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

Optimal. Leaf size=62 \[ \frac {2 i b \log (x)}{(-a+i)^2}-\frac {2 i b \log (-a-b x+i)}{(-a+i)^2}-\frac {a+i}{(-a+i) x} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, 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} \[ \frac {2 i b \log (x)}{(-a+i)^2}-\frac {2 i b \log (-a-b x+i)}{(-a+i)^2}-\frac {a+i}{(-a+i) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.68 \[ \frac {a^2-2 i b x \log (-a-b x+i)+2 i b x \log (x)+1}{(a-i)^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 40, normalized size = 0.65 \[ \frac {2 i \, b x \log \relax (x) - 2 i \, b x \log \left (\frac {b x + a - i}{b}\right ) + a^{2} + 1}{{\left (a^{2} - 2 i \, a - 1\right )} x} \]

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^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.12, size = 106, normalized size = 1.71 \[ -\frac {2 \, b^{2} \log \left (-\frac {a i}{b i x + a i + 1} + \frac {i^{2}}{b i x + a i + 1} + 1\right )}{a^{2} b i + 2 \, a b - b i} - \frac {a b + b i}{{\left (a - i\right )}^{2} {\left (\frac {a i}{b i x + a i + 1} - \frac {i^{2}}{b i x + a i + 1} - 1\right )}} \]

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^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 152, normalized size = 2.45 \[ \frac {a^{2}}{x \left (i-a \right )^{2}}+\frac {1}{x \left (i-a \right )^{2}}-\frac {2 i b \ln \relax (x ) a}{\left (i-a \right )^{3}}-\frac {2 b \ln \relax (x )}{\left (i-a \right )^{3}}+\frac {i b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{\left (i-a \right )^{3}}+\frac {b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{\left (i-a \right )^{3}}-\frac {2 b \arctan \left (b x +a \right ) a}{\left (i-a \right )^{3}}+\frac {2 i b \arctan \left (b x +a \right )}{\left (i-a \right )^{3}} \]

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^2,x)

[Out]

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

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maxima [B]  time = 0.32, size = 113, normalized size = 1.82 \[ -\frac {{\left (2 \, a - 2 i\right )} b \log \left (i \, b x + i \, a + 1\right )}{-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1} + \frac {{\left (2 \, a - 2 i\right )} b \log \relax (x)}{-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1} + \frac {a^{3} + {\left (a^{2} + 1\right )} b x - i \, a^{2} + a - i}{{\left (a^{2} - 2 i \, a - 1\right )} b x^{2} + {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x} \]

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^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.65, size = 100, normalized size = 1.61 \[ \frac {-1+a\,1{}\mathrm {i}}{x\,\left (1+a\,1{}\mathrm {i}\right )}-\frac {4\,b\,\mathrm {atan}\left (\frac {a^2\,1{}\mathrm {i}+2\,a-\mathrm {i}}{{\left (a-\mathrm {i}\right )}^2}+\frac {x\,\left (2\,a^4\,b^2+4\,a^2\,b^2+2\,b^2\right )}{{\left (a-\mathrm {i}\right )}^2\,\left (-1{}\mathrm {i}\,b\,a^3+b\,a^2-1{}\mathrm {i}\,b\,a+b\right )}\right )}{{\left (a-\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*1i - 1)/(x*(a*1i + 1)) - (4*b*atan((2*a + a^2*1i - 1i)/(a - 1i)^2 + (x*(2*b^2 + 4*a^2*b^2 + 2*a^4*b^2))/((a
- 1i)^2*(b - a*b*1i + a^2*b - a^3*b*1i))))/(a - 1i)^2

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sympy [B]  time = 0.74, size = 156, normalized size = 2.52 \[ \frac {2 i b \log {\left (- \frac {2 a^{3} b}{\left (a - i\right )^{2}} + \frac {6 i a^{2} b}{\left (a - i\right )^{2}} + 2 a b + \frac {6 a b}{\left (a - i\right )^{2}} + 4 b^{2} x - 2 i b - \frac {2 i b}{\left (a - i\right )^{2}} \right )}}{\left (a - i\right )^{2}} - \frac {2 i b \log {\left (\frac {2 a^{3} b}{\left (a - i\right )^{2}} - \frac {6 i a^{2} b}{\left (a - i\right )^{2}} + 2 a b - \frac {6 a b}{\left (a - i\right )^{2}} + 4 b^{2} x - 2 i b + \frac {2 i b}{\left (a - i\right )^{2}} \right )}}{\left (a - i\right )^{2}} - \frac {a + i}{x \left (- a + i\right )} \]

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**2,x)

[Out]

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

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