3.33.91 \(\int \frac {1}{2} e^{2 x} (5+2 x+2 \log (2)) \, dx\)

Optimal. Leaf size=14 \[ \frac {1}{2} e^{2 x} (2+x+\log (2)) \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.86, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2176, 2194} \begin {gather*} \frac {1}{4} e^{2 x} (2 x+5+\log (4))-\frac {e^{2 x}}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*E^(2*x) + (E^(2*x)*(5 + 2*x + Log[4]))/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 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} &=\frac {1}{2} \int e^{2 x} (5+2 x+2 \log (2)) \, dx\\ &=\frac {1}{4} e^{2 x} (5+2 x+\log (4))-\frac {1}{2} \int e^{2 x} \, dx\\ &=-\frac {e^{2 x}}{4}+\frac {1}{4} e^{2 x} (5+2 x+\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.36 \begin {gather*} \frac {1}{2} e^{2 x} \left (x+\frac {1}{2} (4+\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(2*x)*(x + (4 + Log[4])/2))/2

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fricas [A]  time = 0.58, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, {\left (x + \log \relax (2) + 2\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x + log(2) + 2)*e^(2*x)

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giac [A]  time = 0.17, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, {\left (x + \log \relax (2) + 2\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x + log(2) + 2)*e^(2*x)

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maple [A]  time = 0.03, size = 12, normalized size = 0.86

method	result	size
gosper	\(\frac {{\mathrm e}^{2 x} \left (\ln \relax (2)+2+x \right )}{2}\)	\(12\)
risch	\(\frac {{\mathrm e}^{2 x} \left (\ln \relax (2)+2+x \right )}{2}\)	\(12\)
norman	\(\left (1+\frac {\ln \relax (2)}{2}\right ) {\mathrm e}^{2 x}+\frac {x \,{\mathrm e}^{2 x}}{2}\)	\(20\)
derivativedivides	\(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \relax (2) {\mathrm e}^{2 x}}{2}\)	\(21\)
default	\(\frac {x \,{\mathrm e}^{2 x}}{2}+{\mathrm e}^{2 x}+\frac {\ln \relax (2) {\mathrm e}^{2 x}}{2}\)	\(21\)
meijerg	\(-1+\frac {5 \,{\mathrm e}^{2 x}}{4}-\frac {\ln \relax (2) \left (1-{\mathrm e}^{2 x}\right )}{2}-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{8}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(2*x)*(ln(2)+2+x)

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maxima [B]  time = 0.34, size = 26, normalized size = 1.86 \begin {gather*} \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, e^{\left (2 \, x\right )} \log \relax (2) + \frac {5}{4} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*x - 1)*e^(2*x) + 1/2*e^(2*x)*log(2) + 5/4*e^(2*x)

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mupad [B]  time = 0.07, size = 13, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,\left (2\,x+\ln \relax (4)+4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*x)*(2*x + log(4) + 4))/4

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sympy [A]  time = 0.13, size = 12, normalized size = 0.86 \begin {gather*} \frac {\left (x + \log {\relax (2 )} + 2\right ) e^{2 x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + log(2) + 2)*exp(2*x)/2

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