\(\int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx\) [177]

Optimal result

Integrand size = 16, antiderivative size = 55 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=-\frac {i-a}{(i+a) x}-\frac {2 i b \log (x)}{(i+a)^2}+\frac {2 i b \log (i+a+b x)}{(i+a)^2} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 55, 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} \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=-\frac {2 i b \log (x)}{(a+i)^2}+\frac {2 i b \log (a+b x+i)}{(a+i)^2}-\frac {-a+i}{(a+i) x} \]

[In]

Int[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 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^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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=\frac {1+a^2-2 i b x \log (x)+2 i b x \log (i+a+b x)}{(i+a)^2 x} \]

[In]

Integrate[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)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (45 ) = 90\).

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.75

method	result	size
parallelrisch	\(-\frac {2 i \ln \left (x \right ) x \,a^{2} b -2 i \ln \left (b x +a +i\right ) x \,a^{2} b +1-2 i b \ln \left (x \right ) x +4 \ln \left (x \right ) x a b +2 i b \ln \left (b x +a +i\right ) x -4 \ln \left (b x +a +i\right ) x a b +2 i a^{3}-a^{4}+2 i a}{\left (a^{2}+1\right )^{2} x}\)	\(96\)
default	\(-\frac {-a^{2}+2 i a +1}{\left (a^{2}+1\right ) x}-\frac {2 b \left (i a^{2}+2 a -i\right ) \ln \left (x \right )}{\left (a^{2}+1\right )^{2}}+\frac {2 b^{2} \left (\frac {\left (i a^{2} b +2 a b -i b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (i a^{3}-3 i a +3 a^{2}-1-\frac {\left (i a^{2} b +2 a b -i b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{\left (a^{2}+1\right )^{2}}\)	\(161\)
risch	\(-\frac {i}{\left (i+a \right ) x}+\frac {a}{\left (i+a \right ) x}-\frac {b \ln \left (4 a^{4} b^{2} x^{2}+8 a^{5} b x +4 a^{6}+8 a^{2} b^{2} x^{2}+16 a^{3} b x +12 a^{4}+4 b^{2} x^{2}+8 a b x +12 a^{2}+4\right )}{i a^{2}-2 a -i}+\frac {2 i b \arctan \left (\frac {\left (2 a^{2} b +2 b \right ) x +2 a^{3}+2 a}{2 a^{2}+2}\right )}{i a^{2}-2 a -i}+\frac {2 b \ln \left (\left (-2 a^{2} b -2 b \right ) x \right )}{i a^{2}-2 a -i}\)	\(189\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=\frac {-2 i \, b x \log \left (x\right ) + 2 i \, b x \log \left (\frac {b x + a + i}{b}\right ) + a^{2} + 1}{{\left (a^{2} + 2 i \, a - 1\right )} x} \]

[In]

integrate((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)

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. 156 vs. \(2 (39) = 78\).

Time = 0.31 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.84 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=- \frac {2 i b \log {\left (- \frac {2 a^{3} b}{\left (a + i\right )^{2}} - \frac {6 i a^{2} b}{\left (a + i\right )^{2}} + 2 a b + \frac {6 a b}{\left (a + i\right )^{2}} + 4 b^{2} x + 2 i b + \frac {2 i b}{\left (a + i\right )^{2}} \right )}}{\left (a + i\right )^{2}} + \frac {2 i b \log {\left (\frac {2 a^{3} b}{\left (a + i\right )^{2}} + \frac {6 i a^{2} b}{\left (a + i\right )^{2}} + 2 a b - \frac {6 a b}{\left (a + i\right )^{2}} + 4 b^{2} x + 2 i b - \frac {2 i b}{\left (a + i\right )^{2}} \right )}}{\left (a + i\right )^{2}} - \frac {- a + i}{x \left (a + i\right )} \]

[In]

integrate((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))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (38) = 76\).

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.29 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=\frac {2 \, {\left (a^{2} - 2 i \, a - 1\right )} b \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {{\left (i \, a^{2} + 2 \, a - i\right )} b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} - \frac {2 \, {\left (i \, a^{2} + 2 \, a - i\right )} b \log \left (x\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {a^{2} - 2 i \, a - 1}{{\left (a^{2} + 1\right )} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=\frac {2 \, b^{2} \log \left (b x + a + i\right )}{-i \, a^{2} b + 2 \, a b + i \, b} + \frac {2 \, b \log \left ({\left | x \right |}\right )}{i \, a^{2} - 2 \, a - i} + \frac {a^{2} + 1}{{\left (a + i\right )}^{2} x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.78 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x^2} \, dx=\frac {a-\mathrm {i}}{x\,\left (a+1{}\mathrm {i}\right )}+\frac {b\,\mathrm {atanh}\left (\frac {a^2+a\,2{}\mathrm {i}-1}{{\left (a+1{}\mathrm {i}\right )}^2}-\frac {x\,\left (2\,a^4\,b^2+4\,a^2\,b^2+2\,b^2\right )}{{\left (a+1{}\mathrm {i}\right )}^2\,\left (-b\,a^3+1{}\mathrm {i}\,b\,a^2-b\,a+b\,1{}\mathrm {i}\right )}\right )\,4{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^2} \]

[In]

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

[Out]

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