3.336 \(\int \frac{e^{-4 i \tan ^{-1}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ \frac{i c (1-i a x)^4}{3 a \left (a^2 c x^2+c\right )^{5/2}}-\frac{i c (1-i a x)^5}{15 a \left (a^2 c x^2+c\right )^{5/2}} \]

[Out]

((I/3)*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))

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Rubi [A]  time = 0.0661462, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {5074, 659, 651} \[ \frac{i c (1-i a x)^4}{3 a \left (a^2 c x^2+c\right )^{5/2}}-\frac{i c (1-i a x)^5}{15 a \left (a^2 c x^2+c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*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 5074

Int[E^(ArcTan[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[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]
) && IGtQ[(I*n)/2, 0]

Rule 659

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 651

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]

Rubi steps

\begin{align*} \int \frac{e^{-4 i \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=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]  time = 0.0392665, size = 77, normalized size = 1.12 \[ \frac{(1-i a x)^{3/2} (a x-4 i) \sqrt{a^2 x^2+1}}{15 a c \sqrt{1+i a x} (a x-i)^2 \sqrt{a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(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])

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Maple [B]  time = 0.072, size = 307, normalized size = 4.5 \begin{align*}{\frac{x}{c}{\frac{1}{\sqrt{{a}^{2}c{x}^{2}+c}}}}-4\,{\frac{1}{{a}^{2}} \left ({\frac{i/5}{ac} \left ( x-{\frac{i}{a}} \right ) ^{-2}{\frac{1}{\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) }}}}+3/5\,ia \left ({\frac{i/3}{ac} \left ( x-{\frac{i}{a}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) }}}}+{\frac{i/3}{a{c}^{2}} \left ( 2\, \left ( x-{\frac{i}{a}} \right ){a}^{2}c+2\,iac \right ){\frac{1}{\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) }}}} \right ) \right ) }+{\frac{4\,i}{a} \left ({\frac{{\frac{i}{3}}}{ac} \left ( x-{\frac{i}{a}} \right ) ^{-1}{\frac{1}{\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) }}}}+{\frac{{\frac{i}{3}}}{a{c}^{2}} \left ( 2\, \left ( x-{\frac{i}{a}} \right ){a}^{2}c+2\,iac \right ){\frac{1}{\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.75508, size = 149, normalized size = 2.16 \begin{align*} -\frac{\sqrt{a^{2} c x^{2} + c}{\left (a^{2} x^{2} - 3 i \, a x + 4\right )}}{15 \, a^{4} c^{2} x^{3} - 45 i \, a^{3} c^{2} x^{2} - 45 \, a^{2} c^{2} x + 15 i \, a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a^2*c*x^2 + c)*(a^2*x^2 - 3*I*a*x + 4)/(15*a^4*c^2*x^3 - 45*I*a^3*c^2*x^2 - 45*a^2*c^2*x + 15*I*a*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.15478, size = 181, normalized size = 2.62 \begin{align*} -\frac{2 \,{\left (5 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{2} c i + 15 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )}^{3} \sqrt{c} + c^{2} i - 5 \,{\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )} c^{\frac{3}{2}}\right )}}{15 \,{\left (\sqrt{c} i - \sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c}\right )}^{5} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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*(5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^2*c*i + 15*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))^3*sqrt(c) + c^
2*i - 5*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*c^(3/2))/((sqrt(c)*i - sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c))^5*a*
c)