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

Optimal. Leaf size=62 \[ \frac{4 a^3}{a x+i}+\frac{8 a^2}{x}-12 i a^3 \log (x)+12 i a^3 \log (a x+i)-\frac{2 i a}{x^2}-\frac{1}{3 x^3} \]

[Out]

-1/(3*x^3) - ((2*I)*a)/x^2 + (8*a^2)/x + (4*a^3)/(I + a*x) - (12*I)*a^3*Log[x] + (12*I)*a^3*Log[I + a*x]

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Rubi [A]  time = 0.0390173, antiderivative size = 62, 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} \[ \frac{4 a^3}{a x+i}+\frac{8 a^2}{x}-12 i a^3 \log (x)+12 i a^3 \log (a x+i)-\frac{2 i a}{x^2}-\frac{1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*x^3) - ((2*I)*a)/x^2 + (8*a^2)/x + (4*a^3)/(I + a*x) - (12*I)*a^3*Log[x] + (12*I)*a^3*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^4} \, dx &=\int \frac{(1+i a x)^2}{x^4 (1-i a x)^2} \, dx\\ &=\int \left (\frac{1}{x^4}+\frac{4 i a}{x^3}-\frac{8 a^2}{x^2}-\frac{12 i a^3}{x}-\frac{4 a^4}{(i+a x)^2}+\frac{12 i a^4}{i+a x}\right ) \, dx\\ &=-\frac{1}{3 x^3}-\frac{2 i a}{x^2}+\frac{8 a^2}{x}+\frac{4 a^3}{i+a x}-12 i a^3 \log (x)+12 i a^3 \log (i+a x)\\ \end{align*}

Mathematica [A]  time = 0.034766, size = 62, normalized size = 1. \[ \frac{4 a^3}{a x+i}+\frac{8 a^2}{x}-12 i a^3 \log (x)+12 i a^3 \log (a x+i)-\frac{2 i a}{x^2}-\frac{1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*x^3) - ((2*I)*a)/x^2 + (8*a^2)/x + (4*a^3)/(I + a*x) - (12*I)*a^3*Log[x] + (12*I)*a^3*Log[I + a*x]

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Maple [A]  time = 0.052, size = 68, normalized size = 1.1 \begin{align*} 4\,{\frac{{a}^{3}}{ax+i}}+6\,i{a}^{3}\ln \left ({a}^{2}{x}^{2}+1 \right ) +12\,{a}^{3}\arctan \left ( ax \right ) -{\frac{1}{3\,{x}^{3}}}-12\,i{a}^{3}\ln \left ( x \right ) -{\frac{2\,ia}{{x}^{2}}}+8\,{\frac{{a}^{2}}{x}} \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^4,x)

[Out]

4*a^3/(a*x+I)+6*I*a^3*ln(a^2*x^2+1)+12*a^3*arctan(a*x)-1/3/x^3-12*I*a^3*ln(x)-2*I*a/x^2+8*a^2/x

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Maxima [A]  time = 1.51194, size = 104, normalized size = 1.68 \begin{align*} 12 \, a^{3} \arctan \left (a x\right ) + 6 i \, a^{3} \log \left (a^{2} x^{2} + 1\right ) - 12 i \, a^{3} \log \left (x\right ) + \frac{72 \, a^{4} x^{4} - 36 i \, a^{3} x^{3} + 46 \, a^{2} x^{2} - 12 i \, a x - 2}{6 \,{\left (a^{2} x^{5} + x^{3}\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^4,x, algorithm="maxima")

[Out]

12*a^3*arctan(a*x) + 6*I*a^3*log(a^2*x^2 + 1) - 12*I*a^3*log(x) + 1/6*(72*a^4*x^4 - 36*I*a^3*x^3 + 46*a^2*x^2
- 12*I*a*x - 2)/(a^2*x^5 + x^3)

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Fricas [A]  time = 1.78426, size = 189, normalized size = 3.05 \begin{align*} \frac{36 \, a^{3} x^{3} + 18 i \, a^{2} x^{2} + 5 \, a x - 36 \,{\left (i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (x\right ) - 36 \,{\left (-i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\frac{a x + i}{a}\right ) - i}{3 \,{\left (a x^{4} + i \, x^{3}\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^4,x, algorithm="fricas")

[Out]

1/3*(36*a^3*x^3 + 18*I*a^2*x^2 + 5*a*x - 36*(I*a^4*x^4 - a^3*x^3)*log(x) - 36*(-I*a^4*x^4 + a^3*x^3)*log((a*x
+ I)/a) - I)/(a*x^4 + I*x^3)

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Sympy [A]  time = 0.758275, size = 63, normalized size = 1.02 \begin{align*} 12 a^{3} \left (- i \log{\left (x \right )} + i \log{\left (x + \frac{i}{a} \right )}\right ) + \frac{36 a^{4} x^{3} + 18 i a^{3} x^{2} + 5 a^{2} x - i a}{3 a^{2} x^{4} + 3 i a x^{3}} \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**4,x)

[Out]

12*a**3*(-I*log(x) + I*log(x + I/a)) + (36*a**4*x**3 + 18*I*a**3*x**2 + 5*a**2*x - I*a)/(3*a**2*x**4 + 3*I*a*x
**3)

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Giac [A]  time = 1.10326, size = 80, normalized size = 1.29 \begin{align*} 12 \, a^{3} i \log \left (a x + i\right ) - 12 \, a^{3} i \log \left ({\left | x \right |}\right ) + \frac{36 \, a^{3} x^{3} + 18 \, a^{2} i x^{2} + 5 \, a x - i}{3 \,{\left (a x + i\right )} x^{3}} \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^4,x, algorithm="giac")

[Out]

12*a^3*i*log(a*x + i) - 12*a^3*i*log(abs(x)) + 1/3*(36*a^3*x^3 + 18*a^2*i*x^2 + 5*a*x - i)/((a*x + i)*x^3)