3.25 \(\int \frac{\sec ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\left (1-2 a^2\right ) b^2 \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}-\frac{\sec ^{-1}(a+b x)}{2 x^2} \]

[Out]

(b*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(2*a*(1 - a^2)*x) + (b^2*ArcSec[a + b*x])/(2*a^2) - ArcSec[a + b*x]/(2*
x^2) - ((1 - 2*a^2)*b^2*ArcTan[(Sqrt[1 + a]*Tan[ArcSec[a + b*x]/2])/Sqrt[1 - a]])/(a^2*(1 - a^2)^(3/2))

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Rubi [A]  time = 0.19247, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5258, 4426, 3785, 3919, 3831, 2659, 205} \[ \frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\left (1-2 a^2\right ) b^2 \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}-\frac{\sec ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]/x^3,x]

[Out]

(b*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(2*a*(1 - a^2)*x) + (b^2*ArcSec[a + b*x])/(2*a^2) - ArcSec[a + b*x]/(2*
x^2) - ((1 - 2*a^2)*b^2*ArcTan[(Sqrt[1 + a]*Tan[ArcSec[a + b*x]/2])/Sqrt[1 - a]])/(a^2*(1 - a^2)^(3/2))

Rule 5258

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d^(m + 1),
Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 4426

Int[((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)])^(n_.)*Tan[(c_.)
+ (d_.)*(x_)], x_Symbol] :> Simp[((e + f*x)^m*(a + b*Sec[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*
d*(n + 1)), Int[(e + f*x)^(m - 1)*(a + b*Sec[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && I
GtQ[m, 0] && NeQ[n, -1]

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}(a+b x)}{x^3} \, dx &=b^2 \operatorname{Subst}\left (\int \frac{x \sec (x) \tan (x)}{(-a+\sec (x))^3} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=-\frac{\sec ^{-1}(a+b x)}{2 x^2}+\frac{1}{2} b^2 \operatorname{Subst}\left (\int \frac{1}{(-a+\sec (x))^2} \, dx,x,\sec ^{-1}(a+b x)\right )\\ &=\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}-\frac{\sec ^{-1}(a+b x)}{2 x^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1-a^2-a \sec (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{2 a \left (1-a^2\right )}\\ &=\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}+\frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\sec ^{-1}(a+b x)}{2 x^2}-\frac{\left (\left (1-2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{\sec (x)}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{2 a^2 \left (1-a^2\right )}\\ &=\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}+\frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\sec ^{-1}(a+b x)}{2 x^2}-\frac{\left (\left (1-2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{2 a^2 \left (1-a^2\right )}\\ &=\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}+\frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\sec ^{-1}(a+b x)}{2 x^2}-\frac{\left (\left (1-2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a+(1+a) x^2} \, dx,x,\tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )\right )}{a^2 \left (1-a^2\right )}\\ &=\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{2 a \left (1-a^2\right ) x}+\frac{b^2 \sec ^{-1}(a+b x)}{2 a^2}-\frac{\sec ^{-1}(a+b x)}{2 x^2}-\frac{\left (1-2 a^2\right ) b^2 \tan ^{-1}\left (\frac{\sqrt{1+a} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a^2 \left (1-a^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 1.04029, size = 198, normalized size = 1.58 \[ -\frac{\frac{b x (a+b x) \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}}{a \left (a^2-1\right )}+\frac{i \left (2 a^2-1\right ) b^2 x^2 \log \left (\frac{4 (a-1) a^2 (a+1) \left (-\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)-\frac{i \left (a^2+a b x-1\right )}{\sqrt{1-a^2}}\right )}{\left (2 a^2-1\right ) b^2 x}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac{b^2 x^2 \sin ^{-1}\left (\frac{1}{a+b x}\right )}{a^2}+\sec ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a + b*x]/x^3,x]

[Out]

-((b*x*(a + b*x)*Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])/(a*(-1 + a^2)) + ArcSec[a + b*x] + (b^2*x^2
*ArcSin[(a + b*x)^(-1)])/a^2 + (I*(-1 + 2*a^2)*b^2*x^2*Log[(4*(-1 + a)*a^2*(1 + a)*(((-I)*(-1 + a^2 + a*b*x))/
Sqrt[1 - a^2] - (a + b*x)*Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]))/((-1 + 2*a^2)*b^2*x)])/(a^2*(1 -
a^2)^(3/2)))/(2*x^2)

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Maple [B]  time = 0.243, size = 452, normalized size = 3.6 \begin{align*} -{\frac{{\rm arcsec} \left (bx+a\right )}{2\,{x}^{2}}}-{\frac{{b}^{2}}{ \left ( 2\,bx+2\,a \right ) \left ({a}^{2}-1 \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{-1+ \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}+{\frac{{a}^{2}{b}^{2}}{bx+a}\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{-1+ \left ( bx+a \right ) ^{2}}+a \left ( bx+a \right ) -1}{bx}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}} \left ({a}^{2}-1 \right ) ^{-{\frac{5}{2}}}}+{\frac{{b}^{2}}{ \left ( 2\,bx+2\,a \right ){a}^{2} \left ({a}^{2}-1 \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{-1+ \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{b \left ( -1+ \left ( bx+a \right ) ^{2} \right ) }{ \left ( 2\,bx+2\,a \right ) a \left ({a}^{2}-1 \right ) x}{\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{3\,{b}^{2}}{2\,bx+2\,a}\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{-1+ \left ( bx+a \right ) ^{2}}+a \left ( bx+a \right ) -1}{bx}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}} \left ({a}^{2}-1 \right ) ^{-{\frac{5}{2}}}}+{\frac{{b}^{2}}{ \left ( 2\,bx+2\,a \right ){a}^{2}}\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{-1+ \left ( bx+a \right ) ^{2}}+a \left ( bx+a \right ) -1}{bx}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}} \left ({a}^{2}-1 \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(b*x+a)/x^3,x)

[Out]

-1/2*arcsec(b*x+a)/x^2-1/2*b^2*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/(a^2-1)*arctan(1/
(-1+(b*x+a)^2)^(1/2))+b^2*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^2/(a^2-1)^(5/2)*ln(2
*((a^2-1)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)+1/2*b^2*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)
^(1/2)/(b*x+a)/a^2/(a^2-1)*arctan(1/(-1+(b*x+a)^2)^(1/2))-1/2*b*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2
)/(b*x+a)/a/(a^2-1)/x-3/2*b^2*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/(a^2-1)^(5/2)*ln(2
*((a^2-1)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)+1/2*b^2*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)
^(1/2)/(b*x+a)/a^2/(a^2-1)^(5/2)*ln(2*((a^2-1)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{2} \int \frac{{\left (b^{2} x + a b\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (b x + a - 1\right )\right )}}{b^{2} x^{4} + 2 \, a b x^{3} +{\left (b^{2} x^{4} + 2 \, a b x^{3} +{\left (a^{2} - 1\right )} x^{2}\right )}{\left (b x + a + 1\right )}{\left (b x + a - 1\right )} +{\left (a^{2} - 1\right )} x^{2}}\,{d x} - \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)/x^3,x, algorithm="maxima")

[Out]

1/2*(2*x^2*integrate(1/2*(b^2*x + a*b)*e^(1/2*log(b*x + a + 1) + 1/2*log(b*x + a - 1))/(b^2*x^4 + 2*a*b*x^3 +
(a^2 - 1)*x^2 + (b^2*x^4 + 2*a*b*x^3 + (a^2 - 1)*x^2)*e^(log(b*x + a + 1) + log(b*x + a - 1))), x) - arctan(sq
rt(b*x + a + 1)*sqrt(b*x + a - 1)))/x^2

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Fricas [A]  time = 2.68186, size = 999, normalized size = 7.99 \begin{align*} \left [\frac{{\left (2 \, a^{2} - 1\right )} \sqrt{a^{2} - 1} b^{2} x^{2} \log \left (\frac{a^{2} b x + a^{3} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (a^{2} + \sqrt{a^{2} - 1} a - 1\right )} +{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1} - a}{x}\right ) + 2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} b^{2} x^{2} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (a^{3} - a\right )} b^{2} x^{2} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (a^{3} - a\right )} b x -{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} \operatorname{arcsec}\left (b x + a\right )}{2 \,{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} x^{2}}, -\frac{2 \,{\left (2 \, a^{2} - 1\right )} \sqrt{-a^{2} + 1} b^{2} x^{2} \arctan \left (-\frac{\sqrt{-a^{2} + 1} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt{-a^{2} + 1}}{a^{2} - 1}\right ) - 2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} b^{2} x^{2} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (a^{3} - a\right )} b^{2} x^{2} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (a^{3} - a\right )} b x +{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} \operatorname{arcsec}\left (b x + a\right )}{2 \,{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)/x^3,x, algorithm="fricas")

[Out]

[1/2*((2*a^2 - 1)*sqrt(a^2 - 1)*b^2*x^2*log((a^2*b*x + a^3 + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^2 + sqrt(a^2
- 1)*a - 1) + (a*b*x + a^2 - 1)*sqrt(a^2 - 1) - a)/x) + 2*(a^4 - 2*a^2 + 1)*b^2*x^2*arctan(-b*x - a + sqrt(b^
2*x^2 + 2*a*b*x + a^2 - 1)) - (a^3 - a)*b^2*x^2 - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^3 - a)*b*x - (a^6 - 2*a
^4 + a^2)*arcsec(b*x + a))/((a^6 - 2*a^4 + a^2)*x^2), -1/2*(2*(2*a^2 - 1)*sqrt(-a^2 + 1)*b^2*x^2*arctan(-(sqrt
(-a^2 + 1)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*sqrt(-a^2 + 1))/(a^2 - 1)) - 2*(a^4 - 2*a^2 + 1)*b^2*x^2*ar
ctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + (a^3 - a)*b^2*x^2 + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^
3 - a)*b*x + (a^6 - 2*a^4 + a^2)*arcsec(b*x + a))/((a^6 - 2*a^4 + a^2)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}{\left (a + b x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(b*x+a)/x**3,x)

[Out]

Integral(asec(a + b*x)/x**3, x)

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Giac [B]  time = 1.66548, size = 375, normalized size = 3. \begin{align*} -b{\left (\frac{{\left (2 \, a^{2} b^{5} - b^{5}\right )} \arctan \left (-\frac{x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{\sqrt{-a^{2} + 1}}\right )}{{\left (a^{4} b^{4} \mathrm{sgn}\left (b x + a\right ) - a^{2} b^{4} \mathrm{sgn}\left (b x + a\right )\right )} \sqrt{-a^{2} + 1}} - \frac{{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a b^{5} + a^{2} b^{4}{\left | b \right |} - b^{4}{\left | b \right |}}{{\left (a^{3} b^{4} \mathrm{sgn}\left (b x + a\right ) - a b^{4} \mathrm{sgn}\left (b x + a\right )\right )}{\left ({\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} - a^{2} + 1\right )}} - \frac{b \arctan \left (-\frac{{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} b + a{\left | b \right |}}{b}\right )}{a^{2} \mathrm{sgn}\left (b x + a\right )}\right )} - \frac{\arccos \left (\frac{1}{b x + a}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)/x^3,x, algorithm="giac")

[Out]

-b*((2*a^2*b^5 - b^5)*arctan(-(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/sqrt(-a^2 + 1))/((a^4*b^4*sgn(b*x
+ a) - a^2*b^4*sgn(b*x + a))*sqrt(-a^2 + 1)) - ((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a*b^5 + a^2*b^
4*abs(b) - b^4*abs(b))/((a^3*b^4*sgn(b*x + a) - a*b^4*sgn(b*x + a))*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2
- 1))^2 - a^2 + 1)) - b*arctan(-((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*b + a*abs(b))/b)/(a^2*sgn(b*x
+ a))) - 1/2*arccos(1/(b*x + a))/x^2