\(\int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx\) [126]

Optimal result

Integrand size = 16, antiderivative size = 121 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-\frac {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:

-10*I*a*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)-(1-I*a*x)^(5/4)/x/(1+I*a*x)^(1/4)- 
5*I*a*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+5*I*a*arctanh((1+I*a*x)^(1/4 
)/(1-I*a*x)^(1/4))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {i \sqrt [4]{1-i a x} \left (i-9 a x+10 a x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{x \sqrt [4]{1+i a x}} \] Input:

Integrate[1/(E^(((5*I)/2)*ArcTan[a*x])*x^2),x]
 

Output:

(I*(1 - I*a*x)^(1/4)*(I - 9*a*x + 10*a*x*Hypergeometric2F1[1/4, 1, 5/4, (I 
 + a*x)/(I - a*x)]))/(x*(1 + I*a*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5585, 105, 105, 104, 25, 827, 216, 219}

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.

\(\displaystyle \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {(1-i a x)^{5/4}}{x^2 (1+i a x)^{5/4}}dx\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {5}{2} i a \int \frac {\sqrt [4]{1-i a x}}{x (i a x+1)^{5/4}}dx-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 105

\(\displaystyle -\frac {5}{2} i a \left (\int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx+\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {5}{2} i a \left (4 \int -\frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {5}{2} i a \left (\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-4 \int \frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {5}{2} i a \left (4 \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )+\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {5}{2} i a \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )+\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {5}{2} i a \left (4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )+\frac {4 \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}\)

Input:

Int[1/(E^(((5*I)/2)*ArcTan[a*x])*x^2),x]
 

Output:

-((1 - I*a*x)^(5/4)/(x*(1 + I*a*x)^(1/4))) - ((5*I)/2)*a*((4*(1 - I*a*x)^( 
1/4))/(1 + I*a*x)^(1/4) + 4*(ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2 
 - ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 5585

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a 
*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))), x] /; FreeQ[{a, m, n}, x] &&  !Intege 
rQ[(I*n - 1)/2]
 

Maple [F]

\[\int \frac {1}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{2}}d x\]

Input:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
 

Output:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
 

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. 212 vs. \(2 (83) = 166\).

Time = 0.13 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-\frac {2 \, \sqrt {a^{2} x^{2} + 1} {\left (9 \, a x - i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 5 \, {\left (-i \, a^{2} x^{2} - a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 5 \, {\left (a^{2} x^{2} - i \, a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 5 \, {\left (a^{2} x^{2} - i \, a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 5 \, {\left (i \, a^{2} x^{2} + a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right )}{2 \, {\left (a x^{2} - i \, x\right )}} \] Input:

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="fricas")
 

Output:

-1/2*(2*sqrt(a^2*x^2 + 1)*(9*a*x - I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) 
+ 5*(-I*a^2*x^2 - a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - 5*(a 
^2*x^2 - I*a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + 5*(a^2*x^2 
- I*a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 5*(I*a^2*x^2 + a*x 
)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1))/(a*x^2 - I*x)
 

Sympy [F]

\[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int \frac {1}{x^{2} \left (\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)/x**2,x)
 

Output:

Integral(1/(x**2*(I*(a*x - I)/sqrt(a**2*x**2 + 1))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int { \frac {1}{x^{2} \left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="maxima")
 

Output:

integrate(1/(x^2*((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2)), x)
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.The choice 
 was done
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int \frac {1}{x^2\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \] Input:

int(1/(x^2*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)),x)
 

Output:

int(1/(x^2*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-\left (\int \frac {\left (a^{2} x^{2}+1\right )^{\frac {1}{4}}}{\sqrt {a i x +1}\, a^{2} x^{4}-2 \sqrt {a i x +1}\, a i \,x^{3}-\sqrt {a i x +1}\, x^{2}}d x \right )-\left (\int \frac {\left (a^{2} x^{2}+1\right )^{\frac {1}{4}}}{\sqrt {a i x +1}\, a^{2} x^{2}-2 \sqrt {a i x +1}\, a i x -\sqrt {a i x +1}}d x \right ) a^{2} \] Input:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
 

Output:

 - (int((a**2*x**2 + 1)**(1/4)/(sqrt(a*i*x + 1)*a**2*x**4 - 2*sqrt(a*i*x + 
 1)*a*i*x**3 - sqrt(a*i*x + 1)*x**2),x) + int((a**2*x**2 + 1)**(1/4)/(sqrt 
(a*i*x + 1)*a**2*x**2 - 2*sqrt(a*i*x + 1)*a*i*x - sqrt(a*i*x + 1)),x)*a**2 
)