3.101 \(\int \frac{e^{6 x}}{1+e^{4 x}} \, dx\)

Optimal. Leaf size=20 \[ \frac{e^{2 x}}{2}-\frac{1}{2} \tan ^{-1}\left (e^{2 x}\right ) \]

[Out]

E^(2*x)/2 - ArcTan[E^(2*x)]/2

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Rubi [A]  time = 0.0239091, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2248, 321, 203} \[ \frac{e^{2 x}}{2}-\frac{1}{2} \tan ^{-1}\left (e^{2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(6*x)/(1 + E^(4*x)),x]

[Out]

E^(2*x)/2 - ArcTan[E^(2*x)]/2

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]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.007745, size = 18, normalized size = 0.9 \[ \frac{1}{2} \left (e^{2 x}-\tan ^{-1}\left (e^{2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(6*x)/(1 + E^(4*x)),x]

[Out]

(E^(2*x) - ArcTan[E^(2*x)])/2

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Maple [A]  time = 0.003, size = 15, normalized size = 0.8 \begin{align*}{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}}{2}}-{\frac{\arctan \left ( \left ({{\rm e}^{x}} \right ) ^{2} \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(6*x)/(1+exp(4*x)),x)

[Out]

1/2*exp(x)^2-1/2*arctan(exp(x)^2)

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Maxima [A]  time = 1.41559, size = 19, normalized size = 0.95 \begin{align*} -\frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(6*x)/(1+exp(4*x)),x, algorithm="maxima")

[Out]

-1/2*arctan(e^(2*x)) + 1/2*e^(2*x)

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Fricas [A]  time = 1.91873, size = 49, normalized size = 2.45 \begin{align*} -\frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(6*x)/(1+exp(4*x)),x, algorithm="fricas")

[Out]

-1/2*arctan(e^(2*x)) + 1/2*e^(2*x)

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Sympy [A]  time = 0.110742, size = 24, normalized size = 1.2 \begin{align*} \frac{e^{2 x}}{2} + \operatorname{RootSum}{\left (16 z^{2} + 1, \left ( i \mapsto i \log{\left (- 4 i + e^{2 x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(6*x)/(1+exp(4*x)),x)

[Out]

exp(2*x)/2 + RootSum(16*_z**2 + 1, Lambda(_i, _i*log(-4*_i + exp(2*x))))

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Giac [A]  time = 1.04763, size = 19, normalized size = 0.95 \begin{align*} -\frac{1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac{1}{2} \, e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(6*x)/(1+exp(4*x)),x, algorithm="giac")

[Out]

-1/2*arctan(e^(2*x)) + 1/2*e^(2*x)