3.178 \(\int \frac{e^{2 i \tan ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=76 \[ -\frac{2 b^2 \log (x)}{(1-i a)^3}+\frac{2 b^2 \log (a+b x+i)}{(1-i a)^3}+\frac{2 i b}{(a+i)^2 x}-\frac{-a+i}{2 (a+i) x^2} \]

[Out]

-(I - a)/(2*(I + a)*x^2) + ((2*I)*b)/((I + a)^2*x) - (2*b^2*Log[x])/(1 - I*a)^3 + (2*b^2*Log[I + a + b*x])/(1
- I*a)^3

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Rubi [A]  time = 0.0492248, antiderivative size = 76, 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, 77} \[ -\frac{2 b^2 \log (x)}{(1-i a)^3}+\frac{2 b^2 \log (a+b x+i)}{(1-i a)^3}+\frac{2 i b}{(a+i)^2 x}-\frac{-a+i}{2 (a+i) x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^((2*I)*ArcTan[a + b*x])/x^3,x]

[Out]

-(I - a)/(2*(I + a)*x^2) + ((2*I)*b)/((I + a)^2*x) - (2*b^2*Log[x])/(1 - I*a)^3 + (2*b^2*Log[I + a + b*x])/(1
- I*a)^3

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 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]))))

Rubi steps

\begin{align*} \int \frac{e^{2 i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac{1+i a+i b x}{x^3 (1-i a-i b x)} \, dx\\ &=\int \left (\frac{i-a}{(i+a) x^3}-\frac{2 i b}{(i+a)^2 x^2}+\frac{2 i b^2}{(i+a)^3 x}-\frac{2 i b^3}{(i+a)^3 (i+a+b x)}\right ) \, dx\\ &=-\frac{i-a}{2 (i+a) x^2}+\frac{2 i b}{(i+a)^2 x}-\frac{2 b^2 \log (x)}{(1-i a)^3}+\frac{2 b^2 \log (i+a+b x)}{(1-i a)^3}\\ \end{align*}

Mathematica [A]  time = 0.0324949, size = 63, normalized size = 0.83 \[ \frac{(a+i) \left (a^2+4 i b x+1\right )-4 i b^2 x^2 \log (a+b x+i)+4 i b^2 x^2 \log (x)}{2 (a+i)^3 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^((2*I)*ArcTan[a + b*x])/x^3,x]

[Out]

((I + a)*(1 + a^2 + (4*I)*b*x) + (4*I)*b^2*x^2*Log[x] - (4*I)*b^2*x^2*Log[I + a + b*x])/(2*(I + a)^3*x^2)

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Maple [B]  time = 0.048, size = 406, normalized size = 5.3 \begin{align*}{\frac{-ia}{ \left ({a}^{2}+1 \right ){x}^{2}}}-{\frac{2\,i{b}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}} \right ) }-3\,{\frac{{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ){a}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{ \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{3\,i{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{ \left ({a}^{2}+1 \right ) ^{3}}}-2\,{\frac{{b}^{2}{a}^{3}}{ \left ({a}^{2}+1 \right ) ^{3}}\arctan \left ( 1/2\,{\frac{2\,{b}^{2}x+2\,ab}{b}} \right ) }+{\frac{6\,i{b}^{2}{a}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}}\arctan \left ({\frac{2\,{b}^{2}x+2\,ab}{2\,b}} \right ) }+6\,{\frac{{b}^{2}a}{ \left ({a}^{2}+1 \right ) ^{3}}\arctan \left ( 1/2\,{\frac{2\,{b}^{2}x+2\,ab}{b}} \right ) }-{\frac{i{b}^{2}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ){a}^{3}}{ \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{{a}^{2}}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}}}-{\frac{1}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}}}+{\frac{2\,i{b}^{2}\ln \left ( x \right ){a}^{3}}{ \left ({a}^{2}+1 \right ) ^{3}}}+{\frac{2\,i{a}^{2}b}{ \left ({a}^{2}+1 \right ) ^{2}x}}+4\,{\frac{ab}{ \left ({a}^{2}+1 \right ) ^{2}x}}-{\frac{2\,ib}{ \left ({a}^{2}+1 \right ) ^{2}x}}-{\frac{6\,i{b}^{2}\ln \left ( x \right ) a}{ \left ({a}^{2}+1 \right ) ^{3}}}+6\,{\frac{{b}^{2}\ln \left ( x \right ){a}^{2}}{ \left ({a}^{2}+1 \right ) ^{3}}}-2\,{\frac{{b}^{2}\ln \left ( x \right ) }{ \left ({a}^{2}+1 \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+I*(b*x+a))^2/(1+(b*x+a)^2)/x^3,x)

[Out]

-I/(a^2+1)/x^2*a-2*I*b^2/(a^2+1)^3*arctan(1/2*(2*b^2*x+2*a*b)/b)-3*b^2/(a^2+1)^3*ln(b^2*x^2+2*a*b*x+a^2+1)*a^2
+b^2/(a^2+1)^3*ln(b^2*x^2+2*a*b*x+a^2+1)+3*I*b^2/(a^2+1)^3*ln(b^2*x^2+2*a*b*x+a^2+1)*a-2*b^2/(a^2+1)^3*arctan(
1/2*(2*b^2*x+2*a*b)/b)*a^3+6*I*b^2/(a^2+1)^3*arctan(1/2*(2*b^2*x+2*a*b)/b)*a^2+6*b^2/(a^2+1)^3*arctan(1/2*(2*b
^2*x+2*a*b)/b)*a-I*b^2/(a^2+1)^3*ln(b^2*x^2+2*a*b*x+a^2+1)*a^3+1/2/(a^2+1)/x^2*a^2-1/2/(a^2+1)/x^2+2*I*b^2/(a^
2+1)^3*ln(x)*a^3+2*I*b/(a^2+1)^2/x*a^2+4*b/(a^2+1)^2/x*a-2*I*b/(a^2+1)^2/x-6*I*b^2/(a^2+1)^3*ln(x)*a+6*b^2/(a^
2+1)^3*ln(x)*a^2-2*b^2/(a^2+1)^3*ln(x)

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Maxima [B]  time = 1.49249, size = 254, normalized size = 3.34 \begin{align*} -\frac{{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} b^{2} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac{{\left (-2 i \, a^{3} - 6 \, a^{2} + 6 i \, a + 2\right )} b^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )}} + \frac{{\left (2 i \, a^{3} + 6 \, a^{2} - 6 i \, a - 2\right )} b^{2} \log \left (x\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac{a^{4} - 2 i \, a^{3} +{\left (4 i \, a^{2} + 8 \, a - 4 i\right )} b x - 2 i \, a - 1}{2 \,{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)/x^3,x, algorithm="maxima")

[Out]

-(2*a^3 - 6*I*a^2 - 6*a + 2*I)*b^2*arctan((b^2*x + a*b)/b)/(a^6 + 3*a^4 + 3*a^2 + 1) + 1/2*(-2*I*a^3 - 6*a^2 +
6*I*a + 2)*b^2*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^6 + 3*a^4 + 3*a^2 + 1) + (2*I*a^3 + 6*a^2 - 6*I*a - 2)*b^2
*log(x)/(a^6 + 3*a^4 + 3*a^2 + 1) + 1/2*(a^4 - 2*I*a^3 + (4*I*a^2 + 8*a - 4*I)*b*x - 2*I*a - 1)/((a^4 + 2*a^2
+ 1)*x^2)

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Fricas [A]  time = 1.93852, size = 181, normalized size = 2.38 \begin{align*} \frac{4 i \, b^{2} x^{2} \log \left (x\right ) - 4 i \, b^{2} x^{2} \log \left (\frac{b x + a + i}{b}\right ) + a^{3} - 4 \,{\left (-i \, a + 1\right )} b x + i \, a^{2} + a + i}{{\left (2 \, a^{3} + 6 i \, a^{2} - 6 \, a - 2 i\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)/x^3,x, algorithm="fricas")

[Out]

(4*I*b^2*x^2*log(x) - 4*I*b^2*x^2*log((b*x + a + I)/b) + a^3 - 4*(-I*a + 1)*b*x + I*a^2 + a + I)/((2*a^3 + 6*I
*a^2 - 6*a - 2*I)*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))**2/(1+(b*x+a)**2)/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.13591, size = 132, normalized size = 1.74 \begin{align*} \frac{2 \, b^{3} \log \left (b x + a + i\right )}{a^{3} b i - 3 \, a^{2} b - 3 \, a b i + b} - \frac{2 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{3} i - 3 \, a^{2} - 3 \, a i + 1} + \frac{a^{3} i - a^{2} + a i - 4 \,{\left (a b + b i\right )} x - 1}{2 \,{\left (a + i\right )}^{3} i x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+I*(b*x+a))^2/(1+(b*x+a)^2)/x^3,x, algorithm="giac")

[Out]

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