3.202 \(\int e^{-2 i \tan ^{-1}(a+b x)} \, dx\)

Optimal. Leaf size=23 \[ -x-\frac {2 i \log (-a-b x+i)}{b} \]

[Out]

-x-2*I*ln(I-a-b*x)/b

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5093, 43} \[ -x-\frac {2 i \log (-a-b x+i)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^((-2*I)*ArcTan[a + b*x]),x]

[Out]

-x - ((2*I)*Log[I - a - b*x])/b

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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^{-2 i \tan ^{-1}(a+b x)} \, dx &=\int \frac {1-i a-i b x}{1+i a+i b x} \, dx\\ &=\int \left (-1-\frac {2 i}{-i+a+b x}\right ) \, dx\\ &=-x-\frac {2 i \log (i-a-b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.39 \[ -\frac {i \log \left ((a+b x)^2+1\right )}{b}+\frac {2 \tan ^{-1}(a+b x)}{b}-x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((-2*I)*ArcTan[a + b*x]),x]

[Out]

-x + (2*ArcTan[a + b*x])/b - (I*Log[1 + (a + b*x)^2])/b

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fricas [A]  time = 0.51, size = 22, normalized size = 0.96 \[ -\frac {b x + 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + 2*I*log((b*x + a - I)/b))/b

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giac [A]  time = 0.13, size = 38, normalized size = 1.65 \[ \frac {{\left (b i x + a i + 1\right )} i}{b} + \frac {2 \, i \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 40, normalized size = 1.74 \[ -x -\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 19, normalized size = 0.83 \[ -x - \frac {2 i \, \log \left (i \, b x + i \, a + 1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*I*log(I*b*x + I*a + 1)/b

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mupad [B]  time = 0.06, size = 21, normalized size = 0.91 \[ -x-\frac {\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,2{}\mathrm {i}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- x - (log(x + (a - 1i)/b)*2i)/b

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sympy [A]  time = 0.22, size = 19, normalized size = 0.83 \[ - x - \frac {2 i \log {\left (i a + i b x + 1 \right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - 2*I*log(I*a + I*b*x + 1)/b

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