\(\int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx\) [154]

Optimal result

Integrand size = 26, antiderivative size = 41 \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{1-a x}-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {871, 12, 272, 65, 214} \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{1-a x}-\text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \]

[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 214

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

Rule 272

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 871

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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {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 {\text {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] (verified)

Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {\sqrt {1-a^2 x^2}}{-1+a x}-\log (x)+\log \left (-1+\sqrt {1-a^2 x^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41

method	result	size
default	\(-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=\frac {a x + {\left (a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

\[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=\int { -\frac {1}{\sqrt {-a^{2} x^{2} + 1} {\left (a x - 1\right )} x} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.80 \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=-\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {2 \, a}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

[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))

Mupad [B] (verification not implemented)

Time = 11.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x (1-a x) \sqrt {1-a^2 x^2}} \, dx=\frac {a\,\sqrt {1-a^2\,x^2}}{\sqrt {-a^2}\,\left (\frac {a}{\sqrt {-a^2}}+x\,\sqrt {-a^2}\right )}-\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right ) \]

[In]

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

[Out]

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