3.16.34 \(\int \frac {1}{4} e^{-4+2 x} (128+40320 e^{2-x}) \, dx\)

Optimal. Leaf size=13 \[ 4 \left (630+2 e^{-2+x}\right )^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.69, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {12, 2248, 37} \begin {gather*} 16 e^{2 x-4} \left (315 e^{2-x}+1\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 + 2*x)*(128 + 40320*E^(2 - x)))/4,x]

[Out]

16*E^(-4 + 2*x)*(1 + 315*E^(2 - x))^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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2248

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[(g*h*Log[G])/(d*e*Log[F])]}, Dist[(Denominator[m]*G^(f*h - (c*g*h)/d))/(d*e*Log[F]), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^((e*(c + d*x))/Denominator[m])], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{-4+2 x} \left (128+40320 e^{2-x}\right ) \, dx\\ &=-\left (\frac {1}{4} \operatorname {Subst}\left (\int \frac {128+40320 x}{x^3} \, dx,x,e^{2-x}\right )\right )\\ &=16 e^{-4+2 x} \left (1+315 e^{2-x}\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.62 \begin {gather*} 32 \left (315 e^{-2+x}+\frac {1}{2} e^{-4+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + 2*x)*(128 + 40320*E^(2 - x)))/4,x]

[Out]

32*(315*E^(-2 + x) + E^(-4 + 2*x)/2)

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fricas [A]  time = 0.55, size = 24, normalized size = 1.85 \begin {gather*} 64 \, {\left (315 \, e^{\left (-x + \log \relax (2) + 2\right )} + 1\right )} e^{\left (2 \, x - 2 \, \log \relax (2) - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20160*exp(log(2)+2-x)+128)/exp(log(2)+2-x)^2,x, algorithm="fricas")

[Out]

64*(315*e^(-x + log(2) + 2) + 1)*e^(2*x - 2*log(2) - 4)

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giac [A]  time = 0.21, size = 15, normalized size = 1.15 \begin {gather*} 16 \, {\left (e^{\left (2 \, x\right )} + 630 \, e^{\left (x + 2\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20160*exp(log(2)+2-x)+128)/exp(log(2)+2-x)^2,x, algorithm="giac")

[Out]

16*(e^(2*x) + 630*e^(x + 2))*e^(-4)

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maple [A]  time = 0.02, size = 16, normalized size = 1.23

method	result	size
risch	\(16 \,{\mathrm e}^{2 x -4}+10080 \,{\mathrm e}^{x -2}\)	\(16\)
norman	\(\frac {\left (64+20160 \,{\mathrm e}^{\ln \relax (2)+2-x}\right ) {\mathrm e}^{2 x -4}}{4}\)	\(24\)
derivativedivides	\(16 \,{\mathrm e}^{2 x -4}+10080 \,{\mathrm e}^{x -2}\)	\(26\)
default	\(16 \,{\mathrm e}^{2 x -4}+10080 \,{\mathrm e}^{x -2}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20160*exp(ln(2)+2-x)+128)/exp(ln(2)+2-x)^2,x,method=_RETURNVERBOSE)

[Out]

16*exp(2*x-4)+10080*exp(x-2)

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maxima [A]  time = 0.58, size = 15, normalized size = 1.15 \begin {gather*} 16 \, e^{\left (2 \, x - 4\right )} + 10080 \, e^{\left (x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20160*exp(log(2)+2-x)+128)/exp(log(2)+2-x)^2,x, algorithm="maxima")

[Out]

16*e^(2*x - 4) + 10080*e^(x - 2)

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mupad [B]  time = 0.06, size = 13, normalized size = 1.00 \begin {gather*} 16\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x\,\left (630\,{\mathrm {e}}^2+{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x - 2*log(2) - 4)*(20160*exp(log(2) - x + 2) + 128),x)

[Out]

16*exp(-4)*exp(x)*(630*exp(2) + exp(x))

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sympy [A]  time = 0.09, size = 14, normalized size = 1.08 \begin {gather*} 10080 e^{x - 2} + 16 e^{2 x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20160*exp(ln(2)+2-x)+128)/exp(ln(2)+2-x)**2,x)

[Out]

10080*exp(x - 2) + 16*exp(2*x - 4)

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