\(\int e^{-i \arctan (a x)} x \, dx\) [37]

Optimal result

Integrand size = 12, antiderivative size = 42 \[ \int e^{-i \arctan (a x)} x \, dx=\frac {(2-i a x) \sqrt {1+a^2 x^2}}{2 a^2}+\frac {i \text {arcsinh}(a x)}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5168, 794, 221} \[ \int e^{-i \arctan (a x)} x \, dx=\frac {i \text {arcsinh}(a x)}{2 a^2}+\frac {\sqrt {a^2 x^2+1} (2-i a x)}{2 a^2} \]

[In]

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

[Out]

((2 - I*a*x)*Sqrt[1 + a^2*x^2])/(2*a^2) + ((I/2)*ArcSinh[a*x])/a^2

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 5168

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n
- 1)/2)*Sqrt[1 + a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1-i a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = \frac {(2-i a x) \sqrt {1+a^2 x^2}}{2 a^2}+\frac {i \int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{2 a} \\ & = \frac {(2-i a x) \sqrt {1+a^2 x^2}}{2 a^2}+\frac {i \text {arcsinh}(a x)}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int e^{-i \arctan (a x)} x \, dx=\frac {(2-i a x) \sqrt {1+a^2 x^2}+i \text {arcsinh}(a x)}{2 a^2} \]

[In]

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

[Out]

((2 - I*a*x)*Sqrt[1 + a^2*x^2] + I*ArcSinh[a*x])/(2*a^2)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40

method	result	size
risch	\(-\frac {i \left (a x +2 i\right ) \sqrt {a^{2} x^{2}+1}}{2 a^{2}}+\frac {i \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 a \sqrt {a^{2}}}\)	\(59\)
default	\(-\frac {i \left (\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}\right )}{a}+\frac {\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}}{a^{2}}\)	\(150\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02 \[ \int e^{-i \arctan (a x)} x \, dx=\frac {\sqrt {a^{2} x^{2} + 1} {\left (-i \, a x + 2\right )} - i \, \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right )}{2 \, a^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int e^{-i \arctan (a x)} x \, dx=- i \int \frac {x \sqrt {a^{2} x^{2} + 1}}{a x - i}\, dx \]

[In]

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

[Out]

-I*Integral(x*sqrt(a**2*x**2 + 1)/(a*x - I), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int e^{-i \arctan (a x)} x \, dx=-\frac {i \, \sqrt {a^{2} x^{2} + 1} x}{2 \, a} + \frac {i \, \operatorname {arsinh}\left (a x\right )}{2 \, a^{2}} + \frac {\sqrt {a^{2} x^{2} + 1}}{a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int e^{-i \arctan (a x)} x \, dx=-\frac {1}{2} \, \sqrt {a^{2} x^{2} + 1} {\left (\frac {i \, x}{a} - \frac {2}{a^{2}}\right )} - \frac {i \, \log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right )}{2 \, a {\left | a \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21 \[ \int e^{-i \arctan (a x)} x \, dx=\frac {\left (\frac {1}{\sqrt {a^2}}-\frac {x\,\sqrt {a^2}\,1{}\mathrm {i}}{2\,a}\right )\,\sqrt {a^2\,x^2+1}+\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )\,1{}\mathrm {i}}{2\,a}}{\sqrt {a^2}} \]

[In]

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

[Out]

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