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)*Log[x])/(I - a) - (2*Log[I - a - b*x])/(1 + I*a)
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Rubi [A] time = 0.0356451, antiderivative size = 41, normalized size of antiderivative = 1.,
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 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]
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]
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 = 0.0187367, 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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Maple [A] time = 0.063, size = 74, normalized size = 1.8 \begin{align*}{\frac{-i\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{i-a}}+2\,{\frac{\arctan \left ( bx+a \right ) }{i-a}}-{\frac{{a}^{2}\ln \left ( x \right ) }{ \left ( i-a \right ) ^{2}}}-{\frac{\ln \left ( x \right ) }{ \left ( i-a \right ) ^{2}}} \end{align*}
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]
-I/(I-a)*ln(b^2*x^2+2*a*b*x+a^2+1)+2/(I-a)*arctan(b*x+a)-1/(I-a)^2*ln(x)*a^2-1/(I-a)^2*ln(x)
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Maxima [A] time = 1.02132, size = 63, normalized size = 1.54 \begin{align*} -\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{align*}
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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Fricas [A] time = 2.22017, size = 73, normalized size = 1.78 \begin{align*} -\frac{{\left (a + i\right )} \log \left (x\right ) - 2 i \, \log \left (\frac{b x + a - i}{b}\right )}{a - i} \end{align*}
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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Sympy [B] time = 2.54326, size = 1538, normalized size = 37.51 \begin{align*} \text{result too large to display} \end{align*}
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]
(-sqrt((-a**8 - 12*I*a**7 + 60*a**6 + 164*I*a**5 - 270*a**4 - 276*I*a**3 + 172*a**2 + 60*I*a - 9)*(-a**8 - 4*I
*a**7 + 4*a**6 - 4*I*a**5 + 10*a**4 + 4*I*a**3 + 4*a**2 + 4*I*a - 1))/(2*(a**8 + 4*I*a**7 - 4*a**6 + 4*I*a**5
- 10*a**4 - 4*I*a**3 - 4*a**2 - 4*I*a + 1)) - 1/2)*log(x + (-sqrt((-a**8 - 12*I*a**7 + 60*a**6 + 164*I*a**5 -
270*a**4 - 276*I*a**3 + 172*a**2 + 60*I*a - 9)*(-a**8 - 4*I*a**7 + 4*a**6 - 4*I*a**5 + 10*a**4 + 4*I*a**3 + 4*
a**2 + 4*I*a - 1))/(2*(a**8 + 4*I*a**7 - 4*a**6 + 4*I*a**5 - 10*a**4 - 4*I*a**3 - 4*a**2 - 4*I*a + 1)) - 1/2)*
(a**11 + 9*I*a**10 - 31*a**9 - 47*I*a**8 + 10*a**7 - 70*I*a**6 + 98*a**5 + 34*I*a**4 + 37*a**3 + 45*I*a**2 - 1
9*a - 3*I)/(a**10*b + 14*I*a**9*b - 85*a**8*b - 296*I*a**7*b + 658*a**6*b + 980*I*a**5*b - 994*a**4*b - 680*I*
a**3*b + 301*a**2*b + 78*I*a*b - 9*b) + (a**25 + 29*I*a**24 - 396*a**23 - 3388*I*a**22 + 20378*a**21 + 91602*I
*a**20 - 319116*a**19 - 880764*I*a**18 + 1948887*a**17 + 3465467*I*a**16 - 4901848*a**15 - 5325624*I*a**14 + 3
970316*a**13 + 967708*I*a**12 + 2488392*a**11 + 4876008*I*a**10 - 5388609*a**9 - 4348701*I*a**8 + 2724068*a**7
+ 1346548*I*a**6 - 523782*a**5 - 157646*I*a**4 + 35524*a**3 + 5652*I*a**2 - 567*a - 27*I)/(a**24*b + 32*I*a**
23*b - 484*a**22*b - 4608*I*a**21*b + 31026*a**20*b + 157344*I*a**19*b - 624948*a**18*b - 1995456*I*a**17*b +
5216127*a**16*b + 11307584*I*a**15*b - 20514376*a**14*b - 31338752*I*a**13*b + 40461564*a**12*b + 44217408*I*a
**11*b - 40876296*a**10*b - 31876224*I*a**9*b + 20859663*a**8*b + 11361696*I*a**7*b - 5089492*a**6*b - 1842944
*I*a**5*b + 526066*a**4*b + 113952*I*a**3*b - 17604*a**2*b - 1728*I*a*b + 81*b)) + (sqrt((-a**8 - 12*I*a**7 +
60*a**6 + 164*I*a**5 - 270*a**4 - 276*I*a**3 + 172*a**2 + 60*I*a - 9)*(-a**8 - 4*I*a**7 + 4*a**6 - 4*I*a**5 +
10*a**4 + 4*I*a**3 + 4*a**2 + 4*I*a - 1))/(2*(a**8 + 4*I*a**7 - 4*a**6 + 4*I*a**5 - 10*a**4 - 4*I*a**3 - 4*a**
2 - 4*I*a + 1)) - 1/2)*log(x + (sqrt((-a**8 - 12*I*a**7 + 60*a**6 + 164*I*a**5 - 270*a**4 - 276*I*a**3 + 172*a
**2 + 60*I*a - 9)*(-a**8 - 4*I*a**7 + 4*a**6 - 4*I*a**5 + 10*a**4 + 4*I*a**3 + 4*a**2 + 4*I*a - 1))/(2*(a**8 +
4*I*a**7 - 4*a**6 + 4*I*a**5 - 10*a**4 - 4*I*a**3 - 4*a**2 - 4*I*a + 1)) - 1/2)*(a**11 + 9*I*a**10 - 31*a**9
- 47*I*a**8 + 10*a**7 - 70*I*a**6 + 98*a**5 + 34*I*a**4 + 37*a**3 + 45*I*a**2 - 19*a - 3*I)/(a**10*b + 14*I*a*
*9*b - 85*a**8*b - 296*I*a**7*b + 658*a**6*b + 980*I*a**5*b - 994*a**4*b - 680*I*a**3*b + 301*a**2*b + 78*I*a*
b - 9*b) + (a**25 + 29*I*a**24 - 396*a**23 - 3388*I*a**22 + 20378*a**21 + 91602*I*a**20 - 319116*a**19 - 88076
4*I*a**18 + 1948887*a**17 + 3465467*I*a**16 - 4901848*a**15 - 5325624*I*a**14 + 3970316*a**13 + 967708*I*a**12
+ 2488392*a**11 + 4876008*I*a**10 - 5388609*a**9 - 4348701*I*a**8 + 2724068*a**7 + 1346548*I*a**6 - 523782*a*
*5 - 157646*I*a**4 + 35524*a**3 + 5652*I*a**2 - 567*a - 27*I)/(a**24*b + 32*I*a**23*b - 484*a**22*b - 4608*I*a
**21*b + 31026*a**20*b + 157344*I*a**19*b - 624948*a**18*b - 1995456*I*a**17*b + 5216127*a**16*b + 11307584*I*
a**15*b - 20514376*a**14*b - 31338752*I*a**13*b + 40461564*a**12*b + 44217408*I*a**11*b - 40876296*a**10*b - 3
1876224*I*a**9*b + 20859663*a**8*b + 11361696*I*a**7*b - 5089492*a**6*b - 1842944*I*a**5*b + 526066*a**4*b + 1
13952*I*a**3*b - 17604*a**2*b - 1728*I*a*b + 81*b))
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Giac [B] time = 1.09915, size = 104, normalized size = 2.54 \begin{align*} 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 )} \end{align*}
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)