3.4.36 \(\int \frac {e^{-4 i \text {ArcTan}(a x)}}{(c+a^2 c x^2)^{3/2}} \, dx\) [336]

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

[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)

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Rubi [A]
time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5182, 673, 665} \begin {gather*} \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}} \end {gather*}

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 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 5182

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]

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*}

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Mathematica [A]
time = 0.03, size = 77, normalized size = 1.12 \begin {gather*} \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}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (57 ) = 114\).
time = 0.07, size = 307, normalized size = 4.45

method	result	size
default	\(\frac {x}{c \sqrt {a^{2} c \,x^{2}+c}}-\frac {4 \left (\frac {i}{5 a c \left (x -\frac {i}{a}\right )^{2} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}+\frac {3 i a \left (\frac {i}{3 a c \left (x -\frac {i}{a}\right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}+\frac {i \left (2 a^{2} c \left (x -\frac {i}{a}\right )+2 i a c \right )}{3 a \,c^{2} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}\right )}{5}\right )}{a^{2}}+\frac {4 i \left (\frac {i}{3 a c \left (x -\frac {i}{a}\right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}+\frac {i \left (2 a^{2} c \left (x -\frac {i}{a}\right )+2 i a c \right )}{3 a \,c^{2} \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2} c +2 i a c \left (x -\frac {i}{a}\right )}}\right )}{a}\)	\(307\)

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,method=_RETURNVERBOSE)

[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*a^2*c*(x-I/a)+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*a^2*c*(x-I/a
)+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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (53) = 106\).
time = 0.27, size = 119, normalized size = 1.72 \begin {gather*} -\frac {x}{15 \, \sqrt {a^{2} c x^{2} + c} c} - \frac {4 i}{5 \, {\left (\sqrt {a^{2} c x^{2} + c} a^{3} c x^{2} - 2 i \, \sqrt {a^{2} c x^{2} + c} a^{2} c x - \sqrt {a^{2} c x^{2} + c} a c\right )}} - \frac {8 i}{15 i \, \sqrt {a^{2} c x^{2} + c} a^{2} c x + 15 \, \sqrt {a^{2} c x^{2} + c} a c} \end {gather*}

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]

-1/15*x/(sqrt(a^2*c*x^2 + c)*c) - 4/5*I/(sqrt(a^2*c*x^2 + c)*a^3*c*x^2 - 2*I*sqrt(a^2*c*x^2 + c)*a^2*c*x - sqr
t(a^2*c*x^2 + c)*a*c) - 8*I/(15*I*sqrt(a^2*c*x^2 + c)*a^2*c*x + 15*sqrt(a^2*c*x^2 + c)*a*c)

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Fricas [A]
time = 2.91, size = 66, normalized size = 0.96 \begin {gather*} -\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 )}} \end {gather*}

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]

-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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^{2} x^{2} + 1\right )^{2}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a x - i\right )^{4}}\, dx \end {gather*}

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]

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

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Giac [B] 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 = 0.47, size = 134, normalized size = 1.94 \begin {gather*} \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} \end {gather*}

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*(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
)

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Mupad [B]
time = 1.00, size = 45, normalized size = 0.65 \begin {gather*} \frac {\sqrt {c\,\left (a^2\,x^2+1\right )}\,\left (a^2\,x^2-a\,x\,3{}\mathrm {i}+4\right )\,1{}\mathrm {i}}{15\,a\,c^2\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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