3.2.55 \(\int \frac {1}{x^2 (1-a x) \sqrt {1-a^2 x^2}} \, dx\) [155]

Optimal. Leaf size=64 \[ -\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 = {871, 821, 272, 65, 214} \begin {gather*} -\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

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*} \int \frac {1}{x^2 (1-a x) \sqrt {1-a^2 x^2}} \, dx &=\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}-\frac {\int \frac {-2 a^2-a^3 x}{x^2 \sqrt {1-a^2 x^2}} \, dx}{a^2}\\ &=-\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}+a \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}+\frac {1}{2} a \text {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (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}\\ &=-\frac {2 \sqrt {1-a^2 x^2}}{x}+\frac {\sqrt {1-a^2 x^2}}{x (1-a x)}-a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.16, size = 64, normalized size = 1.00 \begin {gather*} \frac {(1-2 a x) \sqrt {1-a^2 x^2}}{x (-1+a x)}+2 a \tanh ^{-1}\left (\sqrt {-a^2} x-\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 73, normalized size = 1.14

method	result	size
default	\(-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}-\frac {\sqrt {-a^{2} x^{2}+1}}{x}-a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\)	\(73\)
risch	\(\frac {a^{2} x^{2}-1}{x \sqrt {-a^{2} x^{2}+1}}-a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}+\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(-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^2), x)

________________________________________________________________________________________

Fricas [A]
time = 2.47, size = 76, normalized size = 1.19 \begin {gather*} \frac {a^{2} x^{2} - a x + {\left (a^{2} x^{2} - a x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x - 1\right )}}{a x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)), x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

Mupad [B]
time = 2.59, size = 81, normalized size = 1.27 \begin {gather*} \frac {a^2\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{x}-a\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________