3.66 \(\int \frac{1}{\sqrt{-1+a^{2 x}}} \, dx\)

Optimal. Leaf size=17 \[ \frac{\tan ^{-1}\left (\sqrt{a^{2 x}-1}\right )}{\log (a)} \]

[Out]

ArcTan[Sqrt[-1 + a^(2*x)]]/Log[a]

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Rubi [A]  time = 0.0129283, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {2282, 63, 203} \[ \frac{\tan ^{-1}\left (\sqrt{a^{2 x}-1}\right )}{\log (a)} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 + a^(2*x)],x]

[Out]

ArcTan[Sqrt[-1 + a^(2*x)]]/Log[a]

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]]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0044183, size = 17, normalized size = 1. \[ \frac{\tan ^{-1}\left (\sqrt{a^{2 x}-1}\right )}{\log (a)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 + a^(2*x)],x]

[Out]

ArcTan[Sqrt[-1 + a^(2*x)]]/Log[a]

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Maple [A]  time = 0.007, size = 16, normalized size = 0.9 \begin{align*}{\frac{1}{\ln \left ( a \right ) }\arctan \left ( \sqrt{-1+{a}^{2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan((-1+a^(2*x))^(1/2))/ln(a)

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Maxima [A]  time = 1.445, size = 20, normalized size = 1.18 \begin{align*} \frac{\arctan \left (\sqrt{a^{2 \, x} - 1}\right )}{\log \left (a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(a^(2*x) - 1))/log(a)

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Fricas [A]  time = 2.09896, size = 46, normalized size = 2.71 \begin{align*} \frac{\arctan \left (\sqrt{a^{2 \, x} - 1}\right )}{\log \left (a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(a^(2*x) - 1))/log(a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a^{2 x} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(a**(2*x) - 1), x)

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Giac [A]  time = 1.08796, size = 20, normalized size = 1.18 \begin{align*} \frac{\arctan \left (\sqrt{a^{2 \, x} - 1}\right )}{\log \left (a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(a^(2*x) - 1))/log(a)