\(\int e^{-\frac {5}{2} i \arctan (a x)} \, dx\) [124]

Optimal result

Integrand size = 12, antiderivative size = 226 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=\frac {4 i (1-i a x)^{5/4}}{a \sqrt [4]{1+i a x}}+\frac {5 i \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{a}+\frac {5 i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {5 i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {5 i \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{\sqrt {2} a} \] Output:

4*I*(1-I*a*x)^(5/4)/a/(1+I*a*x)^(1/4)+5*I*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/ 
a+5/2*I*arctan(1-2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4))*2^(1/2)/a-5/2*I* 
arctan(1+2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4))*2^(1/2)/a-5/2*I*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) 
))*2^(1/2)/a
 

Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.17 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=\frac {8 i e^{-\frac {1}{2} i \arctan (a x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},2,\frac {3}{4},-e^{2 i \arctan (a x)}\right )}{a} \] Input:

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

Output:

((8*I)*Hypergeometric2F1[-1/4, 2, 3/4, -E^((2*I)*ArcTan[a*x])])/(a*E^((I/2 
)*ArcTan[a*x]))
 

Rubi [A] (warning: unable to verify)

Time = 0.67 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.21, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {5584, 57, 60, 73, 770, 755, 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.

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

\(\Big \downarrow \) 5584

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 770

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((4*I)*(1 - I*a*x)^(5/4))/(a*(1 + I*a*x)^(1/4)) - 5*(((-I)*(1 - I*a*x)^(1/ 
4)*(1 + I*a*x)^(3/4))/a + ((2*I)*((-(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] + (Sqr 
t[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4)]/(2*Sqrt[2]))/2))/a)
 

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 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 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 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 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 5584

Int[E^(ArcTan[(a_.)*(x_)]*(n_.)), x_Symbol] :> Int[(1 - I*a*x)^(I*(n/2))/(1 
 + I*a*x)^(I*(n/2)), x] /; FreeQ[{a, n}, x] &&  !IntegerQ[(I*n - 1)/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.15 \[ \int e^{-\frac {5}{2} i \arctan (a x)} \, dx=-\frac {{\left (a^{2} x - i \, a\right )} \sqrt {\frac {25 i}{a^{2}}} \log \left (\frac {1}{5} \, a \sqrt {\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (a^{2} x - i \, a\right )} \sqrt {\frac {25 i}{a^{2}}} \log \left (-\frac {1}{5} \, a \sqrt {\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (a^{2} x - i \, a\right )} \sqrt {-\frac {25 i}{a^{2}}} \log \left (\frac {1}{5} \, a \sqrt {-\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + {\left (a^{2} x - i \, a\right )} \sqrt {-\frac {25 i}{a^{2}}} \log \left (-\frac {1}{5} \, a \sqrt {-\frac {25 i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-i \, a x - 9\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, {\left (a^{2} x - i \, a\right )}} \] Input:

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

Output:

-1/2*((a^2*x - I*a)*sqrt(25*I/a^2)*log(1/5*a*sqrt(25*I/a^2) + sqrt(I*sqrt( 
a^2*x^2 + 1)/(a*x + I))) - (a^2*x - I*a)*sqrt(25*I/a^2)*log(-1/5*a*sqrt(25 
*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) - (a^2*x - I*a)*sqrt(-25*I/ 
a^2)*log(1/5*a*sqrt(-25*I/a^2) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + (a 
^2*x - I*a)*sqrt(-25*I/a^2)*log(-1/5*a*sqrt(-25*I/a^2) + sqrt(I*sqrt(a^2*x 
^2 + 1)/(a*x + I))) + 2*sqrt(a^2*x^2 + 1)*(-I*a*x - 9)*sqrt(I*sqrt(a^2*x^2 
 + 1)/(a*x + I)))/(a^2*x - I*a)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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