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

Optimal. Leaf size=264 \[ -\frac{(i a+i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt{-i a-i b x+1}}+\frac{3 (-2 a+3 i) b^2 \sqrt{i a+i b x+1}}{(1+i a) (a+i)^3 \sqrt{-i a-i b x+1}}+\frac{3 (3+2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i} (a+i)^{7/2}}+\frac{(-2 a+3 i) b (i a+i b x+1)^{3/2}}{2 (1+i a) (a+i)^2 x \sqrt{-i a-i b x+1}} \]

[Out]

(3*(3*I - 2*a)*b^2*Sqrt[1 + I*a + I*b*x])/((1 + I*a)*(I + a)^3*Sqrt[1 - I*a - I*b*x]) + ((3*I - 2*a)*b*(1 + I*
a + I*b*x)^(3/2))/(2*(1 + I*a)*(I + a)^2*x*Sqrt[1 - I*a - I*b*x]) - (1 + I*a + I*b*x)^(5/2)/(2*(1 + a^2)*x^2*S
qrt[1 - I*a - I*b*x]) + (3*(3 + (2*I)*a)*b^2*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 -
I*a - I*b*x])])/(Sqrt[I - a]*(I + a)^(7/2))

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Rubi [A]  time = 0.157934, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5095, 96, 94, 93, 208} \[ -\frac{(i a+i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt{-i a-i b x+1}}+\frac{3 (-2 a+3 i) b^2 \sqrt{i a+i b x+1}}{(1+i a) (a+i)^3 \sqrt{-i a-i b x+1}}+\frac{3 (3+2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{\sqrt{-a+i} (a+i)^{7/2}}+\frac{(-2 a+3 i) b (i a+i b x+1)^{3/2}}{2 (1+i a) (a+i)^2 x \sqrt{-i a-i b x+1}} \]

Antiderivative was successfully verified.

[In]

Int[E^((3*I)*ArcTan[a + b*x])/x^3,x]

[Out]

(3*(3*I - 2*a)*b^2*Sqrt[1 + I*a + I*b*x])/((1 + I*a)*(I + a)^3*Sqrt[1 - I*a - I*b*x]) + ((3*I - 2*a)*b*(1 + I*
a + I*b*x)^(3/2))/(2*(1 + I*a)*(I + a)^2*x*Sqrt[1 - I*a - I*b*x]) - (1 + I*a + I*b*x)^(5/2)/(2*(1 + a^2)*x^2*S
qrt[1 - I*a - I*b*x]) + (3*(3 + (2*I)*a)*b^2*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sqrt[1 -
I*a - I*b*x])])/(Sqrt[I - a]*(I + a)^(7/2))

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 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 i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{(1+i a+i b x)^{3/2}}{x^3 (1-i a-i b x)^{3/2}} \, dx\\ &=-\frac{(1+i a+i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt{1-i a-i b x}}+\frac{((3 i-2 a) b) \int \frac{(1+i a+i b x)^{3/2}}{x^2 (1-i a-i b x)^{3/2}} \, dx}{2 \left (1+a^2\right )}\\ &=-\frac{(3 i-2 a) b (1+i a+i b x)^{3/2}}{2 (1-i a) \left (1+a^2\right ) x \sqrt{1-i a-i b x}}-\frac{(1+i a+i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt{1-i a-i b x}}-\frac{\left (3 (3 i-2 a) b^2\right ) \int \frac{\sqrt{1+i a+i b x}}{x (1-i a-i b x)^{3/2}} \, dx}{2 (i+a) \left (1+a^2\right )}\\ &=-\frac{3 (3 i-2 a) b^2 \sqrt{1+i a+i b x}}{(i-a) (1-i a)^3 \sqrt{1-i a-i b x}}-\frac{(3 i-2 a) b (1+i a+i b x)^{3/2}}{2 (1-i a) \left (1+a^2\right ) x \sqrt{1-i a-i b x}}-\frac{(1+i a+i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt{1-i a-i b x}}+\frac{\left (3 (3 i-2 a) b^2\right ) \int \frac{1}{x \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx}{2 (i+a)^3}\\ &=-\frac{3 (3 i-2 a) b^2 \sqrt{1+i a+i b x}}{(i-a) (1-i a)^3 \sqrt{1-i a-i b x}}-\frac{(3 i-2 a) b (1+i a+i b x)^{3/2}}{2 (1-i a) \left (1+a^2\right ) x \sqrt{1-i a-i b x}}-\frac{(1+i a+i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt{1-i a-i b x}}+\frac{\left (3 (3 i-2 a) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-i a-(-1+i a) x^2} \, dx,x,\frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}}\right )}{(i+a)^3}\\ &=-\frac{3 (3 i-2 a) b^2 \sqrt{1+i a+i b x}}{(i-a) (1-i a)^3 \sqrt{1-i a-i b x}}-\frac{(3 i-2 a) b (1+i a+i b x)^{3/2}}{2 (1-i a) \left (1+a^2\right ) x \sqrt{1-i a-i b x}}-\frac{(1+i a+i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt{1-i a-i b x}}+\frac{3 (3+2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{i+a} \sqrt{1+i a+i b x}}{\sqrt{i-a} \sqrt{1-i a-i b x}}\right )}{\sqrt{i-a} (i+a)^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.206453, size = 180, normalized size = 0.68 \[ \frac{\frac{\sqrt{i a+i b x+1} \left (a^3+i a^2-a b^2 x^2+5 i a b x+a+14 i b^2 x^2-5 b x+i\right )}{x^2 \sqrt{-i (a+b x+i)}}-\frac{6 (2 a-3 i) b^2 \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{\sqrt{-1+i a} \sqrt{1+i a}}}{2 (a+i)^3} \]

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]*(I + a + I*a^2 + a^3 - 5*b*x + (5*I)*a*b*x + (14*I)*b^2*x^2 - a*b^2*x^2))/(x^2*Sqrt[(-
I)*(I + a + b*x)]) - (6*(-3*I + 2*a)*b^2*ArcTan[Sqrt[(-I)*(I + a + b*x)]/(Sqrt[(I + a)/(-I + a)]*Sqrt[1 + I*a
+ I*b*x])])/(Sqrt[-1 + I*a]*Sqrt[1 + I*a]))/(2*(I + a)^3)

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Maple [B]  time = 0.116, size = 1955, normalized size = 7.4 \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]

9*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^3*a^2+3*I*b/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^2-5/2*I*a^
4*b/(a^2+1)^2/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+15/2*I*a^2*b/(a^2+1)^2/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+15/2*I*a^
6*b^3/(a^2+1)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-45/2*I*a^4*b^3/(a^2+1)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-31/2*
I*a^4*b^3/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+57/2*I*a^2*b^3/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-1
5/2*a^3*b/(a^2+1)^2/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-75/2*a^3*b^3/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+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-2*I*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)-21/2*I*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)*
a^3-27/2*I*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a+39*I*a^3*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+
27/2*I*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)*a-9*I/(a^2+1)/(
b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a*b^2+9*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^3*b^2+3*I*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)*a-3*I*b/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(
1/2)-6*I/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^3+6*b/(a^2+1)/x/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a+15/(a^2+1)/
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x*b^3*a+45/2*a^5*b^3/(a^2+1)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+15/2*I*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)-45/2*I*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)+1/2*I/(a^2+1)/x^2/(b^2*x^2+2*a*b*x+a^2+
1)^(1/2)*a^3-3/2*I/(a^2+1)/x^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a-30*I*a^5*b^2/(a^2+1)^3/(b^2*x^2+2*a*b*x+a^2+1)^
(1/2)+45/2*I*a^3*b^2/(a^2+1)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+15/2*I*a^7*b^2/(a^2+1)^3/(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)-3/2*b^2/(a^2+1)^2/(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^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*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)-31/2
*I*a^5*b^2/(a^2+1)^2/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+15/(a^2+1)/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*a^2*b^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)-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)+3/2/(a^2+1)/x^2/(b^2*x^2+2*a*b*x+a^2
+1)^(1/2)*a^2+45/2*a^6*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)-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)+29*a^2*b^
2/(a^2+1)^2/(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 [B]  time = 2.39899, size = 1462, normalized size = 5.54 \begin{align*} \frac{{\left (-i \, a - 14\right )} b^{3} x^{3} +{\left (-i \, a^{2} - 13 \, a - 14 i\right )} b^{2} x^{2} - 3 \,{\left ({\left (a^{3} + 3 i \, a^{2} - 3 \, a - i\right )} b x^{3} +{\left (a^{4} + 4 i \, a^{3} - 6 \, a^{2} - 4 i \, a + 1\right )} x^{2}\right )} \sqrt{\frac{{\left (4 \, a^{2} - 12 i \, a - 9\right )} b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}} \log \left (-\frac{{\left (6 \, a - 9 i\right )} b^{3} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (6 \, a - 9 i\right )} b^{2} + 3 \,{\left (a^{5} + 3 i \, a^{4} - 2 \, a^{3} + 2 i \, a^{2} - 3 \, a - i\right )} \sqrt{\frac{{\left (4 \, a^{2} - 12 i \, a - 9\right )} b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}}}{{\left (6 \, a - 9 i\right )} b^{2}}\right ) + 3 \,{\left ({\left (a^{3} + 3 i \, a^{2} - 3 \, a - i\right )} b x^{3} +{\left (a^{4} + 4 i \, a^{3} - 6 \, a^{2} - 4 i \, a + 1\right )} x^{2}\right )} \sqrt{\frac{{\left (4 \, a^{2} - 12 i \, a - 9\right )} b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}} \log \left (-\frac{{\left (6 \, a - 9 i\right )} b^{3} x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (6 \, a - 9 i\right )} b^{2} - 3 \,{\left (a^{5} + 3 i \, a^{4} - 2 \, a^{3} + 2 i \, a^{2} - 3 \, a - i\right )} \sqrt{\frac{{\left (4 \, a^{2} - 12 i \, a - 9\right )} b^{4}}{a^{8} + 6 i \, a^{7} - 14 \, a^{6} - 14 i \, a^{5} - 14 i \, a^{3} + 14 \, a^{2} + 6 i \, a - 1}}}{{\left (6 \, a - 9 i\right )} b^{2}}\right ) +{\left ({\left (-i \, a - 14\right )} b^{2} x^{2} + i \, a^{3} -{\left (5 \, a + 5 i\right )} b x - a^{2} + i \, a - 1\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (2 \, a^{3} + 6 i \, a^{2} - 6 \, a - 2 i\right )} b x^{3} +{\left (2 \, a^{4} + 8 i \, a^{3} - 12 \, a^{2} - 8 i \, a + 2\right )} x^{2}} \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]

((-I*a - 14)*b^3*x^3 + (-I*a^2 - 13*a - 14*I)*b^2*x^2 - 3*((a^3 + 3*I*a^2 - 3*a - I)*b*x^3 + (a^4 + 4*I*a^3 -
6*a^2 - 4*I*a + 1)*x^2)*sqrt((4*a^2 - 12*I*a - 9)*b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 +
6*I*a - 1))*log(-((6*a - 9*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(6*a - 9*I)*b^2 + 3*(a^5 + 3*I*a^4 -
2*a^3 + 2*I*a^2 - 3*a - I)*sqrt((4*a^2 - 12*I*a - 9)*b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^
2 + 6*I*a - 1)))/((6*a - 9*I)*b^2)) + 3*((a^3 + 3*I*a^2 - 3*a - I)*b*x^3 + (a^4 + 4*I*a^3 - 6*a^2 - 4*I*a + 1)
*x^2)*sqrt((4*a^2 - 12*I*a - 9)*b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*I*a - 1))*log(-
((6*a - 9*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(6*a - 9*I)*b^2 - 3*(a^5 + 3*I*a^4 - 2*a^3 + 2*I*a^2 -
3*a - I)*sqrt((4*a^2 - 12*I*a - 9)*b^4/(a^8 + 6*I*a^7 - 14*a^6 - 14*I*a^5 - 14*I*a^3 + 14*a^2 + 6*I*a - 1)))/(
(6*a - 9*I)*b^2)) + ((-I*a - 14)*b^2*x^2 + I*a^3 - (5*a + 5*I)*b*x - a^2 + I*a - 1)*sqrt(b^2*x^2 + 2*a*b*x + a
^2 + 1))/((2*a^3 + 6*I*a^2 - 6*a - 2*I)*b*x^3 + (2*a^4 + 8*I*a^3 - 12*a^2 - 8*I*a + 2)*x^2)

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \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]

undef