\(\int \frac {e^{4 i \arctan (a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [329]

Optimal result

Integrand size = 25, antiderivative size = 69 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}+\frac {i c (1+i a x)^5}{15 a \left (c+a^2 c x^2\right )^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 69, 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, 673, 665} \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {i c (1+i a x)^5}{15 a \left (a^2 c x^2+c\right )^{5/2}}-\frac {i c (1+i a x)^4}{3 a \left (a^2 c x^2+c\right )^{5/2}} \]

[In]

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

[Out]

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

Rule 665

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

Rule 673

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

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^2 \int \frac {(1+i a x)^4}{\left (c+a^2 c x^2\right )^{7/2}} \, dx \\ & = -\frac {i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}-\frac {1}{3} c^2 \int \frac {(1+i a x)^5}{\left (c+a^2 c x^2\right )^{7/2}} \, dx \\ & = -\frac {i c (1+i a x)^4}{3 a \left (c+a^2 c x^2\right )^{5/2}}+\frac {i c (1+i a x)^5}{15 a \left (c+a^2 c x^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {(1+i a x)^{3/2} (4 i+a x) \sqrt {1+a^2 x^2}}{15 a c \sqrt {1-i a x} (i+a x)^2 \sqrt {c+a^2 c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {\left (-a x +i\right ) \left (a x +i\right ) \left (a x +4 i\right ) \left (i a x +1\right )^{4}}{15 a \left (a^{2} x^{2}+1\right )^{2} \left (a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\)	\(57\)
trager	\(\frac {\left (-a^{5} x^{5}-10 a^{3} x^{3}+20 i x^{2} a^{2}+15 a x -4 i\right ) \sqrt {a^{2} c \,x^{2}+c}}{15 c^{2} \left (a^{2} x^{2}+1\right )^{3} a}\)	\(64\)
default	\(\frac {x}{c \sqrt {a^{2} c \,x^{2}+c}}+\frac {2 \left (i \sqrt {-a^{2}}-a \right ) \left (\frac {1}{5 c \sqrt {-a^{2}}\, {\left (x +\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{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 )}}+\frac {3 a^{2} \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 )}{5 \sqrt {-a^{2}}}\right )}{a^{3}}-\frac {2 \left (i \sqrt {-a^{2}}+a \right ) \left (-\frac {1}{5 c \sqrt {-a^{2}}\, {\left (x -\frac {\sqrt {-a^{2}}}{a^{2}}\right )}^{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 )}}-\frac {3 a^{2} \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 )}{5 \sqrt {-a^{2}}}\right )}{a^{3}}-\frac {2 \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 {2 \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}}}\)	\(940\)

[In]

int((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=-\frac {\sqrt {a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 i \, a x + 4\right )}}{15 \, {\left (a^{4} c^{2} x^{3} + 3 i \, a^{3} c^{2} x^{2} - 3 \, a^{2} c^{2} x - i \, a c^{2}\right )}} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((1+I*a*x)**4/(a**2*x**2+1)**2/(a**2*c*x**2+c)**(3/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

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. 134 vs. \(2 (53) = 106\).

Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.94 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{3} \sqrt {c} - 5 i \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )}^{2} c - 5 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c}\right )} c^{\frac {3}{2}} - i \, c^{2}\right )}}{15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} + c} + i \, \sqrt {c}\right )}^{5} a c} \]

[In]

integrate((1+I*a*x)^4/(a^2*x^2+1)^2/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.67 \[ \int \frac {e^{4 i \arctan (a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,\left (a^2\,x^2\,1{}\mathrm {i}-3\,a\,x+4{}\mathrm {i}\right )}{15\,a\,c^2\,{\left (-1+a\,x\,1{}\mathrm {i}\right )}^3} \]

[In]

int((a*x*1i + 1)^4/((c + a^2*c*x^2)^(3/2)*(a^2*x^2 + 1)^2),x)

[Out]

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