3.4.7 \(\int \frac {e^{-2 i \text {ArcTan}(a x)}}{\sqrt {1+a^2 x^2}} \, dx\) [307]

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

[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 = 0.03, 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 = {5181, 49, 41, 221} \begin {gather*} -\frac {\sinh ^{-1}(a x)}{a}+\frac {2 i \sqrt {1-i a x}}{a \sqrt {1+i a x}} \end {gather*}

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 49

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 221

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 5181

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 = 0.04, size = 56, normalized size = 1.37 \begin {gather*} \frac {2 \left (\sqrt {1+a^2 x^2}+(-1-i a x) \text {ArcSin}\left (\frac {\sqrt {1-i a x}}{\sqrt {2}}\right )\right )}{a (-i+a x)} \end {gather*}

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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (34 ) = 68\).
time = 0.08, size = 149, normalized size = 3.63

method	result	size
default	\(-\frac {\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}-i a \left (\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}\right )}{a^{2}}\)	\(149\)

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,method=_RETURNVERBOSE)

[Out]

-1/a^2*(I/a/(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)+I*a*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 = 0.47, size = 33, normalized size = 0.80 \begin {gather*} -\frac {\operatorname {arsinh}\left (a x\right )}{a} + \frac {2 i \, \sqrt {a^{2} x^{2} + 1}}{i \, a^{2} x + a} \end {gather*}

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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Fricas [A]
time = 3.30, size = 54, normalized size = 1.32 \begin {gather*} \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} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [B]
time = 0.49, size = 56, normalized size = 1.37 \begin {gather*} -\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}} \end {gather*}

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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