3.29.46 \(\int \frac {1}{2} (22+4 e^{2 x}-4 e^{x^2} x+\log (5)) \, dx\)

Optimal. Leaf size=23 \[ e^{2 x}-e^{x^2}+x \left (11+\frac {\log (5)}{2}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 2194, 2209} \begin {gather*} -e^{x^2}+e^{2 x}+\frac {1}{2} x (22+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(22 + 4*E^(2*x) - 4*E^x^2*x + Log[5])/2,x]

[Out]

E^(2*x) - E^x^2 + (x*(22 + Log[5]))/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 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]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (22+4 e^{2 x}-4 e^{x^2} x+\log (5)\right ) \, dx\\ &=\frac {1}{2} x (22+\log (5))+2 \int e^{2 x} \, dx-2 \int e^{x^2} x \, dx\\ &=e^{2 x}-e^{x^2}+\frac {1}{2} x (22+\log (5))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.13 \begin {gather*} \frac {1}{2} \left (2 e^{2 x}-2 e^{x^2}+22 x+x \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(22 + 4*E^(2*x) - 4*E^x^2*x + Log[5])/2,x]

[Out]

(2*E^(2*x) - 2*E^x^2 + 22*x + x*Log[5])/2

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fricas [A]  time = 0.42, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, x \log \relax (5) + 11 \, x - e^{\left (x^{2}\right )} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(x^2)*x+2*exp(2*x)+1/2*log(5)+11,x, algorithm="fricas")

[Out]

1/2*x*log(5) + 11*x - e^(x^2) + e^(2*x)

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giac [A]  time = 0.27, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, x \log \relax (5) + 11 \, x - e^{\left (x^{2}\right )} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(x^2)*x+2*exp(2*x)+1/2*log(5)+11,x, algorithm="giac")

[Out]

1/2*x*log(5) + 11*x - e^(x^2) + e^(2*x)

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maple [A]  time = 0.04, size = 20, normalized size = 0.87




method result size



default \(11 x -{\mathrm e}^{x^{2}}+{\mathrm e}^{2 x}+\frac {x \ln \relax (5)}{2}\) \(20\)
norman \({\mathrm e}^{2 x}-{\mathrm e}^{x^{2}}+x \left (11+\frac {\ln \relax (5)}{2}\right )\) \(20\)
risch \(11 x -{\mathrm e}^{x^{2}}+{\mathrm e}^{2 x}+\frac {x \ln \relax (5)}{2}\) \(20\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*exp(x^2)*x+2*exp(2*x)+1/2*ln(5)+11,x,method=_RETURNVERBOSE)

[Out]

11*x-exp(x^2)+exp(2*x)+1/2*x*ln(5)

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maxima [A]  time = 0.34, size = 19, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, x \log \relax (5) + 11 \, x - e^{\left (x^{2}\right )} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(x^2)*x+2*exp(2*x)+1/2*log(5)+11,x, algorithm="maxima")

[Out]

1/2*x*log(5) + 11*x - e^(x^2) + e^(2*x)

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mupad [B]  time = 1.64, size = 19, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{2\,x}-{\mathrm {e}}^{x^2}+x\,\left (\frac {\ln \relax (5)}{2}+11\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(2*x) + log(5)/2 - 2*x*exp(x^2) + 11,x)

[Out]

exp(2*x) - exp(x^2) + x*(log(5)/2 + 11)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.74 \begin {gather*} x \left (\frac {\log {\relax (5 )}}{2} + 11\right ) + e^{2 x} - e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*exp(x**2)*x+2*exp(2*x)+1/2*ln(5)+11,x)

[Out]

x*(log(5)/2 + 11) + exp(2*x) - exp(x**2)

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