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

Optimal. Leaf size=38 \[ \frac {(-a+i) \log (x)}{a+i}-\frac {2 \log (a+b x+i)}{1-i a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 38, 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, 72} \[ \frac {(-a+i) \log (x)}{a+i}-\frac {2 \log (a+b x+i)}{1-i a} \]

Antiderivative was successfully verified.

[In]

Int[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 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]

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} \, 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*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 0.82 \[ -\frac {2 i \log (a+b x+i)+(a-i) \log (x)}{a+i} \]

Antiderivative was successfully verified.

[In]

Integrate[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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fricas [A]  time = 0.63, size = 27, normalized size = 0.71 \[ -\frac {{\left (a - i\right )} \log \relax (x) + 2 i \, \log \left (\frac {b x + a + i}{b}\right )}{a + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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giac [A]  time = 0.12, size = 36, normalized size = 0.95 \[ -\frac {2 \, b i \log \left (b x + a + i\right )}{a b + b i} - \frac {{\left (a - i\right )} \log \left ({\left | x \right |}\right )}{a + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 149, normalized size = 3.92 \[ \frac {2 i \ln \relax (x ) a}{a^{2}+1}-\frac {\ln \relax (x ) a^{2}}{a^{2}+1}+\frac {\ln \relax (x )}{a^{2}+1}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{a^{2}+1}-\frac {\ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{a^{2}+1}+\frac {2 i \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{a^{2}+1}-\frac {2 \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{a^{2}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 80, normalized size = 2.11 \[ -\frac {{\left (2 \, a - 2 i\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} - \frac {{\left (i \, a + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {{\left (a^{2} - 2 i \, a - 1\right )} \log \relax (x)}{a^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.70, size = 32, normalized size = 0.84 \[ \ln \relax (x)\,\left (-1+\frac {2{}\mathrm {i}}{a+1{}\mathrm {i}}\right )-\frac {\ln \left (a+b\,x+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{a+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*(2i/(a + 1i) - 1) - (log(a + b*x + 1i)*2i)/(a + 1i)

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sympy [B]  time = 0.80, size = 107, normalized size = 2.82 \[ - \frac {\left (a - i\right ) \log {\left (\frac {i a^{2} \left (a - i\right )}{a + i} - i a^{2} - \frac {2 a \left (a - i\right )}{a + i} + x \left (- i a b - 3 b\right ) - \frac {i \left (a - i\right )}{a + i} - i \right )}}{a + i} - \frac {2 i \log {\left (- i a^{2} - \frac {2 a^{2}}{a + i} - \frac {4 i a}{a + i} + x \left (- i a b - 3 b\right ) - i + \frac {2}{a + i} \right )}}{a + i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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