3.367 \(\int \sqrt {-1+e^{2 x}} \, dx\)

Optimal. Leaf size=26 \[ \sqrt {e^{2 x}-1}-\tan ^{-1}\left (\sqrt {e^{2 x}-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2282, 50, 63, 203} \[ \sqrt {e^{2 x}-1}-\tan ^{-1}\left (\sqrt {e^{2 x}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + E^(2*x)],x]

[Out]

Sqrt[-1 + E^(2*x)] - ArcTan[Sqrt[-1 + E^(2*x)]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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 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]]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.00 \[ \sqrt {e^{2 x}-1}-\tan ^{-1}\left (\sqrt {e^{2 x}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + E^(2*x)],x]

[Out]

Sqrt[-1 + E^(2*x)] - ArcTan[Sqrt[-1 + E^(2*x)]]

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fricas [A]  time = 0.42, size = 20, normalized size = 0.77 \[ \sqrt {e^{\left (2 \, x\right )} - 1} - \arctan \left (\sqrt {e^{\left (2 \, x\right )} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(e^(2*x) - 1) - arctan(sqrt(e^(2*x) - 1))

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giac [A]  time = 0.87, size = 20, normalized size = 0.77 \[ \sqrt {e^{\left (2 \, x\right )} - 1} - \arctan \left (\sqrt {e^{\left (2 \, x\right )} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(e^(2*x) - 1) - arctan(sqrt(e^(2*x) - 1))

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maple [A]  time = 0.01, size = 21, normalized size = 0.81 \[ -\arctan \left (\sqrt {{\mathrm e}^{2 x}-1}\right )+\sqrt {{\mathrm e}^{2 x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+exp(2*x))^(1/2),x)

[Out]

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

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maxima [A]  time = 0.96, size = 20, normalized size = 0.77 \[ \sqrt {e^{\left (2 \, x\right )} - 1} - \arctan \left (\sqrt {e^{\left (2 \, x\right )} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(e^(2*x) - 1) - arctan(sqrt(e^(2*x) - 1))

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mupad [B]  time = 0.26, size = 31, normalized size = 1.19 \[ \sqrt {{\mathrm {e}}^{2\,x}-1}\,\left (\frac {{\mathrm {e}}^{-x}\,\mathrm {asin}\left ({\mathrm {e}}^{-x}\right )}{\sqrt {1-{\mathrm {e}}^{-2\,x}}}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x) - 1)^(1/2),x)

[Out]

(exp(2*x) - 1)^(1/2)*((exp(-x)*asin(exp(-x)))/(1 - exp(-2*x))^(1/2) + 1)

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sympy [A]  time = 1.39, size = 19, normalized size = 0.73 \[ \begin {cases} \sqrt {e^{2 x} - 1} - \operatorname {acos}{\left (e^{- x} \right )} & \text {for}\: e^{x} < 0 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(exp(2*x) - 1) - acos(exp(-x)), exp(x) < 0))

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