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

Optimal. Leaf size=10 \[ e^x-\tan ^{-1}\left (e^x\right ) \]

[Out]

exp(x)-arctan(exp(x))

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Rubi [A]  time = 0.02, antiderivative size = 10, 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 = {2248, 321, 203} \[ e^x-\tan ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^(3*x)/(1 + E^(2*x)),x]

[Out]

E^x - ArcTan[E^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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \[ e^x-\tan ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*x)/(1 + E^(2*x)),x]

[Out]

E^x - ArcTan[E^x]

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fricas [A]  time = 0.41, size = 8, normalized size = 0.80 \[ -\arctan \left (e^{x}\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(e^x) + e^x

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giac [A]  time = 0.21, size = 8, normalized size = 0.80 \[ -\arctan \left (e^{x}\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(e^x) + e^x

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maple [A]  time = 0.03, size = 9, normalized size = 0.90 \[ -\arctan \left ({\mathrm e}^{x}\right )+{\mathrm e}^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)-arctan(exp(x))

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maxima [A]  time = 1.93, size = 8, normalized size = 0.80 \[ -\arctan \left (e^{x}\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(e^x) + e^x

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mupad [B]  time = 0.08, size = 8, normalized size = 0.80 \[ {\mathrm {e}}^x-\mathrm {atan}\left ({\mathrm {e}}^x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3*x)/(exp(2*x) + 1),x)

[Out]

exp(x) - atan(exp(x))

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sympy [B]  time = 0.12, size = 19, normalized size = 1.90 \[ e^{x} + \operatorname {RootSum} {\left (4 z^{2} + 1, \left (i \mapsto i \log {\left (- 2 i + e^{x} \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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