3.205 \(\int \frac{e^{-2 i \tan ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac{2 b^2 \log (x)}{(1+i a)^3}+\frac{2 b^2 \log (-a-b x+i)}{(1+i a)^3}-\frac{2 i b}{(-a+i)^2 x}-\frac{a+i}{2 (-a+i) x^2} \]

[Out]

-(I + a)/(2*(I - a)*x^2) - ((2*I)*b)/((I - a)^2*x) - (2*b^2*Log[x])/(1 + I*a)^3 + (2*b^2*Log[I - a - b*x])/(1
+ I*a)^3

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Rubi [A]  time = 0.0516425, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5095, 77} \[ -\frac{2 b^2 \log (x)}{(1+i a)^3}+\frac{2 b^2 \log (-a-b x+i)}{(1+i a)^3}-\frac{2 i b}{(-a+i)^2 x}-\frac{a+i}{2 (-a+i) x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]

[Out]

-(I + a)/(2*(I - a)*x^2) - ((2*I)*b)/((I - a)^2*x) - (2*b^2*Log[x])/(1 + I*a)^3 + (2*b^2*Log[I - a - b*x])/(1
+ I*a)^3

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]

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]))))

Rubi steps

\begin{align*} \int \frac{e^{-2 i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{1-i a-i b x}{x^3 (1+i a+i b x)} \, dx\\ &=\int \left (\frac{-i-a}{(-i+a) x^3}+\frac{2 i b}{(-i+a)^2 x^2}-\frac{2 i b^2}{(-i+a)^3 x}+\frac{2 i b^3}{(-i+a)^3 (-i+a+b x)}\right ) \, dx\\ &=-\frac{i+a}{2 (i-a) x^2}-\frac{2 i b}{(i-a)^2 x}-\frac{2 b^2 \log (x)}{(1+i a)^3}+\frac{2 b^2 \log (i-a-b x)}{(1+i a)^3}\\ \end{align*}

Mathematica [A]  time = 0.0361532, size = 66, normalized size = 0.81 \[ \frac{(a-i) \left (a^2-4 i b x+1\right )+4 i b^2 x^2 \log (-a-b x+i)-4 i b^2 x^2 \log (x)}{2 (a-i)^3 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]

[Out]

((-I + a)*(1 + a^2 - (4*I)*b*x) - (4*I)*b^2*x^2*Log[x] + (4*I)*b^2*x^2*Log[I - a - b*x])/(2*(-I + a)^3*x^2)

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Maple [B]  time = 0.053, size = 246, normalized size = 3. \begin{align*}{\frac{i{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{ \left ( i-a \right ) ^{4}}}+{\frac{{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{ \left ( i-a \right ) ^{4}}}-2\,{\frac{{b}^{2}\arctan \left ( bx+a \right ) a}{ \left ( i-a \right ) ^{4}}}+{\frac{2\,i{b}^{2}\arctan \left ( bx+a \right ) }{ \left ( i-a \right ) ^{4}}}-{\frac{i{a}^{3}}{ \left ( i-a \right ) ^{4}{x}^{2}}}+{\frac{{a}^{4}}{2\, \left ( i-a \right ) ^{4}{x}^{2}}}-{\frac{ia}{ \left ( i-a \right ) ^{4}{x}^{2}}}-{\frac{1}{2\, \left ( i-a \right ) ^{4}{x}^{2}}}-{\frac{2\,ib{a}^{2}}{ \left ( i-a \right ) ^{4}x}}+{\frac{2\,ib}{ \left ( i-a \right ) ^{4}x}}-4\,{\frac{ab}{ \left ( i-a \right ) ^{4}x}}-{\frac{2\,i{b}^{2}\ln \left ( x \right ) a}{ \left ( i-a \right ) ^{4}}}-2\,{\frac{{b}^{2}\ln \left ( x \right ) }{ \left ( i-a \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x)

[Out]

I*b^2/(I-a)^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a+b^2/(I-a)^4*ln(b^2*x^2+2*a*b*x+a^2+1)-2*b^2/(I-a)^4*arctan(b*x+a)*a+
2*I*b^2/(I-a)^4*arctan(b*x+a)-I/(I-a)^4/x^2*a^3+1/2/(I-a)^4/x^2*a^4-I/(I-a)^4/x^2*a-1/2/(I-a)^4/x^2-2*I*b/(I-a
)^4/x*a^2+2*I*b/(I-a)^4/x-4*b/(I-a)^4/x*a-2*I*b^2/(I-a)^4*ln(x)*a-2*b^2/(I-a)^4*ln(x)

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Maxima [B]  time = 1.03118, size = 220, normalized size = 2.72 \begin{align*} -\frac{2 \,{\left (-i \, a - 1\right )} b^{2} \log \left (i \, b x + i \, a + 1\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} - \frac{2 \,{\left (i \, a + 1\right )} b^{2} \log \left (x\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} + \frac{4 \,{\left (-i \, a - 1\right )} b^{2} x^{2} + a^{4} - 2 i \, a^{3} +{\left (a^{3} - 5 i \, a^{2} - 7 \, a + 3 i\right )} b x - 2 i \, a - 1}{{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} b x^{3} +{\left (2 \, a^{4} - 8 i \, a^{3} - 12 \, a^{2} + 8 i \, a + 2\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="maxima")

[Out]

-2*(-I*a - 1)*b^2*log(I*b*x + I*a + 1)/(a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1) - 2*(I*a + 1)*b^2*log(x)/(a^4 - 4*I
*a^3 - 6*a^2 + 4*I*a + 1) + (4*(-I*a - 1)*b^2*x^2 + a^4 - 2*I*a^3 + (a^3 - 5*I*a^2 - 7*a + 3*I)*b*x - 2*I*a -
1)/((2*a^3 - 6*I*a^2 - 6*a + 2*I)*b*x^3 + (2*a^4 - 8*I*a^3 - 12*a^2 + 8*I*a + 2)*x^2)

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Fricas [A]  time = 2.18259, size = 181, normalized size = 2.23 \begin{align*} \frac{-4 i \, b^{2} x^{2} \log \left (x\right ) + 4 i \, b^{2} x^{2} \log \left (\frac{b x + a - i}{b}\right ) + a^{3} - 4 \,{\left (i \, a + 1\right )} b x - i \, a^{2} + a - i}{{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="fricas")

[Out]

(-4*I*b^2*x^2*log(x) + 4*I*b^2*x^2*log((b*x + a - I)/b) + a^3 - 4*(I*a + 1)*b*x - I*a^2 + a - I)/((2*a^3 - 6*I
*a^2 - 6*a + 2*I)*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))**2*(1+(b*x+a)**2)/x**3,x)

[Out]

Timed out

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Giac [B]  time = 1.10766, size = 212, normalized size = 2.62 \begin{align*} \frac{2 \, b^{3} \log \left (-\frac{a i}{b i x + a i + 1} + \frac{i^{2}}{b i x + a i + 1} + 1\right )}{a^{3} b i + 3 \, a^{2} b - 3 \, a b i - b} - \frac{\frac{2 \,{\left (a b^{3} i - 3 \, b^{3}\right )} i^{2}}{{\left (b i x + a i + 1\right )} b} + \frac{a b^{2} i - 5 \, b^{2}}{a i + 1}}{2 \,{\left (a - i\right )}^{2}{\left (\frac{a i}{b i x + a i + 1} - \frac{i^{2}}{b i x + a i + 1} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="giac")

[Out]

2*b^3*log(-a*i/(b*i*x + a*i + 1) + i^2/(b*i*x + a*i + 1) + 1)/(a^3*b*i + 3*a^2*b - 3*a*b*i - b) - 1/2*(2*(a*b^
3*i - 3*b^3)*i^2/((b*i*x + a*i + 1)*b) + (a*b^2*i - 5*b^2)/(a*i + 1))/((a - i)^2*(a*i/(b*i*x + a*i + 1) - i^2/
(b*i*x + a*i + 1) - 1)^2)