\(\int e^{2 i \arctan (a+b x)} \, dx\) [175]

Optimal result

Integrand size = 12, antiderivative size = 20 \[ \int e^{2 i \arctan (a+b x)} \, dx=-x+\frac {2 i \log (i+a+b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5201, 45} \[ \int e^{2 i \arctan (a+b x)} \, dx=-x+\frac {2 i \log (a+b x+i)}{b} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int e^{2 i \arctan (a+b x)} \, dx=-x+\frac {2 \arctan (a+b x)}{b}+\frac {i \log \left (1+(a+b x)^2\right )}{b} \]

[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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
parallelrisch	\(\frac {2 i \ln \left (b x +a +i\right )-b x}{b}\)	\(21\)
risch	\(-x +\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (b x +a \right )}{b}\)	\(40\)
default	\(-x +\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int e^{2 i \arctan (a+b x)} \, dx=-\frac {b x - 2 i \, \log \left (\frac {b x + a + i}{b}\right )}{b} \]

[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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int e^{2 i \arctan (a+b x)} \, dx=- x + \frac {2 i \log {\left (a + b x + i \right )}}{b} \]

[In]

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

[Out]

-x + 2*I*log(a + b*x + I)/b

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (16) = 32\).

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.30 \[ \int e^{2 i \arctan (a+b x)} \, dx=-x + \frac {2 \, \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b} + \frac {i \, \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int e^{2 i \arctan (a+b x)} \, dx=-x + \frac {2 i \, \log \left (b x + a + i\right )}{b} \]

[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

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int e^{2 i \arctan (a+b x)} \, dx=-x+\frac {\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,2{}\mathrm {i}}{b} \]

[In]

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

[Out]

(log(x + (a + 1i)/b)*2i)/b - x