\(\int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx\) [214]

Optimal result

Integrand size = 16, antiderivative size = 264 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=-\frac {3 (3 i+2 a) b^2 \sqrt {1-i a-i b x}}{(1+i a)^3 (i+a) \sqrt {1+i a+i b x}}+\frac {(3-2 i a) b (1-i a-i b x)^{3/2}}{2 (i-a)^2 (i+a) x \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{5/2}}{2 \left (1+a^2\right ) x^2 \sqrt {1+i a+i b x}}+\frac {3 (3-2 i a) b^2 \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{7/2} \sqrt {i+a}} \]

[Out]

3*(3-2*I*a)*b^2*arctanh((I+a)^(1/2)*(1+I*a+I*b*x)^(1/2)/(I-a)^(1/2)/(1-I*a-I*b*x)^(1/2))/(I-a)^(7/2)/(I+a)^(1/
2)+1/2*(3-2*I*a)*b*(1-I*a-I*b*x)^(3/2)/(I-a)^2/(I+a)/x/(1+I*a+I*b*x)^(1/2)-1/2*(1-I*a-I*b*x)^(5/2)/(a^2+1)/x^2
/(1+I*a+I*b*x)^(1/2)-3*(3*I+2*a)*b^2*(1-I*a-I*b*x)^(1/2)/(1+I*a)^3/(I+a)/(1+I*a+I*b*x)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5203, 98, 96, 95, 214} \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}+\frac {3 (3-2 i a) b^2 \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{7/2} \sqrt {a+i}}-\frac {3 (2 a+3 i) b^2 \sqrt {-i a-i b x+1}}{(1+i a)^3 (a+i) \sqrt {i a+i b x+1}}+\frac {(3-2 i a) b (-i a-i b x+1)^{3/2}}{2 (-a+i)^2 (a+i) x \sqrt {i a+i b x+1}} \]

[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 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 214

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

Rule 5203

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 \text {arctanh}\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] (verified)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\frac {\frac {\sqrt {-i (i+a+b x)} \left (-i+a-i a^2+a^3-5 b x-5 i a b x-14 i b^2 x^2-a b^2 x^2\right )}{x^2 \sqrt {1+i a+i b x}}+\frac {6 i \sqrt {-1+i a} (3 i+2 a) b^2 \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a} (i+a)}}{2 (-i+a)^3} \]

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

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {i \left (-a \,b^{3} x^{3}-6 i b^{3} x^{3}-a^{2} b^{2} x^{2}-12 i a \,b^{2} x^{2}+a^{3} b x -6 i a^{2} b x +a^{4}+b^{2} x^{2}+a b x -6 b x i+2 a^{2}+1\right )}{2 x^{2} \left (a -i\right )^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}+\frac {b^{2} \left (-\frac {\left (6 a^{2}+3 i a +9\right ) \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{\left (i-a \right ) \sqrt {a^{2}+1}}-\frac {8 i \left (i a +1\right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b \left (i-a \right ) \left (x -\frac {i-a}{b}\right )}\right )}{2 a^{3}-6 i a^{2}-6 a +2 i}\)	\(282\)
default	\(\text {Expression too large to display}\)	\(2328\)

[In]

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

[Out]

-1/2*I*(-a*b^3*x^3-a^2*b^2*x^2+a^3*b*x-6*I*b^3*x^3+a^4+b^2*x^2-12*I*a*b^2*x^2+a*b*x-6*I*a^2*b*x+2*a^2-6*I*b*x+
1)/x^2/(a-I)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2/(-3*I*a^2+a^3+I-3*a)*b^2*(-(3*I*a+6*a^2+9)/(I-a)/(a^2+1)^(1/2
)*ln((2*a^2+2+2*a*b*x+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x)-8*I*(1+I*a)/b/(I-a)/(x-(I-a)/b)*((x-(I
-a)/b)^2*b^2+2*I*b*(x-(I-a)/b))^(1/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 574 vs. \(2 (180) = 360\).

Time = 0.28 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\frac {{\left (i \, a - 14\right )} b^{3} x^{3} + {\left (i \, a^{2} - 13 \, a + 14 i\right )} b^{2} x^{2} - 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (2 \, a + 3 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 3 i\right )} b^{2} + {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (2 \, a + 3 i\right )} b^{2}}\right ) + 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (2 \, a + 3 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 3 i\right )} b^{2} - {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (2 \, a + 3 i\right )} b^{2}}\right ) + {\left ({\left (i \, a - 14\right )} b^{2} x^{2} - i \, a^{3} - 5 \, {\left (a - i\right )} b x - a^{2} - i \, a - 1\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{2 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )}} \]

[In]

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

[Out]

1/2*((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(-((2*a + 3*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + 3*I)*b^2 + (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)))/((2*a + 3*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(-
((2*a + 3*I)*b^3*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(2*a + 3*I)*b^2 - (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)))/((2
*a + 3*I)*b^2)) + ((I*a - 14)*b^2*x^2 - I*a^3 - 5*(a - I)*b*x - a^2 - I*a - 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 +
1))/((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)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\int { \frac {{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\int { \frac {{\left ({\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, b x + i \, a + 1\right )}^{3} x^{3}} \,d x } \]

[In]

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

[Out]

undef

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\int \frac {{\left ({\left (a+b\,x\right )}^2+1\right )}^{3/2}}{x^3\,{\left (1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

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

[Out]

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