3.32 \(\int \frac{e^{4 i \tan ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=38 \[ -\frac{4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac{1}{x} \]

[Out]

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

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Rubi [A]  time = 0.030845, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5062, 88} \[ -\frac{4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/x^2,x]

[Out]

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

Rule 5062

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 - I*a*x)^((I*n)/2))/(1 + I*a*x)^((I*n)/2
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.0242619, size = 38, normalized size = 1. \[ -\frac{4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.049, size = 45, normalized size = 1.2 \begin{align*} -4\,{\frac{a}{ax+i}}-2\,ia\ln \left ({a}^{2}{x}^{2}+1 \right ) -4\,a\arctan \left ( ax \right ) -{x}^{-1}+4\,ia\ln \left ( x \right ) \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^2,x)

[Out]

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

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Maxima [A]  time = 1.53394, size = 72, normalized size = 1.89 \begin{align*} -4 \, a \arctan \left (a x\right ) - 2 i \, a \log \left (a^{2} x^{2} + 1\right ) + 4 i \, a \log \left (x\right ) - \frac{10 \, a^{2} x^{2} - 8 i \, a x + 2}{2 \,{\left (a^{2} x^{3} + x\right )}} \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^2,x, algorithm="maxima")

[Out]

-4*a*arctan(a*x) - 2*I*a*log(a^2*x^2 + 1) + 4*I*a*log(x) - 1/2*(10*a^2*x^2 - 8*I*a*x + 2)/(a^2*x^3 + x)

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Fricas [B]  time = 1.66007, size = 131, normalized size = 3.45 \begin{align*} -\frac{5 \, a x + 4 \,{\left (-i \, a^{2} x^{2} + a x\right )} \log \left (x\right ) + 4 \,{\left (i \, a^{2} x^{2} - a x\right )} \log \left (\frac{a x + i}{a}\right ) + i}{a x^{2} + i \, x} \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^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.608917, size = 37, normalized size = 0.97 \begin{align*} 4 a \left (i \log{\left (x \right )} - i \log{\left (x + \frac{i}{a} \right )}\right ) - \frac{5 a^{2} x + i a}{a^{2} x^{2} + i a x} \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**2,x)

[Out]

4*a*(I*log(x) - I*log(x + I/a)) - (5*a**2*x + I*a)/(a**2*x**2 + I*a*x)

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Giac [A]  time = 1.10653, size = 50, normalized size = 1.32 \begin{align*} -4 \, a i \log \left (a x + i\right ) + 4 \, a i \log \left ({\left | x \right |}\right ) - \frac{5 \, a x + i}{a x^{2} + i x} \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^2,x, algorithm="giac")

[Out]

-4*a*i*log(a*x + i) + 4*a*i*log(abs(x)) - (5*a*x + i)/(a*x^2 + i*x)