\(\int \frac {e^{-3 i \arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [335]

Optimal result

Integrand size = 25, antiderivative size = 49 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {i \sqrt {1+a^2 x^2}}{2 a c (1+i a x)^2 \sqrt {c+a^2 c x^2}} \]

[Out]

1/2*I*(a^2*x^2+1)^(1/2)/a/c/(1+I*a*x)^2/(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5184, 5181, 32} \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {i \sqrt {a^2 x^2+1}}{2 a c (1+i a x)^2 \sqrt {a^2 c x^2+c}} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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])

Rule 5184

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[c^IntPart[p]*((c + d*x^2)^FracP
art[p]/(1 + a^2*x^2)^FracPart[p]), Int[(1 + a^2*x^2)^p*E^(n*ArcTan[a*x]), 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}& = \frac {\sqrt {1+a^2 x^2} \int \frac {e^{-3 i \arctan (a x)}}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {\sqrt {1+a^2 x^2} \int \frac {1}{(1+i a x)^3} \, dx}{c \sqrt {c+a^2 c x^2}} \\ & = \frac {i \sqrt {1+a^2 x^2}}{2 a c (1+i a x)^2 \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {i \sqrt {1+a^2 x^2}}{2 a c (-i+a x)^2 \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {i \sqrt {a^{2} x^{2}+1}}{2 c \sqrt {c \left (a^{2} x^{2}+1\right )}\, a \left (a x -i\right )^{2}}\)	\(42\)
default	\(\frac {i \sqrt {c \left (a^{2} x^{2}+1\right )}}{2 \sqrt {a^{2} x^{2}+1}\, c^{2} a \left (i a x +1\right )^{2}}\)	\(43\)
gosper	\(\frac {\left (-a x +i\right ) \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 a \left (i a x +1\right )^{3} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} \sqrt {a^{2} x^{2} + 1} {\left (-i \, a x^{2} - 2 \, x\right )}}{2 \, {\left (a^{4} c^{2} x^{4} - 2 i \, a^{3} c^{2} x^{3} - 2 i \, a c^{2} x - c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{5} c x^{5} \sqrt {a^{2} c x^{2} + c} - 3 i a^{4} c x^{4} \sqrt {a^{2} c x^{2} + c} - 2 a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - 2 i a^{2} c x^{2} \sqrt {a^{2} c x^{2} + c} - 3 a c x \sqrt {a^{2} c x^{2} + c} + i c \sqrt {a^{2} c x^{2} + c}}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{5} c x^{5} \sqrt {a^{2} c x^{2} + c} - 3 i a^{4} c x^{4} \sqrt {a^{2} c x^{2} + c} - 2 a^{3} c x^{3} \sqrt {a^{2} c x^{2} + c} - 2 i a^{2} c x^{2} \sqrt {a^{2} c x^{2} + c} - 3 a c x \sqrt {a^{2} c x^{2} + c} + i c \sqrt {a^{2} c x^{2} + c}}\, dx\right ) \]

[In]

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

[Out]

I*(Integral(sqrt(a**2*x**2 + 1)/(a**5*c*x**5*sqrt(a**2*c*x**2 + c) - 3*I*a**4*c*x**4*sqrt(a**2*c*x**2 + c) - 2
*a**3*c*x**3*sqrt(a**2*c*x**2 + c) - 2*I*a**2*c*x**2*sqrt(a**2*c*x**2 + c) - 3*a*c*x*sqrt(a**2*c*x**2 + c) + I
*c*sqrt(a**2*c*x**2 + c)), x) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**5*c*x**5*sqrt(a**2*c*x**2 + c) - 3*
I*a**4*c*x**4*sqrt(a**2*c*x**2 + c) - 2*a**3*c*x**3*sqrt(a**2*c*x**2 + c) - 2*I*a**2*c*x**2*sqrt(a**2*c*x**2 +
c) - 3*a*c*x*sqrt(a**2*c*x**2 + c) + I*c*sqrt(a**2*c*x**2 + c)), x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {1}{2 i \, a^{3} c^{\frac {3}{2}} x^{2} + 4 \, a^{2} c^{\frac {3}{2}} x - 2 i \, a c^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,\sqrt {a^2\,x^2+1}}{2\,a\,c^2\,\left (a\,x+1{}\mathrm {i}\right )\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3} \]

[In]

int((a^2*x^2 + 1)^(3/2)/((c + a^2*c*x^2)^(3/2)*(a*x*1i + 1)^3),x)

[Out]

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