(1335\) |
parts | \(\text {Expression too large to display}\) | \(1785\) |
int((((-1+x)*exp(5)+x^3-3*x^2+x-1)*exp(x-147)+x^2*exp(5)^2+(2*x^4+2*x^2)*e xp(5)+x^6+2*x^4-2*x^3+x^2)/(x^2*exp(5)^2+(2*x^4+2*x^2)*exp(5)+x^6+2*x^4+x^ 2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=\frac {x^{4} + x^{2} e^{5} + x^{2} + x + e^{\left (x - 147\right )}}{x^{3} + x e^{5} + x} \]
integrate((((-1+x)*exp(5)+x^3-3*x^2+x-1)*exp(x-147)+x^2*exp(5)^2+(2*x^4+2* x^2)*exp(5)+x^6+2*x^4-2*x^3+x^2)/(x^2*exp(5)^2+(2*x^4+2*x^2)*exp(5)+x^6+2* x^4+x^2),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=x + \frac {e^{x - 147}}{x^{3} + x + x e^{5}} + \frac {1}{x^{2} + 1 + e^{5}} \]
integrate((((-1+x)*exp(5)+x**3-3*x**2+x-1)*exp(x-147)+x**2*exp(5)**2+(2*x* *4+2*x**2)*exp(5)+x**6+2*x**4-2*x**3+x**2)/(x**2*exp(5)**2+(2*x**4+2*x**2) *exp(5)+x**6+2*x**4+x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (21) = 42\).
Time = 0.35 (sec) , antiderivative size = 242, normalized size of antiderivative = 10.52 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=\frac {1}{2} \, {\left (\frac {x}{x^{2} {\left (e^{5} + 1\right )} + e^{10} + 2 \, e^{5} + 1} + \frac {\arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right )}{{\left (e^{5} + 1\right )}^{\frac {3}{2}}}\right )} e^{10} + {\left (\frac {\arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right )}{\sqrt {e^{5} + 1}} - \frac {x}{x^{2} + e^{5} + 1}\right )} e^{5} + {\left (\frac {x}{x^{2} {\left (e^{5} + 1\right )} + e^{10} + 2 \, e^{5} + 1} + \frac {\arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right )}{{\left (e^{5} + 1\right )}^{\frac {3}{2}}}\right )} e^{5} - \frac {3}{2} \, \sqrt {e^{5} + 1} \arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right ) + x + \frac {x {\left (e^{5} + 1\right )}}{2 \, {\left (x^{2} + e^{5} + 1\right )}} + \frac {\arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right )}{\sqrt {e^{5} + 1}} + \frac {x}{2 \, {\left (x^{2} {\left (e^{5} + 1\right )} + e^{10} + 2 \, e^{5} + 1\right )}} - \frac {x}{x^{2} + e^{5} + 1} + \frac {e^{x}}{x^{3} e^{147} + x {\left (e^{152} + e^{147}\right )}} + \frac {\arctan \left (\frac {x}{\sqrt {e^{5} + 1}}\right )}{2 \, {\left (e^{5} + 1\right )}^{\frac {3}{2}}} + \frac {1}{x^{2} + e^{5} + 1} \]
integrate((((-1+x)*exp(5)+x^3-3*x^2+x-1)*exp(x-147)+x^2*exp(5)^2+(2*x^4+2* x^2)*exp(5)+x^6+2*x^4-2*x^3+x^2)/(x^2*exp(5)^2+(2*x^4+2*x^2)*exp(5)+x^6+2* x^4+x^2),x, algorithm=\
1/2*(x/(x^2*(e^5 + 1) + e^10 + 2*e^5 + 1) + arctan(x/sqrt(e^5 + 1))/(e^5 + 1)^(3/2))*e^10 + (arctan(x/sqrt(e^5 + 1))/sqrt(e^5 + 1) - x/(x^2 + e^5 + 1))*e^5 + (x/(x^2*(e^5 + 1) + e^10 + 2*e^5 + 1) + arctan(x/sqrt(e^5 + 1))/ (e^5 + 1)^(3/2))*e^5 - 3/2*sqrt(e^5 + 1)*arctan(x/sqrt(e^5 + 1)) + x + 1/2 *x*(e^5 + 1)/(x^2 + e^5 + 1) + arctan(x/sqrt(e^5 + 1))/sqrt(e^5 + 1) + 1/2 *x/(x^2*(e^5 + 1) + e^10 + 2*e^5 + 1) - x/(x^2 + e^5 + 1) + e^x/(x^3*e^147 + x*(e^152 + e^147)) + 1/2*arctan(x/sqrt(e^5 + 1))/(e^5 + 1)^(3/2) + 1/(x ^2 + e^5 + 1)
Timed out. \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=\text {Timed out} \]
integrate((((-1+x)*exp(5)+x^3-3*x^2+x-1)*exp(x-147)+x^2*exp(5)^2+(2*x^4+2* x^2)*exp(5)+x^6+2*x^4-2*x^3+x^2)/(x^2*exp(5)^2+(2*x^4+2*x^2)*exp(5)+x^6+2* x^4+x^2),x, algorithm=\
Time = 12.49 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {x^2+e^{10} x^2-2 x^3+2 x^4+x^6+e^{-147+x} \left (-1+e^5 (-1+x)+x-3 x^2+x^3\right )+e^5 \left (2 x^2+2 x^4\right )}{x^2+e^{10} x^2+2 x^4+x^6+e^5 \left (2 x^2+2 x^4\right )} \, dx=x+\frac {x+{\mathrm {e}}^{x-147}}{x\,\left (x^2+{\mathrm {e}}^5+1\right )} \]