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]

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

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.44, size = 60, normalized size = 1.15 \[ \frac {i \, a + 2 i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \, b} \]

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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giac [A]  time = 0.48, size = 51, normalized size = 0.98 \[ \frac {\sqrt {{\left (b x + a\right )}^{2} + 1} i}{b} - \frac {\log \left (-a b - {\left (x {\left | b \right |} - \sqrt {{\left (b x + a\right )}^{2} + 1}\right )} {\left | b \right |}\right )}{{\left | b \right |}} \]

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)

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maple [A]  time = 0.11, size = 69, normalized size = 1.33 \[ \frac {i \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b}+\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}} \]

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 [A]  time = 0.33, size = 62, normalized size = 1.19 \[ \frac {\operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b} + \frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b} \]

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]

arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b + I*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/b

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mupad [B]  time = 1.09, size = 97, normalized size = 1.87 \[ \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}\,1{}\mathrm {i}}{b}+\frac {\mathrm {asinh}\left (a+b\,x\right )}{b}+\frac {a\,\mathrm {asinh}\left (a+b\,x\right )\,1{}\mathrm {i}}{b}-\frac {a\,b^2\,\ln \left (\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}+\frac {x\,b^2+a\,b}{\sqrt {b^2}}\right )\,1{}\mathrm {i}}{{\left (b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2)*1i)/b + asinh(a + b*x)/b + (a*asinh(a + b*x)*1i)/b - (a*b^2*log((a^2 + b^
2*x^2 + 2*a*b*x + 1)^(1/2) + (a*b + b^2*x)/(b^2)^(1/2))*1i)/(b^2)^(3/2)

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sympy [A]  time = 3.41, size = 36, normalized size = 0.69 \[ \begin {cases} \frac {i \sqrt {\left (a + b x\right )^{2} + 1} + \operatorname {asinh}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\\frac {x \left (i a + 1\right )}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]

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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