Integrand size = 130, antiderivative size = 35 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=-\frac {e^{-4+\frac {5}{1-x}} \left (-4+\frac {x \left (1+x^2\right )}{e}\right )^2}{x}+\log (x) \]
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=\frac {e^{-6-\frac {5}{-1+x}} \left (-\left (-4 e+x+x^3\right )^2+e^{6+\frac {5}{-1+x}} x \log (x)\right )}{x} \]
Integrate[(E^(-2 - (1 + 4*x)/(-1 + x))*(-x^2 - 3*x^3 - 7*x^4 + 2*x^5 - 11* x^6 + 5*x^7 - 5*x^8 + E^2*(16 - 112*x + 16*x^2) + E^(2 + (1 + 4*x)/(-1 + x ))*(x - 2*x^2 + x^3) + E*(40*x^2 + 16*x^3 + 8*x^4 + 16*x^5)))/(x^2 - 2*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.
| \(\displaystyle \int \frac {e^{-\frac {4 x+1}{x-1}-2} \left (-5 x^8+5 x^7-11 x^6+2 x^5-7 x^4-3 x^3-x^2+e^2 \left (16 x^2-112 x+16\right )+e^{\frac {4 x+1}{x-1}+2} \left (x^3-2 x^2+x\right )+e \left (16 x^5+8 x^4+16 x^3+40 x^2\right )\right )}{x^4-2 x^3+x^2} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{-\frac {4 x+1}{x-1}-2} \left (-5 x^8+5 x^7-11 x^6+2 x^5-7 x^4-3 x^3-x^2+e^2 \left (16 x^2-112 x+16\right )+e^{\frac {4 x+1}{x-1}+2} \left (x^3-2 x^2+x\right )+e \left (16 x^5+8 x^4+16 x^3+40 x^2\right )\right )}{x^2 \left (x^2-2 x+1\right )}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int -\frac {e^{\frac {4 x+1}{1-x}-2} \left (5 x^8-5 x^7+11 x^6-2 x^5+7 x^4+3 x^3+x^2-16 e^2 \left (x^2-7 x+1\right )-e^{2-\frac {4 x+1}{1-x}} \left (x^3-2 x^2+x\right )-8 e \left (2 x^5+x^4+2 x^3+5 x^2\right )\right )}{4 (1-x)^2 x^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {e^{\frac {4 x+1}{1-x}-2} \left (5 x^8-5 x^7+11 x^6-2 x^5+7 x^4+3 x^3+x^2-16 e^2 \left (x^2-7 x+1\right )-e^{2-\frac {4 x+1}{1-x}} \left (x^3-2 x^2+x\right )-8 e \left (2 x^5+x^4+2 x^3+5 x^2\right )\right )}{(1-x)^2 x^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\int \frac {e^{\frac {1-6 x}{x-1}} \left (5 x^8-5 x^7+11 x^6-2 x^5+7 x^4+3 x^3+x^2-16 e^2 \left (x^2-7 x+1\right )-e^{2-\frac {4 x+1}{1-x}} \left (x^3-2 x^2+x\right )-8 e \left (2 x^5+x^4+2 x^3+5 x^2\right )\right )}{(1-x)^2 x^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\int \left (\frac {5 e^{\frac {1-6 x}{x-1}} x^6}{(x-1)^2}-\frac {5 e^{\frac {1-6 x}{x-1}} x^5}{(x-1)^2}+\frac {11 e^{\frac {1-6 x}{x-1}} x^4}{(x-1)^2}-\frac {2 e^{\frac {1-6 x}{x-1}} x^3}{(x-1)^2}+\frac {7 e^{\frac {1-6 x}{x-1}} x^2}{(x-1)^2}+\frac {3 e^{\frac {1-6 x}{x-1}} x}{(x-1)^2}-\frac {8 e^{\frac {1-6 x}{x-1}+1} \left (2 x^3+x^2+2 x+5\right )}{(x-1)^2}+\frac {e^{\frac {1-6 x}{x-1}}}{(x-1)^2}-\frac {e^{\frac {1-6 x}{x-1}+\frac {6 x-1}{x-1}}}{x}-\frac {16 e^{\frac {1-6 x}{x-1}+2} \left (x^2-7 x+1\right )}{(x-1)^2 x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -5 \int e^{\frac {1-6 x}{x-1}} x^4dx-5 \int e^{\frac {1-6 x}{x-1}} x^3dx-16 \int e^{\frac {1-6 x}{x-1}} x^2dx-41 \int e^{\frac {1-6 x}{x-1}}dx+40 \int e^{-\frac {5 x}{x-1}}dx-25 \int e^{\frac {1-6 x}{x-1}} xdx+16 \int e^{-\frac {5 x}{x-1}} xdx-\frac {80 \operatorname {ExpIntegralEi}\left (\frac {5}{1-x}\right )}{e^5}+\frac {60 \operatorname {ExpIntegralEi}\left (\frac {5}{1-x}\right )}{e^6}-4 e^{\frac {5}{1-x}-6}+16 e^{\frac {5}{1-x}-5}-\frac {16 e^{\frac {4 x+1}{1-x}}}{x}+\log (x)\) |
Int[(E^(-2 - (1 + 4*x)/(-1 + x))*(-x^2 - 3*x^3 - 7*x^4 + 2*x^5 - 11*x^6 + 5*x^7 - 5*x^8 + E^2*(16 - 112*x + 16*x^2) + E^(2 + (1 + 4*x)/(-1 + x))*(x - 2*x^2 + x^3) + E*(40*x^2 + 16*x^3 + 8*x^4 + 16*x^5)))/(x^2 - 2*x^3 + x^4 ),x]
3.9.83.3.1 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 = 3.51 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.43
| method | result | size |
| risch | \(\ln \left (x \right )-\frac {\left (x^{6}-8 x^{3} {\mathrm e}+2 x^{4}+16 \,{\mathrm e}^{2}-8 x \,{\mathrm e}+x^{2}\right ) {\mathrm e}^{-\frac {-1+6 x}{-1+x}}}{x}\) | \(50\) |
| norman | \(\frac {\left (\left (-16 \,{\mathrm e}-8\right ) x +{\mathrm e}^{-1} x^{6}+\left (1+8 \,{\mathrm e}\right ) {\mathrm e}^{-1} x^{2}-2 \,{\mathrm e}^{-1} x^{5}-{\mathrm e}^{-1} x^{7}-\left (1+8 \,{\mathrm e}\right ) {\mathrm e}^{-1} x^{3}+2 \left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} x^{4}+16 \,{\mathrm e}\right ) {\mathrm e}^{-1} {\mathrm e}^{-\frac {1+4 x}{-1+x}}}{x \left (-1+x \right )}+\ln \left (x \right )\) | \(114\) |
| parallelrisch | \(\frac {{\mathrm e}^{-2} \left (-x^{7}+{\mathrm e}^{2} \ln \left (x \right ) x^{2} {\mathrm e}^{\frac {1+4 x}{-1+x}}+x^{6}-{\mathrm e}^{2} \ln \left (x \right ) {\mathrm e}^{\frac {1+4 x}{-1+x}} x +8 x^{4} {\mathrm e}-2 x^{5}-8 x^{3} {\mathrm e}+2 x^{4}-16 \,{\mathrm e}^{2} x +8 x^{2} {\mathrm e}-x^{3}+16 \,{\mathrm e}^{2}-8 x \,{\mathrm e}+x^{2}\right ) {\mathrm e}^{-\frac {1+4 x}{-1+x}}}{x \left (-1+x \right )}\) | \(137\) |
| parts | \(\text {Expression too large to display}\) | \(847\) |
| default | \(\text {Expression too large to display}\) | \(907\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(908\) |
int(((x^3-2*x^2+x)*exp(1)^2*exp((1+4*x)/(-1+x))+(16*x^2-112*x+16)*exp(1)^2 +(16*x^5+8*x^4+16*x^3+40*x^2)*exp(1)-5*x^8+5*x^7-11*x^6+2*x^5-7*x^4-3*x^3- x^2)/(x^4-2*x^3+x^2)/exp(1)^2/exp((1+4*x)/(-1+x)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=-\frac {{\left (x^{6} + 2 \, x^{4} - x e^{\left (\frac {6 \, x - 1}{x - 1}\right )} \log \left (x\right ) + x^{2} - 8 \, {\left (x^{3} + x\right )} e + 16 \, e^{2}\right )} e^{\left (-\frac {6 \, x - 1}{x - 1}\right )}}{x} \]
integrate(((x^3-2*x^2+x)*exp(1)^2*exp((1+4*x)/(-1+x))+(16*x^2-112*x+16)*ex p(1)^2+(16*x^5+8*x^4+16*x^3+40*x^2)*exp(1)-5*x^8+5*x^7-11*x^6+2*x^5-7*x^4- 3*x^3-x^2)/(x^4-2*x^3+x^2)/exp(1)^2/exp((1+4*x)/(-1+x)),x, algorithm=\
-(x^6 + 2*x^4 - x*e^((6*x - 1)/(x - 1))*log(x) + x^2 - 8*(x^3 + x)*e + 16* e^2)*e^(-(6*x - 1)/(x - 1))/x
Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=\log {\left (x \right )} + \frac {\left (- x^{6} - 2 x^{4} + 8 e x^{3} - x^{2} + 8 e x - 16 e^{2}\right ) e^{- \frac {4 x + 1}{x - 1}}}{x e^{2}} \]
integrate(((x**3-2*x**2+x)*exp(1)**2*exp((1+4*x)/(-1+x))+(16*x**2-112*x+16 )*exp(1)**2+(16*x**5+8*x**4+16*x**3+40*x**2)*exp(1)-5*x**8+5*x**7-11*x**6+ 2*x**5-7*x**4-3*x**3-x**2)/(x**4-2*x**3+x**2)/exp(1)**2/exp((1+4*x)/(-1+x) ),x)
log(x) + (-x**6 - 2*x**4 + 8*E*x**3 - x**2 + 8*E*x - 16*exp(2))*exp(-2)*ex p(-(4*x + 1)/(x - 1))/x
\[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=\int { -\frac {{\left (5 \, x^{8} - 5 \, x^{7} + 11 \, x^{6} - 2 \, x^{5} + 7 \, x^{4} + 3 \, x^{3} + x^{2} - 16 \, {\left (x^{2} - 7 \, x + 1\right )} e^{2} - 8 \, {\left (2 \, x^{5} + x^{4} + 2 \, x^{3} + 5 \, x^{2}\right )} e - {\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (\frac {4 \, x + 1}{x - 1} + 2\right )}\right )} e^{\left (-\frac {4 \, x + 1}{x - 1} - 2\right )}}{x^{4} - 2 \, x^{3} + x^{2}} \,d x } \]
integrate(((x^3-2*x^2+x)*exp(1)^2*exp((1+4*x)/(-1+x))+(16*x^2-112*x+16)*ex p(1)^2+(16*x^5+8*x^4+16*x^3+40*x^2)*exp(1)-5*x^8+5*x^7-11*x^6+2*x^5-7*x^4- 3*x^3-x^2)/(x^4-2*x^3+x^2)/exp(1)^2/exp((1+4*x)/(-1+x)),x, algorithm=\
-(x^6 + 2*x^4 - 8*x^3*e + x^2 + 16*e^2 + 40*e + 1)*e^(-5/(x - 1) - 6)/x - 1/5*e^(-5/(x - 1) - 6) + integrate((7*x*(40*e + 1) - 40*e - 1)*e^(-5/(x - 1))/(x^4*e^6 - 2*x^3*e^6 + x^2*e^6), x) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (31) = 62\).
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.80 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=-\frac {{\left (x^{6} e^{\left (-\frac {5 \, x}{x - 1}\right )} + 2 \, x^{4} e^{\left (-\frac {5 \, x}{x - 1}\right )} - 8 \, x^{3} e^{\left (-\frac {5 \, x}{x - 1} + 1\right )} + x^{2} e^{\left (-\frac {5 \, x}{x - 1}\right )} - x e \log \left (x\right ) - 8 \, x e^{\left (-\frac {5 \, x}{x - 1} + 1\right )} + 16 \, e^{\left (-\frac {5 \, x}{x - 1} + 2\right )}\right )} e^{\left (-1\right )}}{x} \]
integrate(((x^3-2*x^2+x)*exp(1)^2*exp((1+4*x)/(-1+x))+(16*x^2-112*x+16)*ex p(1)^2+(16*x^5+8*x^4+16*x^3+40*x^2)*exp(1)-5*x^8+5*x^7-11*x^6+2*x^5-7*x^4- 3*x^3-x^2)/(x^4-2*x^3+x^2)/exp(1)^2/exp((1+4*x)/(-1+x)),x, algorithm=\
-(x^6*e^(-5*x/(x - 1)) + 2*x^4*e^(-5*x/(x - 1)) - 8*x^3*e^(-5*x/(x - 1) + 1) + x^2*e^(-5*x/(x - 1)) - x*e*log(x) - 8*x*e^(-5*x/(x - 1) + 1) + 16*e^( -5*x/(x - 1) + 2))*e^(-1)/x
Time = 9.62 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.31 \[ \int \frac {e^{-2-\frac {1+4 x}{-1+x}} \left (-x^2-3 x^3-7 x^4+2 x^5-11 x^6+5 x^7-5 x^8+e^2 \left (16-112 x+16 x^2\right )+e^{2+\frac {1+4 x}{-1+x}} \left (x-2 x^2+x^3\right )+e \left (40 x^2+16 x^3+8 x^4+16 x^5\right )\right )}{x^2-2 x^3+x^4} \, dx=-\frac {16\,{\mathrm {e}}^{\frac {1}{x-1}-\frac {6\,x}{x-1}+2}}{x}-{\mathrm {e}}^{\frac {1}{x-1}-\frac {6\,x}{x-1}}\,\left (x-8\,\mathrm {e}-8\,x^2\,\mathrm {e}-{\mathrm {e}}^{\frac {6\,x}{x-1}-\frac {1}{x-1}}\,\ln \left (x\right )+2\,x^3+x^5\right ) \]
int((exp(-(4*x + 1)/(x - 1))*exp(-2)*(exp(2)*(16*x^2 - 112*x + 16) - x^2 - 3*x^3 - 7*x^4 + 2*x^5 - 11*x^6 + 5*x^7 - 5*x^8 + exp(1)*(40*x^2 + 16*x^3 + 8*x^4 + 16*x^5) + exp((4*x + 1)/(x - 1))*exp(2)*(x - 2*x^2 + x^3)))/(x^2 - 2*x^3 + x^4),x)