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

Optimal. Leaf size=249 \[ -\frac{i \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-22 i a^2-2 (11-10 i a) b x+54 a+29 i\right )}{8 b^4}+\frac{3 \left (8 i a^3-36 a^2-44 i a+17\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}-\frac{3 \left (-8 a^3-36 i a^2+44 a+17 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}-\frac{9 x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b^2}-\frac{2 i x^3 (i a+i b x+1)^{3/2}}{b \sqrt{-i a-i b x+1}} \]

[Out]

(3*(17 - (44*I)*a - 36*a^2 + (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) - ((2*I)*x^3*(1 +
I*a + I*b*x)^(3/2))/(b*Sqrt[1 - I*a - I*b*x]) - (9*x^2*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(4*b^2)
- ((I/8)*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(29*I + 54*a - (22*I)*a^2 - 2*(11 - (10*I)*a)*b*x))/b^
4 - (3*(17*I + 44*a - (36*I)*a^2 - 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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Rubi [A]  time = 0.242333, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5095, 97, 153, 147, 50, 53, 619, 215} \[ -\frac{i \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2} \left (-22 i a^2-2 (11-10 i a) b x+54 a+29 i\right )}{8 b^4}+\frac{3 \left (8 i a^3-36 a^2-44 i a+17\right ) \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{8 b^4}-\frac{3 \left (-8 a^3-36 i a^2+44 a+17 i\right ) \sinh ^{-1}(a+b x)}{8 b^4}-\frac{9 x^2 \sqrt{-i a-i b x+1} (i a+i b x+1)^{3/2}}{4 b^2}-\frac{2 i x^3 (i a+i b x+1)^{3/2}}{b \sqrt{-i a-i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a + b*x])*x^3,x]

[Out]

(3*(17 - (44*I)*a - 36*a^2 + (8*I)*a^3)*Sqrt[1 - I*a - I*b*x]*Sqrt[1 + I*a + I*b*x])/(8*b^4) - ((2*I)*x^3*(1 +
I*a + I*b*x)^(3/2))/(b*Sqrt[1 - I*a - I*b*x]) - (9*x^2*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2))/(4*b^2)
- ((I/8)*Sqrt[1 - I*a - I*b*x]*(1 + I*a + I*b*x)^(3/2)*(29*I + 54*a - (22*I)*a^2 - 2*(11 - (10*I)*a)*b*x))/b^
4 - (3*(17*I + 44*a - (36*I)*a^2 - 8*a^3)*ArcSinh[a + b*x])/(8*b^4)

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 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^3 \, dx &=\int \frac{x^3 (1+i a+i b x)^{3/2}}{(1-i a-i b x)^{3/2}} \, dx\\ &=-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}+\frac{(2 i) \int \frac{x^2 \sqrt{1+i a+i b x} \left (3 (1+i a)+\frac{9 i b x}{2}\right )}{\sqrt{1-i a-i b x}} \, dx}{b}\\ &=-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}+\frac{i \int \frac{x \sqrt{1+i a+i b x} \left (-9 i \left (1+a^2\right ) b+\frac{3}{2} (11-10 i a) b^2 x\right )}{\sqrt{1-i a-i b x}} \, dx}{2 b^3}\\ &=-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{i \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (29 i+54 a-22 i a^2-2 (11-10 i a) b x\right )}{8 b^4}-\frac{\left (3 \left (17 i+44 a-36 i a^2-8 a^3\right )\right ) \int \frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 i a-36 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{i \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (29 i+54 a-22 i a^2-2 (11-10 i a) b x\right )}{8 b^4}-\frac{\left (3 \left (17 i+44 a-36 i a^2-8 a^3\right )\right ) \int \frac{1}{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 i a-36 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{i \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (29 i+54 a-22 i a^2-2 (11-10 i a) b x\right )}{8 b^4}-\frac{\left (3 \left (17 i+44 a-36 i a^2-8 a^3\right )\right ) \int \frac{1}{\sqrt{(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx}{8 b^3}\\ &=\frac{3 \left (17-44 i a-36 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{i \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (29 i+54 a-22 i a^2-2 (11-10 i a) b x\right )}{8 b^4}-\frac{\left (3 \left (17 i+44 a-36 i a^2-8 a^3\right )\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 )}{16 b^5}\\ &=\frac{3 \left (17-44 i a-36 a^2+8 i a^3\right ) \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{8 b^4}-\frac{2 i x^3 (1+i a+i b x)^{3/2}}{b \sqrt{1-i a-i b x}}-\frac{9 x^2 \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2}}{4 b^2}-\frac{i \sqrt{1-i a-i b x} (1+i a+i b x)^{3/2} \left (29 i+54 a-22 i a^2-2 (11-10 i a) b x\right )}{8 b^4}-\frac{3 \left (17 i+44 a-36 i a^2-8 a^3\right ) \sinh ^{-1}(a+b x)}{8 b^4}\\ \end{align*}

Mathematica [A]  time = 0.261418, size = 201, normalized size = 0.81 \[ \frac{\sqrt{i a+i b x+1} \left (a^2 (-233+22 i b x)+2 a^4+78 i a^3-i a \left (10 b^2 x^2-54 i b x+237\right )-2 b^4 x^4+6 i b^3 x^3+11 b^2 x^2-29 i b x+80\right )}{8 b^4 \sqrt{-i (a+b x+i)}}+\frac{3 \sqrt [4]{-1} \left (8 a^3+36 i a^2-44 a-17 i\right ) \sqrt{-i b} \sinh ^{-1}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{b} \sqrt{-i (a+b x+i)}}{\sqrt{-i b}}\right )}{4 b^{9/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + I*a + I*b*x]*(80 + (78*I)*a^3 + 2*a^4 - (29*I)*b*x + 11*b^2*x^2 + (6*I)*b^3*x^3 - 2*b^4*x^4 + a^2*(-
233 + (22*I)*b*x) - I*a*(237 - (54*I)*b*x + 10*b^2*x^2)))/(8*b^4*Sqrt[(-I)*(I + a + b*x)]) + (3*(-1)^(1/4)*(-1
7*I - 44*a + (36*I)*a^2 + 8*a^3)*Sqrt[(-I)*b]*ArcSinh[((1/2 + I/2)*Sqrt[b]*Sqrt[(-I)*(I + a + b*x)])/Sqrt[(-I)
*b]])/(4*b^(9/2))

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Maple [B]  time = 0.129, size = 711, normalized size = 2.9 \begin{align*} \text{result too large to display} \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^3,x)

[Out]

1/2*a^2/b^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-19/2*a^4/b^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+53/2*a/b^3*x/(b^2*x^2+2*a
*b*x+a^2+1)^(1/2)-25/2*a^3/b^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-33/2*a/b^3*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)+5*x^2/b^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2/b*a*x^3/(b^2*x^2+2*a*b*x+a^2+1)^
(1/2)-3/2*x^2/b^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^2+3*a^3/b^3*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)+17/8*I/b*x^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/4*I/b^4*a^5/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-2
65/8*I/b^3*a^2*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-155/8*I/b^4*a^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-x^4/(b^2*x^2+2*a*
b*x+a^2+1)^(1/2)-1/4*I*b*x^5/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/4*I/b^3*a^4*x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+27/2*
I/b^3*a^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)-53/8*I/b^2*a*x^2/(b^2*x^2+2*a*
b*x+a^2+1)^(1/2)-157/8*I/b^4*a/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-51/8*I/b^3*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)-1/4*I*a*x^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+51/8*I/b^3*x/(b^2*x^2+2*a*b*x+a^2+1)
^(1/2)+10/b^4/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07708, size = 633, normalized size = 2.54 \begin{align*} \frac{15 i \, a^{5} - 495 \, a^{4} - 1664 i \, a^{3} +{\left (15 i \, a^{4} - 480 \, a^{3} - 1184 i \, a^{2} + 968 \, a + 256 i\right )} b x + 2152 \, a^{2} -{\left (192 \, a^{4} + 1056 i \, a^{3} +{\left (192 \, a^{3} + 864 i \, a^{2} - 1056 \, a - 408 i\right )} b x - 1920 \, a^{2} - 1464 i \, a + 408\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (-16 i \, b^{4} x^{4} - 48 \, b^{3} x^{3} +{\left (80 \, a + 88 i\right )} b^{2} x^{2} + 16 i \, a^{4} - 624 \, a^{3} - 8 \,{\left (22 \, a^{2} + 54 i \, a - 29\right )} b x - 1864 i \, a^{2} + 1896 \, a + 640 i\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1224 i \, a - 256}{64 \, b^{5} x +{\left (64 \, a + 64 i\right )} 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^3,x, algorithm="fricas")

[Out]

(15*I*a^5 - 495*a^4 - 1664*I*a^3 + (15*I*a^4 - 480*a^3 - 1184*I*a^2 + 968*a + 256*I)*b*x + 2152*a^2 - (192*a^4
+ 1056*I*a^3 + (192*a^3 + 864*I*a^2 - 1056*a - 408*I)*b*x - 1920*a^2 - 1464*I*a + 408)*log(-b*x - a + sqrt(b^
2*x^2 + 2*a*b*x + a^2 + 1)) + (-16*I*b^4*x^4 - 48*b^3*x^3 + (80*a + 88*I)*b^2*x^2 + 16*I*a^4 - 624*a^3 - 8*(22
*a^2 + 54*I*a - 29)*b*x - 1864*I*a^2 + 1896*a + 640*I)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 1224*I*a - 256)/(64
*b^5*x + (64*a + 64*I)*b^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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**3,x)

[Out]

Integral(x**3*(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 = 1.18536, size = 446, normalized size = 1.79 \begin{align*} -\frac{1}{8} \, \sqrt{{\left (b x + a\right )}^{2} + 1}{\left ({\left (2 \,{\left (\frac{i x}{b} - \frac{a b^{11} i - 4 \, b^{11}}{b^{13}}\right )} x + \frac{2 \, a^{2} b^{10} i - 20 \, a b^{10} - 19 \, b^{10} i}{b^{13}}\right )} x - \frac{2 \, a^{3} b^{9} i - 44 \, a^{2} b^{9} - 93 \, a b^{9} i + 48 \, b^{9}}{b^{13}}\right )} - \frac{{\left (8 \, a^{3} + 36 \, a^{2} i - 44 \, a - 17 \, i\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 )}{8 \, b^{3}{\left | b \right |}} - \frac{{\left (8 \, a^{3}{\left | b \right |} + 36 \, a^{2} i{\left | b \right |} - 44 \, a{\left | b \right |} - 17 \, i{\left | b \right |}\right )} \log \left (96 \, b^{4}\right )}{4 \, b^{5}} \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^3,x, algorithm="giac")

[Out]

-1/8*sqrt((b*x + a)^2 + 1)*((2*(i*x/b - (a*b^11*i - 4*b^11)/b^13)*x + (2*a^2*b^10*i - 20*a*b^10 - 19*b^10*i)/b
^13)*x - (2*a^3*b^9*i - 44*a^2*b^9 - 93*a*b^9*i + 48*b^9)/b^13) - 1/8*(8*a^3 + 36*a^2*i - 44*a - 17*i)*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) - sqr
t((b*x + a)^2 + 1))*a*i*abs(b) - a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/(b^3*abs(b)) - 1/4*(8*a^3*ab
s(b) + 36*a^2*i*abs(b) - 44*a*abs(b) - 17*i*abs(b))*log(96*b^4)/b^5