3.18.54 \(\int \frac {2 e^{-11+x-x^2}+e^{-11+x-x^2} (2 x-3 x^2-2 x^3) \log (\frac {x}{4+2 x})}{e^2 (2 x+x^2)} \, dx\)

Optimal. Leaf size=21 \[ e^{-13+x-x^2} \log \left (\frac {x}{4+2 x}\right ) \]

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Rubi [B]  time = 0.58, antiderivative size = 44, normalized size of antiderivative = 2.10, number of steps used = 4, number of rules used = 4, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 1593, 6688, 2288} \begin {gather*} \frac {e^{-x^2+x-13} \left (-2 x^2-3 x+2\right ) \log \left (\frac {x}{2 (x+2)}\right )}{(1-2 x) (x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^(-11 + x - x^2) + E^(-11 + x - x^2)*(2*x - 3*x^2 - 2*x^3)*Log[x/(4 + 2*x)])/(E^2*(2*x + x^2)),x]

[Out]

(E^(-13 + x - x^2)*(2 - 3*x - 2*x^2)*Log[x/(2*(2 + x))])/((1 - 2*x)*(2 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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} &=\frac {\int \frac {2 e^{-11+x-x^2}+e^{-11+x-x^2} \left (2 x-3 x^2-2 x^3\right ) \log \left (\frac {x}{4+2 x}\right )}{2 x+x^2} \, dx}{e^2}\\ &=\frac {\int \frac {2 e^{-11+x-x^2}+e^{-11+x-x^2} \left (2 x-3 x^2-2 x^3\right ) \log \left (\frac {x}{4+2 x}\right )}{x (2+x)} \, dx}{e^2}\\ &=\frac {\int \frac {e^{-11+x-x^2} \left (2+x \left (2-3 x-2 x^2\right ) \log \left (\frac {x}{4+2 x}\right )\right )}{x (2+x)} \, dx}{e^2}\\ &=\frac {e^{-13+x-x^2} \left (2-3 x-2 x^2\right ) \log \left (\frac {x}{2 (2+x)}\right )}{(1-2 x) (2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.00 \begin {gather*} e^{-13+x-x^2} \log \left (\frac {x}{4+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^(-11 + x - x^2) + E^(-11 + x - x^2)*(2*x - 3*x^2 - 2*x^3)*Log[x/(4 + 2*x)])/(E^2*(2*x + x^2)),x
]

[Out]

E^(-13 + x - x^2)*Log[x/(4 + 2*x)]

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fricas [A]  time = 1.30, size = 19, normalized size = 0.90 \begin {gather*} e^{\left (-x^{2} + x - 13\right )} \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-3*x^2+2*x)*exp(-x^2+x-11)*log(x/(2*x+4))+2*exp(-x^2+x-11))/(x^2+2*x)/exp(2),x, algorithm="f
ricas")

[Out]

e^(-x^2 + x - 13)*log(1/2*x/(x + 2))

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giac [A]  time = 0.25, size = 19, normalized size = 0.90 \begin {gather*} e^{\left (-x^{2} + x - 13\right )} \log \left (\frac {x}{2 \, {\left (x + 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-3*x^2+2*x)*exp(-x^2+x-11)*log(x/(2*x+4))+2*exp(-x^2+x-11))/(x^2+2*x)/exp(2),x, algorithm="g
iac")

[Out]

e^(-x^2 + x - 13)*log(1/2*x/(x + 2))

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maple [A]  time = 0.37, size = 25, normalized size = 1.19

method	result	size
norman	\(\ln \left (\frac {x}{2 x +4}\right ) {\mathrm e}^{-2} {\mathrm e}^{-x^{2}+x -11}\)	\(25\)
risch	\(-\ln \left (2+x \right ) {\mathrm e}^{-x^{2}+x -13}+\ln \relax (x ) {\mathrm e}^{-x^{2}+x -13}-\frac {i \mathrm {csgn}\left (\frac {i x}{2+x}\right ) \mathrm {csgn}\left (\frac {i}{2+x}\right ) \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{-x^{2}+x -13}}{2}+\frac {i \mathrm {csgn}\left (\frac {i x}{2+x}\right )^{2} \mathrm {csgn}\left (i x \right ) \pi \,{\mathrm e}^{-x^{2}+x -13}}{2}+\frac {i \mathrm {csgn}\left (\frac {i x}{2+x}\right )^{2} \mathrm {csgn}\left (\frac {i}{2+x}\right ) \pi \,{\mathrm e}^{-x^{2}+x -13}}{2}-\frac {i \mathrm {csgn}\left (\frac {i x}{2+x}\right )^{3} \pi \,{\mathrm e}^{-x^{2}+x -13}}{2}-\ln \relax (2) {\mathrm e}^{-x^{2}+x -13}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3-3*x^2+2*x)*exp(-x^2+x-11)*ln(x/(2*x+4))+2*exp(-x^2+x-11))/(x^2+2*x)/exp(2),x,method=_RETURNVERBOS
E)

[Out]

ln(x/(2*x+4))/exp(2)*exp(-x^2+x-11)

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maxima [A]  time = 0.84, size = 28, normalized size = 1.33 \begin {gather*} -{\left ({\left (\log \relax (2) - \log \relax (x)\right )} e^{x} + e^{x} \log \left (x + 2\right )\right )} e^{\left (-x^{2} - 13\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-3*x^2+2*x)*exp(-x^2+x-11)*log(x/(2*x+4))+2*exp(-x^2+x-11))/(x^2+2*x)/exp(2),x, algorithm="m
axima")

[Out]

-((log(2) - log(x))*e^x + e^x*log(x + 2))*e^(-x^2 - 13)

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mupad [B]  time = 1.42, size = 21, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-13}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^x\,\ln \left (\frac {x}{2\,x+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(2*exp(x - x^2 - 11) - exp(x - x^2 - 11)*log(x/(2*x + 4))*(3*x^2 - 2*x + 2*x^3)))/(2*x + x^2),x)

[Out]

exp(-13)*exp(-x^2)*exp(x)*log(x/(2*x + 4))

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sympy [A]  time = 0.57, size = 19, normalized size = 0.90 \begin {gather*} \frac {e^{- x^{2} + x - 11} \log {\left (\frac {x}{2 x + 4} \right )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3-3*x**2+2*x)*exp(-x**2+x-11)*ln(x/(2*x+4))+2*exp(-x**2+x-11))/(x**2+2*x)/exp(2),x)

[Out]

exp(-2)*exp(-x**2 + x - 11)*log(x/(2*x + 4))

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