3.206 \(\int \frac{e^{-2 i \tan ^{-1}(a+b x)}}{x^4} \, dx\)

Optimal. Leaf size=102 \[ \frac{2 b^2}{(1+i a)^3 x}+\frac{2 i b^3 \log (x)}{(-a+i)^4}-\frac{2 i b^3 \log (-a-b x+i)}{(-a+i)^4}-\frac{i b}{(-a+i)^2 x^2}-\frac{a+i}{3 (-a+i) x^3} \]

[Out]

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

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Rubi [A]  time = 0.0606169, antiderivative size = 102, 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} \[ \frac{2 b^2}{(1+i a)^3 x}+\frac{2 i b^3 \log (x)}{(-a+i)^4}-\frac{2 i b^3 \log (-a-b x+i)}{(-a+i)^4}-\frac{i b}{(-a+i)^2 x^2}-\frac{a+i}{3 (-a+i) x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^((2*I)*ArcTan[a + b*x])*x^4),x]

[Out]

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

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^4} \, dx &=\int \frac{1-i a-i b x}{x^4 (1+i a+i b x)} \, dx\\ &=\int \left (\frac{-i-a}{(-i+a) x^4}+\frac{2 i b}{(-i+a)^2 x^3}-\frac{2 i b^2}{(-i+a)^3 x^2}+\frac{2 i b^3}{(-i+a)^4 x}-\frac{2 i b^4}{(-i+a)^4 (-i+a+b x)}\right ) \, dx\\ &=-\frac{i+a}{3 (i-a) x^3}-\frac{i b}{(i-a)^2 x^2}+\frac{2 b^2}{(1+i a)^3 x}+\frac{2 i b^3 \log (x)}{(i-a)^4}-\frac{2 i b^3 \log (i-a-b x)}{(i-a)^4}\\ \end{align*}

Mathematica [A]  time = 0.0462582, size = 91, normalized size = 0.89 \[ \frac{(a-i) \left (a^3-i a^2-3 i a b x+a+6 i b^2 x^2-3 b x-i\right )-6 i b^3 x^3 \log (-a-b x+i)+6 i b^3 x^3 \log (x)}{3 (a-i)^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((2*I)*ArcTan[a + b*x])*x^4),x]

[Out]

((-I + a)*(-I + a - I*a^2 + a^3 - 3*b*x - (3*I)*a*b*x + (6*I)*b^2*x^2) + (6*I)*b^3*x^3*Log[x] - (6*I)*b^3*x^3*
Log[I - a - b*x])/(3*(-I + a)^4*x^3)

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Maple [B]  time = 0.054, size = 349, normalized size = 3.4 \begin{align*}{\frac{-2\,i{b}^{3}\ln \left ( x \right ) a}{ \left ( i-a \right ) ^{5}}}+{\frac{{b}^{3}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) }{ \left ( i-a \right ) ^{5}}}-2\,{\frac{{b}^{3}\arctan \left ( bx+a \right ) a}{ \left ( i-a \right ) ^{5}}}+{\frac{{\frac{2\,i}{3}}{a}^{2}}{ \left ( i-a \right ) ^{5}{x}^{3}}}+{\frac{i{a}^{4}}{ \left ( i-a \right ) ^{5}{x}^{3}}}+{\frac{2\,i{b}^{3}\arctan \left ( bx+a \right ) }{ \left ( i-a \right ) ^{5}}}+3\,{\frac{{a}^{2}b}{ \left ( i-a \right ) ^{5}{x}^{2}}}-{\frac{b}{ \left ( i-a \right ) ^{5}{x}^{2}}}+{\frac{i{a}^{3}b}{ \left ( i-a \right ) ^{5}{x}^{2}}}-{\frac{{\frac{i}{3}}}{ \left ( i-a \right ) ^{5}{x}^{3}}}-4\,{\frac{a{b}^{2}}{ \left ( i-a \right ) ^{5}x}}+{\frac{i{b}^{3}\ln \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) a}{ \left ( i-a \right ) ^{5}}}-{\frac{{a}^{5}}{3\, \left ( i-a \right ) ^{5}{x}^{3}}}+{\frac{2\,i{b}^{2}}{ \left ( i-a \right ) ^{5}x}}+{\frac{2\,{a}^{3}}{3\, \left ( i-a \right ) ^{5}{x}^{3}}}-{\frac{3\,iba}{ \left ( i-a \right ) ^{5}{x}^{2}}}+{\frac{a}{ \left ( i-a \right ) ^{5}{x}^{3}}}-{\frac{2\,i{a}^{2}{b}^{2}}{ \left ( i-a \right ) ^{5}x}}-2\,{\frac{{b}^{3}\ln \left ( x \right ) }{ \left ( i-a \right ) ^{5}}} \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^4,x)

[Out]

-2*I*b^3/(I-a)^5*ln(x)*a+b^3/(I-a)^5*ln(b^2*x^2+2*a*b*x+a^2+1)-2*b^3/(I-a)^5*arctan(b*x+a)*a+2/3*I/(I-a)^5/x^3
*a^2+I/(I-a)^5/x^3*a^4+2*I*b^3/(I-a)^5*arctan(b*x+a)+3*b/(I-a)^5/x^2*a^2-b/(I-a)^5/x^2+I*b/(I-a)^5/x^2*a^3-1/3
*I/(I-a)^5/x^3-4*b^2/(I-a)^5/x*a+I*b^3/(I-a)^5*ln(b^2*x^2+2*a*b*x+a^2+1)*a-1/3/(I-a)^5/x^3*a^5+2*I*b^2/(I-a)^5
/x+2/3/(I-a)^5/x^3*a^3-3*I*b/(I-a)^5/x^2*a+1/(I-a)^5/x^3*a-2*I*b^2/(I-a)^5/x*a^2-2*b^3/(I-a)^5*ln(x)

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Maxima [B]  time = 1.05809, size = 300, normalized size = 2.94 \begin{align*} \frac{{\left (2 \, a - 2 i\right )} b^{3} \log \left (i \, b x + i \, a + 1\right )}{i \, a^{5} + 5 \, a^{4} - 10 i \, a^{3} - 10 \, a^{2} + 5 i \, a + 1} - \frac{{\left (2 \, a - 2 i\right )} b^{3} \log \left (x\right )}{i \, a^{5} + 5 \, a^{4} - 10 i \, a^{3} - 10 \, a^{2} + 5 i \, a + 1} - \frac{{\left (6 \, a - 6 i\right )} b^{3} x^{3} - i \, a^{5} + 3 \,{\left (a^{2} - 2 i \, a - 1\right )} b^{2} x^{2} - 3 \, a^{4} + 2 i \, a^{3} -{\left (i \, a^{4} + 5 \, a^{3} - 9 i \, a^{2} - 7 \, a + 2 i\right )} b x - 2 \, a^{2} + 3 i \, a + 1}{{\left (3 i \, a^{4} + 12 \, a^{3} - 18 i \, a^{2} - 12 \, a + 3 i\right )} b x^{4} +{\left (3 i \, a^{5} + 15 \, a^{4} - 30 i \, a^{3} - 30 \, a^{2} + 15 i \, a + 3\right )} x^{3}} \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^4,x, algorithm="maxima")

[Out]

(2*a - 2*I)*b^3*log(I*b*x + I*a + 1)/(I*a^5 + 5*a^4 - 10*I*a^3 - 10*a^2 + 5*I*a + 1) - (2*a - 2*I)*b^3*log(x)/
(I*a^5 + 5*a^4 - 10*I*a^3 - 10*a^2 + 5*I*a + 1) - ((6*a - 6*I)*b^3*x^3 - I*a^5 + 3*(a^2 - 2*I*a - 1)*b^2*x^2 -
3*a^4 + 2*I*a^3 - (I*a^4 + 5*a^3 - 9*I*a^2 - 7*a + 2*I)*b*x - 2*a^2 + 3*I*a + 1)/((3*I*a^4 + 12*a^3 - 18*I*a^
2 - 12*a + 3*I)*b*x^4 + (3*I*a^5 + 15*a^4 - 30*I*a^3 - 30*a^2 + 15*I*a + 3)*x^3)

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Fricas [A]  time = 2.12128, size = 248, normalized size = 2.43 \begin{align*} \frac{6 i \, b^{3} x^{3} \log \left (x\right ) - 6 i \, b^{3} x^{3} \log \left (\frac{b x + a - i}{b}\right ) - 6 \,{\left (-i \, a - 1\right )} b^{2} x^{2} + a^{4} - 2 i \, a^{3} +{\left (-3 i \, a^{2} - 6 \, a + 3 i\right )} b x - 2 i \, a - 1}{{\left (3 \, a^{4} - 12 i \, a^{3} - 18 \, a^{2} + 12 i \, a + 3\right )} x^{3}} \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^4,x, algorithm="fricas")

[Out]

(6*I*b^3*x^3*log(x) - 6*I*b^3*x^3*log((b*x + a - I)/b) - 6*(-I*a - 1)*b^2*x^2 + a^4 - 2*I*a^3 + (-3*I*a^2 - 6*
a + 3*I)*b*x - 2*I*a - 1)/((3*a^4 - 12*I*a^3 - 18*a^2 + 12*I*a + 3)*x^3)

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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/(1+I*(b*x+a))**2*(1+(b*x+a)**2)/x**4,x)

[Out]

Timed out

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Giac [B]  time = 1.13172, size = 273, normalized size = 2.68 \begin{align*} -\frac{2 \, b^{4} \log \left (-\frac{a i}{b i x + a i + 1} + \frac{i^{2}}{b i x + a i + 1} + 1\right )}{a^{4} b i + 4 \, a^{3} b - 6 \, a^{2} b i - 4 \, a b + b i} - \frac{\frac{3 \,{\left (a b^{4} i - 8 \, b^{4}\right )} i^{2}}{{\left (b i x + a i + 1\right )} b} + \frac{a b^{3} i - 10 \, b^{3}}{a i + 1} + \frac{3 \,{\left (a^{2} b^{5} + 4 \, a b^{5} i + 5 \, b^{5}\right )} i^{2}}{{\left (b i x + a i + 1\right )}^{2} b^{2}}}{3 \,{\left (a - i\right )}^{3}{\left (\frac{a i}{b i x + a i + 1} - \frac{i^{2}}{b i x + a i + 1} - 1\right )}^{3}} \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^4,x, algorithm="giac")

[Out]

-2*b^4*log(-a*i/(b*i*x + a*i + 1) + i^2/(b*i*x + a*i + 1) + 1)/(a^4*b*i + 4*a^3*b - 6*a^2*b*i - 4*a*b + b*i) -
1/3*(3*(a*b^4*i - 8*b^4)*i^2/((b*i*x + a*i + 1)*b) + (a*b^3*i - 10*b^3)/(a*i + 1) + 3*(a^2*b^5 + 4*a*b^5*i +
5*b^5)*i^2/((b*i*x + a*i + 1)^2*b^2))/((a - i)^3*(a*i/(b*i*x + a*i + 1) - i^2/(b*i*x + a*i + 1) - 1)^3)