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

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 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 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 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]

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 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]]]

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 = 0.02, 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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{4 x}}{1-2 e^{2 x}+3 e^{4 x}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

________________________________________________________________________________________

fricas [A]  time = 1.01, size = 39, normalized size = 0.83 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {3}{2} \, \sqrt {2} e^{\left (2 \, x\right )} - \frac {1}{2} \, \sqrt {2}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \]

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)

________________________________________________________________________________________

giac [A]  time = 0.60, size = 37, normalized size = 0.79 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \]

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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 38, normalized size = 0.81

method	result	size
default	\(\frac {\ln \left (1-2 \,{\mathrm e}^{2 x}+3 \,{\mathrm e}^{4 x}\right )}{12}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (6 \,{\mathrm e}^{2 x}-2\right ) \sqrt {2}}{4}\right )}{12}\)	\(38\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{3}+\frac {i \sqrt {2}}{3}\right )}{12}+\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{3}+\frac {i \sqrt {2}}{3}\right ) \sqrt {2}}{24}+\frac {\ln \left ({\mathrm e}^{2 x}-\frac {1}{3}-\frac {i \sqrt {2}}{3}\right )}{12}-\frac {i \ln \left ({\mathrm e}^{2 x}-\frac {1}{3}-\frac {i \sqrt {2}}{3}\right ) \sqrt {2}}{24}\)	\(70\)

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,method=_RETURNVERBOSE)

[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))

________________________________________________________________________________________

maxima [A]  time = 1.34, size = 37, normalized size = 0.79 \[ \frac {1}{12} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, e^{\left (2 \, x\right )} - 1\right )}\right ) + \frac {1}{12} \, \log \left (3 \, e^{\left (4 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 1\right ) \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.33, size = 39, normalized size = 0.83 \[ \frac {\ln \left (3\,{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1\right )}{12}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}}{2}-\frac {3\,\sqrt {2}\,{\mathrm {e}}^{2\,x}}{2}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*exp(4*x) - 2*exp(2*x) + 1)/12 - (2^(1/2)*atan(2^(1/2)/2 - (3*2^(1/2)*exp(2*x))/2))/12

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 22, normalized size = 0.47 \[ \operatorname {RootSum} {\left (96 z^{2} - 16 z + 1, \left (i \mapsto i \log {\left (8 i + e^{2 x} - 1 \right )} \right )\right )} \]

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)))

________________________________________________________________________________________