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

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

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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} $$\begin{array}{c}\frac{1}{4}{e}^{2x}(2x+5+\log (4))-\frac{e^{2x}}{4}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int e^{2x}(5+2x+2\log (2))dx\\ & =\frac{1}{4}{e}^{2x}(5+2x+\log (4))-\frac{1}{2}\int e^{2x}dx\\ & =-\frac{e^{2x}}{4}+\frac{1}{4}{e}^{2x}(5+2x+\log (4))\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{1}{2}(x+\log (2)+2)e^{\left(2x\right)}\end{array}$$

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 $$\begin{array}{c}\frac{1}{2}(x+\log (2)+2)e^{\left(2x\right)}\end{array}$$

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	$\frac{{\mathrm{e}}^{2x}(\ln (2)+2+x)}{2}$	$12$
risch	$\frac{{\mathrm{e}}^{2x}(\ln (2)+2+x)}{2}$	$12$
norman	$(1+\frac{\ln (2)}{2}){\mathrm{e}}^{2x}+\frac{x{\mathrm{e}}^{2x}}{2}$	$20$
derivativedivides	$\frac{x{\mathrm{e}}^{2x}}{2}+{\mathrm{e}}^{2x}+\frac{\ln (2){\mathrm{e}}^{2x}}{2}$	$21$
default	$\frac{x{\mathrm{e}}^{2x}}{2}+{\mathrm{e}}^{2x}+\frac{\ln (2){\mathrm{e}}^{2x}}{2}$	$21$
meijerg	$-1+\frac{5{\mathrm{e}}^{2x}}{4}-\frac{\ln (2)(1-{\mathrm{e}}^{2x})}{2}-\frac{(-4x+2){\mathrm{e}}^{2x}}{8}$	$32$

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 $$\begin{array}{c}\frac{1}{4}(2x-1)e^{\left(2x\right)}+\frac{1}{2}{e}^{\left(2x\right)}\log (2)+\frac{5}{4}{e}^{\left(2x\right)}\end{array}$$

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 $$\begin{array}{c}\frac{{\mathrm{e}}^{2x}(2x+\ln (4)+4)}{4}\end{array}$$

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 $$\begin{array}{c}\frac{(x+\log (2)+2)e^{2x}}{2}\end{array}$$

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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