\(\int e^{\frac {1}{2} i \arctan (a x)} \, dx\) [78]

Optimal result

Integrand size = 12, antiderivative size = 195 \[ \int e^{\frac {1}{2} i \arctan (a x)} \, dx=\frac {i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{a}-\frac {i \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}+\frac {i \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}\right )}{\sqrt {2} a}-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x} \left (1+\frac {\sqrt {1-i a x}}{\sqrt {1+i a x}}\right )}\right )}{\sqrt {2} a} \] Output:

I*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/a-1/2*I*arctan(1-2^(1/2)*(1-I*a*x)^(1/4) 
/(1+I*a*x)^(1/4))*2^(1/2)/a+1/2*I*arctan(1+2^(1/2)*(1-I*a*x)^(1/4)/(1+I*a* 
x)^(1/4))*2^(1/2)/a-1/2*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 = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.21 \[ \int e^{\frac {1}{2} i \arctan (a x)} \, dx=-\frac {8 i e^{\frac {5}{2} i \arctan (a x)} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},2,\frac {9}{4},-e^{2 i \arctan (a x)}\right )}{5 a} \] Input:

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

Output:

(((-8*I)/5)*E^(((5*I)/2)*ArcTan[a*x])*Hypergeometric2F1[5/4, 2, 9/4, -E^(( 
2*I)*ArcTan[a*x])])/a
 

Rubi [A] (warning: unable to verify)

Time = 0.62 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.23, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5584, 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.

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

\(\Big \downarrow \) 5584

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(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
 

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 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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07 \[ \int e^{\frac {1}{2} i \arctan (a x)} \, dx=\frac {a \sqrt {\frac {i}{a^{2}}} \log \left (i \, a \sqrt {\frac {i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - a \sqrt {\frac {i}{a^{2}}} \log \left (-i \, a \sqrt {\frac {i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + a \sqrt {-\frac {i}{a^{2}}} \log \left (i \, a \sqrt {-\frac {i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) - a \sqrt {-\frac {i}{a^{2}}} \log \left (-i \, a \sqrt {-\frac {i}{a^{2}}} + \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}\right ) + 2 \, {\left (a x + i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, a} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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