3.1.32 \(\int \frac {e^{4 i \text {ArcTan}(a x)}}{x^2} \, dx\) [32]

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

[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 = 0.02, 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 = {5170, 90} \begin {gather*} -\frac {4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac {1}{x} \end {gather*}

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 90

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 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^{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 = 0.02, size = 38, normalized size = 1.00 \begin {gather*} -\frac {1}{x}-\frac {4 a}{i+a x}+4 i a \log (x)-4 i a \log (i+a x) \end {gather*}

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 = 0.09, size = 44, normalized size = 1.16

method	result	size
default	\(-4 a^{2} \left (\frac {1}{a \left (a x +i\right )}+\frac {i \ln \left (a x +i\right )}{a}\right )-\frac {1}{x}+4 i a \ln \left (x \right )\)	\(44\)
risch	\(\frac {-5 a x -i}{\left (a x +i\right ) x}+4 i a \ln \left (x \right )-4 a \arctan \left (a x \right )-2 i a \ln \left (a^{2} x^{2}+1\right )\)	\(48\)
meijerg	\(\frac {a^{2} \left (-\frac {2 \left (3 a^{2} x^{2}+2\right )}{x \sqrt {a^{2}}\, \left (2 a^{2} x^{2}+2\right )}-\frac {3 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{2 \sqrt {a^{2}}}+2 i a \left (-\frac {2 a^{2} x^{2}}{2 a^{2} x^{2}+2}-\ln \left (a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-\frac {3 a^{2} \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{\sqrt {a^{2}}}-\frac {2 i a^{3} x^{2}}{a^{2} x^{2}+1}+\frac {a^{2} \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{2 \sqrt {a^{2}}}\)	\(213\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 53, normalized size = 1.39 \begin {gather*} -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 {5 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{2} x^{3} + x} \end {gather*}

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

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Fricas [A]
time = 1.65, size = 60, normalized size = 1.58 \begin {gather*} -\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 {gather*}

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 = 0.16, size = 44, normalized size = 1.16 \begin {gather*} 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} \end {gather*}

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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Giac [A]
time = 0.42, size = 35, normalized size = 0.92 \begin {gather*} -4 i \, a \log \left (a x + i\right ) + 4 i \, a \log \left ({\left | x \right |}\right ) - \frac {5 \, a x + i}{a x^{2} + i \, x} \end {gather*}

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

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Mupad [B]
time = 0.45, size = 37, normalized size = 0.97 \begin {gather*} -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}} \end {gather*}

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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