\(\int \frac {e^{i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [323]

Optimal result

Integrand size = 24, antiderivative size = 28 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {1}{2 a (i+a x)}+\frac {\arctan (a x)}{2 a} \]

[Out]

1/2/a/(I+a*x)+1/2*arctan(a*x)/a

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5181, 46, 209} \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {\arctan (a x)}{2 a}+\frac {1}{2 a (a x+i)} \]

[In]

Int[E^(I*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

1/(2*a*(I + a*x)) + ArcTan[a*x]/(2*a)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 209

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

Rule 5181

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + I*(n
/2))*(1 + I*a*x)^(p - I*(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(1-i a x)^2 (1+i a x)} \, dx \\ & = \int \left (-\frac {1}{2 (i+a x)^2}+\frac {1}{2 \left (1+a^2 x^2\right )}\right ) \, dx \\ & = \frac {1}{2 a (i+a x)}+\frac {1}{2} \int \frac {1}{1+a^2 x^2} \, dx \\ & = \frac {1}{2 a (i+a x)}+\frac {\arctan (a x)}{2 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {\frac {1}{i+a x}+\arctan (a x)}{2 a} \]

[In]

Integrate[E^(I*ArcTan[a*x])/(1 + a^2*x^2)^(3/2),x]

[Out]

((I + a*x)^(-1) + ArcTan[a*x])/(2*a)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {1}{2 a \left (a x +i\right )}+\frac {\arctan \left (a x \right )}{2 a}\)	\(24\)
default	\(\frac {2 a^{2} x -2 i a}{4 a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{2 a}\)	\(38\)
meijerg	\(\frac {\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}}{2 \sqrt {a^{2}}}+\frac {i a \,x^{2}}{2 a^{2} x^{2}+2}\)	\(61\)
parallelrisch	\(-\frac {i \ln \left (a x -i\right ) x^{2} a^{2}-i \ln \left (a x +i\right ) x^{2} a^{2}-2 i x^{2} a^{2}+i \ln \left (a x -i\right )-i \ln \left (a x +i\right )-2 a x}{4 \left (a^{2} x^{2}+1\right ) a}\)	\(83\)

[In]

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

[Out]

1/2/a/(I+a*x)+1/2*arctan(a*x)/a

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {{\left (i \, a x - 1\right )} \log \left (\frac {a x + i}{a}\right ) + {\left (-i \, a x + 1\right )} \log \left (\frac {a x - i}{a}\right ) + 2}{4 \, {\left (a^{2} x + i \, a\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=i \left (- \frac {i}{2 a^{2} x + 2 i a} + \frac {- \frac {\log {\left (x - \frac {i}{a} \right )}}{4} + \frac {\log {\left (x + \frac {i}{a} \right )}}{4}}{a}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {a x - i}{2 \, {\left (a^{3} x^{2} + a\right )}} + \frac {\arctan \left (a x\right )}{2 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i \, \log \left (a x + i\right )}{4 \, a} - \frac {i \, \log \left (a x - i\right )}{4 \, a} + \frac {1}{2 \, {\left (a x + i\right )} a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {e^{i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {1}{2\,\left (x\,a^2+a\,1{}\mathrm {i}\right )}+\frac {\mathrm {atan}\left (a\,x\right )}{2\,a} \]

[In]

int((a*x*1i + 1)/(a^2*x^2 + 1)^2,x)

[Out]

1/(2*(a*1i + a^2*x)) + atan(a*x)/(2*a)