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

Optimal. Leaf size=20 \[ -x+\frac{2 i \log (a+b x+i)}{b} \]

[Out]

-x + ((2*I)*Log[I + a + b*x])/b

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Rubi [A]  time = 0.0118928, antiderivative size = 20, normalized size of antiderivative = 1., 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 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 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])

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*}

Mathematica [A]  time = 0.0112383, size = 32, normalized size = 1.6 \[ \frac{i \log \left ((a+b x)^2+1\right )}{b}+\frac{2 \tan ^{-1}(a+b x)}{b}-x \]

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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Maple [B]  time = 0.039, size = 51, normalized size = 2.6 \begin{align*} -x+{\frac{i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{b}}+2\,{\frac{1}{b}\arctan \left ( 1/2\,{\frac{2\,{b}^{2}x+2\,ab}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [B]  time = 1.45703, size = 62, normalized size = 3.1 \begin{align*} -x + \frac{2 \, \arctan \left (\frac{b^{2} x + a b}{b}\right )}{b} + \frac{i \, \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.73152, size = 50, normalized size = 2.5 \begin{align*} -\frac{b x - 2 i \, \log \left (\frac{b x + a + i}{b}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.372951, size = 14, normalized size = 0.7 \begin{align*} - x + \frac{2 i \log{\left (a + b x + i \right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 1.11232, size = 23, normalized size = 1.15 \begin{align*} -x + \frac{2 \, i \log \left (b x + a + i\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*i*log(b*x + a + i)/b