\(\int \frac {1}{2} (-3 e^{3 x}+e^x (2+x+\log (2))) \, dx\) [3240]

Optimal result

Integrand size = 21, antiderivative size = 19 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {1}{2} e^x \left (1-e^{2 x}+x+\log (2)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2225, 2207} \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=-\frac {e^x}{2}-\frac {e^{3 x}}{2}+\frac {1}{2} e^x (x+2+\log (2)) \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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{align*} \text {integral}& = \frac {1}{2} \int \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx \\ & = \frac {1}{2} \int e^x (2+x+\log (2)) \, dx-\frac {3}{2} \int e^{3 x} \, dx \\ & = -\frac {e^{3 x}}{2}+\frac {1}{2} e^x (2+x+\log (2))-\frac {\int e^x \, dx}{2} \\ & = -\frac {e^x}{2}-\frac {e^{3 x}}{2}+\frac {1}{2} e^x (2+x+\log (2)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {1}{2} \left (-e^{3 x}+e^x (1+x+\log (2))\right ) \]

[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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {{\mathrm e}^{3 x}}{2}+\frac {\left (\ln \left (2\right )+x +1\right ) {\mathrm e}^{x}}{2}\)	\(17\)
norman	\(\left (\frac {\ln \left (2\right )}{2}+\frac {1}{2}\right ) {\mathrm e}^{x}-\frac {{\mathrm e}^{3 x}}{2}+\frac {{\mathrm e}^{x} x}{2}\)	\(22\)
default	\(-\frac {{\mathrm e}^{3 x}}{2}+\frac {{\mathrm e}^{x} \ln \left (2\right )}{2}+\frac {{\mathrm e}^{x} x}{2}+\frac {{\mathrm e}^{x}}{2}\)	\(23\)
parallelrisch	\(\frac {{\mathrm e}^{x} x}{2}-\frac {{\mathrm e}^{x} {\mathrm e}^{2 x}}{2}+\frac {{\mathrm e}^{x} \ln \left (2\right )}{2}+\frac {{\mathrm e}^{x}}{2}\)	\(25\)
meijerg	\(1-\frac {{\mathrm e}^{3 x}}{2}-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{4}-\left (1+\frac {\ln \left (2\right )}{2}\right ) \left (1-{\mathrm e}^{x}\right )\)	\(32\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {1}{2} \, {\left (x + \log \left (2\right ) + 1\right )} e^{x} - \frac {1}{2} \, e^{\left (3 \, x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {\left (2 x + 2 \log {\left (2 \right )} + 2\right ) e^{x}}{4} - \frac {e^{3 x}}{2} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {1}{2} \, {\left (x - 1\right )} e^{x} + \frac {1}{2} \, e^{x} \log \left (2\right ) - \frac {1}{2} \, e^{\left (3 \, x\right )} + e^{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {1}{2} \, {\left (x + \log \left (2\right ) + 1\right )} e^{x} - \frac {1}{2} \, e^{\left (3 \, x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.68 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{2} \left (-3 e^{3 x}+e^x (2+x+\log (2))\right ) \, dx=\frac {{\mathrm {e}}^x\,\left (x-{\mathrm {e}}^{2\,x}+\ln \left (2\right )+1\right )}{2} \]

[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