3.861 \(\int e^{-3 \tanh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=167 \[ -\frac{\left (6 a^2+18 a+11\right ) \sqrt{a+b x+1} (-a-b x+1)^{3/2}}{6 b^3}-\frac{\left (6 a^2+18 a+11\right ) \sqrt{a+b x+1} \sqrt{-a-b x+1}}{2 b^3}-\frac{\left (6 a^2+18 a+11\right ) \sin ^{-1}(a+b x)}{2 b^3}-\frac{\sqrt{a+b x+1} (-a-b x+1)^{5/2}}{3 b^3}-\frac{(a+1)^2 (-a-b x+1)^{5/2}}{b^3 \sqrt{a+b x+1}} \]

[Out]

-(((1 + a)^2*(1 - a - b*x)^(5/2))/(b^3*Sqrt[1 + a + b*x])) - ((11 + 18*a + 6*a^2)*Sqrt[1 - a - b*x]*Sqrt[1 + a
+ b*x])/(2*b^3) - ((11 + 18*a + 6*a^2)*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/(6*b^3) - ((1 - a - b*x)^(5/2)*
Sqrt[1 + a + b*x])/(3*b^3) - ((11 + 18*a + 6*a^2)*ArcSin[a + b*x])/(2*b^3)

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Rubi [A]  time = 0.200624, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6163, 89, 80, 50, 53, 619, 216} \[ -\frac{\left (6 a^2+18 a+11\right ) \sqrt{a+b x+1} (-a-b x+1)^{3/2}}{6 b^3}-\frac{\left (6 a^2+18 a+11\right ) \sqrt{a+b x+1} \sqrt{-a-b x+1}}{2 b^3}-\frac{\left (6 a^2+18 a+11\right ) \sin ^{-1}(a+b x)}{2 b^3}-\frac{\sqrt{a+b x+1} (-a-b x+1)^{5/2}}{3 b^3}-\frac{(a+1)^2 (-a-b x+1)^{5/2}}{b^3 \sqrt{a+b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/E^(3*ArcTanh[a + b*x]),x]

[Out]

-(((1 + a)^2*(1 - a - b*x)^(5/2))/(b^3*Sqrt[1 + a + b*x])) - ((11 + 18*a + 6*a^2)*Sqrt[1 - a - b*x]*Sqrt[1 + a
+ b*x])/(2*b^3) - ((11 + 18*a + 6*a^2)*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/(6*b^3) - ((1 - a - b*x)^(5/2)*
Sqrt[1 + a + b*x])/(3*b^3) - ((11 + 18*a + 6*a^2)*ArcSin[a + b*x])/(2*b^3)

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a+b x)} x^2 \, dx &=\int \frac{x^2 (1-a-b x)^{3/2}}{(1+a+b x)^{3/2}} \, dx\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}+\frac{\int \frac{(1-a-b x)^{3/2} \left (-(1+a) (3+2 a) b+b^2 x\right )}{\sqrt{1+a+b x}} \, dx}{b^3}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}-\frac{\left (11+18 a+6 a^2\right ) \int \frac{(1-a-b x)^{3/2}}{\sqrt{1+a+b x}} \, dx}{3 b^2}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{\left (11+18 a+6 a^2\right ) (1-a-b x)^{3/2} \sqrt{1+a+b x}}{6 b^3}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}-\frac{\left (11+18 a+6 a^2\right ) \int \frac{\sqrt{1-a-b x}}{\sqrt{1+a+b x}} \, dx}{2 b^2}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{\left (11+18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}-\frac{\left (11+18 a+6 a^2\right ) (1-a-b x)^{3/2} \sqrt{1+a+b x}}{6 b^3}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}-\frac{\left (11+18 a+6 a^2\right ) \int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{2 b^2}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{\left (11+18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}-\frac{\left (11+18 a+6 a^2\right ) (1-a-b x)^{3/2} \sqrt{1+a+b x}}{6 b^3}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}-\frac{\left (11+18 a+6 a^2\right ) \int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{2 b^2}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{\left (11+18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}-\frac{\left (11+18 a+6 a^2\right ) (1-a-b x)^{3/2} \sqrt{1+a+b x}}{6 b^3}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}+\frac{\left (11+18 a+6 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{4 b^4}\\ &=-\frac{(1+a)^2 (1-a-b x)^{5/2}}{b^3 \sqrt{1+a+b x}}-\frac{\left (11+18 a+6 a^2\right ) \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 b^3}-\frac{\left (11+18 a+6 a^2\right ) (1-a-b x)^{3/2} \sqrt{1+a+b x}}{6 b^3}-\frac{(1-a-b x)^{5/2} \sqrt{1+a+b x}}{3 b^3}-\frac{\left (11+18 a+6 a^2\right ) \sin ^{-1}(a+b x)}{2 b^3}\\ \end{align*}

Mathematica [A]  time = 0.169403, size = 190, normalized size = 1.14 \[ -\frac{\sqrt{-b} \left (a^3 (2 b x+51)+a^2 (69 b x+50)+2 a^4+a \left (2 b^3 x^3+9 b^2 x^2+106 b x-51\right )+2 b^4 x^4-9 b^3 x^3+26 b^2 x^2+33 b x-52\right )+6 \left (6 a^2+18 a+11\right ) \sqrt{b} \sqrt{-a^2-2 a b x-b^2 x^2+1} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{6 (-b)^{7/2} \sqrt{-(a+b x-1) (a+b x+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/E^(3*ArcTanh[a + b*x]),x]

[Out]

-(Sqrt[-b]*(-52 + 2*a^4 + 33*b*x + 26*b^2*x^2 - 9*b^3*x^3 + 2*b^4*x^4 + a^3*(51 + 2*b*x) + a^2*(50 + 69*b*x) +
a*(-51 + 106*b*x + 9*b^2*x^2 + 2*b^3*x^3)) + 6*(11 + 18*a + 6*a^2)*Sqrt[b]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*
ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/(6*(-b)^(7/2)*Sqrt[-((-1 + a + b*x)*(1 + a + b*x))])

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Maple [B]  time = 0.048, size = 830, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^3-1/b^6/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))
^(5/2)-2/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)*a^2-6/b^3*a*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)
-4/b^5/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-11/2/b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^
(1/2)*x-9/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a^2-11/2/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/
2)*a-11/2/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))-6/b^5
*a/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-9/b^2*a*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)
*x-9/b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*a-3/b^2/(b
^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))*a^2-1/b^6/(x+1/b+a/b)
^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)*a^2-2/b^6/(x+1/b+a/b)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/
2)*a-2/b^5/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)*a^2-3/b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)
/b))^(1/2)*x*a^2-11/3/b^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)

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Maxima [C]  time = 1.49084, size = 840, normalized size = 5.03 \begin{align*} \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}} a^{2}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} + 2 \, b^{4} x + 2 \, a b^{3} + b^{3}} + \frac{2 \,{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}} a}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} + 2 \, b^{4} x + 2 \, a b^{3} + b^{3}} - \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}} a}{b^{4} x + a b^{3} + b^{3}} - \frac{6 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{b^{4} x + a b^{3} + b^{3}} + \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3} + 2 \, b^{4} x + 2 \, a b^{3} + b^{3}} - \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{b^{4} x + a b^{3} + b^{3}} - \frac{12 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{4} x + a b^{3} + b^{3}} - \frac{3 \, a^{2} \arcsin \left (b x + a\right )}{b^{3}} - \frac{6 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4} x + a b^{3} + b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} x}{2 \, b^{2}} - \frac{9 \, a \arcsin \left (b x + a\right )}{b^{3}} + \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{3 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3} a}{2 \, b^{3}} - \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{3}} - \frac{i \, \arcsin \left (b x + a + 2\right )}{2 \, b^{3}} - \frac{6 \, \arcsin \left (b x + a\right )}{b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 4 \, b x + 4 \, a + 3}}{b^{3}} - \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a^2/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 + 2*b^4*x + 2*a*b^3 + b^3) + 2*(-b^2*x
^2 - 2*a*b*x - a^2 + 1)^(3/2)*a/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 + 2*b^4*x + 2*a*b^3 + b^3) - (-b^2*x^2 - 2*a*b*
x - a^2 + 1)^(3/2)*a/(b^4*x + a*b^3 + b^3) - 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2/(b^4*x + a*b^3 + b^3) +
(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3 + 2*b^4*x + 2*a*b^3 + b^3) - (-b^2*x^2 - 2
*a*b*x - a^2 + 1)^(3/2)/(b^4*x + a*b^3 + b^3) - 12*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/(b^4*x + a*b^3 + b^3)
- 3*a^2*arcsin(b*x + a)/b^3 - 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/(b^4*x + a*b^3 + b^3) + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2 + 4*b*x + 4*a + 3)*x/b^2 - 9*a*arcsin(b*x + a)/b^3 + 1/3*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/b^
3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 4*b*x + 4*a + 3)*a/b^3 - 3*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/b^3 - 1
/2*I*arcsin(b*x + a + 2)/b^3 - 6*arcsin(b*x + a)/b^3 + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 4*b*x + 4*a + 3)/b^3 - 3
*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b^3

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Fricas [A]  time = 2.01453, size = 375, normalized size = 2.25 \begin{align*} \frac{3 \,{\left (6 \, a^{3} +{\left (6 \, a^{2} + 18 \, a + 11\right )} b x + 24 \, a^{2} + 29 \, a + 11\right )} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (2 \, b^{3} x^{3} - 7 \, b^{2} x^{2} + 2 \, a^{3} +{\left (16 \, a + 19\right )} b x + 53 \, a^{2} + 103 \, a + 52\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \,{\left (b^{4} x +{\left (a + 1\right )} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(6*a^3 + (6*a^2 + 18*a + 11)*b*x + 24*a^2 + 29*a + 11)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x +
a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) - (2*b^3*x^3 - 7*b^2*x^2 + 2*a^3 + (16*a + 19)*b*x + 53*a^2 + 103*a + 52)*s
qrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^4*x + (a + 1)*b^3)

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

[Out]

Timed out

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Giac [A]  time = 1.19749, size = 224, normalized size = 1.34 \begin{align*} -\frac{1}{6} \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (x{\left (\frac{2 \, x}{b} - \frac{2 \, a b^{6} + 9 \, b^{6}}{b^{8}}\right )} + \frac{2 \, a^{2} b^{5} + 27 \, a b^{5} + 28 \, b^{5}}{b^{8}}\right )} + \frac{{\left (6 \, a^{2} + 18 \, a + 11\right )} \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{2 \, b^{2}{\left | b \right |}} + \frac{8 \,{\left (a^{2} + 2 \, a + 1\right )}}{b^{2}{\left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b}{b^{2} x + a b} + 1\right )}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x*(2*x/b - (2*a*b^6 + 9*b^6)/b^8) + (2*a^2*b^5 + 27*a*b^5 + 28*b^5)/b
^8) + 1/2*(6*a^2 + 18*a + 11)*arcsin(-b*x - a)*sgn(b)/(b^2*abs(b)) + 8*(a^2 + 2*a + 1)/(b^2*((sqrt(-b^2*x^2 -
2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) + 1)*abs(b))