3.2.1 \(\int \frac {e^{6 x}}{1+e^{4 x}} \, dx\) [101]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2280, 327, 209} \begin {gather*} \frac {e^{2 x}}{2}-\frac {1}{2} \text {ArcTan}\left (e^{2 x}\right ) \end {gather*}

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 209

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

Rule 327

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 2280

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]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 18, normalized size = 0.90 \begin {gather*} \frac {1}{2} \left (e^{2 x}-\tan ^{-1}\left (e^{2 x}\right )\right ) \end {gather*}

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 = 0.02, size = 15, normalized size = 0.75

method	result	size
default	\(\frac {{\mathrm e}^{2 x}}{2}-\frac {\arctan \left ({\mathrm e}^{2 x}\right )}{2}\)	\(15\)
risch	\(\frac {{\mathrm e}^{2 x}}{2}+\frac {i \ln \left ({\mathrm e}^{2 x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{2 x}+i\right )}{4}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.02, size = 14, normalized size = 0.70 \begin {gather*} -\frac {1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} \end {gather*}

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 = 0.41, size = 14, normalized size = 0.70 \begin {gather*} -\frac {1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} \end {gather*}

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 = 0.04, size = 24, normalized size = 1.20 \begin {gather*} \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 {gather*}

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 = 1.11, size = 14, normalized size = 0.70 \begin {gather*} -\frac {1}{2} \, \arctan \left (e^{\left (2 \, x\right )}\right ) + \frac {1}{2} \, e^{\left (2 \, x\right )} \end {gather*}

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)

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Mupad [B]
time = 0.07, size = 14, normalized size = 0.70 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{2}-\frac {\mathrm {atan}\left ({\mathrm {e}}^{2\,x}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(6*x)/(exp(4*x) + 1),x)

[Out]

exp(2*x)/2 - atan(exp(2*x))/2

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