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

Optimal. Leaf size=37 \[ \frac {2 (1-i a) \log (a+b x+i)}{b^2}+\frac {2 i x}{b}-\frac {x^2}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5095, 77} \[ \frac {2 (1-i a) \log (a+b x+i)}{b^2}+\frac {2 i x}{b}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[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 77

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 5095

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*} \int e^{2 i \tan ^{-1}(a+b x)} x \, dx &=\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]  time = 0.02, size = 37, normalized size = 1.00 \[ \frac {2 (1-i a) \log (a+b x+i)}{b^2}+\frac {2 i x}{b}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 35, normalized size = 0.95 \[ -\frac {b^{2} x^{2} - 4 i \, b x + 4 \, {\left (i \, a - 1\right )} \log \left (\frac {b x + a + i}{b}\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [A]  time = 0.13, size = 36, normalized size = 0.97 \[ -\frac {2 \, {\left (a i - 1\right )} \log \left (b x + a + i\right )}{b^{2}} - \frac {b^{2} x^{2} - 4 \, b i x}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.04, size = 107, normalized size = 2.89 \[ -\frac {x^{2}}{2}+\frac {2 i x}{b}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{2}}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}}-\frac {2 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b^{2}}-\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 66, normalized size = 1.78 \[ -\frac {b x^{2} - 4 i \, x}{2 \, b} - \frac {{\left (2 \, a + 2 i\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{2}} + \frac {{\left (-i \, a + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 + 2*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

________________________________________________________________________________________

mupad [B]  time = 0.13, size = 60, normalized size = 1.62 \[ -\ln \left (x+\frac {a+1{}\mathrm {i}}{b}\right )\,\left (-\frac {2}{b^2}+\frac {a\,2{}\mathrm {i}}{b^2}\right )-x\,\left (\frac {\left (-1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {\left (1+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}\right )-\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 0.22, size = 32, normalized size = 0.86 \[ - \frac {x^{2}}{2} + \frac {2 i x}{b} - \frac {2 i \left (a + i\right ) \log {\left (i a + i b x - 1 \right )}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(I*a + I*b*x - 1)/b**2

________________________________________________________________________________________