\(\int e^{2 i \arctan (a+b x)} x \, dx\) [174]

Optimal result

Integrand size = 14, antiderivative size = 37 \[ \int e^{2 i \arctan (a+b x)} x \, dx=\frac {2 i x}{b}-\frac {x^2}{2}+\frac {2 (1-i a) \log (i+a+b x)}{b^2} \]

[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 = 37, 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[E^((2*I)*ArcTan[a + b*x])*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 = 37, 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[E^((2*I)*ArcTan[a + b*x])*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 = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11

method	result	size
parallelrisch	\(-\frac {b^{2} x^{2}-4 \ln \left (b x +a +i\right )+4 i \ln \left (b x +a +i\right ) a -4 b x i}{2 b^{2}}\)	\(41\)
risch	\(-\frac {x^{2}}{2}+\frac {2 i x}{b}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}}-\frac {2 i \arctan \left (b x +a \right )}{b^{2}}-\frac {i a \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}}-\frac {2 a \arctan \left (b x +a \right )}{b^{2}}\)	\(85\)
default	\(\frac {-\frac {1}{2} x^{2} b +2 i x}{b}+\frac {\frac {\left (-2 i a b +2 b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (-2 i a^{2}-2 i-\frac {\left (-2 i a b +2 b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}}{b}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a+b x)} x \, dx=-\frac {b^{2} x^{2} - 4 i \, b x + 4 \, {\left (i \, a - 1\right )} \log \left (\frac {b x + a + i}{b}\right )}{2 \, b^{2}} \]

[In]

integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)*x,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 = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int e^{2 i \arctan (a+b x)} x \, dx=- \frac {x^{2}}{2} + \frac {2 i x}{b} - \frac {2 i \left (a + i\right ) \log {\left (a + b x + i \right )}}{b^{2}} \]

[In]

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

[Out]

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

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. 64 vs. \(2 (29) = 58\).

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.73 \[ \int e^{2 i \arctan (a+b x)} x \, dx=-\frac {b x^{2} - 4 i \, x}{2 \, b} - \frac {2 \, {\left (a + i\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{2}} + \frac {{\left (-i \, a + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a+b x)} x \, dx=-\frac {2 \, {\left (i \, a - 1\right )} \log \left (b x + a + i\right )}{b^{2}} - \frac {b^{2} x^{2} - 4 i \, b x}{2 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.62 \[ \int e^{2 i \arctan (a+b x)} x \, dx=-\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (-\frac {2}{b^2}+\frac {a\,2{}\mathrm {i}}{b^2}\right )-x\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )-\frac {x^2}{2} \]

[In]

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

[Out]

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