3.92.35 \(\int \frac {e^{\frac {-x-e^{2 x} x+\log (e^2 x)}{-4+x}} (-4+5 x+e^{2 x} (4 x+8 x^2-2 x^3)-x \log (e^2 x))}{16 x-8 x^2+x^3} \, dx\)

Optimal. Leaf size=28 \[ 2+e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \]

________________________________________________________________________________________

Rubi [F]  time = 8.33, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{16 x-8 x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))*(-4 + 5*x + E^(2*x)*(4*x + 8*x^2 - 2*x^3) - x*Log[E^2*x]))/(16
*x - 8*x^2 + x^3),x]

[Out]

-2*Defer[Int][E^((2 - 9*x - E^(2*x)*x + 2*x^2 + Log[x])/(-4 + x)), x] + 4*Defer[Int][E^((2 - 9*x - E^(2*x)*x +
2*x^2 + Log[x])/(-4 + x))/(-4 + x)^2, x] + 2*Defer[Int][E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))/(-4 + x)^2
, x] - 8*Defer[Int][E^((2 - 9*x - E^(2*x)*x + 2*x^2 + Log[x])/(-4 + x))/(-4 + x), x] + Defer[Int][E^((-x - E^(
2*x)*x + Log[E^2*x])/(-4 + x))/(-4 + x), x]/4 - Defer[Int][E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))/x, x]/4
- Defer[Int][(E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))*Log[x])/(-4 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{x \left (16-8 x+x^2\right )} \, dx\\ &=\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-4+5 x+e^{2 x} \left (4 x+8 x^2-2 x^3\right )-x \log \left (e^2 x\right )\right )}{(-4+x)^2 x} \, dx\\ &=\int \left (-\frac {2 e^{2 x+\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-2-4 x+x^2\right )}{(-4+x)^2}+\frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} (-4+3 x-x \log (x))}{(-4+x)^2 x}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 x+\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \left (-2-4 x+x^2\right )}{(-4+x)^2} \, dx\right )+\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} (-4+3 x-x \log (x))}{(-4+x)^2 x} \, dx\\ &=-\left (2 \int \frac {e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}} \left (-2-4 x+x^2\right )}{(4-x)^2} \, dx\right )+\int \left (\frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} (-4+3 x)}{(-4+x)^2 x}-\frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \log (x)}{(-4+x)^2}\right ) \, dx\\ &=-\left (2 \int \left (e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}-\frac {2 e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{(-4+x)^2}+\frac {4 e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{-4+x}\right ) \, dx\right )+\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} (-4+3 x)}{(-4+x)^2 x} \, dx-\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \log (x)}{(-4+x)^2} \, dx\\ &=-\left (2 \int e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}} \, dx\right )+4 \int \frac {e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{(-4+x)^2} \, dx-8 \int \frac {e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{-4+x} \, dx+\int \left (\frac {2 e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{(-4+x)^2}+\frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{4 (-4+x)}-\frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{4 x}\right ) \, dx-\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \log (x)}{(-4+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{-4+x} \, dx-\frac {1}{4} \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{x} \, dx-2 \int e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}} \, dx+2 \int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}}}{(-4+x)^2} \, dx+4 \int \frac {e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{(-4+x)^2} \, dx-8 \int \frac {e^{\frac {2-9 x-e^{2 x} x+2 x^2+\log (x)}{-4+x}}}{-4+x} \, dx-\int \frac {e^{\frac {-x-e^{2 x} x+\log \left (e^2 x\right )}{-4+x}} \log (x)}{(-4+x)^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.71, size = 27, normalized size = 0.96 \begin {gather*} e^{-\frac {-2+x+e^{2 x} x}{-4+x}} x^{\frac {1}{-4+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-x - E^(2*x)*x + Log[E^2*x])/(-4 + x))*(-4 + 5*x + E^(2*x)*(4*x + 8*x^2 - 2*x^3) - x*Log[E^2*x]
))/(16*x - 8*x^2 + x^3),x]

[Out]

x^(-4 + x)^(-1)/E^((-2 + x + E^(2*x)*x)/(-4 + x))

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 23, normalized size = 0.82 \begin {gather*} e^{\left (-\frac {x e^{\left (2 \, x\right )} + x - \log \left (x e^{2}\right )}{x - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(exp(2)*x)-x*exp(2*x)-x)/(x-4))/(x^3-8*
x^2+16*x),x, algorithm="fricas")

[Out]

e^(-(x*e^(2*x) + x - log(x*e^2))/(x - 4))

________________________________________________________________________________________

giac [A]  time = 0.25, size = 33, normalized size = 1.18 \begin {gather*} e^{\left (-\frac {x e^{\left (2 \, x\right )}}{x - 4} - \frac {x}{x - 4} + \frac {\log \left (x e^{2}\right )}{x - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(exp(2)*x)-x*exp(2*x)-x)/(x-4))/(x^3-8*
x^2+16*x),x, algorithm="giac")

[Out]

e^(-x*e^(2*x)/(x - 4) - x/(x - 4) + log(x*e^2)/(x - 4))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 24, normalized size = 0.86

method	result	size
risch	\({\mathrm e}^{-\frac {-\ln \left ({\mathrm e}^{2} x \right )+x \,{\mathrm e}^{2 x}+x}{x -4}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((ln(exp(2)*x)-x*exp(2*x)-x)/(x-4))/(x^3-8*x^2+16*x
),x,method=_RETURNVERBOSE)

[Out]

exp(-(-ln(exp(2)*x)+x*exp(2*x)+x)/(x-4))

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 35, normalized size = 1.25 \begin {gather*} e^{\left (-\frac {4 \, e^{\left (2 \, x\right )}}{x - 4} + \frac {\log \relax (x)}{x - 4} - \frac {2}{x - 4} - e^{\left (2 \, x\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(exp(2)*x)+(-2*x^3+8*x^2+4*x)*exp(2*x)+5*x-4)*exp((log(exp(2)*x)-x*exp(2*x)-x)/(x-4))/(x^3-8*
x^2+16*x),x, algorithm="maxima")

[Out]

e^(-4*e^(2*x)/(x - 4) + log(x)/(x - 4) - 2/(x - 4) - e^(2*x) - 1)

________________________________________________________________________________________

mupad [B]  time = 7.62, size = 38, normalized size = 1.36 \begin {gather*} x^{\frac {1}{x-4}}\,{\mathrm {e}}^{-\frac {x}{x-4}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{2\,x}}{x-4}}\,{\mathrm {e}}^{\frac {2}{x-4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(x - log(x*exp(2)) + x*exp(2*x))/(x - 4))*(5*x + exp(2*x)*(4*x + 8*x^2 - 2*x^3) - x*log(x*exp(2)) -
4))/(16*x - 8*x^2 + x^3),x)

[Out]

x^(1/(x - 4))*exp(-x/(x - 4))*exp(-(x*exp(2*x))/(x - 4))*exp(2/(x - 4))

________________________________________________________________________________________

sympy [A]  time = 0.80, size = 19, normalized size = 0.68 \begin {gather*} e^{\frac {- x e^{2 x} - x + \log {\left (x e^{2} \right )}}{x - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(exp(2)*x)+(-2*x**3+8*x**2+4*x)*exp(2*x)+5*x-4)*exp((ln(exp(2)*x)-x*exp(2*x)-x)/(x-4))/(x**3-8
*x**2+16*x),x)

[Out]

exp((-x*exp(2*x) - x + log(x*exp(2)))/(x - 4))

________________________________________________________________________________________