3.40.14 \(\int \frac {2+e^{-5+2 x} (-16+24 x-12 x^2+2 x^3)-2 \log (2)}{-8+12 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ e^{-5+2 x}+\frac {-1+\log (2)}{(-2+x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 23, normalized size of antiderivative = 1.28, number of steps used = 3, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 2194} \begin {gather*} e^{2 x-5}-\frac {1-\log (2)}{(2-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + E^(-5 + 2*x)*(-16 + 24*x - 12*x^2 + 2*x^3) - 2*Log[2])/(-8 + 12*x - 6*x^2 + x^3),x]

[Out]

E^(-5 + 2*x) - (1 - Log[2])/(2 - x)^2

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} e^{-5+2 x}+\frac {-1+\log (2)}{(-2+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^(-5 + 2*x)*(-16 + 24*x - 12*x^2 + 2*x^3) - 2*Log[2])/(-8 + 12*x - 6*x^2 + x^3),x]

[Out]

E^(-5 + 2*x) + (-1 + Log[2])/(-2 + x)^2

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 30, normalized size = 1.67 \begin {gather*} \frac {{\left (x^{2} - 4 \, x + 4\right )} e^{\left (2 \, x - 5\right )} + \log \relax (2) - 1}{x^{2} - 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+24*x-16)*exp(2*x-5)-2*log(2)+2)/(x^3-6*x^2+12*x-8),x, algorithm="fricas")

[Out]

((x^2 - 4*x + 4)*e^(2*x - 5) + log(2) - 1)/(x^2 - 4*x + 4)

________________________________________________________________________________________

giac [B]  time = 0.14, size = 50, normalized size = 2.78 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} - 4 \, x e^{\left (2 \, x\right )} + e^{5} \log \relax (2) - e^{5} + 4 \, e^{\left (2 \, x\right )}}{x^{2} e^{5} - 4 \, x e^{5} + 4 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+24*x-16)*exp(2*x-5)-2*log(2)+2)/(x^3-6*x^2+12*x-8),x, algorithm="giac")

[Out]

(x^2*e^(2*x) - 4*x*e^(2*x) + e^5*log(2) - e^5 + 4*e^(2*x))/(x^2*e^5 - 4*x*e^5 + 4*e^5)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 28, normalized size = 1.56

method	result	size
derivativedivides	\(-\frac {4}{\left (2 x -4\right )^{2}}+\frac {4 \ln \relax (2)}{\left (2 x -4\right )^{2}}+{\mathrm e}^{2 x -5}\)	\(28\)
default	\(-\frac {4}{\left (2 x -4\right )^{2}}+\frac {4 \ln \relax (2)}{\left (2 x -4\right )^{2}}+{\mathrm e}^{2 x -5}\)	\(28\)
risch	\(-\frac {1}{x^{2}-4 x +4}+\frac {\ln \relax (2)}{x^{2}-4 x +4}+{\mathrm e}^{2 x -5}\)	\(33\)
norman	\(\frac {{\mathrm e}^{2 x -5} x^{2}-1-4 \,{\mathrm e}^{2 x -5} x +4 \,{\mathrm e}^{2 x -5}+\ln \relax (2)}{\left (x -2\right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-12*x^2+24*x-16)*exp(2*x-5)-2*ln(2)+2)/(x^3-6*x^2+12*x-8),x,method=_RETURNVERBOSE)

[Out]

-4/(2*x-4)^2+4*ln(2)/(2*x-4)^2+exp(2*x-5)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left (x^{3} - 6 \, x^{2} + 12 \, x\right )} e^{\left (2 \, x\right )}}{x^{3} e^{5} - 6 \, x^{2} e^{5} + 12 \, x e^{5} - 8 \, e^{5}} + \frac {16 \, e^{\left (-1\right )} E_{3}\left (-2 \, x + 4\right )}{{\left (x - 2\right )}^{2}} + \frac {\log \relax (2)}{x^{2} - 4 \, x + 4} - \frac {1}{x^{2} - 4 \, x + 4} + 24 \, \int \frac {e^{\left (2 \, x\right )}}{x^{4} e^{5} - 8 \, x^{3} e^{5} + 24 \, x^{2} e^{5} - 32 \, x e^{5} + 16 \, e^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+24*x-16)*exp(2*x-5)-2*log(2)+2)/(x^3-6*x^2+12*x-8),x, algorithm="maxima")

[Out]

(x^3 - 6*x^2 + 12*x)*e^(2*x)/(x^3*e^5 - 6*x^2*e^5 + 12*x*e^5 - 8*e^5) + 16*e^(-1)*exp_integral_e(3, -2*x + 4)/
(x - 2)^2 + log(2)/(x^2 - 4*x + 4) - 1/(x^2 - 4*x + 4) + 24*integrate(e^(2*x)/(x^4*e^5 - 8*x^3*e^5 + 24*x^2*e^
5 - 32*x*e^5 + 16*e^5), x)

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 22, normalized size = 1.22 \begin {gather*} {\mathrm {e}}^{2\,x-5}+\frac {\ln \relax (2)-1}{x^2-4\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x - 5)*(24*x - 12*x^2 + 2*x^3 - 16) - 2*log(2) + 2)/(12*x - 6*x^2 + x^3 - 8),x)

[Out]

exp(2*x - 5) + (log(2) - 1)/(x^2 - 4*x + 4)

________________________________________________________________________________________

sympy [A]  time = 0.16, size = 22, normalized size = 1.22 \begin {gather*} e^{2 x - 5} - \frac {2 - 2 \log {\relax (2 )}}{2 x^{2} - 8 x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-12*x**2+24*x-16)*exp(2*x-5)-2*ln(2)+2)/(x**3-6*x**2+12*x-8),x)

[Out]

exp(2*x - 5) - (2 - 2*log(2))/(2*x**2 - 8*x + 8)

________________________________________________________________________________________