\(\int e^{2 i \arctan (a x)} x \, dx\) [13]

Optimal result

Integrand size = 12, antiderivative size = 29 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i+a x)}{a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5170, 78} \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 \log (a x+i)}{a^2}+\frac {2 i x}{a}-\frac {x^2}{2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i+a x)}{a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {-a^{2} x^{2}+4 i a x +4 \ln \left (a x +i\right )}{2 a^{2}}\)	\(29\)
risch	\(-\frac {x^{2}}{2}+\frac {2 i x}{a}+\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{2}}-\frac {2 i \arctan \left (a x \right )}{a^{2}}\)	\(38\)
default	\(\frac {-\frac {1}{2} a \,x^{2}+2 i x}{a}+\frac {\frac {\ln \left (a^{2} x^{2}+1\right )}{a}-\frac {2 i \arctan \left (a x \right )}{a}}{a}\)	\(46\)
meijerg	\(\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}+\frac {i \left (\frac {2 x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2}}-\frac {2 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{a \sqrt {a^{2}}}-\frac {a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a^{2} x^{2} - 4 i \, a x - 4 \, \log \left (\frac {a x + i}{a}\right )}{2 \, a^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int e^{2 i \arctan (a x)} x \, dx=- \frac {x^{2}}{2} + \frac {2 i x}{a} + \frac {2 \log {\left (a x + i \right )}}{a^{2}} \]

[In]

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

[Out]

-x**2/2 + 2*I*x/a + 2*log(a*x + I)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a x^{2} - 4 i \, x}{2 \, a} - \frac {2 i \, \arctan \left (a x\right )}{a^{2}} + \frac {\log \left (a^{2} x^{2} + 1\right )}{a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a^{2} x^{2} - 4 i \, a x}{2 \, a^{2}} + \frac {2 \, \log \left (a x + i\right )}{a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {x^2}{2}+\frac {x\,2{}\mathrm {i}}{a} \]

[In]

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

[Out]

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