3.525 \(\int \frac{e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx\)

Optimal. Leaf size=47 \[ \frac{1}{12} \log \left (-2 e^{2 x}+3 e^{4 x}+1\right )-\frac{\tan ^{-1}\left (\frac{1-3 e^{2 x}}{\sqrt{2}}\right )}{6 \sqrt{2}} \]

[Out]

-ArcTan[(1 - 3*E^(2*x))/Sqrt[2]]/(6*Sqrt[2]) + Log[1 - 2*E^(2*x) + 3*E^(4*x)]/12

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Rubi [A]  time = 0.0591705, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2282, 634, 618, 204, 628} \[ \frac{1}{12} \log \left (-2 e^{2 x}+3 e^{4 x}+1\right )-\frac{\tan ^{-1}\left (\frac{1-3 e^{2 x}}{\sqrt{2}}\right )}{6 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(1 - 3*E^(2*x))/Sqrt[2]]/(6*Sqrt[2]) + Log[1 - 2*E^(2*x) + 3*E^(4*x)]/12

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{1-2 x+3 x^2} \, dx,x,e^{2 x}\right )\\ &=\frac{1}{12} \operatorname{Subst}\left (\int \frac{-2+6 x}{1-2 x+3 x^2} \, dx,x,e^{2 x}\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-2 x+3 x^2} \, dx,x,e^{2 x}\right )\\ &=\frac{1}{12} \log \left (1-2 e^{2 x}+3 e^{4 x}\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-8-x^2} \, dx,x,-2+6 e^{2 x}\right )\\ &=-\frac{\tan ^{-1}\left (\frac{1-3 e^{2 x}}{\sqrt{2}}\right )}{6 \sqrt{2}}+\frac{1}{12} \log \left (1-2 e^{2 x}+3 e^{4 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.020503, size = 44, normalized size = 0.94 \[ \frac{1}{12} \left (\log \left (-2 e^{2 x}+3 e^{4 x}+1\right )+\sqrt{2} \tan ^{-1}\left (\frac{3 e^{2 x}-1}{\sqrt{2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*ArcTan[(-1 + 3*E^(2*x))/Sqrt[2]] + Log[1 - 2*E^(2*x) + 3*E^(4*x)])/12

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Maple [A]  time = 0.007, size = 38, normalized size = 0.8 \begin{align*}{\frac{\ln \left ( 1-2\, \left ({{\rm e}^{x}} \right ) ^{2}+3\, \left ({{\rm e}^{x}} \right ) ^{4} \right ) }{12}}+{\frac{\sqrt{2}}{12}\arctan \left ({\frac{ \left ( 6\, \left ({{\rm e}^{x}} \right ) ^{2}-2 \right ) \sqrt{2}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*ln(1-2*exp(x)^2+3*exp(x)^4)+1/12*2^(1/2)*arctan(1/4*(6*exp(x)^2-2)*2^(1/2))

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Maxima [A]  time = 1.41344, size = 50, normalized size = 1.06 \begin{align*} \frac{1}{12} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac{1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="maxima")

[Out]

1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*e^(2*x) - 1)) + 1/12*log(3*e^(4*x) - 2*e^(2*x) + 1)

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Fricas [A]  time = 1.78919, size = 127, normalized size = 2.7 \begin{align*} \frac{1}{12} \, \sqrt{2} \arctan \left (\frac{3}{2} \, \sqrt{2} e^{\left (2 \, x\right )} - \frac{1}{2} \, \sqrt{2}\right ) + \frac{1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="fricas")

[Out]

1/12*sqrt(2)*arctan(3/2*sqrt(2)*e^(2*x) - 1/2*sqrt(2)) + 1/12*log(3*e^(4*x) - 2*e^(2*x) + 1)

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Sympy [A]  time = 0.134202, size = 22, normalized size = 0.47 \begin{align*} \operatorname{RootSum}{\left (96 z^{2} - 16 z + 1, \left ( i \mapsto i \log{\left (8 i + e^{2 x} - 1 \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(96*_z**2 - 16*_z + 1, Lambda(_i, _i*log(8*_i + exp(2*x) - 1)))

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Giac [A]  time = 1.13426, size = 50, normalized size = 1.06 \begin{align*} \frac{1}{12} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac{1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(4*x)/(1-2*exp(2*x)+3*exp(4*x)),x, algorithm="giac")

[Out]

1/12*sqrt(2)*arctan(1/2*sqrt(2)*(3*e^(2*x) - 1)) + 1/12*log(3*e^(4*x) - 2*e^(2*x) + 1)