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

Optimal. Leaf size=189 \[ \frac{a b^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right )^2 x}-\frac{a b^3 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\left (1-a^2\right )^{5/2}}+\frac{(a+b x-1)^{3/2} (a+b x+1)^{3/2}}{3 \left (1-a^2\right ) x^3}-\frac{a b \sqrt{a+b x-1} (a+b x+1)^{3/2}}{2 (1-a) (a+1)^2 x^2}-\frac{a}{3 x^3}-\frac{b}{2 x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.156407, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5909, 14, 96, 94, 93, 205} \[ \frac{a b^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right )^2 x}-\frac{a b^3 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\left (1-a^2\right )^{5/2}}+\frac{(a+b x-1)^{3/2} (a+b x+1)^{3/2}}{3 \left (1-a^2\right ) x^3}-\frac{a b \sqrt{a+b x-1} (a+b x+1)^{3/2}}{2 (1-a) (a+1)^2 x^2}-\frac{a}{3 x^3}-\frac{b}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]/x^4,x]

[Out]

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

Rule 5909

Int[E^(ArcCosh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[-1 + u]*Sqrt[1 + u])^n, x] /; RationalQ[m
] && IntegerQ[n] && PolynomialQ[u, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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

Mathematica [C]  time = 0.156376, size = 179, normalized size = 0.95 \[ \frac{1}{6} \left (\frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (a^2 \left (b^2 x^2+4\right )-a^3 b x-2 a^4+a b x+2 b^2 x^2-2\right )}{\left (a^2-1\right )^2 x^3}-\frac{3 i a b^3 \log \left (\frac{4 \left (1-a^2\right )^{3/2} \left (\sqrt{1-a^2} \sqrt{a+b x-1} \sqrt{a+b x+1}+i a^2+i a b x-i\right )}{a b^3 x}\right )}{\left (1-a^2\right )^{5/2}}-\frac{2 a}{x^3}-\frac{3 b}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]/x^4,x]

[Out]

((-2*a)/x^3 - (3*b)/x^2 + (Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(-2 - 2*a^4 + a*b*x - a^3*b*x + 2*b^2*x^2 + a^
2*(4 + b^2*x^2)))/((-1 + a^2)^2*x^3) - ((3*I)*a*b^3*Log[(4*(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]))/(a*b^3*x)])/(1 - a^2)^(5/2))/6

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 374, normalized size = 2. \begin{align*} -{\frac{1}{6\, \left ({a}^{2}-1 \right ) ^{3}{x}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1} \left ( 3\,\sqrt{{a}^{2}-1}\ln \left ( 2\,{\frac{xab+\sqrt{{a}^{2}-1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+{a}^{2}-1}{x}} \right ){x}^{3}a{b}^{3}-{x}^{2}{a}^{4}{b}^{2}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+x{a}^{5}b\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{x}^{2}{a}^{2}{b}^{2}+2\,{a}^{6}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}-2\,x{a}^{3}b\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{x}^{2}{b}^{2}-6\,{a}^{4}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}xab+6\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{a}^{2}-2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1} \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}-{\frac{b}{2\,{x}^{2}}}-{\frac{a}{3\,{x}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^4,x)

[Out]

-1/6*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*(3*(a^2-1)^(1/2)*ln(2*(x*a*b+(a^2-1)^(1/2)*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+
a^2-1)/x)*x^3*a*b^3-x^2*a^4*b^2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+x*a^5*b*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)-(b^2*x^2+2
*a*b*x+a^2-1)^(1/2)*x^2*a^2*b^2+2*a^6*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)-2*x*a^3*b*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+2*
(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*x^2*b^2-6*a^4*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)+(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*x*a*b+
6*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*a^2-2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2))/(b^2*x^2+2*a*b*x+a^2-1)^(1/2)/(a^2-1)^3/x
^3-1/2*b/x^2-1/3*a/x^3

________________________________________________________________________________________

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((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.04357, size = 1035, normalized size = 5.48 \begin{align*} \left [\frac{3 \, \sqrt{a^{2} - 1} a b^{3} x^{3} \log \left (\frac{a^{2} b x + a^{3} +{\left (a^{2} - \sqrt{a^{2} - 1} a - 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} -{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1} - a}{x}\right ) - 2 \, a^{7} +{\left (a^{4} + a^{2} - 2\right )} b^{3} x^{3} + 6 \, a^{5} - 6 \, a^{3} - 3 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b x -{\left (2 \, a^{6} -{\left (a^{4} + a^{2} - 2\right )} b^{2} x^{2} - 6 \, a^{4} +{\left (a^{5} - 2 \, a^{3} + a\right )} b x + 6 \, a^{2} - 2\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + 2 \, a}{6 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}, \frac{6 \, \sqrt{-a^{2} + 1} a b^{3} x^{3} \arctan \left (-\frac{\sqrt{-a^{2} + 1} b x - \sqrt{-a^{2} + 1} \sqrt{b x + a + 1} \sqrt{b x + a - 1}}{a^{2} - 1}\right ) - 2 \, a^{7} +{\left (a^{4} + a^{2} - 2\right )} b^{3} x^{3} + 6 \, a^{5} - 6 \, a^{3} - 3 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b x -{\left (2 \, a^{6} -{\left (a^{4} + a^{2} - 2\right )} b^{2} x^{2} - 6 \, a^{4} +{\left (a^{5} - 2 \, a^{3} + a\right )} b x + 6 \, a^{2} - 2\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + 2 \, a}{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((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(b*x+a-1)**(1/2)*(b*x+a+1)**(1/2))/x**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.64251, size = 695, normalized size = 3.68 \begin{align*} -\frac{\frac{6 \, a b^{4} \arctan \left (\frac{{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 2 \, a}{2 \, \sqrt{-a^{2} + 1}}\right )}{{\left (a^{4} - 2 \, a^{2} + 1\right )} \sqrt{-a^{2} + 1}} - \frac{a b^{4} + 3 \, b^{4}}{a^{3} + 3 \, a^{2} + 3 \, a + 1} - \frac{4 \,{\left (12 \, a^{4} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{8} - 16 \, a^{5} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{6} - 3 \, a b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{10} + 6 \, a^{2} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{8} - 56 \, a^{3} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{6} + 48 \, a^{4} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} + 12 \, b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{8} - 48 \, a b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{6} + 192 \, a^{2} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} - 96 \, a^{3} b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 144 \, a b^{4}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} + 32 \, a^{2} b^{4} + 64 \, b^{4}\right )}}{{\left (a^{4} - 2 \, a^{2} + 1\right )}{\left ({\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} - 4 \, a{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} + 4\right )}^{3}} + \frac{3 \,{\left (b x + a + 1\right )} b^{4} - a b^{4} - 3 \, b^{4}}{b^{3} x^{3}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))/x^4,x, algorithm="giac")

[Out]

-1/6*(6*a*b^4*arctan(1/2*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2*a)/sqrt(-a^2 + 1))/((a^4 - 2*a^2 + 1)*
sqrt(-a^2 + 1)) - (a*b^4 + 3*b^4)/(a^3 + 3*a^2 + 3*a + 1) - 4*(12*a^4*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a -
1))^8 - 16*a^5*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^6 - 3*a*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))
^10 + 6*a^2*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^8 - 56*a^3*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))
^6 + 48*a^4*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 + 12*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^8 -
48*a*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^6 + 192*a^2*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 -
96*a^3*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 144*a*b^4*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 + 3
2*a^2*b^4 + 64*b^4)/((a^4 - 2*a^2 + 1)*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 - 4*a*(sqrt(b*x + a + 1) - s
qrt(b*x + a - 1))^2 + 4)^3) + (3*(b*x + a + 1)*b^4 - a*b^4 - 3*b^4)/(b^3*x^3))/b