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

Optimal. Leaf size=63 \[ -\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a \sqrt{c}}+\frac{2 i (1-i a x)}{a \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.0611335, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {5074, 653, 217, 206} \[ -\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{a \sqrt{c}}+\frac{2 i (1-i a x)}{a \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 653

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{-2 i \tan ^{-1}(a x)}}{\sqrt{c+a^2 c x^2}} \, dx &=c \int \frac{(1-i a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac{2 i (1-i a x)}{a \sqrt{c+a^2 c x^2}}-\int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx\\ &=\frac{2 i (1-i a x)}{a \sqrt{c+a^2 c x^2}}-\operatorname{Subst}\left (\int \frac{1}{1-a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c+a^2 c x^2}}\right )\\ &=\frac{2 i (1-i a x)}{a \sqrt{c+a^2 c x^2}}-\frac{\tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c+a^2 c x^2}}\right )}{a \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0577683, size = 117, normalized size = 1.86 \[ \frac{2 \sqrt{a^2 x^2+1} \left ((1-i a x) \sqrt{1+i a x}-i \sqrt{1-i a x} (a x-i) \sin ^{-1}\left (\frac{\sqrt{1-i a x}}{\sqrt{2}}\right )\right )}{a \sqrt{1-i a x} (a x-i) \sqrt{a^2 c x^2+c}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^((2*I)*ArcTan[a*x])*Sqrt[c + a^2*c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.201, size = 87, normalized size = 1.4 \begin{align*} -{\ln \left ({{a}^{2}cx{\frac{1}{\sqrt{{a}^{2}c}}}}+\sqrt{{a}^{2}c{x}^{2}+c} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+2\,{\frac{1}{{a}^{2}c}\sqrt{ \left ( x-{\frac{i}{a}} \right ) ^{2}{a}^{2}c+2\,iac \left ( x-{\frac{i}{a}} \right ) } \left ( x-{\frac{i}{a}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99296, size = 332, normalized size = 5.27 \begin{align*} -\frac{{\left (a^{2} c x - i \, a c\right )} \sqrt{\frac{1}{a^{2} c}} \log \left (\frac{2 \,{\left (a^{2} c x + \sqrt{a^{2} c x^{2} + c} a^{2} c \sqrt{\frac{1}{a^{2} c}}\right )}}{x}\right ) -{\left (a^{2} c x - i \, a c\right )} \sqrt{\frac{1}{a^{2} c}} \log \left (\frac{2 \,{\left (a^{2} c x - \sqrt{a^{2} c x^{2} + c} a^{2} c \sqrt{\frac{1}{a^{2} c}}\right )}}{x}\right ) - 4 \, \sqrt{a^{2} c x^{2} + c}}{2 \, a^{2} c x - 2 i \, a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32336, size = 101, normalized size = 1.6 \begin{align*} -i^{2}{\left (\frac{\log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{a \sqrt{c}} + \frac{4}{{\left ({\left (\sqrt{a^{2} c} x - \sqrt{a^{2} c x^{2} + c}\right )} i + \sqrt{c}\right )} a}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-i^2*(log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/(a*sqrt(c)) + 4/(((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 + c))*i
+ sqrt(c))*a))