\(\int \frac {e^{-2 x} (1-2 x+(3-6 x^2) \log (4))}{3 \log (4)} \, dx\) [4103]

Optimal result

Integrand size = 28, antiderivative size = 20 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=e^{-2 x} \left (x+x^2+\frac {x}{3 \log (4)}\right ) \]

[Out]

exp(ln(1/6*x/ln(2)+x+x^2)-2*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=e^{-2 x} x^2+e^{-2 x} x+\frac {e^{-2 x} x}{3 \log (4)} \]

[In]

Int[(1 - 2*x + (3 - 6*x^2)*Log[4])/(3*E^(2*x)*Log[4]),x]

[Out]

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

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right ) \, dx}{3 \log (4)} \\ & = \frac {\int \left (e^{-2 x}-2 e^{-2 x} x-3 e^{-2 x} \left (-1+2 x^2\right ) \log (4)\right ) \, dx}{3 \log (4)} \\ & = \frac {\int e^{-2 x} \, dx}{3 \log (4)}-\frac {2 \int e^{-2 x} x \, dx}{3 \log (4)}-\int e^{-2 x} \left (-1+2 x^2\right ) \, dx \\ & = -\frac {e^{-2 x}}{6 \log (4)}+\frac {e^{-2 x} x}{3 \log (4)}-\frac {\int e^{-2 x} \, dx}{3 \log (4)}-\int \left (-e^{-2 x}+2 e^{-2 x} x^2\right ) \, dx \\ & = \frac {e^{-2 x} x}{3 \log (4)}-2 \int e^{-2 x} x^2 \, dx+\int e^{-2 x} \, dx \\ & = -\frac {1}{2} e^{-2 x}+e^{-2 x} x^2+\frac {e^{-2 x} x}{3 \log (4)}-2 \int e^{-2 x} x \, dx \\ & = -\frac {1}{2} e^{-2 x}+e^{-2 x} x+e^{-2 x} x^2+\frac {e^{-2 x} x}{3 \log (4)}-\int e^{-2 x} \, dx \\ & = e^{-2 x} x+e^{-2 x} x^2+\frac {e^{-2 x} x}{3 \log (4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=\frac {e^{-2 x} x (1+\log (64)+x \log (64))}{\log (64)} \]

[In]

Integrate[(1 - 2*x + (3 - 6*x^2)*Log[4])/(3*E^(2*x)*Log[4]),x]

[Out]

(x*(1 + Log[64] + x*Log[64]))/(E^(2*x)*Log[64])

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25

method	result	size
gosper	\({\mathrm e}^{\ln \left (\frac {x \left (6 x \ln \left (2\right )+6 \ln \left (2\right )+1\right )}{6 \ln \left (2\right )}\right )-2 x}\)	\(25\)
risch	\(\frac {\left (2 \left (3 x^{2}+3 x \right ) \ln \left (2\right )+x \right ) {\mathrm e}^{-2 x}}{6 \ln \left (2\right )}\)	\(26\)
norman	\({\mathrm e}^{\ln \left (\frac {2 \left (3 x^{2}+3 x \right ) \ln \left (2\right )+x}{6 \ln \left (2\right )}\right )-2 x}\)	\(28\)
parallelrisch	\({\mathrm e}^{\ln \left (\frac {2 \left (3 x^{2}+3 x \right ) \ln \left (2\right )+x}{6 \ln \left (2\right )}\right )-2 x}\)	\(28\)
default	\(\frac {x \,{\mathrm e}^{-2 x}-3 \,{\mathrm e}^{-2 x} \ln \left (2\right )+\frac {3 \ln \left (2\right ) \left (4 x^{2} {\mathrm e}^{-2 x}+4 x \,{\mathrm e}^{-2 x}+2 \,{\mathrm e}^{-2 x}\right )}{2}}{6 \ln \left (2\right )}\)	\(49\)

[In]

int((2*(-6*x^2+3)*ln(2)+1-2*x)*exp(ln(1/6*(2*(3*x^2+3*x)*ln(2)+x)/ln(2))-2*x)/(2*(3*x^2+3*x)*ln(2)+x),x,method
=_RETURNVERBOSE)

[Out]

exp(ln(1/6*x*(6*x*ln(2)+6*ln(2)+1)/ln(2))-2*x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=e^{\left (-2 \, x + \log \left (\frac {6 \, {\left (x^{2} + x\right )} \log \left (2\right ) + x}{6 \, \log \left (2\right )}\right )\right )} \]

[In]

integrate((2*(-6*x^2+3)*log(2)+1-2*x)*exp(log(1/6*(2*(3*x^2+3*x)*log(2)+x)/log(2))-2*x)/(2*(3*x^2+3*x)*log(2)+
x),x, algorithm="fricas")

[Out]

e^(-2*x + log(1/6*(6*(x^2 + x)*log(2) + x)/log(2)))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=\frac {\left (6 x^{2} \log {\left (2 \right )} + x + 6 x \log {\left (2 \right )}\right ) e^{- 2 x}}{6 \log {\left (2 \right )}} \]

[In]

integrate((2*(-6*x**2+3)*ln(2)+1-2*x)*exp(ln(1/6*(2*(3*x**2+3*x)*ln(2)+x)/ln(2))-2*x)/(2*(3*x**2+3*x)*ln(2)+x)
,x)

[Out]

(6*x**2*log(2) + x + 6*x*log(2))*exp(-2*x)/(6*log(2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (18) = 36\).

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.45 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=\frac {6 \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} \log \left (2\right ) + {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} - 6 \, e^{\left (-2 \, x\right )} \log \left (2\right ) - e^{\left (-2 \, x\right )}}{12 \, \log \left (2\right )} \]

[In]

integrate((2*(-6*x^2+3)*log(2)+1-2*x)*exp(log(1/6*(2*(3*x^2+3*x)*log(2)+x)/log(2))-2*x)/(2*(3*x^2+3*x)*log(2)+
x),x, algorithm="maxima")

[Out]

1/12*(6*(2*x^2 + 2*x + 1)*e^(-2*x)*log(2) + (2*x + 1)*e^(-2*x) - 6*e^(-2*x)*log(2) - e^(-2*x))/log(2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=\frac {6 \, x^{2} e^{\left (-2 \, x\right )} \log \left (2\right ) + 6 \, x e^{\left (-2 \, x\right )} \log \left (2\right ) + x e^{\left (-2 \, x\right )}}{6 \, \log \left (2\right )} \]

[In]

integrate((2*(-6*x^2+3)*log(2)+1-2*x)*exp(log(1/6*(2*(3*x^2+3*x)*log(2)+x)/log(2))-2*x)/(2*(3*x^2+3*x)*log(2)+
x),x, algorithm="giac")

[Out]

1/6*(6*x^2*e^(-2*x)*log(2) + 6*x*e^(-2*x)*log(2) + x*e^(-2*x))/log(2)

Mupad [B] (verification not implemented)

Time = 16.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.90 \[ \int \frac {e^{-2 x} \left (1-2 x+\left (3-6 x^2\right ) \log (4)\right )}{3 \log (4)} \, dx=\frac {{\mathrm {e}}^{-2\,x}\,\left (x-3\,\ln \left (2\right )+\ln \left (8\right )+72\,x^2\,{\ln \left (2\right )}^2+36\,x^3\,{\ln \left (2\right )}^2+3\,x\,\ln \left (2\right )+x\,\ln \left (512\right )+36\,x\,{\ln \left (2\right )}^2+6\,x^2\,\ln \left (2\right )+x^2\,\ln \left (64\right )\right )}{6\,\ln \left (2\right )\,\left (6\,\ln \left (2\right )+6\,x\,\ln \left (2\right )+1\right )} \]

[In]

int(-(exp(log((x/6 + (log(2)*(3*x + 3*x^2))/3)/log(2)) - 2*x)*(2*x + 2*log(2)*(6*x^2 - 3) - 1))/(x + 2*log(2)*
(3*x + 3*x^2)),x)

[Out]

(exp(-2*x)*(x - 3*log(2) + log(8) + 72*x^2*log(2)^2 + 36*x^3*log(2)^2 + 3*x*log(2) + x*log(512) + 36*x*log(2)^
2 + 6*x^2*log(2) + x^2*log(64)))/(6*log(2)*(6*log(2) + 6*x*log(2) + 1))