\(\int \frac {e^{4 i \arctan (a x)}}{x} \, dx\) [31]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\frac {4 i}{i+a x}+\log (x) \]

[Out]

4*I/(I+a*x)+ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 90} \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\log (x)+\frac {4 i}{a x+i} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\frac {4 i}{i+a x}+\log (x) \]

[In]

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

[Out]

(4*I)/(I + a*x) + Log[x]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {4 i}{a x +i}+\ln \left (x \right )\)	\(15\)
risch	\(\frac {4 i}{a x +i}+\ln \left (-x \right )\)	\(17\)
norman	\(\frac {-4 a^{2} x^{2}+4 i a x}{a^{2} x^{2}+1}+\ln \left (x \right )\)	\(30\)
parallelrisch	\(\frac {a^{2} \ln \left (x \right ) x^{2}-4 a^{2} x^{2}+4 i a x +\ln \left (x \right )}{a^{2} x^{2}+1}\)	\(38\)
meijerg	\(-\frac {a^{2} x^{2}}{2 a^{2} x^{2}+2}+\frac {1}{2}+\ln \left (x \right )+\frac {\ln \left (a^{2}\right )}{2}+\frac {2 i a \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{\sqrt {a^{2}}}-\frac {7 a^{2} x^{2}}{2 \left (a^{2} x^{2}+1\right )}-\frac {2 i a \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{\sqrt {a^{2}}}\)	\(138\)

[In]

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

[Out]

4*I/(I+a*x)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\frac {{\left (a x + i\right )} \log \left (x\right ) + 4 i}{a x + i} \]

[In]

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

[Out]

((a*x + I)*log(x) + 4*I)/(a*x + I)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\log {\left (x \right )} + \frac {4 i}{a x + i} \]

[In]

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

[Out]

log(x) + 4*I/(a*x + I)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=-\frac {4 \, {\left (-i \, a x - 1\right )}}{a^{2} x^{2} + 1} + \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\frac {4 i}{a x + i} + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

4*I/(a*x + I) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4 i \arctan (a x)}}{x} \, dx=\ln \left (x\right )+\frac {4{}\mathrm {i}}{a\,x+1{}\mathrm {i}} \]

[In]

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

[Out]

log(x) + 4i/(a*x + 1i)