\(\int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx\) [110]

Optimal result

Integrand size = 16, antiderivative size = 170 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}+\frac {5 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac {11 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}+\frac {3}{8} i a^3 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {3}{8} i a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:

-1/3*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x^3+5/12*I*a*(1-I*a*x)^(1/4)*(1+I*a*x 
)^(3/4)/x^2+11/24*a^2*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x+3/8*I*a^3*arctan(( 
1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))-3/8*I*a^3*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 = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\frac {\sqrt [4]{1-i a x} \left (-8+2 i a x+a^2 x^2+11 i a^3 x^3-18 i a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{24 x^3 \sqrt [4]{1+i a x}} \] Input:

Integrate[1/(E^((I/2)*ArcTan[a*x])*x^4),x]
 

Output:

((1 - I*a*x)^(1/4)*(-8 + (2*I)*a*x + a^2*x^2 + (11*I)*a^3*x^3 - (18*I)*a^3 
*x^3*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x)]))/(24*x^3*(1 + I* 
a*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5585, 110, 27, 168, 27, 168, 27, 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.

\(\displaystyle \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \int \frac {4 a x+5 i}{x^3 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{2} \int -\frac {a (11-10 i a x)}{2 x^2 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \int \frac {11-10 i a x}{x^2 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-\int \frac {9 i a}{2 x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-\frac {9}{2} i a \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-18 i a \int -\frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (18 i a \int \frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-18 i a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-18 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {1}{6} a \left (\frac {1}{4} a \left (-18 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {11 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

Input:

Int[1/(E^((I/2)*ArcTan[a*x])*x^4),x]
 

Output:

-1/3*((1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))/x^3 - (a*((((-5*I)/2)*(1 - I*a* 
x)^(1/4)*(1 + I*a*x)^(3/4))/x^2 + (a*((-11*(1 - I*a*x)^(1/4)*(1 + I*a*x)^( 
3/4))/x - (18*I)*a*(ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2 - ArcTan 
h[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2)))/4))/6
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 110

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[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[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]

\[\int \frac {1}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{4}}d x\]

Input:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
 

Output:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\frac {-9 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 9 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 9 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 9 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) + 2 \, {\left (11 \, a^{2} x^{2} + 10 i \, a x - 8\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{48 \, x^{3}} \] Input:

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x, algorithm="fricas")
 

Output:

1/48*(-9*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - 9*a^3*x^ 
3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + 9*a^3*x^3*log(sqrt(I*sqrt 
(a^2*x^2 + 1)/(a*x + I)) - I) + 9*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/( 
a*x + I)) - 1) + 2*(11*a^2*x^2 + 10*I*a*x - 8)*sqrt(a^2*x^2 + 1)*sqrt(I*sq 
rt(a^2*x^2 + 1)/(a*x + I)))/x^3
 

Sympy [F]

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

integrate(1/((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/2)/x**4,x)
 

Output:

Integral(1/(x**4*sqrt(I*(a*x - I)/sqrt(a**2*x**2 + 1))), x)
 

Maxima [F]

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

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(1/(x^4*sqrt((I*a*x + 1)/sqrt(a^2*x^2 + 1))), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,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. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\int \frac {1}{x^4\,\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}} \,d x \] Input:

int(1/(x^4*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/2)),x)
 

Output:

int(1/(x^4*((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\frac {22 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}} a^{2} x^{2}+20 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}} a i x -16 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}}+9 \left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {1}{4}}}{a^{2} x^{3}+x}d x \right ) a^{3} i \,x^{3}}{48 x^{3}} \] Input:

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
 

Output:

(22*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(1/4)*a**2*x**2 + 20*sqrt(a*i*x + 1)* 
(a**2*x**2 + 1)**(1/4)*a*i*x - 16*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(1/4) + 
 9*int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(1/4))/(a**2*x**3 + x),x)*a**3*i* 
x**3)/(48*x**3)