3.3.2 \(\int e^{-2 i \text {ArcTan}(a+b x)} \, dx\) [202]

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

[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 = {5201, 45} \begin {gather*} -x-\frac {2 i \log (-a-b x+i)}{b} \end {gather*}

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 45

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 5201

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 \begin {gather*} -x+\frac {2 \text {ArcTan}(a+b x)}{b}-\frac {i \log \left (1+(a+b x)^2\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.07, size = 22, normalized size = 0.96

method	result	size
default	\(-x -\frac {2 i \ln \left (-b x -a +i\right )}{b}\)	\(22\)
risch	\(-x -\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (b x +a \right )}{b}\)	\(40\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 19, normalized size = 0.83 \begin {gather*} -x - \frac {2 i \, \log \left (i \, b x + i \, a + 1\right )}{b} \end {gather*}

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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Fricas [A]
time = 2.55, size = 22, normalized size = 0.96 \begin {gather*} -\frac {b x + 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{b} \end {gather*}

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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Sympy [A]
time = 0.07, size = 15, normalized size = 0.65 \begin {gather*} - x - \frac {2 i \log {\left (a + b x - i \right )}}{b} \end {gather*}

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(a + b*x - I)/b

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Giac [A]
time = 0.45, size = 37, normalized size = 1.61 \begin {gather*} \frac {i \, {\left (i \, b x + i \, a + 1\right )}}{b} + \frac {2 i \, \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b} \end {gather*}

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]

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

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

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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