\(\int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx\) [203]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=\frac {(i+a) \log (x)}{i-a}-\frac {2 \log (i-a-b x)}{1+i a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5203, 78} \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=\frac {(a+i) \log (x)}{-a+i}-\frac {2 \log (-a-b x+i)}{1+i a} \]

[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 78

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 5203

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=\frac {-((i+a) \log (x))+2 i \log (i-a-b x)}{-i+a} \]

[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)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\left (-a^{2}-1\right ) \ln \left (x \right )}{\left (i-a \right )^{2}}-\frac {2 i \ln \left (-b x -a +i\right )}{i-a}\)	\(42\)
risch	\(-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{i-a}+\frac {2 \arctan \left (b x +a \right )}{i-a}+\frac {i \ln \left (x \right )}{i-a}+\frac {\ln \left (x \right ) a}{i-a}\)	\(72\)
parallelrisch	\(-\frac {252 \ln \left (b x +a -i\right ) a^{4}-72 \ln \left (b x +a -i\right ) a^{2}-168 i \ln \left (b x +a -i\right ) a^{3}+18 i \ln \left (b x +a -i\right ) a +252 i \ln \left (b x +a -i\right ) a^{5}+112 \ln \left (b x +a -i\right ) x \,a^{3} b -16 \ln \left (b x +a -i\right ) x a b +2 i \ln \left (b x +a -i\right ) x b +140 i \ln \left (b x +a -i\right ) x \,a^{4} b -56 i \ln \left (b x +a -i\right ) x \,a^{2} b -168 \ln \left (b x +a -i\right ) a^{6}+18 \ln \left (b x +a -i\right ) a^{8}-\ln \left (x \right )+2 \ln \left (b x +a -i\right )+16 \ln \left (b x +a -i\right ) x \,a^{7} b -112 \ln \left (b x +a -i\right ) x \,a^{5} b -28 \ln \left (x \right ) x \,a^{3} b +20 \ln \left (x \right ) x \,a^{7} b -\ln \left (x \right ) x \,a^{9} b -14 \ln \left (x \right ) x \,a^{5} b +7 \ln \left (x \right ) x a b -i b \ln \left (x \right ) x +27 a^{2} \ln \left (x \right )-8 i a \ln \left (x \right )-14 i \ln \left (x \right ) x \,a^{4} b +7 i \ln \left (x \right ) x \,a^{8} b +2 i \ln \left (b x +a -i\right ) x \,a^{8} b -28 i \ln \left (x \right ) x \,a^{6} b -56 i \ln \left (b x +a -i\right ) x \,a^{6} b -42 \ln \left (x \right ) a^{6}-42 \ln \left (x \right ) a^{4}+27 \ln \left (x \right ) a^{8}-\ln \left (x \right ) a^{10}+20 i \ln \left (x \right ) x \,a^{2} b +48 i \ln \left (x \right ) a^{3}-48 i \ln \left (x \right ) a^{7}-72 i \ln \left (b x +a -i\right ) a^{7}+8 i \ln \left (x \right ) a^{9}+2 i \ln \left (b x +a -i\right ) a^{9}}{\left (-a^{5}+5 i a^{4}+10 a^{3}-10 i a^{2}-5 a +i\right ) \left (-b x -a +i\right ) \left (-a^{4}+4 i a^{3}+6 a^{2}-4 i a -1\right )}\)	\(494\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.66 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=-\frac {{\left (a + i\right )} \log \left (x\right ) - 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{a - i} \]

[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)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (24) = 48\).

Time = 0.41 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.41 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=- \frac {\left (a + i\right ) \log {\left (a^{2} - \frac {a^{2} \left (a + i\right )}{a - i} + \frac {2 i a \left (a + i\right )}{a - i} + x \left (a b + 3 i b\right ) + 1 + \frac {a + i}{a - i} \right )}}{a - i} + \frac {2 i \log {\left (a^{2} + \frac {2 i a^{2}}{a - i} + \frac {4 a}{a - i} + x \left (a b + 3 i b\right ) + 1 - \frac {2 i}{a - i} \right )}}{a - i} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=-\frac {2 \, {\left (-i \, a - 1\right )} \log \left (i \, b x + i \, a + 1\right )}{a^{2} - 2 i \, a - 1} - \frac {{\left (a^{2} + 1\right )} \log \left (x\right )}{a^{2} - 2 i \, a - 1} \]

[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)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (32) = 64\).

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.66 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=i \, b {\left (\frac {{\left (a + i\right )} \log \left (\frac {i \, a}{i \, b x + i \, a + 1} + \frac {1}{i \, b x + i \, a + 1} - 1\right )}{-i \, a b - b} - \frac {i \, \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x} \, dx=-\frac {2\,\ln \left (a+b\,x-\mathrm {i}\right )}{1+a\,1{}\mathrm {i}}+\ln \left (x\right )\,\left (\frac {2}{1+a\,1{}\mathrm {i}}-1\right ) \]

[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)