3.55.85 \(\int \frac {1}{2} (2+e^{x/2}+2 e^4 \log (\frac {9}{4})) \, dx\)

Optimal. Leaf size=22 \[ 4+e^{x/2}+x+e^4 \left (2+x \log \left (\frac {9}{4}\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2194} \begin {gather*} e^{x/2}+x \left (1+e^4 \log \left (\frac {9}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + E^(x/2) + 2*E^4*Log[9/4])/2,x]

[Out]

E^(x/2) + x*(1 + E^4*Log[9/4])

Rule 12

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

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 \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx\\ &=x \left (1+e^4 \log \left (\frac {9}{4}\right )\right )+\frac {1}{2} \int e^{x/2} \, dx\\ &=e^{x/2}+x \left (1+e^4 \log \left (\frac {9}{4}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.82 \begin {gather*} e^{x/2}+x+e^4 x \log \left (\frac {9}{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^(x/2) + 2*E^4*Log[9/4])/2,x]

[Out]

E^(x/2) + x + E^4*x*Log[9/4]

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fricas [A]  time = 0.44, size = 13, normalized size = 0.59 \begin {gather*} -x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*exp(1/2*x)-exp(4)*log(4/9)+1,x, algorithm="fricas")

[Out]

-x*e^4*log(4/9) + x + e^(1/2*x)

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giac [A]  time = 0.29, size = 13, normalized size = 0.59 \begin {gather*} -x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*exp(1/2*x)-exp(4)*log(4/9)+1,x, algorithm="giac")

[Out]

-x*e^4*log(4/9) + x + e^(1/2*x)

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maple [A]  time = 0.06, size = 14, normalized size = 0.64

method	result	size
default	\(x -{\mathrm e}^{4} \ln \left (\frac {4}{9}\right ) x +{\mathrm e}^{\frac {x}{2}}\)	\(14\)
derivativedivides	\({\mathrm e}^{\frac {x}{2}}+\left (-2 \,{\mathrm e}^{4} \ln \left (\frac {4}{9}\right )+2\right ) \ln \left ({\mathrm e}^{\frac {x}{2}}\right )\)	\(20\)
risch	\(-2 x \,{\mathrm e}^{4} \ln \relax (2)+2 \,{\mathrm e}^{4} x \ln \relax (3)+{\mathrm e}^{\frac {x}{2}}+x\)	\(21\)
norman	\(\left (-2 \,{\mathrm e}^{4} \ln \relax (2)+2 \,{\mathrm e}^{4} \ln \relax (3)+1\right ) x +{\mathrm e}^{\frac {x}{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*exp(1/2*x)-exp(4)*ln(4/9)+1,x,method=_RETURNVERBOSE)

[Out]

x-exp(4)*ln(4/9)*x+exp(1/2*x)

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maxima [A]  time = 0.38, size = 13, normalized size = 0.59 \begin {gather*} -x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*exp(1/2*x)-exp(4)*log(4/9)+1,x, algorithm="maxima")

[Out]

-x*e^4*log(4/9) + x + e^(1/2*x)

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mupad [B]  time = 0.08, size = 15, normalized size = 0.68 \begin {gather*} {\mathrm {e}}^{x/2}-x\,\left ({\mathrm {e}}^4\,\ln \left (\frac {4}{9}\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x/2)/2 - exp(4)*log(4/9) + 1,x)

[Out]

exp(x/2) - x*(exp(4)*log(4/9) - 1)

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sympy [A]  time = 0.09, size = 24, normalized size = 1.09 \begin {gather*} x \left (- 2 e^{4} \log {\relax (2 )} + 1 + 2 e^{4} \log {\relax (3 )}\right ) + e^{\frac {x}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*exp(1/2*x)-exp(4)*ln(4/9)+1,x)

[Out]

x*(-2*exp(4)*log(2) + 1 + 2*exp(4)*log(3)) + exp(x/2)

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