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

Optimal. Leaf size=30 \[ \frac{2}{a (a x+i)}-\frac{i \log (a x+i)}{a} \]

[Out]

2/(a*(I + a*x)) - (I*Log[I + a*x])/a

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Rubi [A]  time = 0.0382867, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5073, 43} \[ \frac{2}{a (a x+i)}-\frac{i \log (a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(a*(I + a*x)) - (I*Log[I + a*x])/a

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])

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0141234, size = 30, normalized size = 1. \[ \frac{2}{a (a x+i)}-\frac{i \log (a x+i)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(a*(I + a*x)) - (I*Log[I + a*x])/a

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Maple [A]  time = 0.044, size = 40, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{a \left ( ax+i \right ) }}-{\frac{{\frac{i}{2}}\ln \left ({a}^{2}{x}^{2}+1 \right ) }{a}}-{\frac{\arctan \left ( ax \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^3/(a^2*x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.47707, size = 59, normalized size = 1.97 \begin{align*} \frac{4 \, a x - 4 i}{2 \,{\left (a^{3} x^{2} + a\right )}} - \frac{\arctan \left (a x\right )}{a} - \frac{i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77536, size = 69, normalized size = 2.3 \begin{align*} \frac{{\left (-i \, a x + 1\right )} \log \left (\frac{a x + i}{a}\right ) + 2}{a^{2} x + i \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.458285, size = 22, normalized size = 0.73 \begin{align*} \frac{2 a}{a^{3} x + i a^{2}} - \frac{i \log{\left (a x + i \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**3/(a**2*x**2+1)**2,x)

[Out]

2*a/(a**3*x + I*a**2) - I*log(a*x + I)/a

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Giac [A]  time = 1.1164, size = 34, normalized size = 1.13 \begin{align*} -\frac{i \log \left (a x + i\right )}{a} + \frac{2}{{\left (a x + i\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^3/(a^2*x^2+1)^2,x, algorithm="giac")

[Out]

-i*log(a*x + i)/a + 2/((a*x + i)*a)