\(\int \frac {e^{2 i \arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [331]

Optimal result

Integrand size = 25, antiderivative size = 54 \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {2 i (1+i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {x}{3 c \sqrt {c+a^2 c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, 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 = {5183, 667, 197} \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {x}{3 c \sqrt {a^2 c x^2+c}}-\frac {2 i (1+i a x)}{3 a \left (a^2 c x^2+c\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 667

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)*((a + c*x^2)^(p + 1)/(c*(p
+ 1))), x] - Dist[e^2*((p + 2)/(c*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 5183

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

Rubi steps \begin{align*} \text {integral}& = c \int \frac {(1+i a x)^2}{\left (c+a^2 c x^2\right )^{5/2}} \, dx \\ & = -\frac {2 i (1+i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {1}{3} \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx \\ & = -\frac {2 i (1+i a x)}{3 a \left (c+a^2 c x^2\right )^{3/2}}+\frac {x}{3 c \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {(2-i a x) \sqrt {1+i a x} \sqrt {1+a^2 x^2}}{3 a c \sqrt {1-i a x} (i+a x) \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85

method	result	size
trager	\(\frac {\left (a^{3} x^{3}+3 a x -2 i\right ) \sqrt {a^{2} c \,x^{2}+c}}{3 c^{2} \left (a^{2} x^{2}+1\right )^{2} a}\)	\(46\)
gosper	\(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (a x +2 i\right ) \left (i a x +1\right )^{2}}{3 a \left (a^{2} x^{2}+1\right ) \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(57\)
default	\(-\frac {x}{c \sqrt {a^{2} c \,x^{2}+c}}+\frac {\left (i \sqrt {-a^{2}}+a \right ) \left (-\frac {1}{3 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}-\frac {2 \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c +2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c +2 c \sqrt {-a^{2}}\, \left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{a \sqrt {-a^{2}}}+\frac {\left (i \sqrt {-a^{2}}-a \right ) \left (\frac {1}{3 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}+\frac {2 \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right ) a^{2} c -2 c \sqrt {-a^{2}}}{3 c^{2} \sqrt {-a^{2}}\, \sqrt {{\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{2} a^{2} c -2 c \sqrt {-a^{2}}\, \left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}}\right )}{a \sqrt {-a^{2}}}\)	\(398\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {a^{2} c x^{2} + c} {\left (a x + 2 i\right )}}{3 \, {\left (a^{3} c^{2} x^{2} + 2 i \, a^{2} c^{2} x - a c^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.41 \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {2 \, \sqrt {a^{2} c} {\left (3 \, \sqrt {a^{2} c} x - 3 \, \sqrt {a^{2} c x^{2} + c} + i \, \sqrt {c}\right )}}{3 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c} + i \, \sqrt {c}\right )}^{3} a^{2} c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.59 \[ \int \frac {e^{2 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {a^3\,x^3+3\,a\,x-2{}\mathrm {i}}{3\,a\,{\left (c\,\left (a^2\,x^2+1\right )\right )}^{3/2}} \]

[In]

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

[Out]

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