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]

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

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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]))

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]

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*}

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Mathematica [A]  time = 0.03, size = 38, normalized size = 1.00 \[ -\frac {4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac {1}{x} \]

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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fricas [A]  time = 0.42, size = 60, normalized size = 1.58 \[ -\frac {5 \, a x + 4 \, {\left (-i \, a^{2} x^{2} + a x\right )} \log \relax (x) + 4 \, {\left (i \, a^{2} x^{2} - a x\right )} \log \left (\frac {a x + i}{a}\right ) + i}{a x^{2} + i \, x} \]

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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giac [A]  time = 0.14, size = 37, normalized size = 0.97 \[ -4 \, a i \log \left (a x + i\right ) + 4 \, a i \log \left ({\left | x \right |}\right ) - \frac {5 \, a x + i}{a x^{2} + i x} \]

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)

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maple [A]  time = 0.06, size = 45, normalized size = 1.18 \[ -\frac {1}{x}+4 i a \ln \relax (x )-\frac {4 a}{a x +i}-2 i a \ln \left (a^{2} x^{2}+1\right )-4 a \arctan \left (a x \right ) \]

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]

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

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maxima [A]  time = 0.42, size = 53, normalized size = 1.39 \[ -4 \, a \arctan \left (a x\right ) - 2 i \, a \log \left (a^{2} x^{2} + 1\right ) + 4 i \, a \log \relax (x) - \frac {10 \, a^{2} x^{2} - 8 i \, a x + 2}{2 \, {\left (a^{2} x^{3} + x\right )}} \]

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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mupad [B]  time = 0.45, size = 37, normalized size = 0.97 \[ -8\,a\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )-\frac {5\,x+\frac {1{}\mathrm {i}}{a}}{x^2+\frac {x\,1{}\mathrm {i}}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 8*a*atan(2*a*x + 1i) - (5*x + 1i/a)/((x*1i)/a + x^2)

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sympy [A]  time = 0.30, size = 44, normalized size = 1.16 \[ 4 a \left (i \log {\left (8 a^{2} x \right )} - i \log {\left (8 a^{2} x + 8 i a \right )}\right ) + \frac {5 a x + i}{- a x^{2} - i x} \]

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(8*a**2*x) - I*log(8*a**2*x + 8*I*a)) + (5*a*x + I)/(-a*x**2 - I*x)

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