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

Optimal. Leaf size=22 \[ \left (5+\frac {1}{9} e^{2-2 x} \log ^2(4)\right ) (-3+\log (5)) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2194} \begin {gather*} -\frac {1}{9} e^{2-2 x} \log ^2(4) (3-\log (5)) \end {gather*}

Antiderivative was successfully verified.

[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 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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.86 \begin {gather*} \frac {1}{9} e^{2-2 x} \log ^2(4) (-3+\log (5)) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.16, size = 22, normalized size = 1.00 \begin {gather*} \frac {4}{9} \, {\left (\log \relax (5) \log \relax (2)^{2} - 3 \, \log \relax (2)^{2}\right )} e^{\left (-2 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 22, normalized size = 1.00 \begin {gather*} \frac {4}{9} \, {\left (\log \relax (5) \log \relax (2)^{2} - 3 \, \log \relax (2)^{2}\right )} e^{\left (-2 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 19, normalized size = 0.86

method	result	size
gosper	\(\frac {4 \ln \relax (2)^{2} \left (\ln \relax (5)-3\right ) {\mathrm e}^{-2 x +2}}{9}\)	\(19\)
norman	\(\left (\frac {4 \ln \relax (2)^{2} \ln \relax (5)}{9}-\frac {4 \ln \relax (2)^{2}}{3}\right ) {\mathrm e}^{-2 x +2}\)	\(25\)
derivativedivides	\(-\frac {\left (-8 \ln \relax (2)^{2} \ln \relax (5)+24 \ln \relax (2)^{2}\right ) {\mathrm e}^{-2 x +2}}{18}\)	\(26\)
default	\(-\frac {\left (-\frac {8 \ln \relax (2)^{2} \ln \relax (5)}{9}+\frac {8 \ln \relax (2)^{2}}{3}\right ) {\mathrm e}^{-2 x +2}}{2}\)	\(26\)
risch	\(\frac {4 \,{\mathrm e}^{-2 x +2} \ln \relax (2)^{2} \ln \relax (5)}{9}-\frac {4 \,{\mathrm e}^{-2 x +2} \ln \relax (2)^{2}}{3}\)	\(28\)
meijerg	\(-\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2} x -2 x} \ln \relax (2)^{2} \ln \relax (5) \left (1-{\mathrm e}^{-2 \,{\mathrm e}^{2} x}\right )}{9}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2} x -2 x} \ln \relax (2)^{2} \left (1-{\mathrm e}^{-2 \,{\mathrm e}^{2} x}\right )}{3}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.39, size = 22, normalized size = 1.00 \begin {gather*} \frac {4}{9} \, {\left (\log \relax (5) \log \relax (2)^{2} - 3 \, \log \relax (2)^{2}\right )} e^{\left (-2 \, x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.16, size = 16, normalized size = 0.73 \begin {gather*} \frac {4\,{\mathrm {e}}^{2-2\,x}\,{\ln \relax (2)}^2\,\left (\ln \relax (5)-3\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.11, size = 24, normalized size = 1.09 \begin {gather*} \frac {\left (- 12 \log {\relax (2 )}^{2} + 4 \log {\relax (2 )}^{2} \log {\relax (5 )}\right ) e^{2 - 2 x}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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