3.30 \(\int e^{4 i \tan ^{-1}(a x)} \, dx\)

Optimal. Leaf size=31 \[ \frac{4}{a (a x+i)}-\frac{4 i \log (a x+i)}{a}+x \]

[Out]

x + 4/(a*(I + a*x)) - ((4*I)*Log[I + a*x])/a

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Rubi [A]  time = 0.0140897, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5061, 43} \[ \frac{4}{a (a x+i)}-\frac{4 i \log (a x+i)}{a}+x \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x]),x]

[Out]

x + 4/(a*(I + a*x)) - ((4*I)*Log[I + a*x])/a

Rule 5061

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]

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 e^{4 i \tan ^{-1}(a x)} \, dx &=\int \frac{(1+i a x)^2}{(1-i a x)^2} \, dx\\ &=\int \left (1-\frac{4}{(i+a x)^2}-\frac{4 i}{i+a x}\right ) \, dx\\ &=x+\frac{4}{a (i+a x)}-\frac{4 i \log (i+a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0204097, size = 42, normalized size = 1.35 \[ -\frac{2 i \log \left (a^2 x^2+1\right )}{a}+\frac{4}{a (a x+i)}-\frac{4 \tan ^{-1}(a x)}{a}+x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((4*I)*ArcTan[a*x]),x]

[Out]

x + 4/(a*(I + a*x)) - (4*ArcTan[a*x])/a - ((2*I)*Log[1 + a^2*x^2])/a

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Maple [A]  time = 0.043, size = 41, normalized size = 1.3 \begin{align*} x+4\,{\frac{1}{a \left ( ax+i \right ) }}-{\frac{2\,i\ln \left ({a}^{2}{x}^{2}+1 \right ) }{a}}-4\,{\frac{\arctan \left ( ax \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.52143, size = 61, normalized size = 1.97 \begin{align*} x + \frac{8 \, a x - 8 i}{2 \,{\left (a^{3} x^{2} + a\right )}} - \frac{4 \, \arctan \left (a x\right )}{a} - \frac{2 i \, \log \left (a^{2} x^{2} + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53931, size = 95, normalized size = 3.06 \begin{align*} \frac{a^{2} x^{2} + i \, a x - 4 \,{\left (i \, a x - 1\right )} \log \left (\frac{a x + i}{a}\right ) + 4}{a^{2} x + i \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.421508, size = 26, normalized size = 0.84 \begin{align*} \frac{4 a}{a^{3} x + i a^{2}} + x - \frac{4 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)**4/(a**2*x**2+1)**2,x)

[Out]

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

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Giac [A]  time = 1.12512, size = 35, normalized size = 1.13 \begin{align*} x - \frac{4 \, i \log \left (a x + i\right )}{a} + \frac{4}{{\left (a x + i\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2,x, algorithm="giac")

[Out]

x - 4*i*log(a*x + i)/a + 4/((a*x + i)*a)