3.33.40 $\int \frac{1}{2}(-3e^{3x}+e^x(2+x+\log (2)))dx$

Optimal. Leaf size=19 $$\frac{1}{2}{e}^x(1-e^{2x}+x+\log (2))$$

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 3, integrand size = 21, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.143, Rules used = {12, 2194, 2176} $$\begin{array}{c}-\frac{e^x}{2}-\frac{e^{3x}}{2}+\frac{1}{2}{e}^x(x+2+\log (2))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-3*E^(3*x) + E^x*(2 + x + Log[2]))/2,x]

[Out]

-1/2*E^x - E^(3*x)/2 + (E^x*(2 + x + Log[2]))/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 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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int (-3e^{3x}+e^x(2+x+\log (2)))dx\\ & =\frac{1}{2}\int e^x(2+x+\log (2))dx-\frac{3}{2}\int e^{3x}dx\\ & =-\frac{e^{3x}}{2}+\frac{1}{2}{e}^x(2+x+\log (2))-\frac{\int e^{x}dx}{2}\\ & =-\frac{e^x}{2}-\frac{e^{3x}}{2}+\frac{1}{2}{e}^x(2+x+\log (2))\end{array}\end{array}$$

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Mathematica [A]  time = 0.05, size = 21, normalized size = 1.11 $$\begin{array}{c}\frac{1}{2}(-e^{3x}+e^x(1+x+\log (2)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^(3*x) + E^x*(2 + x + Log[2]))/2,x]

[Out]

(-E^(3*x) + E^x*(1 + x + Log[2]))/2

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fricas [A]  time = 0.66, size = 16, normalized size = 0.84 $$\begin{array}{c}\frac{1}{2}(x+\log (2)+1)e^x-\frac{1}{2}{e}^{\left(3x\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/2*exp(x)*exp(2*x)+1/2*(log(2)+2+x)*exp(x),x, algorithm="fricas")

[Out]

1/2*(x + log(2) + 1)*e^x - 1/2*e^(3*x)

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giac [A]  time = 0.23, size = 16, normalized size = 0.84 $$\begin{array}{c}\frac{1}{2}(x+\log (2)+1)e^x-\frac{1}{2}{e}^{\left(3x\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/2*exp(x)*exp(2*x)+1/2*(log(2)+2+x)*exp(x),x, algorithm="giac")

[Out]

1/2*(x + log(2) + 1)*e^x - 1/2*e^(3*x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.89

method	result	size
risch	$-\frac{{\mathrm{e}}^{3x}}{2}+\frac{(\ln (2)+x+1){\mathrm{e}}^x}{2}$	$17$
norman	$(\frac{1}{2}+\frac{\ln (2)}{2}){\mathrm{e}}^x-\frac{{\mathrm{e}}^{3x}}{2}+\frac{{\mathrm{e}}^{x}x}{2}$	$22$
default	$\frac{{\mathrm{e}}^{x}x}{2}+\frac{{\mathrm{e}}^x}{2}+\frac{{\mathrm{e}}^x\ln (2)}{2}-\frac{{\mathrm{e}}^{3x}}{2}$	$23$
meijerg	$1-\frac{(-2x+2){\mathrm{e}}^x}{4}-\frac{{\mathrm{e}}^{3x}}{2}-(1+\frac{\ln (2)}{2})(1-{\mathrm{e}}^x)$	$32$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-3/2*exp(x)*exp(2*x)+1/2*(ln(2)+2+x)*exp(x),x,method=_RETURNVERBOSE)

[Out]

-1/2*exp(3*x)+1/2*(ln(2)+x+1)*exp(x)

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maxima [A]  time = 0.38, size = 22, normalized size = 1.16 $$\begin{array}{c}\frac{1}{2}(x-1)e^x+\frac{1}{2}{e}^x\log (2)-\frac{1}{2}{e}^{\left(3x\right)}+e^x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/2*exp(x)*exp(2*x)+1/2*(log(2)+2+x)*exp(x),x, algorithm="maxima")

[Out]

1/2*(x - 1)*e^x + 1/2*e^x*log(2) - 1/2*e^(3*x) + e^x

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mupad [B]  time = 0.05, size = 15, normalized size = 0.79 $$\begin{array}{c}\frac{{\mathrm{e}}^x(x-{\mathrm{e}}^{2x}+\ln (2)+1)}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x + log(2) + 2))/2 - (3*exp(3*x))/2,x)

[Out]

(exp(x)*(x - exp(2*x) + log(2) + 1))/2

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sympy [A]  time = 0.13, size = 20, normalized size = 1.05 $$\begin{array}{c}\frac{(2x+2\log (2)+2)e^x}{4}-\frac{e^{3x}}{2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-3/2*exp(x)*exp(2*x)+1/2*(ln(2)+2+x)*exp(x),x)

[Out]

(2*x + 2*log(2) + 2)*exp(x)/4 - exp(3*x)/2

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