3.323 \(\int \frac {e^{i \tan ^{-1}(a x)}}{(1+a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=28 \[ \frac {\tan ^{-1}(a x)}{2 a}+\frac {1}{2 a (a x+i)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5073, 44, 203} \[ \frac {\tan ^{-1}(a x)}{2 a}+\frac {1}{2 a (a x+i)} \]

Antiderivative was successfully verified.

[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 44

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] && L
tQ[m + n + 2, 0])

Rule 203

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

Rule 5073

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*} \int \frac {e^{i \tan ^{-1}(a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx &=\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 {\tan ^{-1}(a x)}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.75 \[ \frac {\tan ^{-1}(a x)+\frac {1}{a x+i}}{2 a} \]

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.45, size = 49, normalized size = 1.75 \[ \frac {{\left (i \, a x - 1\right )} \log \left (\frac {a x + i}{a}\right ) + {\left (-i \, a x + 1\right )} \log \left (\frac {a x - i}{a}\right ) + 2}{4 \, a^{2} x + 4 i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 41, normalized size = 1.46 \[ \frac {i \log \left (a x + i\right )}{4 \, a} + \frac {\log \left (-a i x - 1\right )}{4 \, a i} + \frac {1}{2 \, {\left (a x + i\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*log(-a*i*x - 1)/(a*i) + 1/2/((a*x + i)*a)

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maple [A]  time = 0.04, size = 38, normalized size = 1.36 \[ \frac {2 a^{2} x -2 i a}{4 a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 28, normalized size = 1.00 \[ \frac {a x - i}{2 \, {\left (a^{3} x^{2} + a\right )}} + \frac {\arctan \left (a x\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.48, size = 23, normalized size = 0.82 \[ \frac {1}{2\,\left (x\,a^2+a\,1{}\mathrm {i}\right )}+\frac {\mathrm {atan}\left (a\,x\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.21, size = 34, normalized size = 1.21 \[ \frac {1}{2 a^{2} x + 2 i a} - \frac {\frac {i \log {\left (x - \frac {i}{a} \right )}}{4} - \frac {i \log {\left (x + \frac {i}{a} \right )}}{4}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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