\(\int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx\) [308]

Optimal result

Integrand size = 24, antiderivative size = 32 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {2}{a (i-a x)}+\frac {i \log (i-a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5181, 45} \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {i \log (-a x+i)}{a}-\frac {2}{a (-a x+i)} \]

[In]

Int[1/(E^((3*I)*ArcTan[a*x])*Sqrt[1 + a^2*x^2]),x]

[Out]

-2/(a*(I - a*x)) + (I*Log[I - a*x])/a

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5181

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - I*a*x)^(p + I*(n
/2))*(1 + I*a*x)^(p - I*(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[d, a^2*c] && (IntegerQ[p] || GtQ[c,
0])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {2}{a (i-a x)}+\frac {i \log (i-a x)}{a} \]

[In]

Integrate[1/(E^((3*I)*ArcTan[a*x])*Sqrt[1 + a^2*x^2]),x]

[Out]

-2/(a*(I - a*x)) + (I*Log[I - a*x])/a

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {2}{a \left (-a x +i\right )}+\frac {i \ln \left (-a x +i\right )}{a}\)	\(30\)
risch	\(\frac {2}{a \left (a x -i\right )}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}-\frac {\arctan \left (a x \right )}{a}\)	\(40\)
meijerg	\(\frac {i \left (-\frac {i x a \left (9 i a x +6\right )}{3 \left (i a x +1\right )^{2}}+2 \ln \left (i a x +1\right )\right )}{2 a}+\frac {x \left (i a x +2\right )}{2 \left (i a x +1\right )^{2}}\)	\(59\)
parallelrisch	\(\frac {i \ln \left (a x -i\right ) x^{2} a^{2}+2 \ln \left (a x -i\right ) x a -2 i x^{2} a^{2}-i \ln \left (a x -i\right )-2 a x}{\left (-a x +i\right )^{2} a}\)	\(65\)

[In]

int(1/(1+I*a*x)^3*(a^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {{\left (i \, a x + 1\right )} \log \left (\frac {a x - i}{a}\right ) + 2}{a^{2} x - i \, a} \]

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{a^{2} x - i a} + \frac {i \log {\left (a x - i \right )}}{a} \]

[In]

integrate(1/(1+I*a*x)**3*(a**2*x**2+1),x)

[Out]

2/(a**2*x - I*a) + I*log(a*x - I)/a

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {4 \, {\left (-i \, a x - 1\right )}}{2 i \, a^{3} x^{2} + 4 \, a^{2} x - 2 i \, a} + \frac {i \, \log \left (i \, a x + 1\right )}{a} \]

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {i \, \log \left (a x - i\right )}{a} + \frac {2}{{\left (a x - i\right )} a} \]

[In]

integrate(1/(1+I*a*x)^3*(a^2*x^2+1),x, algorithm="giac")

[Out]

I*log(a*x - I)/a + 2/((a*x - I)*a)

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {2}{-a^2\,x+a\,1{}\mathrm {i}}+\frac {\ln \left (a\,x-\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]

[In]

int((a^2*x^2 + 1)/(a*x*1i + 1)^3,x)

[Out]

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