\(\int e^{\frac {5}{2} i \arctan (a x)} x \, dx\) [96]

Optimal result

Integrand size = 14, antiderivative size = 255 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=-\frac {25 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{4 a^2}-\frac {5 (1-i a x)^{3/4} (1+i a x)^{5/4}}{2 a^2}-\frac {2 (1+i a x)^{9/4}}{a^2 \sqrt [4]{1-i a x}}+\frac {25 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}-\frac {25 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{4 \sqrt {2} a^2}+\frac {25 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{4 \sqrt {2} a^2} \] Output:

-25/4*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a^2-5/2*(1-I*a*x)^(3/4)*(1+I*a*x)^(5 
/4)/a^2-2*(1+I*a*x)^(9/4)/a^2/(1-I*a*x)^(1/4)+25/8*arctan(1-2^(1/2)*(1-I*a 
*x)^(1/4)/(1+I*a*x)^(1/4))*2^(1/2)/a^2-25/8*arctan(1+2^(1/2)*(1-I*a*x)^(1/ 
4)/(1+I*a*x)^(1/4))*2^(1/2)/a^2+25/8*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^2
 

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.28 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 \left (-3 (1+i a x)^{9/4}+20 i \sqrt [4]{2} (i+a x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {3}{4},\frac {7}{4},\frac {1}{2} (1-i a x)\right )\right )}{3 a^2 \sqrt [4]{1-i a x}} \] Input:

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

Output:

(2*(-3*(1 + I*a*x)^(9/4) + (20*I)*2^(1/4)*(I + a*x)*Hypergeometric2F1[-5/4 
, 3/4, 7/4, (1 - I*a*x)/2]))/(3*a^2*(1 - I*a*x)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.23, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5585, 87, 60, 60, 73, 854, 826, 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.

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

\(\Big \downarrow \) 5585

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(-2*(1 + I*a*x)^(9/4))/(a^2*(1 - I*a*x)^(1/4)) + ((5*I)*(((I/2)*(1 - I*a*x 
)^(3/4)*(1 + I*a*x)^(5/4))/a + (5*((I*(1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4)) 
/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 + (Log[1 + Sqrt[1 - I*a*x] - (Sqrt[2]*(1 - I*a*x)^(1/4))/(1 + I*a*x) 
^(1/4)]/(2*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))/a))/4))/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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.93 \[ \int e^{\frac {5}{2} i \arctan (a x)} x \, dx=\frac {2 \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} \, a^{2} \sqrt {\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (\frac {4}{25} \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - 2 \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} \log \left (-\frac {4}{25} \, a^{2} \sqrt {-\frac {625 i}{16 \, a^{4}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - {\left (2 \, a^{2} x^{2} - 9 i \, a x + 43\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{4 \, a^{2}} \] Input:

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

Output:

1/4*(2*a^2*sqrt(625/16*I/a^4)*log(4/25*a^2*sqrt(625/16*I/a^4) + sqrt(I*sqr 
t(a^2*x^2 + 1)/(a*x + I))) - 2*a^2*sqrt(625/16*I/a^4)*log(-4/25*a^2*sqrt(6 
25/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I))) + 2*a^2*sqrt(-625/16*I 
/a^4)*log(4/25*a^2*sqrt(-625/16*I/a^4) + sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I 
))) - 2*a^2*sqrt(-625/16*I/a^4)*log(-4/25*a^2*sqrt(-625/16*I/a^4) + sqrt(I 
*sqrt(a^2*x^2 + 1)/(a*x + I))) - (2*a^2*x^2 - 9*I*a*x + 43)*sqrt(I*sqrt(a^ 
2*x^2 + 1)/(a*x + I)))/a^2
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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