\(\int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx\) [227]

Optimal result

Integrand size = 16, antiderivative size = 178 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\frac {6 i b \sqrt {1-i a-i b x}}{(i-a)^2 \sqrt {1+i a+i b x}}-\frac {(1-i a-i b x)^{3/2}}{(1+i a) x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {i+a} b \text {arctanh}\left (\frac {\sqrt {i+a} \sqrt {1+i a+i b x}}{\sqrt {i-a} \sqrt {1-i a-i b x}}\right )}{(i-a)^{5/2}} \] Output:

6*I*b*(1-I*a-I*b*x)^(1/2)/(I-a)^2/(1+I*a+I*b*x)^(1/2)-(1-I*a-I*b*x)^(3/2)/ 
(1+I*a)/x/(1+I*a+I*b*x)^(1/2)-6*I*(I+a)^(1/2)*b*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)^(5/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=\frac {\frac {\sqrt {-i (i+a+b x)} \left (1+a^2+5 i b x+a b x\right )}{x \sqrt {1+i a+i b x}}-\frac {6 i \sqrt {-1+i a} b \text {arctanh}\left (\frac {\sqrt {-1-i a} \sqrt {-i (i+a+b x)}}{\sqrt {-1+i a} \sqrt {1+i a+i b x}}\right )}{\sqrt {-1-i a}}}{(-i+a)^2} \] Input:

Integrate[1/(E^((3*I)*ArcTan[a + b*x])*x^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5618, 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.

\(\displaystyle \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx\)

\(\Big \downarrow \) 5618

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(E^((3*I)*ArcTan[a + b*x])*x^2),x]
 

Output:

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

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 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 = 1.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, \left (i+a \right )}{\left (-i+a \right )^{2} x}+\frac {b \left (\frac {3 \sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )}{i-a}+\frac {4 i \left (i a +1\right ) \sqrt {\left (x -\frac {i-a}{b}\right )^{2} b^{2}+2 i b \left (x -\frac {i-a}{b}\right )}}{b \left (i-a \right ) \left (x -\frac {i-a}{b}\right )}\right )}{a^{2}-2 i a -1}\)	\(187\)
default	\(\text {Expression too large to display}\)	\(1543\)

Input:

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

Output:

-I*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*(I+a)/(-I+a)^2/x+1/(-2*I*a+a^2-1)*b*(3*(a 
^2+1)^(1/2)/(I-a)*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)+4*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. 389 vs. \(2 (116) = 232\).

Time = 0.14 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.19 \[ \int \frac {e^{-3 i \arctan (a+b x)}}{x^2} \, dx=-\frac {{\left (i \, a - 5\right )} b^{2} x^{2} + {\left (i \, a^{2} - 4 \, a + 5 i\right )} b x - 3 \, {\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + 3 \, {\left ({\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} \log \left (-\frac {b^{2} x - {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} \sqrt {\frac {{\left (a + i\right )} b^{2}}{a^{5} - 5 i \, a^{4} - 10 \, a^{3} + 10 i \, a^{2} + 5 \, a - i}} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b}{b}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (i \, a - 5\right )} b x + i \, a^{2} + i\right )}}{{\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x} \] Input:

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

Output:

-((I*a - 5)*b^2*x^2 + (I*a^2 - 4*a + 5*I)*b*x - 3*((a^2 - 2*I*a - 1)*b*x^2 
 + (a^3 - 3*I*a^2 - 3*a + I)*x)*sqrt((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 
 10*I*a^2 + 5*a - I))*log(-(b^2*x + (a^3 - 3*I*a^2 - 3*a + I)*sqrt((a + I) 
*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I)) - sqrt(b^2*x^2 + 2*a*b 
*x + a^2 + 1)*b)/b) + 3*((a^2 - 2*I*a - 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + 
I)*x)*sqrt((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a^3 + 10*I*a^2 + 5*a - I))*log( 
-(b^2*x - (a^3 - 3*I*a^2 - 3*a + I)*sqrt((a + I)*b^2/(a^5 - 5*I*a^4 - 10*a 
^3 + 10*I*a^2 + 5*a - I)) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b)/b) + sqrt 
(b^2*x^2 + 2*a*b*x + a^2 + 1)*((I*a - 5)*b*x + I*a^2 + I))/((a^2 - 2*I*a - 
 1)*b*x^2 + (a^3 - 3*I*a^2 - 3*a + I)*x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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