3.214 \(\int \frac{e^{-3 i \tan ^{-1}(a+b x)}}{x^3} \, dx\)
Optimal. Leaf size=264 \[ -\frac{(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt{i a+i b x+1}}-\frac{3 (2 a+3 i) b^2 \sqrt{-i a-i b x+1}}{(1+i a)^3 (a+i) \sqrt{i a+i b x+1}}+\frac{3 (3-2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{7/2} \sqrt{a+i}}+\frac{(3-2 i a) b (-i a-i b x+1)^{3/2}}{2 (-a+i)^2 (a+i) x \sqrt{i a+i b x+1}} \]
[Out]
(-3*(3*I + 2*a)*b^2*Sqrt[1 - I*a - I*b*x])/((1 + I*a)^3*(I + a)*Sqrt[1 + I*a + I*b*x]) + ((3 - (2*I)*a)*b*(1 -
I*a - I*b*x)^(3/2))/(2*(I - a)^2*(I + a)*x*Sqrt[1 + I*a + I*b*x]) - (1 - I*a - I*b*x)^(5/2)/(2*(1 + a^2)*x^2*
Sqrt[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])])/((I - a)^(7/2)*Sqrt[I + a])
________________________________________________________________________________________
Rubi [A] time = 0.160294, 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} \[ -\frac{(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt{i a+i b x+1}}-\frac{3 (2 a+3 i) b^2 \sqrt{-i a-i b x+1}}{(1+i a)^3 (a+i) \sqrt{i a+i b x+1}}+\frac{3 (3-2 i a) b^2 \tanh ^{-1}\left (\frac{\sqrt{a+i} \sqrt{i a+i b x+1}}{\sqrt{-a+i} \sqrt{-i a-i b x+1}}\right )}{(-a+i)^{7/2} \sqrt{a+i}}+\frac{(3-2 i a) b (-i a-i b x+1)^{3/2}}{2 (-a+i)^2 (a+i) x \sqrt{i a+i b x+1}} \]
Antiderivative was successfully verified.
[In]
Int[1/(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)^3*(I + a)*Sqrt[1 + I*a + I*b*x]) + ((3 - (2*I)*a)*b*(1 -
I*a - I*b*x)^(3/2))/(2*(I - a)^2*(I + a)*x*Sqrt[1 + I*a + I*b*x]) - (1 - I*a - I*b*x)^(5/2)/(2*(1 + a^2)*x^2*
Sqrt[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])])/((I - a)^(7/2)*Sqrt[I + a])
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{(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) 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)^2 (i+a)}\\ &=-\frac{3 (3-2 i a) b^2 \sqrt{1-i a-i b x}}{(i-a)^3 (i+a) \sqrt{1+i a+i b x}}+\frac{(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) 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 (3-2 i a) b^2 \sqrt{1-i a-i b x}}{(i-a)^3 (i+a) \sqrt{1+i a+i b x}}+\frac{(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) 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 (3-2 i a) b^2 \sqrt{1-i a-i b x}}{(i-a)^3 (i+a) \sqrt{1+i a+i b x}}+\frac{(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) 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 )}{(i-a)^{7/2} \sqrt{i+a}}\\ \end{align*}
Mathematica [A] time = 0.222158, size = 184, normalized size = 0.7 \[ \frac{\frac{\sqrt{-i (a+b x+i)} \left (a^3-i a^2-a b^2 x^2-5 i a b x+a-14 i b^2 x^2-5 b x-i\right )}{x^2 \sqrt{i a+i b x+1}}+\frac{6 i \sqrt{\frac{a+i}{a-i}} (2 a+3 i) b^2 \tan ^{-1}\left (\frac{\sqrt{-i (a+b x+i)}}{\sqrt{\frac{a+i}{a-i}} \sqrt{i a+i b x+1}}\right )}{a+i}}{2 (a-i)^3} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x^3),x]
[Out]
((Sqrt[(-I)*(I + a + 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*Sqr
t[1 + I*a + I*b*x]) + ((6*I)*Sqrt[(I + a)/(-I + a)]*(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])])/(I + a))/(2*(-I + a)^3)
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Maple [B] time = 0.135, size = 3042, normalized size = 11.5 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^3,x)
[Out]
9/2*I/(I-a)^4*b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x-27/2*I/(I-a)^4*b^3*a^2/(a^2+1)*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)-9*I/(I-a)^4*b^3*a^4/(a^2+1)*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)+1/2*I/(I-a)^3*a*b/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)-3/4*I/(
I-a)^3*a^3*b^3/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/4*I/(I-a)^3*a^3*b^3/(a^2+1)^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*I/(I-a)^3*a^5*b^3/(a^2+1)^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)-1/2*I/(I-a)^3*a*b^3/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)*x-3/4*I/(
I-a)^3*a*b^3/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-3/4*I/(I-a)^3*a*b^3/(a^2+1)^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)-3/4*I/(I-a)^3*b^3/(a^2+1)*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/4*I
/(I-a)^3*b^3/(a^2+1)*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)-3/2*I/(I-a)^3*b^3
/(a^2+1)*a^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)-9/2*I/(I-a)^4*b^3*a^2/(a^2+
1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-3/(I-a)^4/(x-I/b+a/b)^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)+3/(I-a)
^4*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)-2*I/(I-a)^5*b^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-6*I/(I-a)^5*b
^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-3/(I-a)^5*b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*x-3/(I-a)^5*b^2*((x
-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a-3/(I-a)^5*b^3*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b
^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)+2*I/(I-a)^5*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)-1/(I-a)^3
/b/(x-I/b+a/b)^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)-3/(I-a)^3*b^3*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*
b)^(1/2)*x-3/(I-a)^3*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a-2*I/(I-a)^3/(x-I/b+a/b)^2*((x-(I-a)/b)^
2*b^2+2*I*(x-(I-a)/b)*b)^(5/2)+2*I/(I-a)^3*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(3/2)-9/4*I/(I-a)^3*b^2/(
a^2+1)*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-I/(I-a)^3*a^2*b^2/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-3*b^3/(I-a)
^3*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)+9/2*I/(I-a)^4
*b^2*((x-(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2)*a+9/2*I/(I-a)^4*b^3*ln((I*b+(x-(I-a)/b)*b^2)/(b^2)^(1/2)+((x-
(I-a)/b)^2*b^2+2*I*(x-(I-a)/b)*b)^(1/2))/(b^2)^(1/2)+1/2*I/(I-a)^3/(a^2+1)/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(5/2)-1
/2*I/(I-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-3/2*I/(I-a)^3*b^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)
+3/2*I/(I-a)^3*b^2/(a^2+1)^(1/2)*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)-9*I/(I-
a)^5*b^2*a^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+6*I/(I-a)^5*b^2*(a^2+1)^(1/2)*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*I/(I-a)^3*b^2/(a^2+1)^(1/2)*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^2+9*I/(I-a)^4*b^2*a/(a^2+1)^(1/2)*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*I/(I-a)^4*b^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)*x+3*I/(I-a)^4*b/(a^2+1)/x*(b^2*x^2+
2*a*b*x+a^2+1)^(5/2)+6*I/(I-a)^5*b^2*(a^2+1)^(1/2)*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^2-9/2*I/(I-a)^4*b^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9/2*I/(I-a)^4*b^3/(a^2+1)*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)-6*I/(I-a)^4*b^2*a/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(
3/2)-27/2*I/(I-a)^4*b^2*a^3/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-27/2*I/(I-a)^4*b^2*a/(a^2+1)*(b^2*x^2+2*a*b*
x+a^2+1)^(1/2)+9*I/(I-a)^4*b^2*a^3/(a^2+1)^(1/2)*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*I/(I-a)^3*a^2*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*I/(I-a)^5*b^3*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-9*I/(I-a)^5*b^3*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)-6*I/(I-a)^5*b^3*a^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)-9/4*I/(I-a)^3*a^4*b^2/(a^2+1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-9/4*I/(I-a)^3*a^2*b^2/(a^2+1)^2*(b^
2*x^2+2*a*b*x+a^2+1)^(1/2)+3/2*I/(I-a)^3*a^4*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)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^3,x, algorithm="maxima")
[Out]
integrate(((b*x + a)^2 + 1)^(3/2)/((I*b*x + I*a + 1)^3*x^3), x)
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Fricas [B] time = 2.48799, size = 1458, normalized size = 5.52 \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/(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(-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(1/(1+I*(b*x+a))**3*(1+(b*x+a)**2)**(3/2)/x**3,x)
[Out]
Timed out
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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/(1+I*(b*x+a))^3*(1+(b*x+a)^2)^(3/2)/x^3,x, algorithm="giac")
[Out]
undef