3.4.67 \(\int \sqrt {-1+e^{2 x}} \, dx\) [367]

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

[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 = {2320, 52, 65, 209} \begin {gather*} \sqrt {e^{2 x}-1}-\tan ^{-1}\left (\sqrt {e^{2 x}-1}\right ) \end {gather*}

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 52

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 65

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 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 2320

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} \text {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,e^{2 x}\right )\\ &=\sqrt {-1+e^{2 x}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,e^{2 x}\right )\\ &=\sqrt {-1+e^{2 x}}-\text {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 = 0.02, size = 26, normalized size = 1.00 \begin {gather*} \sqrt {-1+e^{2 x}}-\tan ^{-1}\left (\sqrt {-1+e^{2 x}}\right ) \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.53, size = 30, normalized size = 1.15 \begin {gather*} \text {ConditionalExpression}\left [\sqrt {-1+E^{2 x}}-\text {ArcCos}\left [E^{-x}\right ],E^x>-1\text {\&\&}E^x<1\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[-1 + E^(2*x)],x]')

[Out]

ConditionalExpression[Sqrt[-1 + E ^ (2 x)] - ArcCos[E ^ (-x)], E ^ x > -1 && E ^ x < 1]

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Maple [A]
time = 0.04, size = 21, normalized size = 0.81

method	result	size
derivativedivides	\(-\arctan \left (\sqrt {{\mathrm e}^{2 x}-1}\right )+\sqrt {{\mathrm e}^{2 x}-1}\)	\(21\)
default	\(-\arctan \left (\sqrt {{\mathrm e}^{2 x}-1}\right )+\sqrt {{\mathrm e}^{2 x}-1}\)	\(21\)
risch	\(-\arctan \left (\sqrt {{\mathrm e}^{2 x}-1}\right )+\sqrt {{\mathrm e}^{2 x}-1}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.36, size = 20, normalized size = 0.77 \begin {gather*} \sqrt {e^{\left (2 \, x\right )} - 1} - \arctan \left (\sqrt {e^{\left (2 \, x\right )} - 1}\right ) \end {gather*}

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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Fricas [A]
time = 0.34, size = 20, normalized size = 0.77 \begin {gather*} \sqrt {e^{\left (2 \, x\right )} - 1} - \arctan \left (\sqrt {e^{\left (2 \, x\right )} - 1}\right ) \end {gather*}

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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Sympy [A]
time = 0.60, size = 24, normalized size = 0.92 \begin {gather*} \begin {cases} \sqrt {e^{2 x} - 1} - \operatorname {acos}{\left (e^{- x} \right )} & \text {for}\: e^{x} > -1 \wedge e^{x} < 1 \end {cases} \end {gather*}

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) > -1) & (exp(x) < 1)))

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Giac [A]
time = 0.00, size = 23, normalized size = 0.88 \begin {gather*} \sqrt {\mathrm {e}^{2 x}-1}-\arctan \left (\sqrt {\mathrm {e}^{2 x}-1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \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 ) \end {gather*}

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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