\(\int \frac {1}{2} (2+e^{x/2}+2 e^4 \log (\frac {9}{4})) \, dx\) [5485]

Optimal result

Integrand size = 22, antiderivative size = 22 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=4+e^{x/2}+x+e^4 \left (2+x \log \left (\frac {9}{4}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {12, 2225} \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=e^{x/2}+x \left (1+e^4 \log \left (\frac {9}{4}\right )\right ) \]

[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 2225

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=e^{x/2}+x+e^4 x \log \left (\frac {9}{4}\right ) \]

[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]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64

method	result	size
default	\(x -{\mathrm e}^{4} \ln \left (\frac {4}{9}\right ) x +{\mathrm e}^{\frac {x}{2}}\)	\(14\)
parts	\(x -{\mathrm e}^{4} \ln \left (\frac {4}{9}\right ) x +{\mathrm e}^{\frac {x}{2}}\)	\(14\)
parallelrisch	\({\mathrm e}^{\frac {x}{2}}+\left (-{\mathrm e}^{4} \ln \left (\frac {4}{9}\right )+1\right ) x\)	\(16\)
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	\(2 \,{\mathrm e}^{4} x \ln \left (3\right )-2 x \,{\mathrm e}^{4} \ln \left (2\right )+{\mathrm e}^{\frac {x}{2}}+x\)	\(21\)
norman	\(\left (2 \,{\mathrm e}^{4} \ln \left (3\right )-2 \,{\mathrm e}^{4} \ln \left (2\right )+1\right ) x +{\mathrm e}^{\frac {x}{2}}\)	\(22\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=-x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=x \left (- 2 e^{4} \log {\left (2 \right )} + 1 + 2 e^{4} \log {\left (3 \right )}\right ) + e^{\frac {x}{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=-x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx=-x e^{4} \log \left (\frac {4}{9}\right ) + x + e^{\left (\frac {1}{2} \, x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {1}{2} \left (2+e^{x/2}+2 e^4 \log \left (\frac {9}{4}\right )\right ) \, dx={\mathrm {e}}^{x/2}-x\,\left ({\mathrm {e}}^4\,\ln \left (\frac {4}{9}\right )-1\right ) \]

[In]

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

[Out]

exp(x/2) - x*(exp(4)*log(4/9) - 1)