3.74.45 \(\int \frac {1}{2} (-2 e^{e^{\log ^2(10)}}+e^{x/2}) \, dx\)

Optimal. Leaf size=20 \[ 1+e^{x/2}-e^{e^{\log ^2(10)}} x \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 0.95, 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 e^{e^{\log ^2(10)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^E^Log[10]^2 + E^(x/2))/2,x]

[Out]

E^(x/2) - E^E^Log[10]^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 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^{e^{\log ^2(10)}}+e^{x/2}\right ) \, dx\\ &=-e^{e^{\log ^2(10)}} x+\frac {1}{2} \int e^{x/2} \, dx\\ &=e^{x/2}-e^{e^{\log ^2(10)}} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.25 \begin {gather*} \frac {1}{2} \left (2 e^{x/2}-2 e^{e^{\log ^2(10)}} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^E^Log[10]^2 + E^(x/2))/2,x]

[Out]

(2*E^(x/2) - 2*E^E^Log[10]^2*x)/2

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fricas [A]  time = 0.80, size = 14, normalized size = 0.70 \begin {gather*} -x e^{\left (e^{\left (\log \left (10\right )^{2}\right )}\right )} + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(log(10)^2))+1/2*exp(1/2*x),x, algorithm="fricas")

[Out]

-x*e^(e^(log(10)^2)) + e^(1/2*x)

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giac [A]  time = 0.13, size = 14, normalized size = 0.70 \begin {gather*} -x e^{\left (e^{\left (\log \left (10\right )^{2}\right )}\right )} + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(log(10)^2))+1/2*exp(1/2*x),x, algorithm="giac")

[Out]

-x*e^(e^(log(10)^2)) + e^(1/2*x)

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maple [A]  time = 0.06, size = 15, normalized size = 0.75

method	result	size
default	\({\mathrm e}^{\frac {x}{2}}-x \,{\mathrm e}^{{\mathrm e}^{\ln \left (10\right )^{2}}}\)	\(15\)
norman	\({\mathrm e}^{\frac {x}{2}}-x \,{\mathrm e}^{{\mathrm e}^{\ln \left (10\right )^{2}}}\)	\(15\)
risch	\(-{\mathrm e}^{{\mathrm e}^{\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}} x +{\mathrm e}^{\frac {x}{2}}\)	\(18\)
derivativedivides	\({\mathrm e}^{\frac {x}{2}}-2 \,{\mathrm e}^{{\mathrm e}^{\ln \left (10\right )^{2}}} \ln \left ({\mathrm e}^{\frac {x}{2}}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(exp(ln(10)^2))+1/2*exp(1/2*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/2*x)-x*exp(exp(ln(10)^2))

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maxima [A]  time = 0.36, size = 14, normalized size = 0.70 \begin {gather*} -x e^{\left (e^{\left (\log \left (10\right )^{2}\right )}\right )} + e^{\left (\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(log(10)^2))+1/2*exp(1/2*x),x, algorithm="maxima")

[Out]

-x*e^(e^(log(10)^2)) + e^(1/2*x)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.70 \begin {gather*} {\mathrm {e}}^{x/2}-x\,{\mathrm {e}}^{{\mathrm {e}}^{{\ln \left (10\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x/2)/2 - exp(exp(log(10)^2)),x)

[Out]

exp(x/2) - x*exp(exp(log(10)^2))

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sympy [A]  time = 0.08, size = 14, normalized size = 0.70 \begin {gather*} - x e^{e^{\log {\left (10 \right )}^{2}}} + e^{\frac {x}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(exp(ln(10)**2))+1/2*exp(1/2*x),x)

[Out]

-x*exp(exp(log(10)**2)) + exp(x/2)

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