3.7.66 \(\int \frac {e^{5 x}}{1+e^{10 x}} \, dx\) [666]

Optimal. Leaf size=10 \[ \frac {1}{5} \tan ^{-1}\left (e^{5 x}\right ) \]

[Out]

1/5*arctan(exp(5*x))

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Rubi [A]
time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2281, 209} \begin {gather*} \frac {1}{5} \text {ArcTan}\left (e^{5 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(5*x)/(1 + E^(10*x)),x]

[Out]

ArcTan[E^(5*x)]/5

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 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps

\begin {align*} \int \frac {e^{5 x}}{1+e^{10 x}} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{5 x}\right )\\ &=\frac {1}{5} \tan ^{-1}\left (e^{5 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 10, normalized size = 1.00 \begin {gather*} \frac {1}{5} \tan ^{-1}\left (e^{5 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(5*x)/(1 + E^(10*x)),x]

[Out]

ArcTan[E^(5*x)]/5

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Maple [A]
time = 0.02, size = 8, normalized size = 0.80

method	result	size
default	\(\frac {\arctan \left ({\mathrm e}^{5 x}\right )}{5}\)	\(8\)
risch	\(\frac {i \ln \left ({\mathrm e}^{5 x}+i\right )}{10}-\frac {i \ln \left ({\mathrm e}^{5 x}-i\right )}{10}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5*x)/(1+exp(10*x)),x,method=_RETURNVERBOSE)

[Out]

1/5*arctan(exp(x)^5)

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Maxima [A]
time = 0.52, size = 7, normalized size = 0.70 \begin {gather*} \frac {1}{5} \, \arctan \left (e^{\left (5 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(5*x)/(1+exp(10*x)),x, algorithm="maxima")

[Out]

1/5*arctan(e^(5*x))

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Fricas [A]
time = 0.35, size = 7, normalized size = 0.70 \begin {gather*} \frac {1}{5} \, \arctan \left (e^{\left (5 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(5*x)/(1+exp(10*x)),x, algorithm="fricas")

[Out]

1/5*arctan(e^(5*x))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (7) = 14\).
time = 0.04, size = 17, normalized size = 1.70 \begin {gather*} \operatorname {RootSum} {\left (100 z^{2} + 1, \left ( i \mapsto i \log {\left (10 i + e^{5 x} \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(5*x)/(1+exp(10*x)),x)

[Out]

RootSum(100*_z**2 + 1, Lambda(_i, _i*log(10*_i + exp(5*x))))

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Giac [A]
time = 6.46, size = 7, normalized size = 0.70 \begin {gather*} \frac {1}{5} \, \arctan \left (e^{\left (5 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(5*x)/(1+exp(10*x)),x, algorithm="giac")

[Out]

1/5*arctan(e^(5*x))

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Mupad [B]
time = 0.06, size = 7, normalized size = 0.70 \begin {gather*} \frac {\mathrm {atan}\left ({\mathrm {e}}^{5\,x}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5*x)/(exp(10*x) + 1),x)

[Out]

atan(exp(5*x))/5

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