3.33 \(\int \frac {e^{4 i \tan ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=52 \[ -\frac {4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac {4 i a}{x}-\frac {1}{2 x^2} \]

[Out]

-1/2/x^2-4*I*a/x-4*I*a^2/(I+a*x)-8*a^2*ln(x)+8*a^2*ln(I+a*x)

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Rubi [A]  time = 0.04, antiderivative size = 52, 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} \[ -\frac {4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac {4 i a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^((4*I)*ArcTan[a*x])/x^3,x]

[Out]

-1/(2*x^2) - ((4*I)*a)/x - ((4*I)*a^2)/(I + a*x) - 8*a^2*Log[x] + 8*a^2*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^3} \, dx &=\int \frac {(1+i a x)^2}{x^3 (1-i a x)^2} \, dx\\ &=\int \left (\frac {1}{x^3}+\frac {4 i a}{x^2}-\frac {8 a^2}{x}+\frac {4 i a^3}{(i+a x)^2}+\frac {8 a^3}{i+a x}\right ) \, dx\\ &=-\frac {1}{2 x^2}-\frac {4 i a}{x}-\frac {4 i a^2}{i+a x}-8 a^2 \log (x)+8 a^2 \log (i+a x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 52, normalized size = 1.00 \[ -\frac {4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac {4 i a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((4*I)*ArcTan[a*x])/x^3,x]

[Out]

-1/2*1/x^2 - ((4*I)*a)/x - ((4*I)*a^2)/(I + a*x) - 8*a^2*Log[x] + 8*a^2*Log[I + a*x]

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fricas [A]  time = 0.47, size = 78, normalized size = 1.50 \[ \frac {-16 i \, a^{2} x^{2} + 7 \, a x - {\left (16 \, a^{3} x^{3} + 16 i \, a^{2} x^{2}\right )} \log \relax (x) + {\left (16 \, a^{3} x^{3} + 16 i \, a^{2} x^{2}\right )} \log \left (\frac {a x + i}{a}\right ) - i}{2 \, a x^{3} + 2 i \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="fricas")

[Out]

(-16*I*a^2*x^2 + 7*a*x - (16*a^3*x^3 + 16*I*a^2*x^2)*log(x) + (16*a^3*x^3 + 16*I*a^2*x^2)*log((a*x + I)/a) - I
)/(2*a*x^3 + 2*I*x^2)

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giac [A]  time = 0.12, size = 47, normalized size = 0.90 \[ 8 \, a^{2} \log \left (a x + i\right ) - 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {16 \, a^{2} i x^{2} - 7 \, a x + i}{2 \, {\left (a x + i\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="giac")

[Out]

8*a^2*log(a*x + i) - 8*a^2*log(abs(x)) - 1/2*(16*a^2*i*x^2 - 7*a*x + i)/((a*x + i)*x^2)

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maple [A]  time = 0.06, size = 60, normalized size = 1.15 \[ -\frac {1}{2 x^{2}}-\frac {4 i a}{x}-8 a^{2} \ln \relax (x )-\frac {4 i a^{2}}{a x +i}+4 a^{2} \ln \left (a^{2} x^{2}+1\right )-8 i a^{2} \arctan \left (a x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x)

[Out]

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

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maxima [A]  time = 0.42, size = 69, normalized size = 1.33 \[ -8 i \, a^{2} \arctan \left (a x\right ) + 4 \, a^{2} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{2} \log \relax (x) + \frac {-16 i \, a^{3} x^{3} - 9 \, a^{2} x^{2} - 8 i \, a x - 1}{2 \, {\left (a^{2} x^{4} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/x^3,x, algorithm="maxima")

[Out]

-8*I*a^2*arctan(a*x) + 4*a^2*log(a^2*x^2 + 1) - 8*a^2*log(x) + 1/2*(-16*I*a^3*x^3 - 9*a^2*x^2 - 8*I*a*x - 1)/(
a^2*x^4 + x^2)

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mupad [B]  time = 0.47, size = 43, normalized size = 0.83 \[ -a^2\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )\,16{}\mathrm {i}+\frac {8\,a^2\,x^2+\frac {a\,x\,7{}\mathrm {i}}{2}+\frac {1}{2}}{x^2\,\left (-1+a\,x\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x*1i + 1)^4/(x^3*(a^2*x^2 + 1)^2),x)

[Out]

((a*x*7i)/2 + 8*a^2*x^2 + 1/2)/(x^2*(a*x*1i - 1)) - a^2*atan(2*a*x + 1i)*16i

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sympy [A]  time = 0.35, size = 60, normalized size = 1.15 \[ 8 a^{2} \left (- \log {\left (16 a^{3} x \right )} + \log {\left (16 a^{3} x + 16 i a^{2} \right )}\right ) + \frac {16 i a^{2} x^{2} - 7 a x + i}{- 2 a x^{3} - 2 i x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2/x**3,x)

[Out]

8*a**2*(-log(16*a**3*x) + log(16*a**3*x + 16*I*a**2)) + (16*I*a**2*x**2 - 7*a*x + I)/(-2*a*x**3 - 2*I*x**2)

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