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

Optimal. Leaf size=49 \[ \frac{i \sqrt{a^2 x^2+1}}{2 a c (1+i a x)^2 \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.0720498, antiderivative size = 49, 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 = {5076, 5073, 32} \[ \frac{i \sqrt{a^2 x^2+1}}{2 a c (1+i a x)^2 \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[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 5076

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])

Rule 5073

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 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]

Rubi steps

\begin{align*} \int \frac{e^{-3 i \tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=\frac{\sqrt{1+a^2 x^2} \int \frac{e^{-3 i \tan ^{-1}(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]  time = 0.0405942, size = 57, normalized size = 1.16 \[ -\frac{i \sqrt{a^2 x^2+1} \sqrt{a^2 c x^2+c}}{2 a c^2 (a x-i)^3 (a x+i)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.041, size = 45, normalized size = 0.9 \begin{align*}{\frac{-ax+i}{2\,a \left ( 1+iax \right ) ^{3}} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}} \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02254, size = 39, normalized size = 0.8 \begin{align*} \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 2.23682, size = 151, normalized size = 3.08 \begin{align*} \frac{\sqrt{a^{2} c x^{2} + c} \sqrt{a^{2} x^{2} + 1}{\left (-i \, a x^{2} - 2 \, x\right )}}{2 \, a^{4} c^{2} x^{4} - 4 i \, a^{3} c^{2} x^{3} - 4 i \, a c^{2} x - 2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}}{\left (i \, a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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