\(\int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx\) [100]

Optimal result

Integrand size = 16, antiderivative size = 163 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {25 a^2 \sqrt [4]{1+i a x}}{2 \sqrt [4]{1-i a x}}-\frac {5 i a (1+i a x)^{5/4}}{4 x \sqrt [4]{1-i a x}}-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}+\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:

-25/2*a^2*(1+I*a*x)^(1/4)/(1-I*a*x)^(1/4)-5/4*I*a*(1+I*a*x)^(5/4)/x/(1-I*a 
*x)^(1/4)-1/2*(1+I*a*x)^(9/4)/x^2/(1-I*a*x)^(1/4)+25/4*a^2*arctan((1+I*a*x 
)^(1/4)/(1-I*a*x)^(1/4))+25/4*a^2*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 = 99, normalized size of antiderivative = 0.61 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {-6-33 i a x-102 a^2 x^2-129 i a^3 x^3+50 a^2 x^2 (1-i a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )}{12 x^2 \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \] Input:

Integrate[E^(((5*I)/2)*ArcTan[a*x])/x^3,x]
 

Output:

(-6 - (33*I)*a*x - 102*a^2*x^2 - (129*I)*a^3*x^3 + 50*a^2*x^2*(1 - I*a*x)* 
Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)])/(12*x^2*(1 - I*a*x)^( 
1/4)*(1 + I*a*x)^(3/4))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5585, 107, 105, 105, 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.

\(\displaystyle \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx\)

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{4} i a \left (\frac {5}{2} i a \left (4 \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{9/4}}{2 x^2 \sqrt [4]{1-i a x}}\)

Input:

Int[E^(((5*I)/2)*ArcTan[a*x])/x^3,x]
 

Output:

-1/2*(1 + I*a*x)^(9/4)/(x^2*(1 - I*a*x)^(1/4)) + ((5*I)/4)*a*(-((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*(-1/2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] - ArcTan 
h[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2)))
 

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 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(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] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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]
 

Maple [F]

\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}{x^{3}}d x\]

Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
 

Output:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 25 i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 25 \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (43 \, a^{2} x^{2} + 9 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="fricas")
 

Output:

1/8*(25*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 25*I*a^2*x^ 
2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) - 25*I*a^2*x^2*log(sqrt(I*s 
qrt(a^2*x^2 + 1)/(a*x + I)) - I) - 25*a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1) 
/(a*x + I)) - 1) - 2*(43*a^2*x^2 + 9*I*a*x + 2)*sqrt(I*sqrt(a^2*x^2 + 1)/( 
a*x + I)))/x^2
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\text {Timed out} \] Input:

integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)/x**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x, algorithm="maxima")
 

Output:

integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2)/x^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,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. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}}{x^3} \,d x \] Input:

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^3,x)
 

Output:

int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^3, x)
 

Reduce [F]

\[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {-36 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}} a^{2} x^{2}-18 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}} a i x -4 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}-25 \left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}}{a^{4} x^{5}+2 a^{2} x^{3}+x}d x \right ) a^{4} x^{4}-25 \left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}}{a^{4} x^{5}+2 a^{2} x^{3}+x}d x \right ) a^{2} x^{2}}{8 x^{2} \left (a^{2} x^{2}+1\right )} \] Input:

int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^3,x)
 

Output:

( - 36*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4)*a**2*x**2 - 18*sqrt(a*i*x + 
1)*(a**2*x**2 + 1)**(3/4)*a*i*x - 4*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4) 
 - 25*int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4))/(a**4*x**5 + 2*a**2*x** 
3 + x),x)*a**4*x**4 - 25*int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4))/(a** 
4*x**5 + 2*a**2*x**3 + x),x)*a**2*x**2)/(8*x**2*(a**2*x**2 + 1))