3.166 \(\int e^{i \tan ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=52 \[ \frac{\sinh ^{-1}(a+b x)}{b}+\frac{i \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{b} \]

[Out]

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

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Rubi [A]  time = 0.0337835, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5093, 50, 53, 619, 215} \[ \frac{\sinh ^{-1}(a+b x)}{b}+\frac{i \sqrt{-i a-i b x+1} \sqrt{i a+i b x+1}}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(I*ArcTan[a + b*x]),x]

[Out]

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

Rule 5093

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 - I*a*c - I*b*c*x)^((I*n)/2)/(1 + I*a*c +
I*b*c*x)^((I*n)/2), x] /; FreeQ[{a, b, c, n}, x]

Rule 50

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

Rule 53

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

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/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 e^{i \tan ^{-1}(a+b x)} \, dx &=\int \frac{\sqrt{1+i a+i b x}}{\sqrt{1-i a-i b x}} \, dx\\ &=\frac{i \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{b}+\int \frac{1}{\sqrt{1-i a-i b x} \sqrt{1+i a+i b x}} \, dx\\ &=\frac{i \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{b}+\int \frac{1}{\sqrt{(1-i a) (1+i a)+2 a b x+b^2 x^2}} \, dx\\ &=\frac{i \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{b}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{4 b^2}}} \, dx,x,2 a b+2 b^2 x\right )}{2 b^2}\\ &=\frac{i \sqrt{1-i a-i b x} \sqrt{1+i a+i b x}}{b}+\frac{\sinh ^{-1}(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0177386, size = 28, normalized size = 0.54 \[ \frac{\sinh ^{-1}(a+b x)+i \sqrt{(a+b x)^2+1}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(I*ArcTan[a + b*x]),x]

[Out]

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

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Maple [A]  time = 0.088, size = 69, normalized size = 1.3 \begin{align*}{\frac{i}{b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67551, size = 144, normalized size = 2.77 \begin{align*} \frac{i \, a + 2 i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.80244, size = 36, normalized size = 0.69 \begin{align*} \begin{cases} \frac{i \sqrt{\left (a + b x\right )^{2} + 1} + \operatorname{asinh}{\left (a + b x \right )}}{b} & \text{for}\: b \neq 0 \\\frac{x \left (i a + 1\right )}{\sqrt{a^{2} + 1}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((I*sqrt((a + b*x)**2 + 1) + asinh(a + b*x))/b, Ne(b, 0)), (x*(I*a + 1)/sqrt(a**2 + 1), True))

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Giac [A]  time = 1.16162, size = 69, normalized size = 1.33 \begin{align*} \frac{\sqrt{{\left (b x + a\right )}^{2} + 1} i}{b} - \frac{\log \left (-a b -{\left (x{\left | b \right |} - \sqrt{{\left (b x + a\right )}^{2} + 1}\right )}{\left | b \right |}\right )}{{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt((b*x + a)^2 + 1)*i/b - log(-a*b - (x*abs(b) - sqrt((b*x + a)^2 + 1))*abs(b))/abs(b)