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

Optimal. Leaf size=41 \[ -\frac {\sinh ^{-1}(a x)}{a}+\frac {2 i \sqrt {1-i a x}}{a \sqrt {1+i a x}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5073, 47, 41, 215} \[ -\frac {\sinh ^{-1}(a x)}{a}+\frac {2 i \sqrt {1-i a x}}{a \sqrt {1+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*Sqrt[1 - I*a*x])/(a*Sqrt[1 + I*a*x]) - ArcSinh[a*x]/a

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 215

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 56, normalized size = 1.37 \[ \frac {2 \left (\sqrt {a^2 x^2+1}+(-1-i a x) \sin ^{-1}\left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{a (a x-i)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 54, normalized size = 1.32 \[ \frac {2 \, a x + {\left (a x - i\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1}\right ) + 2 \, \sqrt {a^{2} x^{2} + 1} - 2 i}{a^{2} x - i \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [B]  time = 0.18, size = 143, normalized size = 3.49 \[ -\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{a^{3} \left (x -\frac {i}{a}\right )^{2}}+\frac {i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a}-\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 33, normalized size = 0.80 \[ -\frac {\operatorname {arsinh}\left (a x\right )}{a} + \frac {2 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{2} x + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsinh(a*x)/a + 2*I*sqrt(a^2*x^2 + 1)/(I*a^2*x + a)

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mupad [B]  time = 0.49, size = 56, normalized size = 1.37 \[ -\frac {\mathrm {asinh}\left (x\,\sqrt {a^2}\right )}{\sqrt {a^2}}-\frac {2\,\sqrt {a^2\,x^2+1}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- asinh(x*(a^2)^(1/2))/(a^2)^(1/2) - (2*(a^2*x^2 + 1)^(1/2))/((((a^2)^(1/2)*1i)/a - x*(a^2)^(1/2))*(a^2)^(1/2)
)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{2} x^{2} - 2 i a x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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