3.90 \(\int \frac{\cosh ^{-1}(a+b x)}{x^4} \, dx\)

Optimal. Leaf size=154 \[ \frac{a b^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right )^2 x}-\frac{\left (2 a^2+1\right ) b^3 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{3 \left (1-a^2\right )^{5/2}}+\frac{b \sqrt{a+b x-1} \sqrt{a+b x+1}}{6 \left (1-a^2\right ) x^2}-\frac{\cosh ^{-1}(a+b x)}{3 x^3} \]

[Out]

(b*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(6*(1 - a^2)*x^2) + (a*b^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(2*(
1 - a^2)^2*x) - ArcCosh[a + b*x]/(3*x^3) - ((1 + 2*a^2)*b^3*ArcTan[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a
]*Sqrt[-1 + a + b*x])])/(3*(1 - a^2)^(5/2))

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Rubi [A]  time = 0.171842, antiderivative size = 154, 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 = {5866, 5802, 103, 151, 12, 93, 205} \[ \frac{a b^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right )^2 x}-\frac{\left (2 a^2+1\right ) b^3 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{3 \left (1-a^2\right )^{5/2}}+\frac{b \sqrt{a+b x-1} \sqrt{a+b x+1}}{6 \left (1-a^2\right ) x^2}-\frac{\cosh ^{-1}(a+b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a + b*x]/x^4,x]

[Out]

(b*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(6*(1 - a^2)*x^2) + (a*b^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(2*(
1 - a^2)^2*x) - ArcCosh[a + b*x]/(3*x^3) - ((1 + 2*a^2)*b^3*ArcTan[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a
]*Sqrt[-1 + a + b*x])])/(3*(1 - a^2)^(5/2))

Rule 5866

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5802

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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{\cosh ^{-1}(a+b x)}{x^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cosh ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\cosh ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x} \left (-\frac{a}{b}+\frac{x}{b}\right )^3} \, dx,x,a+b x\right )\\ &=\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{6 \left (1-a^2\right ) x^2}-\frac{\cosh ^{-1}(a+b x)}{3 x^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\frac{2 a}{b}+\frac{x}{b}}{\sqrt{-1+x} \sqrt{1+x} \left (-\frac{a}{b}+\frac{x}{b}\right )^2} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )}\\ &=\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{6 \left (1-a^2\right ) x^2}+\frac{a b^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right )^2 x}-\frac{\cosh ^{-1}(a+b x)}{3 x^3}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1+2 a^2}{b^2 \sqrt{-1+x} \sqrt{1+x} \left (-\frac{a}{b}+\frac{x}{b}\right )} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )^2}\\ &=\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{6 \left (1-a^2\right ) x^2}+\frac{a b^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right )^2 x}-\frac{\cosh ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (1+2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x} \left (-\frac{a}{b}+\frac{x}{b}\right )} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )^2}\\ &=\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{6 \left (1-a^2\right ) x^2}+\frac{a b^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right )^2 x}-\frac{\cosh ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (1+2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{b}-\frac{a}{b}-\left (\frac{1}{b}-\frac{a}{b}\right ) x^2} \, dx,x,\frac{\sqrt{1+a+b x}}{\sqrt{-1+a+b x}}\right )}{3 \left (1-a^2\right )^2}\\ &=\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{6 \left (1-a^2\right ) x^2}+\frac{a b^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right )^2 x}-\frac{\cosh ^{-1}(a+b x)}{3 x^3}-\frac{\left (1+2 a^2\right ) b^3 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{-1+a+b x}}\right )}{3 \left (1-a^2\right )^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.296328, size = 162, normalized size = 1.05 \[ \frac{1}{6} \left (-\frac{i \left (2 a^2+1\right ) b^3 \log \left (\frac{12 \left (1-a^2\right )^{3/2} \left (\sqrt{1-a^2} \sqrt{a+b x-1} \sqrt{a+b x+1}+i a^2+i a b x-i\right )}{b^3 \left (2 a^2 x+x\right )}\right )}{\left (1-a^2\right )^{5/2}}+\frac{b \sqrt{a+b x-1} \sqrt{a+b x+1} \left (-a^2+3 a b x+1\right )}{\left (a^2-1\right )^2 x^2}-\frac{2 \cosh ^{-1}(a+b x)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a + b*x]/x^4,x]

[Out]

((b*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(1 - a^2 + 3*a*b*x))/((-1 + a^2)^2*x^2) - (2*ArcCosh[a + b*x])/x^3 -
(I*(1 + 2*a^2)*b^3*Log[(12*(1 - a^2)^(3/2)*(-I + I*a^2 + I*a*b*x + Sqrt[1 - a^2]*Sqrt[-1 + a + b*x]*Sqrt[1 + a
+ b*x]))/(b^3*(x + 2*a^2*x))])/(1 - a^2)^(5/2))/6

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Maple [B]  time = 0.02, size = 397, normalized size = 2.6 \begin{align*} -{\frac{{\rm arccosh} \left (bx+a\right )}{3\,{x}^{3}}}-{\frac{{b}^{3}{a}^{2}}{ \left ( 3+3\,a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{ \left ( bx+a \right ) ^{2}-1}+a \left ( bx+a \right ) -1}{bx}} \right ) \left ({a}^{2}-1 \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}}-{\frac{{b}^{3}}{ \left ( 6+6\,a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{ \left ( bx+a \right ) ^{2}-1}+a \left ( bx+a \right ) -1}{bx}} \right ) \left ({a}^{2}-1 \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}}+{\frac{{b}^{2}{a}^{3}}{2\,x \left ({a}^{2}-1 \right ) ^{2} \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{b{a}^{4}}{6\,{x}^{2} \left ({a}^{2}-1 \right ) ^{2} \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{{b}^{2}a}{2\,x \left ({a}^{2}-1 \right ) ^{2} \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{{a}^{2}b}{3\,{x}^{2} \left ({a}^{2}-1 \right ) ^{2} \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{b}{6\,{x}^{2} \left ({a}^{2}-1 \right ) ^{2} \left ( 1+a \right ) \left ( a-1 \right ) }\sqrt{bx+a-1}\sqrt{bx+a+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(b*x+a)/x^4,x)

[Out]

-1/3*arccosh(b*x+a)/x^3-1/3*b^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/(a^2-1)^(3/2)/(1+a)/(a-1)/((b*x+a)^2-1)^(1/2)*
ln(2*((a^2-1)^(1/2)*((b*x+a)^2-1)^(1/2)+a*(b*x+a)-1)/b/x)*a^2-1/6*b^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/(a^2-1)^
(3/2)/(1+a)/(a-1)/((b*x+a)^2-1)^(1/2)*ln(2*((a^2-1)^(1/2)*((b*x+a)^2-1)^(1/2)+a*(b*x+a)-1)/b/x)+1/2*b^2*(b*x+a
-1)^(1/2)*(b*x+a+1)^(1/2)/x/(a^2-1)^2/(1+a)/(a-1)*a^3-1/6*b*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x^2/(a^2-1)^2/(1+a
)/(a-1)*a^4-1/2*b^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x/(a^2-1)^2/(1+a)/(a-1)*a+1/3*b*(b*x+a-1)^(1/2)*(b*x+a+1)^
(1/2)/x^2/(a^2-1)^2/(1+a)/(a-1)*a^2-1/6*b*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/x^2/(a^2-1)^2/(1+a)/(a-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.71583, size = 1305, normalized size = 8.47 \begin{align*} \left [\frac{{\left (2 \, a^{2} + 1\right )} \sqrt{a^{2} - 1} b^{3} x^{3} \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 ) + 3 \,{\left (a^{3} - a\right )} b^{3} x^{3} + 2 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \,{\left (a^{6} - 3 \, a^{4} -{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} + 3 \, a^{2} - 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (3 \,{\left (a^{3} - a\right )} b^{2} x^{2} -{\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}, \frac{2 \,{\left (2 \, a^{2} + 1\right )} \sqrt{-a^{2} + 1} b^{3} x^{3} \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 ) + 3 \,{\left (a^{3} - a\right )} b^{3} x^{3} + 2 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \,{\left (a^{6} - 3 \, a^{4} -{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3} + 3 \, a^{2} - 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (3 \,{\left (a^{3} - a\right )} b^{2} x^{2} -{\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acosh}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(b*x+a)/x**4,x)

[Out]

Integral(acosh(a + b*x)/x**4, x)

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Giac [B]  time = 1.19736, size = 459, normalized size = 2.98 \begin{align*} \frac{1}{3} \, b{\left (\frac{{\left (2 \, a^{2} b^{2} + b^{2}\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} - 2 \, a^{2} + 1\right )} \sqrt{-a^{2} + 1}} - \frac{2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} a^{2} b^{2} - 6 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{4} b^{2} - 4 \, a^{5} b{\left | b \right |} +{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} b^{2} + 7 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{2} b^{2} + 8 \, a^{3} b{\left | b \right |} -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} b^{2} - 4 \, a b{\left | b \right |}}{{\left (a^{4} - 2 \, a^{2} + 1\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 )}^{2}}\right )} - \frac{\log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} - 1}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/3*b*((2*a^2*b^2 + b^2)*arctan(-(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/sqrt(-a^2 + 1))/((a^4 - 2*a^2
+ 1)*sqrt(-a^2 + 1)) - (2*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))^3*a^2*b^2 - 6*(x*abs(b) - sqrt(b^2*x^
2 + 2*a*b*x + a^2 - 1))*a^4*b^2 - 4*a^5*b*abs(b) + (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))^3*b^2 + 7*(x
*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a^2*b^2 + 8*a^3*b*abs(b) - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a
^2 - 1))*b^2 - 4*a*b*abs(b))/((a^4 - 2*a^2 + 1)*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))^2 - a^2 + 1)^2
)) - 1/3*log(b*x + a + sqrt((b*x + a)^2 - 1))/x^3