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

Optimal. Leaf size=55 \[ -\frac {2 i b \log (x)}{(a+i)^2}+\frac {2 i b \log (a+b x+i)}{(a+i)^2}-\frac {-a+i}{(a+i) x} \]

[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 = 55, 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[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 = 39, normalized size = 0.71 \[ \frac {a^2+2 i b x \log (a+b x+i)-2 i b x \log (x)+1}{(a+i)^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[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 = 0.50, size = 40, normalized size = 0.73 \[ \frac {-2 i \, b x \log \relax (x) + 2 i \, b x \log \left (\frac {b x + a + i}{b}\right ) + a^{2} + 1}{{\left (a^{2} + 2 i \, a - 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+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 [A]  time = 0.12, size = 68, normalized size = 1.24 \[ -\frac {2 \, b^{2} \log \left (b x + a + i\right )}{a^{2} b i - 2 \, a b - b i} + \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{2} i - 2 \, a - i} - \frac {{\left (a^{2} i + i\right )} i}{{\left (a + i\right )}^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [B]  time = 0.05, size = 260, normalized size = 4.73 \[ -\frac {2 i a}{\left (a^{2}+1\right ) x}+\frac {a^{2}}{\left (a^{2}+1\right ) x}-\frac {1}{\left (a^{2}+1\right ) x}-\frac {2 i b \ln \relax (x ) a^{2}}{\left (a^{2}+1\right )^{2}}+\frac {2 i b \ln \relax (x )}{\left (a^{2}+1\right )^{2}}-\frac {4 b \ln \relax (x ) a}{\left (a^{2}+1\right )^{2}}+\frac {i b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{\left (a^{2}+1\right )^{2}}-\frac {i b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{\left (a^{2}+1\right )^{2}}+\frac {2 b \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{\left (a^{2}+1\right )^{2}}-\frac {4 i b \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{\left (a^{2}+1\right )^{2}}+\frac {2 b \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{\left (a^{2}+1\right )^{2}}-\frac {2 b \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{\left (a^{2}+1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 125, normalized size = 2.27 \[ \frac {2 \, {\left (a^{2} - 2 i \, a - 1\right )} b \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {{\left (i \, a^{2} + 2 \, a - i\right )} b \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{4} + 2 \, a^{2} + 1} + \frac {{\left (-2 i \, a^{2} - 4 \, a + 2 i\right )} b \log \relax (x)}{a^{4} + 2 \, a^{2} + 1} + \frac {a^{2} - 2 i \, a - 1}{{\left (a^{2} + 1\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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mupad [B]  time = 0.64, size = 98, normalized size = 1.78 \[ \frac {a-\mathrm {i}}{x\,\left (a+1{}\mathrm {i}\right )}+\frac {b\,\mathrm {atanh}\left (\frac {a^2+a\,2{}\mathrm {i}-1}{{\left (a+1{}\mathrm {i}\right )}^2}-\frac {x\,\left (2\,a^4\,b^2+4\,a^2\,b^2+2\,b^2\right )}{{\left (a+1{}\mathrm {i}\right )}^2\,\left (-b\,a^3+1{}\mathrm {i}\,b\,a^2-b\,a+b\,1{}\mathrm {i}\right )}\right )\,4{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [B]  time = 0.61, size = 158, normalized size = 2.87 \[ - \frac {2 i b \log {\left (- \frac {2 a^{3} b}{\left (a + i\right )^{2}} - \frac {6 i a^{2} b}{\left (a + i\right )^{2}} + 2 a b + \frac {6 a b}{\left (a + i\right )^{2}} + 4 b^{2} x + 2 i b + \frac {2 i b}{\left (a + i\right )^{2}} \right )}}{\left (a + i\right )^{2}} + \frac {2 i b \log {\left (\frac {2 a^{3} b}{\left (a + i\right )^{2}} + \frac {6 i a^{2} b}{\left (a + i\right )^{2}} + 2 a b - \frac {6 a b}{\left (a + i\right )^{2}} + 4 b^{2} x + 2 i b - \frac {2 i b}{\left (a + i\right )^{2}} \right )}}{\left (a + i\right )^{2}} - \frac {a - i}{x \left (- a - i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+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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