3.1.77 \(\int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx\) [77]

3.1.77.1 Optimal result

Integrand size = 16, antiderivative size = 170 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}-\frac {7 i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{12 x^2}+\frac {23 a^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{24 x}-\frac {17}{8} i a^3 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {17}{8} i a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

output

-1/3*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x^3-7/12*I*a*(1-I*a*x)^(1/4)*(1+I*a*x 
)^(3/4)/x^2+23/24*a^2*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x-17/8*I*a^3*arctan( 
(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+17/8*I*a^3*arctanh((1+I*a*x)^(1/4)/(1-I*a 
*x)^(1/4))
 

3.1.77.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx=\frac {\sqrt [4]{1-i a x} \left (-8-22 i a x+37 a^2 x^2+23 i a^3 x^3+102 i a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{24 x^3 \sqrt [4]{1+i a x}} \]

input

Integrate[E^(((3*I)/2)*ArcTan[a*x])/x^4,x]
 

output

((1 - I*a*x)^(1/4)*(-8 - (22*I)*a*x + 37*a^2*x^2 + (23*I)*a^3*x^3 + (102*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))
 

3.1.77.3 Rubi [A] (verified)

Time = 0.29 (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.

\(\displaystyle \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 5585

\(\displaystyle \int \frac {(1+i a x)^{3/4}}{x^4 (1-i a x)^{3/4}}dx\)

\(\Big \downarrow \) 110

\(\displaystyle \frac {1}{3} \int \frac {a (7 i-4 a x)}{2 x^3 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} a \int \frac {7 i-4 a x}{x^3 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{2} \int \frac {a (14 i a x+23)}{2 x^2 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \int \frac {14 i a x+23}{x^2 (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (-\int -\frac {51 i a}{2 x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (\frac {51}{2} i a \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (102 i a \int -\frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (-102 i a \int \frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (102 i a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (102 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (102 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {23 \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {7 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{3 x^3}\)

input

Int[E^(((3*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*((((-7*I)/2)*(1 - I*a* 
x)^(1/4)*(1 + I*a*x)^(3/4))/x^2 - (a*((-23*(1 - I*a*x)^(1/4)*(1 + I*a*x)^( 
3/4))/x + (102*I)*a*(ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2 - ArcTa 
nh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2)))/4))/6
 

3.1.77.3.1 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]
 

3.1.77.4 Maple [F]

\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {3}{2}}}{x^{4}}d x\]

input

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x)
 

output

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x)
 

3.1.77.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx=\frac {51 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 51 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 51 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 51 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) + 2 \, {\left (23 \, a^{2} x^{2} - 14 i \, a x - 8\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{48 \, x^{3}} \]

input

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x, algorithm="fricas")
 

output

1/48*(51*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 51*a^3*x 
^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) - 51*a^3*x^3*log(sqrt(I*sq 
rt(a^2*x^2 + 1)/(a*x + I)) - I) - 51*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1 
)/(a*x + I)) - 1) + 2*(23*a^2*x^2 - 14*I*a*x - 8)*sqrt(a^2*x^2 + 1)*sqrt(I 
*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^3
 

3.1.77.6 Sympy [F]

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

input

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(3/2)/x**4,x)
 

output

Integral((I*(a*x - I)/sqrt(a**2*x**2 + 1))**(3/2)/x**4, x)
 

3.1.77.7 Maxima [F]

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

input

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x, algorithm="maxima")
 

output

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(3/2)/x^4, x)
 

3.1.77.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx=\text {Exception raised: TypeError} \]

input

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(3/2)/x^4,x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 
]Warning, replacing 0 by 23, a substitution variable should perhaps be pur 
ged.Warni
 

3.1.77.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x^4} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{3/2}}{x^4} \,d x \]

input

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

output

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