3.1.68 \(\int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx\) [68]

3.1.68.1 Optimal result

Integrand size = 16, antiderivative size = 202 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}-\frac {7 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x^3}+\frac {29 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{96 x^2}+\frac {83 i a^3 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{192 x}-\frac {11}{64} a^4 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {11}{64} a^4 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

output

-1/4*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/x^4-7/24*I*a*(1-I*a*x)^(3/4)*(1+I*a*x 
)^(1/4)/x^3+29/96*a^2*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/x^2+83/192*I*a^3*(1- 
I*a*x)^(3/4)*(1+I*a*x)^(1/4)/x-11/64*a^4*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^ 
(1/4))-11/64*a^4*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))
 

3.1.68.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.49 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {(1-i a x)^{3/4} \left (48+104 i a x-114 a^2 x^2-141 i a^3 x^3+83 a^4 x^4+22 a^4 x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )\right )}{192 x^4 (1+i a x)^{3/4}} \]

input

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

output

-1/192*((1 - I*a*x)^(3/4)*(48 + (104*I)*a*x - 114*a^2*x^2 - (141*I)*a^3*x^ 
3 + 83*a^4*x^4 + 22*a^4*x^4*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - 
a*x)]))/(x^4*(1 + I*a*x)^(3/4))
 

3.1.68.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {5585, 110, 27, 168, 27, 168, 27, 168, 27, 104, 756, 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.

\(\displaystyle \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx\)

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \int \frac {7 i-6 a x}{x^4 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{3} \int \frac {a (28 i a x+29)}{2 x^3 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \int \frac {28 i a x+29}{x^3 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (-\frac {1}{2} \int -\frac {a (83 i-58 a x)}{2 x^2 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \int \frac {83 i-58 a x}{x^2 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-\int \frac {33 a}{2 x \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-\frac {33}{2} a \int \frac {1}{x \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-66 a \int \frac {1}{\frac {i a x+1}{1-i a x}-1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 756

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-66 a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-66 a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{8} a \left (-\frac {1}{6} a \left (\frac {1}{4} a \left (-66 a \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {83 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {29 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {7 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 x^4}\)

input

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

output

-1/4*((1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x^4 + (a*((((-7*I)/3)*(1 - I*a* 
x)^(3/4)*(1 + I*a*x)^(1/4))/x^3 - (a*((-29*(1 - I*a*x)^(3/4)*(1 + I*a*x)^( 
1/4))/(2*x^2) + (a*(((-83*I)*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x - 66*a 
*(-1/2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] - ArcTanh[(1 + I*a*x)^( 
1/4)/(1 - I*a*x)^(1/4)]/2)))/4))/6))/8
 

3.1.68.3.1 Defintions of rubi rules used

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 756

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

3.1.68.4 Maple [F]

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

input

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

output

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

3.1.68.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx=-\frac {33 \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 33 i \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 33 i \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 33 \, a^{4} x^{4} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (83 \, a^{4} x^{4} + 25 i \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 8 i \, a x - 48\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{384 \, x^{4}} \]

input

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

output

-1/384*(33*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 33*I*a^4 
*x^4*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) - 33*I*a^4*x^4*log(sqrt( 
I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) - 33*a^4*x^4*log(sqrt(I*sqrt(a^2*x^2 + 
 1)/(a*x + I)) - 1) - 2*(83*a^4*x^4 + 25*I*a^3*x^3 + 2*a^2*x^2 - 8*I*a*x - 
 48)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^4
 

3.1.68.6 Sympy [F]

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

input

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

output

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

3.1.68.7 Maxima [F]

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

input

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

output

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

3.1.68.8 Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^5} \, dx=\text {Exception raised: TypeError} \]

input

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^5,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 -28, a substitution variable should perhaps be pu 
rged.Warn
 

3.1.68.9 Mupad [F(-1)]

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

input

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

output

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