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

Optimal. Leaf size=40 \[ \frac {2 (1+i a) \log (-a-b x+i)}{b^2}-\frac {2 i x}{b}-\frac {x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 40, 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[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 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*}

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Mathematica [A]  time = 0.02, size = 40, 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[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

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fricas [A]  time = 0.44, size = 35, normalized size = 0.88 \[ -\frac {b^{2} x^{2} + 4 i \, b x + 4 \, {\left (-i \, a - 1\right )} \log \left (\frac {b x + a - i}{b}\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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

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giac [B]  time = 0.14, size = 80, normalized size = 2.00 \[ -\frac {i {\left (\frac {4 \, {\left (a - i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b} - \frac {{\left (b i x + a i + 1\right )}^{2} {\left (i - \frac {2 \, {\left (a b i + 3 \, b\right )} i}{{\left (b i x + a i + 1\right )} b}\right )}}{b i^{2}}\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 85, normalized size = 2.12 \[ -\frac {x^{2}}{2}-\frac {2 i x}{b}-\frac {2 \arctan \left (b x +a \right ) a}{b^{2}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{2}}+\frac {2 i \arctan \left (b x +a \right )}{b^{2}}+\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 36, normalized size = 0.90 \[ \frac {i \, {\left (i \, b x^{2} - 4 \, x\right )}}{2 \, b} - \frac {2 \, {\left (-i \, a - 1\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.50, size = 51, normalized size = 1.28 \[ \ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {2}{b^2}+\frac {a\,2{}\mathrm {i}}{b^2}\right )-\frac {x^2}{2}+x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.25, size = 32, normalized size = 0.80 \[ - \frac {x^{2}}{2} - \frac {2 i x}{b} + \frac {2 i \left (a - i\right ) \log {\left (i a + i b x + 1 \right )}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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(I*a + I*b*x + 1)/b**2

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