3.71.19 \(\int \frac {e^{-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x} (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4))}{\log (4)} \, dx\)

Optimal. Leaf size=28 \[ 2 e^{x-\left (3+\log \left (e^{2 e^{4+\frac {x}{\log (4)}}}\right )\right )^2} x \]

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Rubi [B]  time = 0.91, antiderivative size = 106, normalized size of antiderivative = 3.79, number of steps used = 4, number of rules used = 3, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {12, 6741, 2288} \begin {gather*} -\frac {2 e^{x-12 e^{\frac {x}{\log (4)}+4}-4 e^{\frac {2 x}{\log (4)}+8}-9} \left (12 x e^{\frac {x}{\log (4)}+4}+8 x e^{\frac {2 x}{\log (4)}+8}-x \log (4)\right )}{\log (4) \left (-\frac {12 e^{\frac {x}{\log (4)}+4}}{\log (4)}-\frac {8 e^{\frac {2 x}{\log (4)}+8}}{\log (4)}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-9 - 12*E^((x + 4*Log[4])/Log[4]) - 4*E^((2*(x + 4*Log[4]))/Log[4]) + x)*(-24*E^((x + 4*Log[4])/Log[4]
)*x - 16*E^((2*(x + 4*Log[4]))/Log[4])*x + (2 + 2*x)*Log[4]))/Log[4],x]

[Out]

(-2*E^(-9 - 12*E^(4 + x/Log[4]) - 4*E^(8 + (2*x)/Log[4]) + x)*(12*E^(4 + x/Log[4])*x + 8*E^(8 + (2*x)/Log[4])*
x - x*Log[4]))/((1 - (12*E^(4 + x/Log[4]))/Log[4] - (8*E^(8 + (2*x)/Log[4]))/Log[4])*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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \exp \left (-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x\right ) \left (-24 e^{\frac {x+4 \log (4)}{\log (4)}} x-16 e^{\frac {2 (x+4 \log (4))}{\log (4)}} x+(2+2 x) \log (4)\right ) \, dx}{\log (4)}\\ &=\frac {\int 2 \exp \left (-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x\right ) \left (-12 e^{4+\frac {x}{\log (4)}} x-8 e^{8+\frac {2 x}{\log (4)}} x+\log (4)+x \log (4)\right ) \, dx}{\log (4)}\\ &=\frac {2 \int \exp \left (-9-12 e^{\frac {x+4 \log (4)}{\log (4)}}-4 e^{\frac {2 (x+4 \log (4))}{\log (4)}}+x\right ) \left (-12 e^{4+\frac {x}{\log (4)}} x-8 e^{8+\frac {2 x}{\log (4)}} x+\log (4)+x \log (4)\right ) \, dx}{\log (4)}\\ &=-\frac {2 e^{-9-12 e^{4+\frac {x}{\log (4)}}-4 e^{8+\frac {2 x}{\log (4)}}+x} \left (12 e^{4+\frac {x}{\log (4)}} x+8 e^{8+\frac {2 x}{\log (4)}} x-x \log (4)\right )}{\left (1-\frac {12 e^{4+\frac {x}{\log (4)}}}{\log (4)}-\frac {8 e^{8+\frac {2 x}{\log (4)}}}{\log (4)}\right ) \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.75, size = 33, normalized size = 1.18 \begin {gather*} 2 e^{-9-12 e^{4+\frac {x}{\log (4)}}-4 e^{8+\frac {2 x}{\log (4)}}+x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-9 - 12*E^((x + 4*Log[4])/Log[4]) - 4*E^((2*(x + 4*Log[4]))/Log[4]) + x)*(-24*E^((x + 4*Log[4])/
Log[4])*x - 16*E^((2*(x + 4*Log[4]))/Log[4])*x + (2 + 2*x)*Log[4]))/Log[4],x]

[Out]

2*E^(-9 - 12*E^(4 + x/Log[4]) - 4*E^(8 + (2*x)/Log[4]) + x)*x

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fricas [A]  time = 0.96, size = 36, normalized size = 1.29 \begin {gather*} 2 \, x e^{\left (x - 4 \, e^{\left (\frac {x + 8 \, \log \relax (2)}{\log \relax (2)}\right )} - 12 \, e^{\left (\frac {x + 8 \, \log \relax (2)}{2 \, \log \relax (2)}\right )} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2)+x)/log(2))+2*(2*x+2)*log(2))*exp(-4
*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp(1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm="fricas")

[Out]

2*x*e^(x - 4*e^((x + 8*log(2))/log(2)) - 12*e^(1/2*(x + 8*log(2))/log(2)) - 9)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (4 \, x e^{\left (\frac {x + 8 \, \log \relax (2)}{\log \relax (2)}\right )} + 6 \, x e^{\left (\frac {x + 8 \, \log \relax (2)}{2 \, \log \relax (2)}\right )} - {\left (x + 1\right )} \log \relax (2)\right )} e^{\left (x - 4 \, e^{\left (\frac {x + 8 \, \log \relax (2)}{\log \relax (2)}\right )} - 12 \, e^{\left (\frac {x + 8 \, \log \relax (2)}{2 \, \log \relax (2)}\right )} - 9\right )}}{\log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2)+x)/log(2))+2*(2*x+2)*log(2))*exp(-4
*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp(1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm="giac")

[Out]

integrate(-2*(4*x*e^((x + 8*log(2))/log(2)) + 6*x*e^(1/2*(x + 8*log(2))/log(2)) - (x + 1)*log(2))*e^(x - 4*e^(
(x + 8*log(2))/log(2)) - 12*e^(1/2*(x + 8*log(2))/log(2)) - 9)/log(2), x)

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maple [A]  time = 0.09, size = 37, normalized size = 1.32

method	result	size
risch	\(2 x \,{\mathrm e}^{-4 \,{\mathrm e}^{\frac {8 \ln \relax (2)+x}{\ln \relax (2)}}-12 \,{\mathrm e}^{\frac {8 \ln \relax (2)+x}{2 \ln \relax (2)}}+x -9}\)	\(37\)
norman	\(2 x \,{\mathrm e}^{-4 \,{\mathrm e}^{\frac {8 \ln \relax (2)+x}{\ln \relax (2)}}-12 \,{\mathrm e}^{\frac {8 \ln \relax (2)+x}{2 \ln \relax (2)}}+x -9}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-16*x*exp(1/2*(8*ln(2)+x)/ln(2))^2-24*x*exp(1/2*(8*ln(2)+x)/ln(2))+2*(2*x+2)*ln(2))*exp(-4*exp(1/2*(8
*ln(2)+x)/ln(2))^2-12*exp(1/2*(8*ln(2)+x)/ln(2))+x-9)/ln(2),x,method=_RETURNVERBOSE)

[Out]

2*x*exp(-4*exp((8*ln(2)+x)/ln(2))-12*exp(1/2*(8*ln(2)+x)/ln(2))+x-9)

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maxima [A]  time = 0.63, size = 30, normalized size = 1.07 \begin {gather*} 2 \, x e^{\left (x - 4 \, e^{\left (\frac {x}{\log \relax (2)} + 8\right )} - 12 \, e^{\left (\frac {x}{2 \, \log \relax (2)} + 4\right )} - 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-16*x*exp(1/2*(8*log(2)+x)/log(2))^2-24*x*exp(1/2*(8*log(2)+x)/log(2))+2*(2*x+2)*log(2))*exp(-4
*exp(1/2*(8*log(2)+x)/log(2))^2-12*exp(1/2*(8*log(2)+x)/log(2))+x-9)/log(2),x, algorithm="maxima")

[Out]

2*x*e^(x - 4*e^(x/log(2) + 8) - 12*e^(1/2*x/log(2) + 4) - 9)

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mupad [B]  time = 4.22, size = 32, normalized size = 1.14 \begin {gather*} 2\,x\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{\frac {x}{\ln \relax (2)}}\,{\mathrm {e}}^8}\,{\mathrm {e}}^{-12\,{\mathrm {e}}^{\frac {x}{2\,\ln \relax (2)}}\,{\mathrm {e}}^4}\,{\mathrm {e}}^{-9}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 12*exp((x/2 + 4*log(2))/log(2)) - 4*exp((2*(x/2 + 4*log(2)))/log(2)) - 9)*(24*x*exp((x/2 + 4*log
(2))/log(2)) - 2*log(2)*(2*x + 2) + 16*x*exp((2*(x/2 + 4*log(2)))/log(2))))/(2*log(2)),x)

[Out]

2*x*exp(-4*exp(x/log(2))*exp(8))*exp(-12*exp(x/(2*log(2)))*exp(4))*exp(-9)*exp(x)

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sympy [A]  time = 0.29, size = 36, normalized size = 1.29 \begin {gather*} 2 x e^{x - 4 e^{\frac {2 \left (\frac {x}{2} + 4 \log {\relax (2 )}\right )}{\log {\relax (2 )}}} - 12 e^{\frac {\frac {x}{2} + 4 \log {\relax (2 )}}{\log {\relax (2 )}}} - 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-16*x*exp(1/2*(8*ln(2)+x)/ln(2))**2-24*x*exp(1/2*(8*ln(2)+x)/ln(2))+2*(2*x+2)*ln(2))*exp(-4*exp
(1/2*(8*ln(2)+x)/ln(2))**2-12*exp(1/2*(8*ln(2)+x)/ln(2))+x-9)/ln(2),x)

[Out]

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

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