Integrand size = 57, antiderivative size = 29 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=e^{1-x}-\frac {-25-\frac {1}{x}+2 x}{4 (4-x)} \] Output:
exp(1)/exp(x)-(2*x-1/x-25)/(-4*x+16)
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=e^{1-x}-\frac {69}{16 (-4+x)}+\frac {1}{16 x} \] Input:
Integrate[(E^x*(-4 + 2*x + 17*x^2) + E*(-64*x^2 + 32*x^3 - 4*x^4))/(E^x*(6 4*x^2 - 32*x^3 + 4*x^4)),x]
Output:
E^(1 - x) - 69/(16*(-4 + x)) + 1/(16*x)
Time = 0.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2026, 7239, 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.
\(\displaystyle \int \frac {e^{-x} \left (e^x \left (17 x^2+2 x-4\right )+e \left (-4 x^4+32 x^3-64 x^2\right )\right )}{4 x^4-32 x^3+64 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (e^x \left (17 x^2+2 x-4\right )+e \left (-4 x^4+32 x^3-64 x^2\right )\right )}{x^2 \left (4 x^2-32 x+64\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {17 x^2+2 x-4}{4 (x-4)^2 x^2}-e^{1-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{1-x}+\frac {1}{16 x}+\frac {69}{16 (4-x)}\) |
Input:
Int[(E^x*(-4 + 2*x + 17*x^2) + E*(-64*x^2 + 32*x^3 - 4*x^4))/(E^x*(64*x^2 - 32*x^3 + 4*x^4)),x]
Output:
E^(1 - x) + 69/(16*(4 - x)) + 1/(16*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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.34 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
parts | \({\mathrm e} \,{\mathrm e}^{-x}+\frac {1}{16 x}-\frac {69}{16 \left (x -4\right )}\) | \(21\) |
risch | \(\frac {-\frac {17 x}{4}-\frac {1}{4}}{\left (x -4\right ) x}+{\mathrm e}^{1-x}\) | \(22\) |
norman | \(\frac {\left (x^{2} {\mathrm e}-\frac {17 \,{\mathrm e}^{x} x}{4}-4 x \,{\mathrm e}-\frac {{\mathrm e}^{x}}{4}\right ) {\mathrm e}^{-x}}{x \left (x -4\right )}\) | \(35\) |
parallelrisch | \(\frac {\left (4 x^{2} {\mathrm e}-16 x \,{\mathrm e}-17 \,{\mathrm e}^{x} x -{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{4 x \left (x -4\right )}\) | \(37\) |
default | \(\frac {1}{16 x}-\frac {69}{16 \left (x -4\right )}-16 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-x}}{x -4}+{\mathrm e}^{-4} \operatorname {expIntegral}_{1}\left (x -4\right )\right )+8 \,{\mathrm e} \left (-\frac {4 \,{\mathrm e}^{-x}}{x -4}+3 \,{\mathrm e}^{-4} \operatorname {expIntegral}_{1}\left (x -4\right )\right )-{\mathrm e} \left (-{\mathrm e}^{-x}-\frac {16 \,{\mathrm e}^{-x}}{x -4}+8 \,{\mathrm e}^{-4} \operatorname {expIntegral}_{1}\left (x -4\right )\right )\) | \(94\) |
orering | \(-\frac {\left (17 x^{5}-135 x^{4}+230 x^{3}+10 x^{2}+24 x -32\right ) \left (\left (17 x^{2}+2 x -4\right ) {\mathrm e}^{x}+\left (-4 x^{4}+32 x^{3}-64 x^{2}\right ) {\mathrm e}\right ) {\mathrm e}^{-x}}{\left (17 x^{4}-100 x^{3}-18 x^{2}+40 x -32\right ) \left (4 x^{4}-32 x^{3}+64 x^{2}\right )}-\frac {\left (17 x^{3}-84 x^{2}-6 x +4\right ) \left (x -4\right ) x \left (\frac {\left (\left (34 x +2\right ) {\mathrm e}^{x}+\left (17 x^{2}+2 x -4\right ) {\mathrm e}^{x}+\left (-16 x^{3}+96 x^{2}-128 x \right ) {\mathrm e}\right ) {\mathrm e}^{-x}}{4 x^{4}-32 x^{3}+64 x^{2}}-\frac {\left (\left (17 x^{2}+2 x -4\right ) {\mathrm e}^{x}+\left (-4 x^{4}+32 x^{3}-64 x^{2}\right ) {\mathrm e}\right ) {\mathrm e}^{-x} \left (16 x^{3}-96 x^{2}+128 x \right )}{\left (4 x^{4}-32 x^{3}+64 x^{2}\right )^{2}}-\frac {\left (\left (17 x^{2}+2 x -4\right ) {\mathrm e}^{x}+\left (-4 x^{4}+32 x^{3}-64 x^{2}\right ) {\mathrm e}\right ) {\mathrm e}^{-x}}{4 x^{4}-32 x^{3}+64 x^{2}}\right )}{17 x^{4}-100 x^{3}-18 x^{2}+40 x -32}\) | \(340\) |
Input:
int(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64 *x^2)/exp(x),x,method=_RETURNVERBOSE)
Output:
exp(1)/exp(x)+1/16/x-69/16/(x-4)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=\frac {{\left (4 \, {\left (x^{2} - 4 \, x\right )} e - {\left (17 \, x + 1\right )} e^{x}\right )} e^{\left (-x\right )}}{4 \, {\left (x^{2} - 4 \, x\right )}} \] Input:
integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32* x^3+64*x^2)/exp(x),x, algorithm="fricas")
Output:
1/4*(4*(x^2 - 4*x)*e - (17*x + 1)*e^x)*e^(-x)/(x^2 - 4*x)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=\frac {- 17 x - 1}{4 x^{2} - 16 x} + e e^{- x} \] Input:
integrate(((17*x**2+2*x-4)*exp(x)+(-4*x**4+32*x**3-64*x**2)*exp(1))/(4*x** 4-32*x**3+64*x**2)/exp(x),x)
Output:
(-17*x - 1)/(4*x**2 - 16*x) + E*exp(-x)
\[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=\int { -\frac {{\left (4 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )} e - {\left (17 \, x^{2} + 2 \, x - 4\right )} e^{x}\right )} e^{\left (-x\right )}}{4 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )}} \,d x } \] Input:
integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32* x^3+64*x^2)/exp(x),x, algorithm="maxima")
Output:
16*e^(-3)*exp_integral_e(2, x - 4)/(x - 4) - 1/4*(17*x^2 - 4*(x^3*e - 8*x^ 2*e)*e^(-x) - 67*x - 4)/(x^3 - 8*x^2 + 16*x) - 32*integrate(e^(-x + 1)/(x^ 3 - 12*x^2 + 48*x - 64), x)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=\frac {4 \, x^{2} e^{\left (-x + 1\right )} - 16 \, x e^{\left (-x + 1\right )} - 17 \, x - 1}{4 \, {\left (x^{2} - 4 \, x\right )}} \] Input:
integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32* x^3+64*x^2)/exp(x),x, algorithm="giac")
Output:
1/4*(4*x^2*e^(-x + 1) - 16*x*e^(-x + 1) - 17*x - 1)/(x^2 - 4*x)
Time = 3.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx={\mathrm {e}}^{-x}\,\mathrm {e}+\frac {17\,x}{4\,\left (4\,x-x^2\right )}+\frac {1}{4\,\left (4\,x-x^2\right )} \] Input:
int((exp(-x)*(exp(x)*(2*x + 17*x^2 - 4) - exp(1)*(64*x^2 - 32*x^3 + 4*x^4) ))/(64*x^2 - 32*x^3 + 4*x^4),x)
Output:
exp(-x)*exp(1) + (17*x)/(4*(4*x - x^2)) + 1/(4*(4*x - x^2))
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{64 x^2-32 x^3+4 x^4} \, dx=\frac {-17 e^{x} x^{2}-4 e^{x}+16 e \,x^{2}-64 e x}{16 e^{x} x \left (x -4\right )} \] Input:
int(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64 *x^2)/exp(x),x)
Output:
( - 17*e**x*x**2 - 4*e**x + 16*e*x**2 - 64*e*x)/(16*e**x*x*(x - 4))