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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {5203, 78} \begin {gather*} \frac {(a+i) \log (x)}{-a+i}-\frac {2 \log (-a-b x+i)}{1+i a} \end {gather*}

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 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*} \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*}

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 42, normalized size = 1.02

method	result	size
default	\(-\frac {2 i \ln \left (-b x -a +i\right )}{i-a}+\frac {\left (-a^{2}-1\right ) \ln \left (x \right )}{\left (i-a \right )^{2}}\)	\(42\)
risch	\(-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{i-a}+\frac {2 \arctan \left (b x +a \right )}{i-a}+\frac {i \ln \left (x \right )}{i-a}+\frac {\ln \left (x \right ) a}{i-a}\)	\(72\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 47, normalized size = 1.15 \begin {gather*} -\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} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 3.91, size = 27, normalized size = 0.66 \begin {gather*} -\frac {{\left (a + i\right )} \log \left (x\right ) - 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{a - i} \end {gather*}

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)

________________________________________________________________________________________

Sympy [B] 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 = 0.47, size = 99, normalized size = 2.41 \begin {gather*} - \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} \end {gather*}

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(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)

________________________________________________________________________________________

Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (32) = 64\).
time = 0.45, size = 70, normalized size = 1.71 \begin {gather*} -i \, b {\left (\frac {{\left (-i \, a + 1\right )} \log \left (\frac {a}{i \, b x + i \, a + 1} - \frac {i}{i \, b x + i \, a + 1} + i\right )}{a b - i \, b} + \frac {i \, \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b}\right )} \end {gather*}

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]

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

________________________________________________________________________________________

Mupad [B]
time = 0.72, size = 34, normalized size = 0.83 \begin {gather*} -\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 ) \end {gather*}

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)

________________________________________________________________________________________