3.1.18 \(\int \frac {e^{2 i \text {ArcTan}(a x)}}{x^4} \, dx\) [18]

Optimal. Leaf size=48 \[ -\frac {1}{3 x^3}-\frac {i a}{x^2}+\frac {2 a^2}{x}-2 i a^3 \log (x)+2 i a^3 \log (i+a x) \]

[Out]

-1/3/x^3-I*a/x^2+2*a^2/x-2*I*a^3*ln(x)+2*I*a^3*ln(I+a*x)

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Rubi [A]
time = 0.02, antiderivative size = 48, 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 = {5170, 78} \begin {gather*} -2 i a^3 \log (x)+2 i a^3 \log (a x+i)+\frac {2 a^2}{x}-\frac {i a}{x^2}-\frac {1}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5170

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^{2 i \tan ^{-1}(a x)}}{x^4} \, dx &=\int \frac {1+i a x}{x^4 (1-i a x)} \, dx\\ &=\int \left (\frac {1}{x^4}+\frac {2 i a}{x^3}-\frac {2 a^2}{x^2}-\frac {2 i a^3}{x}+\frac {2 i a^4}{i+a x}\right ) \, dx\\ &=-\frac {1}{3 x^3}-\frac {i a}{x^2}+\frac {2 a^2}{x}-2 i a^3 \log (x)+2 i a^3 \log (i+a x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 48, normalized size = 1.00 \begin {gather*} -\frac {1}{3 x^3}-\frac {i a}{x^2}+\frac {2 a^2}{x}-2 i a^3 \log (x)+2 i a^3 \log (i+a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 60, normalized size = 1.25

method	result	size
risch	\(\frac {2 a^{2} x^{2}-i a x -\frac {1}{3}}{x^{3}}-2 i a^{3} \ln \left (-x \right )+2 a^{3} \arctan \left (a x \right )+i a^{3} \ln \left (a^{2} x^{2}+1\right )\)	\(56\)
default	\(2 a^{4} \left (\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\right )-\frac {1}{3 x^{3}}-2 i a^{3} \ln \left (x \right )-\frac {i a}{x^{2}}+\frac {2 a^{2}}{x}\)	\(60\)
meijerg	\(\frac {a^{4} \left (\frac {2 a^{2}}{x \left (a^{2}\right )^{\frac {3}{2}}}-\frac {2}{3 x^{3} \left (a^{2}\right )^{\frac {3}{2}}}+\frac {2 a^{3} \arctan \left (a x \right )}{\left (a^{2}\right )^{\frac {3}{2}}}\right )}{2 \sqrt {a^{2}}}+i a^{3} \left (\ln \left (a^{2} x^{2}+1\right )-2 \ln \left (x \right )-\ln \left (a^{2}\right )-\frac {1}{a^{2} x^{2}}\right )-\frac {a^{4} \left (-\frac {2}{x \sqrt {a^{2}}}-\frac {2 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{2 \sqrt {a^{2}}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 51, normalized size = 1.06 \begin {gather*} 2 \, a^{3} \arctan \left (a x\right ) + i \, a^{3} \log \left (a^{2} x^{2} + 1\right ) - 2 i \, a^{3} \log \left (x\right ) + \frac {6 \, a^{2} x^{2} - 3 i \, a x - 1}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*arctan(a*x) + I*a^3*log(a^2*x^2 + 1) - 2*I*a^3*log(x) + 1/3*(6*a^2*x^2 - 3*I*a*x - 1)/x^3

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Fricas [A]
time = 2.62, size = 47, normalized size = 0.98 \begin {gather*} \frac {-6 i \, a^{3} x^{3} \log \left (x\right ) + 6 i \, a^{3} x^{3} \log \left (\frac {a x + i}{a}\right ) + 6 \, a^{2} x^{2} - 3 i \, a x - 1}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-6*I*a^3*x^3*log(x) + 6*I*a^3*x^3*log((a*x + I)/a) + 6*a^2*x^2 - 3*I*a*x - 1)/x^3

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Sympy [A]
time = 0.12, size = 54, normalized size = 1.12 \begin {gather*} - 2 a^{3} \left (i \log {\left (4 a^{4} x \right )} - i \log {\left (4 a^{4} x + 4 i a^{3} \right )}\right ) - \frac {- 6 a^{2} x^{2} + 3 i a x + 1}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**3*(I*log(4*a**4*x) - I*log(4*a**4*x + 4*I*a**3)) - (-6*a**2*x**2 + 3*I*a*x + 1)/(3*x**3)

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Giac [A]
time = 0.39, size = 39, normalized size = 0.81 \begin {gather*} 2 i \, a^{3} \log \left (a x + i\right ) - 2 i \, a^{3} \log \left ({\left | x \right |}\right ) + \frac {6 \, a^{2} x^{2} - 3 i \, a x - 1}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*a^3*log(a*x + I) - 2*I*a^3*log(abs(x)) + 1/3*(6*a^2*x^2 - 3*I*a*x - 1)/x^3

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Mupad [B]
time = 0.07, size = 34, normalized size = 0.71 \begin {gather*} 4\,a^3\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )-\frac {-2\,a^2\,x^2+a\,x\,1{}\mathrm {i}+\frac {1}{3}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a^3*atan(2*a*x + 1i) - (a*x*1i - 2*a^2*x^2 + 1/3)/x^3

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