3.1.33 \(\int \frac {e^{4 i \text {ArcTan}(a x)}}{x^3} \, dx\) [33]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {5170, 90} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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]

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^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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 52, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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]

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 53, normalized size = 1.02

method	result	size
default	\(-4 a^{3} \left (\frac {i}{a \left (a x +i\right )}-\frac {2 \ln \left (a x +i\right )}{a}\right )-\frac {1}{2 x^{2}}-\frac {4 i a}{x}-8 a^{2} \ln \left (x \right )\)	\(53\)
risch	\(\frac {-8 i a^{2} x^{2}+\frac {7}{2} a x -\frac {1}{2} i}{\left (a x +i\right ) x^{2}}-8 a^{2} \ln \left (x \right )-8 i a^{2} \arctan \left (a x \right )+4 a^{2} \ln \left (a^{2} x^{2}+1\right )\)	\(62\)
meijerg	\(\frac {a^{2} \left (\frac {3 a^{2} x^{2}}{3 a^{2} x^{2}+3}+2 \ln \left (a^{2} x^{2}+1\right )-1-4 \ln \left (x \right )-2 \ln \left (a^{2}\right )-\frac {1}{a^{2} x^{2}}\right )}{2}+\frac {2 i a^{3} \left (-\frac {2 \left (3 a^{2} x^{2}+2\right )}{x \sqrt {a^{2}}\, \left (2 a^{2} x^{2}+2\right )}-\frac {3 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{\sqrt {a^{2}}}-3 a^{2} \left (-\frac {2 a^{2} x^{2}}{2 a^{2} x^{2}+2}-\ln \left (a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-\frac {2 i a^{3} \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{\sqrt {a^{2}}}+\frac {a^{4} x^{2}}{2 a^{2} x^{2}+2}\)	\(226\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 69, normalized size = 1.33 \begin {gather*} -8 i \, a^{2} \arctan \left (a x\right ) + 4 \, a^{2} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{2} \log \left (x\right ) + \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 )}} \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^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)

________________________________________________________________________________________

Fricas [A]
time = 2.05, size = 77, normalized size = 1.48 \begin {gather*} \frac {-16 i \, a^{2} x^{2} + 7 \, a x - 16 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (x\right ) + 16 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (\frac {a x + i}{a}\right ) - i}{2 \, {\left (a x^{3} + i \, x^{2}\right )}} \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^3,x, algorithm="fricas")

[Out]

1/2*(-16*I*a^2*x^2 + 7*a*x - 16*(a^3*x^3 + I*a^2*x^2)*log(x) + 16*(a^3*x^3 + I*a^2*x^2)*log((a*x + I)/a) - I)/
(a*x^3 + I*x^2)

________________________________________________________________________________________

Sympy [A]
time = 0.18, size = 58, normalized size = 1.12 \begin {gather*} 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}} \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**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)

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 46, normalized size = 0.88 \begin {gather*} 8 \, a^{2} \log \left (a x + i\right ) - 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {16 i \, a^{2} x^{2} - 7 \, a x + i}{2 \, {\left (a x + i\right )} x^{2}} \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^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.47, size = 43, normalized size = 0.83 \begin {gather*} -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 )} \end {gather*}

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

________________________________________________________________________________________