\(\int e^{-24+2 x} (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)) \, dx\) [4582]

Optimal result

Integrand size = 26, antiderivative size = 18 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=e^{2 x} \left (-\frac {2}{e^{12}}-\log (4)\right )^2 \]

[Out]

exp(x)^2*(-2*ln(2)-2/exp(12))^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2225} \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=e^{2 x-24} \left (2+e^{12} \log (4)\right )^2 \]

[In]

Int[E^(-24 + 2*x)*(8 + 8*E^12*Log[4] + 2*E^24*Log[4]^2),x]

[Out]

E^(-24 + 2*x)*(2 + E^12*Log[4])^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 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}& = \left (2 \left (2+e^{12} \log (4)\right )^2\right ) \int e^{-24+2 x} \, dx \\ & = e^{-24+2 x} \left (2+e^{12} \log (4)\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=e^{-24+2 x} \left (2+e^{12} \log (4)\right )^2 \]

[In]

Integrate[E^(-24 + 2*x)*(8 + 8*E^12*Log[4] + 2*E^24*Log[4]^2),x]

[Out]

E^(-24 + 2*x)*(2 + E^12*Log[4])^2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.56

method	result	size
gosper	\(4 \left ({\mathrm e}^{24} \ln \left (2\right )^{2}+2 \,{\mathrm e}^{12} \ln \left (2\right )+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
default	\(4 \left ({\mathrm e}^{24} \ln \left (2\right )^{2}+2 \,{\mathrm e}^{12} \ln \left (2\right )+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
norman	\(4 \left ({\mathrm e}^{24} \ln \left (2\right )^{2}+2 \,{\mathrm e}^{12} \ln \left (2\right )+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
derivativedivides	\(\frac {\left (8 \,{\mathrm e}^{24} \ln \left (2\right )^{2}+16 \,{\mathrm e}^{12} \ln \left (2\right )+8\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}}{2}\)	\(29\)
parallelrisch	\(\frac {\left (8 \,{\mathrm e}^{24} \ln \left (2\right )^{2}+16 \,{\mathrm e}^{12} \ln \left (2\right )+8\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}}{2}\)	\(29\)
risch	\(4 \ln \left (2\right )^{2} {\mathrm e}^{2 x -24} {\mathrm e}^{24}+8 \ln \left (2\right ) {\mathrm e}^{2 x -24} {\mathrm e}^{12}+4 \,{\mathrm e}^{2 x -24}\)	\(36\)
meijerg	\(-4 \ln \left (2\right )^{2} \left (1-{\mathrm e}^{2 x}\right )-8 \,{\mathrm e}^{-12} \ln \left (2\right ) \left (1-{\mathrm e}^{2 x}\right )-4 \,{\mathrm e}^{-24} \left (1-{\mathrm e}^{2 x}\right )\)	\(42\)

[In]

int((8*exp(12)^2*ln(2)^2+16*exp(12)*ln(2)+8)*exp(x)^2/exp(12)^2,x,method=_RETURNVERBOSE)

[Out]

4*(exp(12)^2*ln(2)^2+2*exp(12)*ln(2)+1)*exp(x)^2/exp(12)^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=4 \, {\left (e^{24} \log \left (2\right )^{2} + 2 \, e^{12} \log \left (2\right ) + 1\right )} e^{\left (2 \, x - 24\right )} \]

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="fricas")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.50 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=\frac {\left (4 + 8 e^{12} \log {\left (2 \right )} + 4 e^{24} \log {\left (2 \right )}^{2}\right ) e^{2 x}}{e^{24}} \]

[In]

integrate((8*exp(12)**2*ln(2)**2+16*exp(12)*ln(2)+8)*exp(x)**2/exp(12)**2,x)

[Out]

(4 + 8*exp(12)*log(2) + 4*exp(24)*log(2)**2)*exp(-24)*exp(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=4 \, {\left (e^{24} \log \left (2\right )^{2} + 2 \, e^{12} \log \left (2\right ) + 1\right )} e^{\left (2 \, x - 24\right )} \]

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="maxima")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx=4 \, {\left (e^{24} \log \left (2\right )^{2} + 2 \, e^{12} \log \left (2\right ) + 1\right )} e^{\left (2 \, x - 24\right )} \]

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="giac")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int e^{-24+2 x} \left (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)\right ) \, dx={\mathrm {e}}^{2\,x-24}\,\left (8\,{\mathrm {e}}^{12}\,\ln \left (2\right )+4\,{\mathrm {e}}^{24}\,{\ln \left (2\right )}^2+4\right ) \]

[In]

int(exp(2*x)*exp(-24)*(16*exp(12)*log(2) + 8*exp(24)*log(2)^2 + 8),x)

[Out]

exp(2*x - 24)*(8*exp(12)*log(2) + 4*exp(24)*log(2)^2 + 4)