3.182 \(\int e^{3 i \tan ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=227 \[ \frac{\left (-6 i a^2+18 a+11 i\right ) \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{6 b^3}+\frac{\left (-6 i a^2+18 a+11 i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 b^3}+\frac{\left (-6 a^2-18 i a+11\right ) \sinh ^{-1}(a+b x)}{2 b^3}+\frac{i \sqrt{-i a-i b x+1} (i a+i b x+1)^{5/2}}{3 b^3}-\frac{i (a+i)^2 (i a+i b x+1)^{5/2}}{b^3 \sqrt{-i a-i b x+1}} \]

[Out]

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

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Rubi [A]  time = 0.169239, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5095, 89, 80, 50, 53, 619, 215} \[ \frac{\left (-6 i a^2+18 a+11 i\right ) \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{6 b^3}+\frac{\left (-6 i a^2+18 a+11 i\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{2 b^3}+\frac{\left (-6 a^2-18 i a+11\right ) \sinh ^{-1}(a+b x)}{2 b^3}+\frac{i \sqrt{-i a-i b x+1} (i a+i b x+1)^{5/2}}{3 b^3}-\frac{i (a+i)^2 (i a+i b x+1)^{5/2}}{b^3 \sqrt{-i a-i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a + b*x])*x^2,x]

[Out]

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

Rule 5095

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1 -
I*a*c - I*b*c*x)^((I*n)/2))/(1 + I*a*c + I*b*c*x)^((I*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 215

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

Rubi steps

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

Mathematica [A]  time = 0.236446, size = 160, normalized size = 0.7 \[ \frac{\sqrt{i a+i b x+1} \left (-2 a^3-53 i a^2+a (103-16 i b x)-2 b^3 x^3+7 i b^2 x^2+19 b x+52 i\right )}{6 b^3 \sqrt{-i (a+b x+i)}}+\frac{(-1)^{3/4} \left (6 a^2+18 i a-11\right ) \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{\sqrt{-i b} b^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((3*I)*ArcTan[a + b*x])*x^2,x]

[Out]

(Sqrt[1 + I*a + I*b*x]*(52*I - (53*I)*a^2 - 2*a^3 + 19*b*x + (7*I)*b^2*x^2 - 2*b^3*x^3 + a*(103 - (16*I)*b*x))
)/(6*b^3*Sqrt[(-I)*(I + a + b*x)]) + ((-1)^(3/4)*(-11 + (18*I)*a + 6*a^2)*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-
I)*(I + a + b*x)])/Sqrt[(-I)*b]])/(Sqrt[(-I)*b]*b^(5/2))

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Maple [B]  time = 0.12, size = 519, normalized size = 2.3 \begin{align*}{\frac{17\,a}{2\,{b}^{3}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{17\,{a}^{3}}{2\,{b}^{3}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{{\frac{i}{3}}a{x}^{3}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{\frac{9\,ia}{{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{{\frac{26\,i}{3}}}{{b}^{3}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{11}{2\,{b}^{2}}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{11\,x}{2\,{b}^{2}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{\frac{{\frac{i}{3}}{a}^{3}x}{{b}^{2}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{{\frac{i}{3}}b{x}^{4}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{{\frac{53\,i}{3}}ax}{{b}^{2}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{{\frac{25\,i}{3}}{a}^{2}}{{b}^{3}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{\frac{{\frac{i}{3}}{a}^{4}}{{b}^{3}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{{\frac{13\,i}{3}}{x}^{2}}{b}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-3\,{\frac{{a}^{2}}{{b}^{2}\sqrt{{b}^{2}}}\ln \left ({\frac{{b}^{2}x+ab}{\sqrt{{b}^{2}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) }+{\frac{3\,a{x}^{2}}{2\,b}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}-{\frac{3\,{x}^{3}}{2}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}}+{\frac{23\,{a}^{2}x}{2\,{b}^{2}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/2*a/b^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+17/2*a^3/b^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/3*I*a*x^3/(b^2*x^2+2*a*b
*x+a^2+1)^(1/2)-9*I/b^2*a*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+26/3*I/b^3/(b^
2*x^2+2*a*b*x+a^2+1)^(1/2)+11/2/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-11/2
*x/b^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/3*I/b^2*a^3*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/3*I*b*x^4/(b^2*x^2+2*a*b*
x+a^2+1)^(1/2)+53/3*I/b^2*a*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+25/3*I/b^3*a^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/3*I
/b^3*a^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+13/3*I/b*x^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3*a^2/b^2*ln((b^2*x+a*b)/(b^
2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)+3/2/b*a*x^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/2*x^3/(b^2*x^2
+2*a*b*x+a^2+1)^(1/2)+23/2*a^2/b^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01836, size = 489, normalized size = 2.15 \begin{align*} \frac{-7 i \, a^{4} + 166 \, a^{3} +{\left (-7 i \, a^{3} + 159 \, a^{2} + 249 i \, a - 96\right )} b x + 408 i \, a^{2} +{\left (72 \, a^{3} + 12 \,{\left (6 \, a^{2} + 18 i \, a - 11\right )} b x + 288 i \, a^{2} - 348 \, a - 132 i\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (-8 i \, b^{3} x^{3} - 28 \, b^{2} x^{2} - 8 i \, a^{3} +{\left (64 \, a + 76 i\right )} b x + 212 \, a^{2} + 412 i \, a - 208\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 345 \, a - 96 i}{24 \, b^{4} x +{\left (24 \, a + 24 i\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-7*I*a^4 + 166*a^3 + (-7*I*a^3 + 159*a^2 + 249*I*a - 96)*b*x + 408*I*a^2 + (72*a^3 + 12*(6*a^2 + 18*I*a - 11)
*b*x + 288*I*a^2 - 348*a - 132*I)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (-8*I*b^3*x^3 - 28*b^2*x
^2 - 8*I*a^3 + (64*a + 76*I)*b*x + 212*a^2 + 412*I*a - 208)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 345*a - 96*I)/
(24*b^4*x + (24*a + 24*I)*b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16247, size = 375, normalized size = 1.65 \begin{align*} -\frac{1}{6} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (\frac{2 \, i x}{b} - \frac{2 \, a b^{6} i - 9 \, b^{6}}{b^{8}}\right )} x + \frac{2 \, a^{2} b^{5} i - 27 \, a b^{5} - 28 \, b^{5} i}{b^{8}}\right )} + \frac{{\left (6 \, a^{2} + 18 \, a i - 11\right )} \log \left (3 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} a b + a^{3} b + 2 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{2} b i + 2 \, a^{2} b i +{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}^{3}{\left | b \right |} + 3 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a^{2}{\left | b \right |} + 4 \,{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )} a i{\left | b \right |} - a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{6 \, b^{2}{\left | b \right |}} + \frac{{\left (6 \, a^{2}{\left | b \right |} + 18 \, a i{\left | b \right |} - 11 \,{\left | b \right |}\right )} \log \left (8 \, b^{3}\right )}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt((b*x + a)^2 + 1)*((2*i*x/b - (2*a*b^6*i - 9*b^6)/b^8)*x + (2*a^2*b^5*i - 27*a*b^5 - 28*b^5*i)/b^8) +
1/6*(6*a^2 + 18*a*i - 11)*log(3*(x*abs(b) - sqrt((b*x + a)^2 + 1))^2*a*b + a^3*b + 2*(x*abs(b) - sqrt((b*x +
a)^2 + 1))^2*b*i + 2*a^2*b*i + (x*abs(b) - sqrt((b*x + a)^2 + 1))^3*abs(b) + 3*(x*abs(b) - sqrt((b*x + a)^2 +
1))*a^2*abs(b) + 4*(x*abs(b) - sqrt((b*x + a)^2 + 1))*a*i*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*ab
s(b))/(b^2*abs(b)) + 1/3*(6*a^2*abs(b) + 18*a*i*abs(b) - 11*abs(b))*log(8*b^3)/b^4