Integrand size = 91, antiderivative size = 25 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=-1+\frac {e^{2 x+\frac {x}{e^8 (5-x)}}}{x}+x \]
Time = 1.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=\frac {e^{\left (2-\frac {1}{e^8 (-5+x)}\right ) x}}{x}+x \]
Integrate[(E^8*(25*x^2 - 10*x^3 + x^4) + E^((-x + E^8*(-10*x + 2*x^2))/(E^ 8*(-5 + x)))*(5*x + E^8*(-25 + 60*x - 21*x^2 + 2*x^3)))/(E^8*(25*x^2 - 10* x^3 + x^4)),x]
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.
\(\displaystyle \int \frac {e^{\frac {e^8 \left (2 x^2-10 x\right )-x}{e^8 (x-5)}} \left (e^8 \left (2 x^3-21 x^2+60 x-25\right )+5 x\right )+e^8 \left (x^4-10 x^3+25 x^2\right )}{e^8 \left (x^4-10 x^3+25 x^2\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^8 \left (x^4-10 x^3+25 x^2\right )+e^{\frac {x+2 e^8 \left (5 x-x^2\right )}{e^8 (5-x)}} \left (5 x-e^8 \left (-2 x^3+21 x^2-60 x+25\right )\right )}{x^4-10 x^3+25 x^2}dx}{e^8}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {e^8 \left (x^4-10 x^3+25 x^2\right )+e^{\frac {x+2 e^8 \left (5 x-x^2\right )}{e^8 (5-x)}} \left (5 x-e^8 \left (-2 x^3+21 x^2-60 x+25\right )\right )}{x^2 \left (x^2-10 x+25\right )}dx}{e^8}\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle \frac {4 \int \frac {e^8 \left (x^4-10 x^3+25 x^2\right )+e^{\frac {x+2 e^8 \left (5 x-x^2\right )}{e^8 (5-x)}} \left (5 x-e^8 \left (-2 x^3+21 x^2-60 x+25\right )\right )}{4 (5-x)^2 x^2}dx}{e^8}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^8 \left (x^4-10 x^3+25 x^2\right )+e^{\frac {x+2 e^8 \left (5 x-x^2\right )}{e^8 (5-x)}} \left (5 x-e^8 \left (-2 x^3+21 x^2-60 x+25\right )\right )}{(5-x)^2 x^2}dx}{e^8}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {e^{2 x-\frac {x}{e^8 (x-5)}} \left (2 e^8 x^3-21 e^8 x^2+5 \left (1+12 e^8\right ) x-25 e^8\right )}{(5-x)^2 x^2}+e^8\right )dx}{e^8}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\int \frac {e^{-\frac {x}{e^8 (x-5)}+2 x+8}}{x^2}dx+\int \frac {e^{2 x-\frac {x}{e^8 (x-5)}}}{(x-5)^2}dx-\frac {1}{5} \int \frac {e^{2 x-\frac {x}{e^8 (x-5)}}}{x-5}dx+\frac {1}{5} \left (1+10 e^8\right ) \int \frac {e^{2 x-\frac {x}{e^8 (x-5)}}}{x}dx+e^8 x}{e^8}\) |
Int[(E^8*(25*x^2 - 10*x^3 + x^4) + E^((-x + E^8*(-10*x + 2*x^2))/(E^8*(-5 + x)))*(5*x + E^8*(-25 + 60*x - 21*x^2 + 2*x^3)))/(E^8*(25*x^2 - 10*x^3 + x^4)),x]
3.28.34.3.1 Defintions of rubi rules used
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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.53 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x +\frac {{\mathrm e}^{\frac {x \left (2 x \,{\mathrm e}^{8}-10 \,{\mathrm e}^{8}-1\right ) {\mathrm e}^{-8}}{-5+x}}}{x}\) | \(28\) |
parallelrisch | \(\frac {{\mathrm e}^{-8} \left (x^{2} {\mathrm e}^{8}+20 x \,{\mathrm e}^{8}+{\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{-5+x}} {\mathrm e}^{8}\right )}{x}\) | \(59\) |
parts | \(x +\frac {\left (x \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{-5+x}}-5 \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{-5+x}}\right ) {\mathrm e}^{-4}}{x \left (-5+x \right )}\) | \(83\) |
norman | \(\frac {\left (x^{3} {\mathrm e}^{4}-25 x \,{\mathrm e}^{4}+x \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{-5+x}}-5 \,{\mathrm e}^{4} {\mathrm e}^{\frac {\left (\left (2 x^{2}-10 x \right ) {\mathrm e}^{8}-x \right ) {\mathrm e}^{-8}}{-5+x}}\right ) {\mathrm e}^{-4}}{x \left (-5+x \right )}\) | \(92\) |
int((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^2-x)/( -5+x)/exp(4)^2)+(x^4-10*x^3+25*x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/exp(4)^2 ,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=\frac {x^{2} + e^{\left (\frac {{\left (2 \, {\left (x^{2} - 5 \, x\right )} e^{8} - x\right )} e^{\left (-8\right )}}{x - 5}\right )}}{x} \]
integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^ 2-x)/(-5+x)/exp(4)^2)+(x^4-10*x^3+25*x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/ex p(4)^2,x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=x + \frac {e^{\frac {- x + \left (2 x^{2} - 10 x\right ) e^{8}}{\left (x - 5\right ) e^{8}}}}{x} \]
integrate((((2*x**3-21*x**2+60*x-25)*exp(4)**2+5*x)*exp(((2*x**2-10*x)*exp (4)**2-x)/(-5+x)/exp(4)**2)+(x**4-10*x**3+25*x**2)*exp(4)**2)/(x**4-10*x** 3+25*x**2)/exp(4)**2,x)
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.04 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx={\left ({\left (x - \frac {25}{x - 5} + 10 \, \log \left (x - 5\right )\right )} e^{8} + 10 \, {\left (\frac {5}{x - 5} - \log \left (x - 5\right )\right )} e^{8} - \frac {25 \, e^{8}}{x - 5} + \frac {e^{\left (2 \, x - \frac {5}{x e^{8} - 5 \, e^{8}} - e^{\left (-8\right )} + 8\right )}}{x}\right )} e^{\left (-8\right )} \]
integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^ 2-x)/(-5+x)/exp(4)^2)+(x^4-10*x^3+25*x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/ex p(4)^2,x, algorithm=\
((x - 25/(x - 5) + 10*log(x - 5))*e^8 + 10*(5/(x - 5) - log(x - 5))*e^8 - 25*e^8/(x - 5) + e^(2*x - 5/(x*e^8 - 5*e^8) - e^(-8) + 8)/x)*e^(-8)
Time = 0.40 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=\frac {{\left (x^{2} e^{8} + e^{\left (\frac {2 \, x^{2} e^{8} - 10 \, x e^{8} - x}{x e^{8} - 5 \, e^{8}} + 8\right )}\right )} e^{\left (-8\right )}}{x} \]
integrate((((2*x^3-21*x^2+60*x-25)*exp(4)^2+5*x)*exp(((2*x^2-10*x)*exp(4)^ 2-x)/(-5+x)/exp(4)^2)+(x^4-10*x^3+25*x^2)*exp(4)^2)/(x^4-10*x^3+25*x^2)/ex p(4)^2,x, algorithm=\
Time = 11.90 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^8 \left (25 x^2-10 x^3+x^4\right )+e^{\frac {-x+e^8 \left (-10 x+2 x^2\right )}{e^8 (-5+x)}} \left (5 x+e^8 \left (-25+60 x-21 x^2+2 x^3\right )\right )}{e^8 \left (25 x^2-10 x^3+x^4\right )} \, dx=x+\frac {{\mathrm {e}}^{-\frac {10\,x}{x-5}}\,{\mathrm {e}}^{\frac {2\,x^2}{x-5}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-8}}{x-5}}}{x} \]