3.74.73 \(\int \frac {e^{-e^{-x} (-4+x)} (e^x (-4+x) (-25 x+5 x^2)+e^{2 x} (40-50 x+10 x^2))}{-12 x^3+3 x^4} \, dx\)

Optimal. Leaf size=23 \[ \frac {5 e^{-e^{-x} (-4+x)+2 x}}{3 x^2} \]

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Rubi [B]  time = 0.26, antiderivative size = 50, normalized size of antiderivative = 2.17, number of steps used = 2, number of rules used = 2, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1593, 2288} \begin {gather*} \frac {5 e^{e^{-x} (4-x)+x} \left (5 x-x^2\right )}{3 \left (e^{-x} (4-x)+e^{-x}\right ) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-4 + x)*(-25*x + 5*x^2) + E^(2*x)*(40 - 50*x + 10*x^2))/(E^((-4 + x)/E^x)*(-12*x^3 + 3*x^4)),x]

[Out]

(5*E^((4 - x)/E^x + x)*(5*x - x^2))/(3*(E^(-x) + (4 - x)/E^x)*x^3)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-e^{-x} (-4+x)} \left (e^x (-4+x) \left (-25 x+5 x^2\right )+e^{2 x} \left (40-50 x+10 x^2\right )\right )}{x^3 (-12+3 x)} \, dx\\ &=\frac {5 e^{e^{-x} (4-x)+x} \left (5 x-x^2\right )}{3 \left (e^{-x}+e^{-x} (4-x)\right ) x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 23, normalized size = 1.00 \begin {gather*} \frac {5 e^{-e^{-x} (-4+x)+2 x}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-4 + x)*(-25*x + 5*x^2) + E^(2*x)*(40 - 50*x + 10*x^2))/(E^((-4 + x)/E^x)*(-12*x^3 + 3*x^4)),x
]

[Out]

(5*E^(-((-4 + x)/E^x) + 2*x))/(3*x^2)

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fricas [B]  time = 0.58, size = 50, normalized size = 2.17 \begin {gather*} \frac {5 \, e^{\left (-{\left (x^{2} - 8 \, x + 16\right )} e^{\left (-x - \log \left (x - 4\right )\right )} + 2 \, x + 2 \, \log \left (x - 4\right )\right )}}{3 \, {\left (x^{4} - 8 \, x^{3} + 16 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-25*x)*exp(x)^2*exp(log(x-4)-x)+(10*x^2-50*x+40)*exp(x)^2)/(3*x^4-12*x^3)/exp(exp(log(x-4)-x)
),x, algorithm="fricas")

[Out]

5/3*e^(-(x^2 - 8*x + 16)*e^(-x - log(x - 4)) + 2*x + 2*log(x - 4))/(x^4 - 8*x^3 + 16*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 \, {\left (2 \, {\left (x^{2} - 5 \, x + 4\right )} e^{\left (2 \, x\right )} + {\left (x^{2} - 5 \, x\right )} e^{\left (x + \log \left (x - 4\right )\right )}\right )} e^{\left (-e^{\left (-x + \log \left (x - 4\right )\right )}\right )}}{3 \, {\left (x^{4} - 4 \, x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-25*x)*exp(x)^2*exp(log(x-4)-x)+(10*x^2-50*x+40)*exp(x)^2)/(3*x^4-12*x^3)/exp(exp(log(x-4)-x)
),x, algorithm="giac")

[Out]

integrate(5/3*(2*(x^2 - 5*x + 4)*e^(2*x) + (x^2 - 5*x)*e^(x + log(x - 4)))*e^(-e^(-x + log(x - 4)))/(x^4 - 4*x
^3), x)

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maple [A]  time = 0.16, size = 21, normalized size = 0.91

method	result	size
norman	\(\frac {5 \,{\mathrm e}^{2 x} {\mathrm e}^{-\left (x -4\right ) {\mathrm e}^{-x}}}{3 x^{2}}\)	\(21\)
risch	\(\frac {5 \,{\mathrm e}^{-x \,{\mathrm e}^{-x}+4 \,{\mathrm e}^{-x}+2 x}}{3 x^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2-25*x)*exp(x)^2*exp(ln(x-4)-x)+(10*x^2-50*x+40)*exp(x)^2)/(3*x^4-12*x^3)/exp(exp(ln(x-4)-x)),x,meth
od=_RETURNVERBOSE)

[Out]

5/3*exp(x)^2/x^2/exp((x-4)/exp(x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{3} \, \int \frac {{\left ({\left (x^{2} - 5 \, x\right )} {\left (x - 4\right )} e^{x} + 2 \, {\left (x^{2} - 5 \, x + 4\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-{\left (x - 4\right )} e^{\left (-x\right )}\right )}}{x^{4} - 4 \, x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-25*x)*exp(x)^2*exp(log(x-4)-x)+(10*x^2-50*x+40)*exp(x)^2)/(3*x^4-12*x^3)/exp(exp(log(x-4)-x)
),x, algorithm="maxima")

[Out]

5/3*integrate(((x^2 - 5*x)*(x - 4)*e^x + 2*(x^2 - 5*x + 4)*e^(2*x))*e^(-(x - 4)*e^(-x))/(x^4 - 4*x^3), x)

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mupad [B]  time = 4.72, size = 24, normalized size = 1.04 \begin {gather*} \frac {5\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-x}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{-x}}}{3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(log(x - 4) - x))*(exp(2*x)*(10*x^2 - 50*x + 40) - exp(2*x)*exp(log(x - 4) - x)*(25*x - 5*x^2)))
/(12*x^3 - 3*x^4),x)

[Out]

(5*exp(4*exp(-x))*exp(2*x)*exp(-x*exp(-x)))/(3*x^2)

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sympy [A]  time = 0.20, size = 19, normalized size = 0.83 \begin {gather*} \frac {5 e^{2 x} e^{- \left (x - 4\right ) e^{- x}}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2-25*x)*exp(x)**2*exp(ln(x-4)-x)+(10*x**2-50*x+40)*exp(x)**2)/(3*x**4-12*x**3)/exp(exp(ln(x-4
)-x)),x)

[Out]

5*exp(2*x)*exp(-(x - 4)*exp(-x))/(3*x**2)

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