\(\int e^{-2 i \arctan (a+b x)} x \, dx\) [201]

Optimal result

Integrand size = 14, antiderivative size = 40 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=-\frac {2 i x}{b}-\frac {x^2}{2}+\frac {2 (1+i a) \log (i-a-b x)}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5203, 78} \[ \int e^{-2 i \arctan (a+b x)} x \, dx=\frac {2 (1+i a) \log (-a-b x+i)}{b^2}-\frac {2 i x}{b}-\frac {x^2}{2} \]

[In]

Int[x/E^((2*I)*ArcTan[a + b*x]),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5203

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((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, d, e, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1-i a-i b x)}{1+i a+i b x} \, dx \\ & = \int \left (-\frac {2 i}{b}-x+\frac {2 (1+i a)}{b (-i+a+b x)}\right ) \, dx \\ & = -\frac {2 i x}{b}-\frac {x^2}{2}+\frac {2 (1+i a) \log (i-a-b x)}{b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=-\frac {2 i x}{b}-\frac {x^2}{2}+\frac {2 (1+i a) \log (i-a-b x)}{b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\frac {1}{2} x^{2} b +2 i x}{b}+\frac {\left (2 i a +2\right ) \ln \left (-b x -a +i\right )}{b^{2}}\)	\(39\)
risch	\(-\frac {x^{2}}{2}-\frac {2 i x}{b}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}}+\frac {2 i \arctan \left (b x +a \right )}{b^{2}}+\frac {i a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}}-\frac {2 a \arctan \left (b x +a \right )}{b^{2}}\)	\(85\)
parallelrisch	\(-\frac {-b^{3} x^{3}+4 i \ln \left (b x +a -i\right ) x a b -3 i x^{2} b^{2}-a \,b^{2} x^{2}+4 i \ln \left (b x +a -i\right ) a^{2}-4 i+4 i a^{2}+4 \ln \left (b x +a -i\right ) x b -4 i \ln \left (b x +a -i\right )+8 \ln \left (b x +a -i\right ) a +8 a}{2 b^{2} \left (-b x -a +i\right )}\)	\(118\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=-\frac {b^{2} x^{2} + 4 i \, b x + 4 \, {\left (-i \, a - 1\right )} \log \left (\frac {b x + a - i}{b}\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.72 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=- \frac {x^{2}}{2} - \frac {2 i x}{b} + \frac {2 i \left (a - i\right ) \log {\left (a + b x - i \right )}}{b^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=\frac {i \, {\left (i \, b x^{2} - 4 \, x\right )}}{2 \, b} - \frac {2 \, {\left (-i \, a - 1\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (32) = 64\).

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.80 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=-\frac {i \, {\left (\frac {{\left (i \, b x + i \, a + 1\right )}^{2} {\left (-\frac {2 i \, {\left (i \, a b + 3 \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} + i\right )}}{b} + \frac {4 \, {\left (a - i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b}\right )}}{2 \, b} \]

[In]

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

[Out]

-1/2*I*((I*b*x + I*a + 1)^2*(-2*I*(I*a*b + 3*b)/((I*b*x + I*a + 1)*b) + I)/b + 4*(a - I)*log(1/(sqrt((b*x + a)
^2 + 1)*abs(b)))/b)/b

Mupad [B] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.28 \[ \int e^{-2 i \arctan (a+b x)} x \, dx=\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {2}{b^2}+\frac {a\,2{}\mathrm {i}}{b^2}\right )-\frac {x^2}{2}+x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right ) \]

[In]

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

[Out]

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