\(\int \frac {1}{9} e^{2-2 x} (6 \log ^2(4)-2 \log ^2(4) \log (5)) \, dx\) [8442]

Optimal result

Integrand size = 26, antiderivative size = 22 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\left (5+\frac {1}{9} e^{2-2 x} \log ^2(4)\right ) (-3+\log (5)) \]

[Out]

(ln(5)-3)*(5+4/9*exp(1-x)^2*ln(2)^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2225} \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=-\frac {1}{9} e^{2-2 x} \log ^2(4) (3-\log (5)) \]

[In]

Int[(E^(2 - 2*x)*(6*Log[4]^2 - 2*Log[4]^2*Log[5]))/9,x]

[Out]

-1/9*(E^(2 - 2*x)*Log[4]^2*(3 - Log[5]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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}{9} \left (2 \log ^2(4) (3-\log (5))\right ) \int e^{2-2 x} \, dx \\ & = -\frac {1}{9} e^{2-2 x} \log ^2(4) (3-\log (5)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {1}{9} e^{2-2 x} \log ^2(4) (-3+\log (5)) \]

[In]

Integrate[(E^(2 - 2*x)*(6*Log[4]^2 - 2*Log[4]^2*Log[5]))/9,x]

[Out]

(E^(2 - 2*x)*Log[4]^2*(-3 + Log[5]))/9

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
gosper	\(\frac {4 \ln \left (2\right )^{2} \left (\ln \left (5\right )-3\right ) {\mathrm e}^{2-2 x}}{9}\)	\(19\)
norman	\(\left (\frac {4 \ln \left (2\right )^{2} \ln \left (5\right )}{9}-\frac {4 \ln \left (2\right )^{2}}{3}\right ) {\mathrm e}^{2-2 x}\)	\(25\)
derivativedivides	\(-\frac {\left (-8 \ln \left (2\right )^{2} \ln \left (5\right )+24 \ln \left (2\right )^{2}\right ) {\mathrm e}^{2-2 x}}{18}\)	\(26\)
default	\(-\frac {\left (-\frac {8 \ln \left (2\right )^{2} \ln \left (5\right )}{9}+\frac {8 \ln \left (2\right )^{2}}{3}\right ) {\mathrm e}^{2-2 x}}{2}\)	\(26\)
parallelrisch	\(-\frac {\left (-\frac {8 \ln \left (2\right )^{2} \ln \left (5\right )}{9}+\frac {8 \ln \left (2\right )^{2}}{3}\right ) {\mathrm e}^{2-2 x}}{2}\)	\(26\)
risch	\(\frac {4 \,{\mathrm e}^{2-2 x} \ln \left (2\right )^{2} \ln \left (5\right )}{9}-\frac {4 \,{\mathrm e}^{2-2 x} \ln \left (2\right )^{2}}{3}\)	\(28\)

[In]

int(1/9*(-8*ln(2)^2*ln(5)+24*ln(2)^2)*exp(1-x)^2,x,method=_RETURNVERBOSE)

[Out]

4/9*ln(2)^2*(ln(5)-3)*exp(1-x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {4}{9} \, {\left (\log \left (5\right ) \log \left (2\right )^{2} - 3 \, \log \left (2\right )^{2}\right )} e^{\left (-2 \, x + 2\right )} \]

[In]

integrate(1/9*(-8*log(2)^2*log(5)+24*log(2)^2)*exp(1-x)^2,x, algorithm="fricas")

[Out]

4/9*(log(5)*log(2)^2 - 3*log(2)^2)*e^(-2*x + 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {\left (- 12 \log {\left (2 \right )}^{2} + 4 \log {\left (2 \right )}^{2} \log {\left (5 \right )}\right ) e^{2 - 2 x}}{9} \]

[In]

integrate(1/9*(-8*ln(2)**2*ln(5)+24*ln(2)**2)*exp(1-x)**2,x)

[Out]

(-12*log(2)**2 + 4*log(2)**2*log(5))*exp(2 - 2*x)/9

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {4}{9} \, {\left (\log \left (5\right ) \log \left (2\right )^{2} - 3 \, \log \left (2\right )^{2}\right )} e^{\left (-2 \, x + 2\right )} \]

[In]

integrate(1/9*(-8*log(2)^2*log(5)+24*log(2)^2)*exp(1-x)^2,x, algorithm="maxima")

[Out]

4/9*(log(5)*log(2)^2 - 3*log(2)^2)*e^(-2*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {4}{9} \, {\left (\log \left (5\right ) \log \left (2\right )^{2} - 3 \, \log \left (2\right )^{2}\right )} e^{\left (-2 \, x + 2\right )} \]

[In]

integrate(1/9*(-8*log(2)^2*log(5)+24*log(2)^2)*exp(1-x)^2,x, algorithm="giac")

[Out]

4/9*(log(5)*log(2)^2 - 3*log(2)^2)*e^(-2*x + 2)

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{9} e^{2-2 x} \left (6 \log ^2(4)-2 \log ^2(4) \log (5)\right ) \, dx=\frac {4\,{\mathrm {e}}^{2-2\,x}\,{\ln \left (2\right )}^2\,\left (\ln \left (5\right )-3\right )}{9} \]

[In]

int(-(exp(2 - 2*x)*(8*log(2)^2*log(5) - 24*log(2)^2))/9,x)

[Out]

(4*exp(2 - 2*x)*log(2)^2*(log(5) - 3))/9