3.56.26 \(\int \frac {6+e^{2 x} (-2+4 x)}{9-6 e^{2 x}+e^{4 x}} \, dx\)

Optimal. Leaf size=14 \[ \frac {2 x}{3-e^{2 x}} \]

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Rubi [B]  time = 0.32, antiderivative size = 90, normalized size of antiderivative = 6.43, number of steps used = 17, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6688, 6742, 2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 31, 29} \begin {gather*} \frac {2 x^2}{3}-\frac {1}{6} (1-2 x)^2+\frac {2 x}{3-e^{2 x}}-\frac {2 x}{3}-\frac {1}{3} (1-2 x) \log \left (1-\frac {e^{2 x}}{3}\right )+\frac {1}{3} \log \left (3-e^{2 x}\right )-\frac {2}{3} x \log \left (1-\frac {e^{2 x}}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + E^(2*x)*(-2 + 4*x))/(9 - 6*E^(2*x) + E^(4*x)),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6+e^{2 x} (-2+4 x)}{\left (3-e^{2 x}\right )^2} \, dx\\ &=\int \left (\frac {12 x}{\left (-3+e^{2 x}\right )^2}+\frac {2 (-1+2 x)}{-3+e^{2 x}}\right ) \, dx\\ &=2 \int \frac {-1+2 x}{-3+e^{2 x}} \, dx+12 \int \frac {x}{\left (-3+e^{2 x}\right )^2} \, dx\\ &=-\frac {1}{6} (1-2 x)^2+\frac {2}{3} \int \frac {e^{2 x} (-1+2 x)}{-3+e^{2 x}} \, dx+4 \int \frac {e^{2 x} x}{\left (-3+e^{2 x}\right )^2} \, dx-4 \int \frac {x}{-3+e^{2 x}} \, dx\\ &=-\frac {1}{6} (1-2 x)^2+\frac {2 x}{3-e^{2 x}}+\frac {2 x^2}{3}-\frac {1}{3} (1-2 x) \log \left (1-\frac {e^{2 x}}{3}\right )-\frac {2}{3} \int \log \left (1-\frac {e^{2 x}}{3}\right ) \, dx-\frac {4}{3} \int \frac {e^{2 x} x}{-3+e^{2 x}} \, dx+2 \int \frac {1}{-3+e^{2 x}} \, dx\\ &=-\frac {1}{6} (1-2 x)^2+\frac {2 x}{3-e^{2 x}}+\frac {2 x^2}{3}-\frac {1}{3} (1-2 x) \log \left (1-\frac {e^{2 x}}{3}\right )-\frac {2}{3} x \log \left (1-\frac {e^{2 x}}{3}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3}\right )}{x} \, dx,x,e^{2 x}\right )+\frac {2}{3} \int \log \left (1-\frac {e^{2 x}}{3}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{(-3+x) x} \, dx,x,e^{2 x}\right )\\ &=-\frac {1}{6} (1-2 x)^2+\frac {2 x}{3-e^{2 x}}+\frac {2 x^2}{3}-\frac {1}{3} (1-2 x) \log \left (1-\frac {e^{2 x}}{3}\right )-\frac {2}{3} x \log \left (1-\frac {e^{2 x}}{3}\right )+\frac {1}{3} \text {Li}_2\left (\frac {e^{2 x}}{3}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-3+x} \, dx,x,e^{2 x}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3}\right )}{x} \, dx,x,e^{2 x}\right )\\ &=-\frac {1}{6} (1-2 x)^2-\frac {2 x}{3}+\frac {2 x}{3-e^{2 x}}+\frac {2 x^2}{3}+\frac {1}{3} \log \left (3-e^{2 x}\right )-\frac {1}{3} (1-2 x) \log \left (1-\frac {e^{2 x}}{3}\right )-\frac {2}{3} x \log \left (1-\frac {e^{2 x}}{3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 12, normalized size = 0.86 \begin {gather*} -\frac {2 x}{-3+e^{2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + E^(2*x)*(-2 + 4*x))/(9 - 6*E^(2*x) + E^(4*x)),x]

[Out]

(-2*x)/(-3 + E^(2*x))

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fricas [A]  time = 0.87, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2 \, x}{e^{\left (2 \, x\right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-2)*exp(x)^2+6)/(exp(x)^4-6*exp(x)^2+9),x, algorithm="fricas")

[Out]

-2*x/(e^(2*x) - 3)

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giac [A]  time = 0.15, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2 \, x}{e^{\left (2 \, x\right )} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-2)*exp(x)^2+6)/(exp(x)^4-6*exp(x)^2+9),x, algorithm="giac")

[Out]

-2*x/(e^(2*x) - 3)

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maple [A]  time = 0.06, size = 12, normalized size = 0.86

method	result	size
norman	\(-\frac {2 x}{{\mathrm e}^{2 x}-3}\)	\(12\)
risch	\(-\frac {2 x}{{\mathrm e}^{2 x}-3}\)	\(12\)
default	\(\frac {2 \ln \left ({\mathrm e}^{x}\right )}{3}-\frac {2 x \,{\mathrm e}^{2 x}}{3 \left ({\mathrm e}^{2 x}-3\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x-2)*exp(x)^2+6)/(exp(x)^4-6*exp(x)^2+9),x,method=_RETURNVERBOSE)

[Out]

-2*x/(exp(x)^2-3)

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maxima [A]  time = 0.38, size = 19, normalized size = 1.36 \begin {gather*} \frac {2}{3} \, x - \frac {2 \, x e^{\left (2 \, x\right )}}{3 \, {\left (e^{\left (2 \, x\right )} - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-2)*exp(x)^2+6)/(exp(x)^4-6*exp(x)^2+9),x, algorithm="maxima")

[Out]

2/3*x - 2/3*x*e^(2*x)/(e^(2*x) - 3)

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mupad [B]  time = 0.13, size = 11, normalized size = 0.79 \begin {gather*} -\frac {2\,x}{{\mathrm {e}}^{2\,x}-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(4*x - 2) + 6)/(exp(4*x) - 6*exp(2*x) + 9),x)

[Out]

-(2*x)/(exp(2*x) - 3)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.71 \begin {gather*} - \frac {2 x}{e^{2 x} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-2)*exp(x)**2+6)/(exp(x)**4-6*exp(x)**2+9),x)

[Out]

-2*x/(exp(2*x) - 3)

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