3.841 \(\int \frac{e^{3 \tanh ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=202 \[ -\frac{3 (2 a+3) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(1-a)^3 \sqrt{1-a^2}}-\frac{(a+b x+1)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{-a-b x+1}}+\frac{3 (2 a+3) b^2 \sqrt{a+b x+1}}{(1-a)^3 (a+1) \sqrt{-a-b x+1}}-\frac{(2 a+3) b (a+b x+1)^{3/2}}{2 (1-a)^2 (a+1) x \sqrt{-a-b x+1}} \]

[Out]

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

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Rubi [A]  time = 0.12217, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6163, 96, 94, 93, 208} \[ -\frac{3 (2 a+3) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(1-a)^3 \sqrt{1-a^2}}-\frac{(a+b x+1)^{5/2}}{2 \left (1-a^2\right ) x^2 \sqrt{-a-b x+1}}+\frac{3 (2 a+3) b^2 \sqrt{a+b x+1}}{(1-a)^3 (a+1) \sqrt{-a-b x+1}}-\frac{(2 a+3) b (a+b x+1)^{3/2}}{2 (1-a)^2 (a+1) x \sqrt{-a-b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a + b*x])/x^3,x]

[Out]

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

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.134306, size = 136, normalized size = 0.67 \[ \frac{\sqrt{a+b x+1} \left (a^3-a^2-a \left (b^2 x^2+5 b x+1\right )-14 b^2 x^2+5 b x+1\right )}{2 (a-1)^3 x^2 \sqrt{-a-b x+1}}-\frac{3 (2 a+3) b^2 \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )}{(a-1)^{7/2} \sqrt{a+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a + b*x])/x^3,x]

[Out]

(Sqrt[1 + a + b*x]*(1 - a^2 + a^3 + 5*b*x - 14*b^2*x^2 - a*(1 + 5*b*x + b^2*x^2)))/(2*(-1 + a)^3*x^2*Sqrt[1 -
a - b*x]) - (3*(3 + 2*a)*b^2*ArcTan[Sqrt[1 - a - b*x]/(Sqrt[(-1 + a)/(1 + a)]*Sqrt[1 + a + b*x])])/((-1 + a)^(
7/2)*Sqrt[1 + a])

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Maple [B]  time = 0.045, size = 2194, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)/x^3,x)

[Out]

-21/2*a^3*b^2/(-a^2+1)^(5/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-3*a*b^2/
(-a^2+1)^(3/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-45/2*a^2*b^2/(-a^2+1)^
(5/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-3*b^2/(-a^2+1)^(3/2)*ln((-2*a^2
+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-3/2/(-a^2+1)/x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)
*a+30*a^5*b^2/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+45/2*a^3*b^2/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)
-45/2*a^4*b^2/(-a^2+1)^(7/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-45/2*a^3
*b^2/(-a^2+1)^(7/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+15/2*a^7*b^2/(-a^
2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+45/2*a^6*b^2/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-15/2*a^2*b^2/(-a^
2+1)^(7/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-3/2*b^2/(-a^2+1)^(5/2)*ln(
(-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+3/2*b^2/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2
+1)^(1/2)+3*b^2/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2/(-a^2+1)/x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3*b/(-
a^2+1)/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+39*a^3*b^2/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+29*a^2*b^2/(-a^2+
1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+27/2*a*b^2/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+31/2*a^5*b^2/(-a^2+1)
^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+75/2*a^4*b^2/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+15/2*a^2*b^2/(-a^2+1)
^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+30*a^4*b^2/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+2*b^3*(-2*b^2*x-2*a*b)/
(-4*b^2*(-a^2+1)-4*a^2*b^2)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-27/2*a*b^2/(-a^2+1)^(5/2)*ln((-2*a^2+2-2*x*a*b+2*(-
a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)+6/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b^3+9/(-a^2+1)/(-b
^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^3*b^2+15/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^2*b^2+9/(-a^2+1)/(-b^2*x^2-2*
a*b*x-a^2+1)^(1/2)*a*b^2-15/2*a^5*b^2/(-a^2+1)^(7/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a
^2+1)^(1/2))/x)-1/2/(-a^2+1)/x^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^3-3/2/(-a^2+1)/x^2/(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2)*a^2+31/2*a^4*b^3/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+75/2*a^3*b^3/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a
^2+1)^(1/2)*x+57/2*a^2*b^3/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+9/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/
2)*x*b^3*a^2+15/(-a^2+1)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x*b^3*a-5/2*a*b/(-a^2+1)^2/x/(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2)+15/2*a^3*b^3/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+13/2*a*b^3/(-a^2+1)^2/(-b^2*x^2-2*a*b*x-a^2+1)^
(1/2)*x-5/2*a^4*b/(-a^2+1)^2/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-15/2*a^3*b/(-a^2+1)^2/x/(-b^2*x^2-2*a*b*x-a^2+1)
^(1/2)-15/2*a^2*b/(-a^2+1)^2/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+15/2*a^6*b^3/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)
^(1/2)*x+45/2*a^5*b^3/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+45/2*a^4*b^3/(-a^2+1)^3/(-b^2*x^2-2*a*b*x-a^
2+1)^(1/2)*x-3*b/(-a^2+1)/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a^2-6*b/(-a^2+1)/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a

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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((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.515, size = 1184, normalized size = 5.86 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, a + 3\right )} b^{3} x^{3} +{\left (2 \, a^{2} + a - 3\right )} b^{2} x^{2}\right )} \sqrt{-a^{2} + 1} \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 2 \,{\left (a^{5} -{\left (a^{3} + 14 \, a^{2} - a - 14\right )} b^{2} x^{2} - a^{4} - 2 \, a^{3} - 5 \,{\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a^{2} + a - 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{4 \,{\left ({\left (a^{5} - 3 \, a^{4} + 2 \, a^{3} + 2 \, a^{2} - 3 \, a + 1\right )} b x^{3} +{\left (a^{6} - 4 \, a^{5} + 5 \, a^{4} - 5 \, a^{2} + 4 \, a - 1\right )} x^{2}\right )}}, -\frac{3 \,{\left ({\left (2 \, a + 3\right )} b^{3} x^{3} +{\left (2 \, a^{2} + a - 3\right )} b^{2} x^{2}\right )} \sqrt{a^{2} - 1} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) +{\left (a^{5} -{\left (a^{3} + 14 \, a^{2} - a - 14\right )} b^{2} x^{2} - a^{4} - 2 \, a^{3} - 5 \,{\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a^{2} + a - 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \,{\left ({\left (a^{5} - 3 \, a^{4} + 2 \, a^{3} + 2 \, a^{2} - 3 \, a + 1\right )} b x^{3} +{\left (a^{6} - 4 \, a^{5} + 5 \, a^{4} - 5 \, a^{2} + 4 \, a - 1\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="fricas")

[Out]

[-1/4*(3*((2*a + 3)*b^3*x^3 + (2*a^2 + a - 3)*b^2*x^2)*sqrt(-a^2 + 1)*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^
3 - a)*b*x - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) + 2*(a^5
- (a^3 + 14*a^2 - a - 14)*b^2*x^2 - a^4 - 2*a^3 - 5*(a^3 - a^2 - a + 1)*b*x + 2*a^2 + a - 1)*sqrt(-b^2*x^2 - 2
*a*b*x - a^2 + 1))/((a^5 - 3*a^4 + 2*a^3 + 2*a^2 - 3*a + 1)*b*x^3 + (a^6 - 4*a^5 + 5*a^4 - 5*a^2 + 4*a - 1)*x^
2), -1/2*(3*((2*a + 3)*b^3*x^3 + (2*a^2 + a - 3)*b^2*x^2)*sqrt(a^2 - 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 +
1)*(a*b*x + a^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) + (a^5 - (a^3 + 1
4*a^2 - a - 14)*b^2*x^2 - a^4 - 2*a^3 - 5*(a^3 - a^2 - a + 1)*b*x + 2*a^2 + a - 1)*sqrt(-b^2*x^2 - 2*a*b*x - a
^2 + 1))/((a^5 - 3*a^4 + 2*a^3 + 2*a^2 - 3*a + 1)*b*x^3 + (a^6 - 4*a^5 + 5*a^4 - 5*a^2 + 4*a - 1)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x + 1\right )^{3}}{x^{3} \left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)**3/(1-(b*x+a)**2)**(3/2)/x**3,x)

[Out]

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

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Giac [B]  time = 1.28377, size = 933, normalized size = 4.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)/x^3,x, algorithm="giac")

[Out]

-8*b^3/((a^3*abs(b) - 3*a^2*abs(b) + 3*a*abs(b) - abs(b))*((sqrt(-(b*x + a)^2 + 1)*abs(b) + b)/(b^2*x + a*b) -
1)) - 3*(2*a*b^3 + 3*b^3)*arctan(((sqrt(-(b*x + a)^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/((a
^3*abs(b) - 3*a^2*abs(b) + 3*a*abs(b) - abs(b))*sqrt(a^2 - 1)) + (2*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)^2*a^4*
b^3/(b^2*x + a*b)^2 + 2*a^4*b^3 - 5*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)*a^3*b^3/(b^2*x + a*b) + 6*(sqrt(-(b*x
+ a)^2 + 1)*abs(b) + b)^2*a^3*b^3/(b^2*x + a*b)^2 - 3*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)^3*a^3*b^3/(b^2*x + a
*b)^3 + 6*a^3*b^3 - 18*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)*a^2*b^3/(b^2*x + a*b) + 3*(sqrt(-(b*x + a)^2 + 1)*a
bs(b) + b)^2*a^2*b^3/(b^2*x + a*b)^2 - 6*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)^3*a^2*b^3/(b^2*x + a*b)^3 - a^2*b
^3 + 2*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)*a*b^3/(b^2*x + a*b) + 12*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)^2*a*b^
3/(b^2*x + a*b)^2 + 2*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)^3*a*b^3/(b^2*x + a*b)^3 - 2*(sqrt(-(b*x + a)^2 + 1)*
abs(b) + b)^2*b^3/(b^2*x + a*b)^2)/((a^5*abs(b) - 3*a^4*abs(b) + 3*a^3*abs(b) - a^2*abs(b))*((sqrt(-(b*x + a)^
2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + a - 2*(sqrt(-(b*x + a)^2 + 1)*abs(b) + b)/(b^2*x + a*b))^2)