\(\int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x} \, dx\) [89]

Optimal result

Integrand size = 16, antiderivative size = 204 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x} \, dx=2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right ) \] Output:

2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+2^(1/2)*arctan(1-2^(1/2)*(1-I*a* 
x)^(1/4)/(1+I*a*x)^(1/4))-2^(1/2)*arctan(1+2^(1/2)*(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^(1/2)*arctanh(2^(1/ 
2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)/(1+(1-I*a*x)^(1/2)/(1+I*a*x)^(1/2)))
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.47 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x} \, dx=-2 2^{3/4} \sqrt [4]{1-i a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-i a x)\right )-\frac {4 \sqrt [4]{1-i a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},-\frac {1-i a x}{-1-i a x}\right )}{\sqrt [4]{1+i a x}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.30, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.062, Rules used = {5585, 140, 73, 104, 25, 770, 755, 827, 216, 219, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 140

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*d^(m + n)*f^p   Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] 
, x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x 
)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 
0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n, -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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[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 755

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

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {3}{2} i \arctan (a x)}}{x} \, dx=\frac {1}{2} \, \sqrt {4 i} \log \left (\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac {1}{2} \, \sqrt {4 i} \log \left (-\frac {1}{2} i \, \sqrt {4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \frac {1}{2} \, \sqrt {-4 i} \log \left (\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + \frac {1}{2} \, \sqrt {-4 i} \log \left (-\frac {1}{2} i \, \sqrt {-4 i} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + i \, \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - i \, \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) \] Input:

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

Output:

1/2*sqrt(4*I)*log(1/2*I*sqrt(4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - 
 1/2*sqrt(4*I)*log(-1/2*I*sqrt(4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) 
 - 1/2*sqrt(-4*I)*log(1/2*I*sqrt(-4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I 
))) + 1/2*sqrt(-4*I)*log(-1/2*I*sqrt(-4*I) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x 
 + I))) - log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + I*log(sqrt(I*sqrt 
(a^2*x^2 + 1)/(a*x + I)) + I) - I*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) 
- I) + log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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