3.56.27 \(\int \frac {e^{1-2 e^3-4 x} (3+12 x) \log (2)}{e^{2-8 x}-8 e^{1-4 x} x+16 x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {3 e^{-2 e^3} x \log (2)}{e^{1-4 x}-4 x} \]

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Rubi [A]  time = 0.29, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6688, 6686} \begin {gather*} \frac {3 e^{1-2 e^3} \log (2)}{4 \left (e-4 e^{4 x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - 2*E^3 - 4*x)*(3 + 12*x)*Log[2])/(E^(2 - 8*x) - 8*E^(1 - 4*x)*x + 16*x^2),x]

[Out]

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

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (2) \int \frac {e^{1-2 e^3-4 x} (3+12 x)}{e^{2-8 x}-8 e^{1-4 x} x+16 x^2} \, dx\\ &=\log (2) \int \frac {3 e^{1-2 e^3+4 x} (1+4 x)}{\left (e-4 e^{4 x} x\right )^2} \, dx\\ &=(3 \log (2)) \int \frac {e^{1-2 e^3+4 x} (1+4 x)}{\left (e-4 e^{4 x} x\right )^2} \, dx\\ &=\frac {3 e^{1-2 e^3} \log (2)}{4 \left (e-4 e^{4 x} x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.08 \begin {gather*} \frac {3 e^{1-2 e^3} \log (2)}{4 \left (e-4 e^{4 x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - 2*E^3 - 4*x)*(3 + 12*x)*Log[2])/(E^(2 - 8*x) - 8*E^(1 - 4*x)*x + 16*x^2),x]

[Out]

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

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fricas [A]  time = 0.62, size = 24, normalized size = 0.96 \begin {gather*} -\frac {3 \, x e^{\left (-2 \, e^{3}\right )} \log \relax (2)}{4 \, x - e^{\left (-4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x+3)*log(2)*exp(-4*x+1)/(exp(-4*x+1)^2-8*x*exp(-4*x+1)+16*x^2)/exp(exp(3))^2,x, algorithm="frica
s")

[Out]

-3*x*e^(-2*e^3)*log(2)/(4*x - e^(-4*x + 1))

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giac [A]  time = 0.17, size = 30, normalized size = 1.20 \begin {gather*} -\frac {3 \, e \log \relax (2)}{4 \, {\left (4 \, x e^{\left (4 \, x + 2 \, e^{3}\right )} - e^{\left (2 \, e^{3} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x+3)*log(2)*exp(-4*x+1)/(exp(-4*x+1)^2-8*x*exp(-4*x+1)+16*x^2)/exp(exp(3))^2,x, algorithm="giac"
)

[Out]

-3/4*e*log(2)/(4*x*e^(4*x + 2*e^3) - e^(2*e^3 + 1))

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maple [A]  time = 0.14, size = 25, normalized size = 1.00

method	result	size
risch	\(-\frac {3 \ln \relax (2) {\mathrm e}^{-2 \,{\mathrm e}^{3}} x}{4 x -{\mathrm e}^{-4 x +1}}\)	\(25\)
norman	\(-\frac {3 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}} \ln \relax (2) {\mathrm e}^{-4 x +1}}{4 \left (4 x -{\mathrm e}^{-4 x +1}\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x+3)*ln(2)*exp(-4*x+1)/(exp(-4*x+1)^2-8*x*exp(-4*x+1)+16*x^2)/exp(exp(3))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 30, normalized size = 1.20 \begin {gather*} -\frac {3 \, e \log \relax (2)}{4 \, {\left (4 \, x e^{\left (4 \, x + 2 \, e^{3}\right )} - e^{\left (2 \, e^{3} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x+3)*log(2)*exp(-4*x+1)/(exp(-4*x+1)^2-8*x*exp(-4*x+1)+16*x^2)/exp(exp(3))^2,x, algorithm="maxim
a")

[Out]

-3/4*e*log(2)/(4*x*e^(4*x + 2*e^3) - e^(2*e^3 + 1))

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mupad [B]  time = 3.54, size = 29, normalized size = 1.16 \begin {gather*} -\frac {3\,x\,\ln \relax (2)}{4\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-4\,x}\,\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*exp(3))*exp(1 - 4*x)*log(2)*(12*x + 3))/(exp(2 - 8*x) - 8*x*exp(1 - 4*x) + 16*x^2),x)

[Out]

-(3*x*log(2))/(4*x*exp(2*exp(3)) - exp(2*exp(3))*exp(-4*x)*exp(1))

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sympy [A]  time = 0.12, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 x \log {\relax (2 )}}{- 4 x e^{2 e^{3}} + e^{1 - 4 x} e^{2 e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x+3)*ln(2)*exp(-4*x+1)/(exp(-4*x+1)**2-8*x*exp(-4*x+1)+16*x**2)/exp(exp(3))**2,x)

[Out]

3*x*log(2)/(-4*x*exp(2*exp(3)) + exp(1 - 4*x)*exp(2*exp(3)))

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