3.5.65 \(\int \frac {1}{2} e^{-5+x} (-3-3 x) \, dx\)

Optimal. Leaf size=16 \[ \frac {3}{2} \left (-\frac {3}{16}-e^{-5+x} x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 2176, 2194} \begin {gather*} \frac {3 e^{x-5}}{2}-\frac {3}{2} e^{x-5} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-5 + x)*(-3 - 3*x))/2,x]

[Out]

(3*E^(-5 + x))/2 - (3*E^(-5 + x)*(1 + x))/2

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.62 \begin {gather*} -\frac {3}{2} e^{-5+x} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*E^(-5 + x)*x)/2

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fricas [A]  time = 0.66, size = 8, normalized size = 0.50 \begin {gather*} -\frac {3}{2} \, e^{\left (x + \log \relax (x) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x-3)*exp(log(x)+x-5)/x,x, algorithm="fricas")

[Out]

-3/2*e^(x + log(x) - 5)

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giac [A]  time = 0.50, size = 7, normalized size = 0.44 \begin {gather*} -\frac {3}{2} \, x e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x-3)*exp(log(x)+x-5)/x,x, algorithm="giac")

[Out]

-3/2*x*e^(x - 5)

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maple [A]  time = 0.02, size = 8, normalized size = 0.50




method result size



risch \(-\frac {3 x \,{\mathrm e}^{x -5}}{2}\) \(8\)
gosper \(-\frac {3 \,{\mathrm e}^{\ln \relax (x )+x -5}}{2}\) \(9\)
norman \(-\frac {3 \,{\mathrm e}^{\ln \relax (x )+x -5}}{2}\) \(9\)
default \(-\frac {3 \,{\mathrm e}^{x -5} \left (x -5\right )}{2}-\frac {15 \,{\mathrm e}^{x -5}}{2}\) \(17\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-3*x-3)*exp(ln(x)+x-5)/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 16, normalized size = 1.00 \begin {gather*} -\frac {3}{2} \, {\left (x - 1\right )} e^{\left (x - 5\right )} - \frac {3}{2} \, e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x-3)*exp(log(x)+x-5)/x,x, algorithm="maxima")

[Out]

-3/2*(x - 1)*e^(x - 5) - 3/2*e^(x - 5)

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mupad [B]  time = 0.44, size = 7, normalized size = 0.44 \begin {gather*} -\frac {3\,x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + log(x) - 5)*(3*x + 3))/(2*x),x)

[Out]

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

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sympy [A]  time = 0.09, size = 10, normalized size = 0.62 \begin {gather*} - \frac {3 x e^{x - 5}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-3*x-3)*exp(ln(x)+x-5)/x,x)

[Out]

-3*x*exp(x - 5)/2

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