3.1.17 \(\int (6+2 x+e^{2-x} (1-x) \log (2)) \, dx\)

Optimal. Leaf size=15 \[ x \left (6+x+e^{2-x} \log (2)\right ) \]

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Rubi [B]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 2.20, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2176, 2194} \begin {gather*} x^2+6 x+e^{2-x} \log (2)-e^{2-x} (1-x) \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[6 + 2*x + E^(2 - x)*(1 - x)*Log[2],x]

[Out]

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

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} &=6 x+x^2+\log (2) \int e^{2-x} (1-x) \, dx\\ &=6 x+x^2-e^{2-x} (1-x) \log (2)-\log (2) \int e^{2-x} \, dx\\ &=6 x+x^2+e^{2-x} \log (2)-e^{2-x} (1-x) \log (2)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 1.20 \begin {gather*} 6 x+x^2+e^{2-x} x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[6 + 2*x + E^(2 - x)*(1 - x)*Log[2],x]

[Out]

6*x + x^2 + E^(2 - x)*x*Log[2]

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fricas [A]  time = 0.61, size = 17, normalized size = 1.13 \begin {gather*} x e^{\left (-x + 2\right )} \log \relax (2) + x^{2} + 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+1)*log(2)*exp(2-x)+2*x+6,x, algorithm="fricas")

[Out]

x*e^(-x + 2)*log(2) + x^2 + 6*x

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giac [A]  time = 0.15, size = 17, normalized size = 1.13 \begin {gather*} x e^{\left (-x + 2\right )} \log \relax (2) + x^{2} + 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+1)*log(2)*exp(2-x)+2*x+6,x, algorithm="giac")

[Out]

x*e^(-x + 2)*log(2) + x^2 + 6*x

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maple [A]  time = 0.04, size = 18, normalized size = 1.20

method	result	size
default	\(6 x +{\mathrm e}^{2-x} \ln \relax (2) x +x^{2}\)	\(18\)
norman	\(6 x +{\mathrm e}^{2-x} \ln \relax (2) x +x^{2}\)	\(18\)
risch	\(6 x +{\mathrm e}^{2-x} \ln \relax (2) x +x^{2}\)	\(18\)
derivativedivides	\(\left (2-x \right )^{2}-20+10 x -\ln \relax (2) {\mathrm e}^{2-x} \left (2-x \right )+2 \,{\mathrm e}^{2-x} \ln \relax (2)\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)*ln(2)*exp(2-x)+2*x+6,x,method=_RETURNVERBOSE)

[Out]

6*x+exp(2-x)*ln(2)*x+x^2

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maxima [A]  time = 0.53, size = 17, normalized size = 1.13 \begin {gather*} x e^{\left (-x + 2\right )} \log \relax (2) + x^{2} + 6 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+1)*log(2)*exp(2-x)+2*x+6,x, algorithm="maxima")

[Out]

x*e^(-x + 2)*log(2) + x^2 + 6*x

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mupad [B]  time = 0.17, size = 14, normalized size = 0.93 \begin {gather*} x\,\left (x+{\mathrm {e}}^{2-x}\,\ln \relax (2)+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - exp(2 - x)*log(2)*(x - 1) + 6,x)

[Out]

x*(x + exp(2 - x)*log(2) + 6)

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sympy [A]  time = 0.09, size = 15, normalized size = 1.00 \begin {gather*} x^{2} + x e^{2 - x} \log {\relax (2 )} + 6 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x+1)*ln(2)*exp(2-x)+2*x+6,x)

[Out]

x**2 + x*exp(2 - x)*log(2) + 6*x

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