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

Optimal. Leaf size=41 \[ \frac {(a+i) \log (x)}{-a+i}-\frac {2 \log (-a-b x+i)}{1+i a} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 41, 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, 72} \[ \frac {(a+i) \log (x)}{-a+i}-\frac {2 \log (-a-b x+i)}{1+i a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.83 \[ \frac {2 i \log (-a-b x+i)-(a+i) \log (x)}{a-i} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 27, normalized size = 0.66 \[ -\frac {{\left (a + i\right )} \log \relax (x) - 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{a - i} \]

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,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.14, size = 77, normalized size = 1.88 \[ b i {\left (\frac {{\left (a i - 1\right )} \log \left (-\frac {a i^{2}}{b i x + a i + 1} + i - \frac {i}{b i x + a i + 1}\right )}{a b - b i} - \frac {i \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b}\right )} \]

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,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 74, normalized size = 1.80 \[ -\frac {\ln \relax (x ) a^{2}}{\left (i-a \right )^{2}}-\frac {\ln \relax (x )}{\left (i-a \right )^{2}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{i-a}+\frac {2 \arctan \left (b x +a \right )}{i-a} \]

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,x)

[Out]

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

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maxima [A]  time = 0.33, size = 47, normalized size = 1.15 \[ -\frac {2 \, {\left (-i \, a - 1\right )} \log \left (i \, b x + i \, a + 1\right )}{a^{2} - 2 i \, a - 1} - \frac {{\left (a^{2} + 1\right )} \log \relax (x)}{a^{2} - 2 i \, a - 1} \]

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,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.72, size = 34, normalized size = 0.83 \[ -\frac {2\,\ln \left (a+b\,x-\mathrm {i}\right )}{1+a\,1{}\mathrm {i}}+\ln \relax (x)\,\left (\frac {2}{1+a\,1{}\mathrm {i}}-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.97, size = 102, normalized size = 2.49 \[ - \frac {\left (a + i\right ) \log {\left (i a^{2} - \frac {i a^{2} \left (a + i\right )}{a - i} - \frac {2 a \left (a + i\right )}{a - i} + x \left (i a b - 3 b\right ) + i + \frac {i \left (a + i\right )}{a - i} \right )}}{a - i} + \frac {2 i \log {\left (i a^{2} - \frac {2 a^{2}}{a - i} + \frac {4 i a}{a - i} + x \left (i a b - 3 b\right ) + i + \frac {2}{a - i} \right )}}{a - i} \]

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,x)

[Out]

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

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