\(\int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx\) [228]

Optimal result

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

-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)+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*(3-2*I*a)*b^2*arcta 
nh((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)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\frac {\frac {\sqrt {-i (i+a+b x)} \left (-i+a-i a^2+a^3-5 b x-5 i a b x-14 i b^2 x^2-a b^2 x^2\right )}{x^2 \sqrt {1+i a+i b x}}+\frac {6 i \sqrt {-1+i a} (3 i+2 a) b^2 \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a} (i+a)}}{2 (-i+a)^3} \] Input:

Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x^3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5618, 107, 105, 105, 104, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 5618

\(\displaystyle \int \frac {(-i a-i b x+1)^{3/2}}{x^3 (i a+i b x+1)^{3/2}}dx\)

\(\Big \downarrow \) 107

\(\displaystyle -\frac {(2 a+3 i) b \int \frac {(-i a-i b x+1)^{3/2}}{x^2 (i a+i b x+1)^{3/2}}dx}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(2 a+3 i) b \left (\frac {3 b \int \frac {\sqrt {-i a-i b x+1}}{x (i a+i b x+1)^{3/2}}dx}{-a+i}-\frac {(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt {i a+i b x+1}}\right )}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {(2 a+3 i) b \left (\frac {3 b \left (\frac {(a+i) \int \frac {1}{x \sqrt {-i a-i b x+1} \sqrt {i a+i b x+1}}dx}{-a+i}+\frac {2 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}\right )}{-a+i}-\frac {(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt {i a+i b x+1}}\right )}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {(2 a+3 i) b \left (\frac {3 b \left (\frac {2 (a+i) \int \frac {1}{-i a+\frac {(1-i a) (i a+i b x+1)}{-i a-i b x+1}-1}d\frac {\sqrt {i a+i b x+1}}{\sqrt {-i a-i b x+1}}}{-a+i}+\frac {2 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}\right )}{-a+i}-\frac {(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt {i a+i b x+1}}\right )}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {(2 a+3 i) b \left (\frac {3 b \left (\frac {2 \sqrt {-i a-i b x+1}}{(1+i a) \sqrt {i a+i b x+1}}-\frac {2 i \sqrt {a+i} \text {arctanh}\left (\frac {\sqrt {a+i} \sqrt {i a+i b x+1}}{\sqrt {-a+i} \sqrt {-i a-i b x+1}}\right )}{(-a+i)^{3/2}}\right )}{-a+i}-\frac {(-i a-i b x+1)^{3/2}}{(1+i a) x \sqrt {i a+i b x+1}}\right )}{2 \left (a^2+1\right )}-\frac {(-i a-i b x+1)^{5/2}}{2 \left (a^2+1\right ) x^2 \sqrt {i a+i b x+1}}\)

Input:

Int[1/(E^((3*I)*ArcTan[a + b*x])*x^3),x]
 

Output:

-1/2*(1 - I*a - I*b*x)^(5/2)/((1 + a^2)*x^2*Sqrt[1 + I*a + I*b*x]) - ((3*I 
 + 2*a)*b*(-((1 - I*a - I*b*x)^(3/2)/((1 + I*a)*x*Sqrt[1 + I*a + I*b*x])) 
+ (3*b*((2*Sqrt[1 - I*a - I*b*x])/((1 + I*a)*Sqrt[1 + I*a + I*b*x]) - ((2* 
I)*Sqrt[I + a]*ArcTanh[(Sqrt[I + a]*Sqrt[1 + I*a + I*b*x])/(Sqrt[I - a]*Sq 
rt[1 - I*a - I*b*x])])/(I - a)^(3/2)))/(I - a)))/(2*(1 + a^2))
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[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 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> 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] + Simp[(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] || SumSimplerQ[m, 1])
 

rule 221

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 5618

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]
 

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.07

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

Input:

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

Output:

-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/(-I+a)^3/(b^2*x^2+2*a*b*x+a^2+1 
)^(1/2)+1/2/(a^3-3*a-3*I*a^2+I)*b^2*(-3*(I*a+2*a^2+3)/(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 = 0.16 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^3} \, dx=\frac {{\left (i \, a - 14\right )} b^{3} x^{3} + {\left (i \, a^{2} - 13 \, a + 14 i\right )} b^{2} x^{2} - 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (2 \, a + 3 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 3 i\right )} b^{2} + {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (2 \, a + 3 i\right )} b^{2}}\right ) + 3 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}} \log \left (-\frac {{\left (2 \, a + 3 i\right )} b^{3} x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (2 \, a + 3 i\right )} b^{2} - {\left (a^{5} - 3 i \, a^{4} - 2 \, a^{3} - 2 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (4 \, a^{2} + 12 i \, a - 9\right )} b^{4}}{a^{8} - 6 i \, a^{7} - 14 \, a^{6} + 14 i \, a^{5} + 14 i \, a^{3} + 14 \, a^{2} - 6 i \, a - 1}}}{{\left (2 \, a + 3 i\right )} b^{2}}\right ) + {\left ({\left (i \, a - 14\right )} b^{2} x^{2} - i \, a^{3} - 5 \, {\left (a - i\right )} b x - a^{2} - i \, a - 1\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{2 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )}} \] Input:

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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