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

Optimal. Leaf size=41 \[ \frac{\sqrt{1-a^2 x^2}}{1-a x}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

Sqrt[1 - a^2*x^2]/(1 - a*x) - ArcTanh[Sqrt[1 - a^2*x^2]]

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Rubi [A]  time = 0.0397202, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {857, 12, 266, 63, 208} \[ \frac{\sqrt{1-a^2 x^2}}{1-a x}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 - a*x)*Sqrt[1 - a^2*x^2]),x]

[Out]

Sqrt[1 - a^2*x^2]/(1 - a*x) - ArcTanh[Sqrt[1 - a^2*x^2]]

Rule 857

Int[(((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(d*(f + g*x)
^(n + 1)*(a + c*x^2)^(p + 1))/(2*a*p*(e*f - d*g)*(d + e*x)), x] + Dist[1/(p*(2*c*d)*(e*f - d*g)), Int[(f + g*x
)^n*(a + c*x^2)^p*(c*e*f*(2*p + 1) - c*d*g*(n + 2*p + 1) + c*e*g*(n + 2*p + 2)*x), x], x] /; FreeQ[{a, c, d, e
, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[n, 0] && ILtQ[n + 2*p, 0] &
&  !IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0301259, size = 41, normalized size = 1. \[ \frac{\sqrt{1-a^2 x^2}}{1-a x}-\tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 - a*x)*Sqrt[1 - a^2*x^2]),x]

[Out]

Sqrt[1 - a^2*x^2]/(1 - a*x) - ArcTanh[Sqrt[1 - a^2*x^2]]

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Maple [A]  time = 0.053, size = 58, normalized size = 1.4 \begin{align*} -{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -{\frac{1}{a}\sqrt{- \left ( x-{a}^{-1} \right ) ^{2}{a}^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctanh(1/(-a^2*x^2+1)^(1/2))-1/a/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{\sqrt{-a^{2} x^{2} + 1}{\left (a x - 1\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58115, size = 116, normalized size = 2.83 \begin{align*} \frac{a x +{\left (a x - 1\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt{-a^{2} x^{2} + 1} - 1}{a x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x + (a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(a*x**2*sqrt(-a**2*x**2 + 1) - x*sqrt(-a**2*x**2 + 1)), x)

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Giac [A]  time = 1.29234, size = 100, normalized size = 2.44 \begin{align*} -\frac{a \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{{\left | a \right |}} + \frac{2 \, a}{{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 2*a/(((sqrt(-a^2*x^2 + 1)*abs(a) + a
)/(a^2*x) - 1)*abs(a))