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

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]

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

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Rubi [A]  time = 0.0346615, antiderivative size = 41, normalized size of antiderivative = 1., 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[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 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 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 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 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]

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*}

Mathematica [A]  time = 0.0296359, size = 52, normalized size = 1.27 \[ -\frac{2 i \left (\frac{\sqrt{1+i a x}}{\sqrt{1-i a x}}+\sin ^{-1}\left (\frac{\sqrt{1-i a x}}{\sqrt{2}}\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.064, size = 63, normalized size = 1.5 \begin{align*} 2\,{\frac{x}{\sqrt{{a}^{2}{x}^{2}+1}}}-{\ln \left ({{a}^{2}x{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{2\,i}{a}{\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.988022, size = 66, normalized size = 1.61 \begin{align*} \frac{2 \, x}{\sqrt{a^{2} x^{2} + 1}} - \frac{\operatorname{arsinh}\left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} - \frac{2 i}{\sqrt{a^{2} x^{2} + 1} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.76655, size = 126, normalized size = 3.07 \begin{align*} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(3/2),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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (i a x + 1\right )^{2}}{\left (a^{2} x^{2} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*a*x)^2/(a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

undef