\(\int \frac {e^{-4 i \arctan (a x)}}{(1+a^2 x^2)^{3/2}} \, dx\) [327]

Optimal result

Integrand size = 24, antiderivative size = 67 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}}+\frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}} \]

[Out]

1/5*I*(1-I*a*x)^(3/2)/a/(1+I*a*x)^(5/2)+1/15*I*(1-I*a*x)^(3/2)/a/(1+I*a*x)^(3/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5181, 47, 37} \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}}+\frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}} \]

[In]

Int[1/(E^((4*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

((I/5)*(1 - I*a*x)^(3/2))/(a*(1 + I*a*x)^(5/2)) + ((I/15)*(1 - I*a*x)^(3/2))/(a*(1 + I*a*x)^(3/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 {\sqrt {1-i a x}}{(1+i a x)^{7/2}} \, dx \\ & = \frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}}+\frac {1}{5} \int \frac {\sqrt {1-i a x}}{(1+i a x)^{5/2}} \, dx \\ & = \frac {i (1-i a x)^{3/2}}{5 a (1+i a x)^{5/2}}+\frac {i (1-i a x)^{3/2}}{15 a (1+i a x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {(1-i a x)^{3/2} (-4 i+a x)}{15 a \sqrt {1+i a x} (-i+a x)^2} \]

[In]

Integrate[1/(E^((4*I)*ArcTan[a*x])*(1 + a^2*x^2)^(3/2)),x]

[Out]

((1 - I*a*x)^(3/2)*(-4*I + a*x))/(15*a*Sqrt[1 + I*a*x]*(-I + a*x)^2)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (-a x +4 i\right ) \sqrt {a^{2} x^{2}+1}}{15 a \left (i a x +1\right )^{4}}\)	\(46\)
default	\(\frac {\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {i}{a}\right )^{4}}-\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {i}{a}\right )^{3}}}{a^{4}}\)	\(92\)

[In]

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

[Out]

-1/15*(I-a*x)*(I+a*x)*(-a*x+4*I)*(a^2*x^2+1)^(1/2)/a/(1+I*a*x)^4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=-\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 3 \, a x + {\left (a^{2} x^{2} - 3 i \, a x + 4\right )} \sqrt {a^{2} x^{2} + 1} + i}{15 \, {\left (a^{4} x^{3} - 3 i \, a^{3} x^{2} - 3 \, a^{2} x + i \, a\right )}} \]

[In]

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

[Out]

-1/15*(a^3*x^3 - 3*I*a^2*x^2 - 3*a*x + (a^2*x^2 - 3*I*a*x + 4)*sqrt(a^2*x^2 + 1) + I)/(a^4*x^3 - 3*I*a^3*x^2 -
3*a^2*x + I*a)

Sympy [F]

\[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {a^{2} x^{2} + 1}}{\left (a x - i\right )^{4}}\, dx \]

[In]

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

[Out]

Integral(sqrt(a**2*x**2 + 1)/(a*x - I)**4, x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (43) = 86\).

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {2 i \, \sqrt {a^{2} x^{2} + 1}}{-5 i \, a^{4} x^{3} - 15 \, a^{3} x^{2} + 15 i \, a^{2} x + 5 \, a} + \frac {i \, \sqrt {a^{2} x^{2} + 1}}{15 \, {\left (a^{3} x^{2} - 2 i \, a^{2} x - a\right )}} - \frac {i \, \sqrt {a^{2} x^{2} + 1}}{15 i \, a^{2} x + 15 \, a} \]

[In]

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

[Out]

2*I*sqrt(a^2*x^2 + 1)/(-5*I*a^4*x^3 - 15*a^3*x^2 + 15*I*a^2*x + 5*a) + 1/15*I*sqrt(a^2*x^2 + 1)/(a^3*x^2 - 2*I
*a^2*x - a) - I*sqrt(a^2*x^2 + 1)/(15*I*a^2*x + 15*a)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (43) = 86\).

Time = 0.30 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.66 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (4 \, a^{4} - 25 \, a^{2} {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{2} - 15 i \, a {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{3} + 15 \, {\left (\sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{4} - 5 \, a^{3} {\left (-i \, \sqrt {a^{2} + \frac {1}{x^{2}}} + \frac {i}{x}\right )}\right )}}{15 \, {\left (-i \, a + \sqrt {a^{2} + \frac {1}{x^{2}}} - \frac {1}{x}\right )}^{5}} \]

[In]

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

[Out]

2/15*(4*a^4 - 25*a^2*(sqrt(a^2 + 1/x^2) - 1/x)^2 - 15*I*a*(sqrt(a^2 + 1/x^2) - 1/x)^3 + 15*(sqrt(a^2 + 1/x^2)
- 1/x)^4 - 5*a^3*(-I*sqrt(a^2 + 1/x^2) + I/x))/(-I*a + sqrt(a^2 + 1/x^2) - 1/x)^5

Mupad [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-4 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^2\,x^2+1}\,\left (a^2\,x^2-a\,x\,3{}\mathrm {i}+4\right )\,1{}\mathrm {i}}{15\,a\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3} \]

[In]

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

[Out]

((a^2*x^2 + 1)^(1/2)*(a^2*x^2 - a*x*3i + 4)*1i)/(15*a*(a*x*1i + 1)^3)