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

Optimal. Leaf size=93 \[ \frac{2 b^2}{(1-i a)^3 x}-\frac{2 i b^3 \log (x)}{(a+i)^4}+\frac{2 i b^3 \log (a+b x+i)}{(a+i)^4}+\frac{i b}{(a+i)^2 x^2}-\frac{-a+i}{3 (a+i) x^3} \]

[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 = 0.0564335, antiderivative size = 93, 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[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 = 0.0460624, size = 88, normalized size = 0.95 \[ \frac{(a+i) \left (a^3+i a^2+3 i a b x+a-6 i b^2 x^2-3 b x+i\right )+6 i b^3 x^3 \log (a+b x+i)-6 i b^3 x^3 \log (x)}{3 (a+i)^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[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*Lo
g[I + a + b*x])/(3*(I + a)^4*x^3)

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Maple [B]  time = 0.052, size = 560, normalized size = 6. \begin{align*} \text{result too large to display} \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^4,x)

[Out]

2*b^2/(a^2+1)^3/x+1/3/(a^2+1)/x^3*a^2+2*b^3/(a^2+1)^4*arctan(1/2*(2*b^2*x+2*a*b)/b)-2*I*b^2/(a^2+1)^3/x*a^3+6*
I*b^2/(a^2+1)^3/x*a+I*b^3/(a^2+1)^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a^4-6*I*b^3/(a^2+1)^4*ln(b^2*x^2+2*a*b*x+a^2+1)*
a^2-8*I*b^3/(a^2+1)^4*arctan(1/2*(2*b^2*x+2*a*b)/b)*a^3+8*I*b^3/(a^2+1)^4*arctan(1/2*(2*b^2*x+2*a*b)/b)*a+I*b/
(a^2+1)^2/x^2*a^2-2*I*b^3/(a^2+1)^4*ln(x)*a^4+12*I*b^3/(a^2+1)^4*ln(x)*a^2-8*b^3/(a^2+1)^4*ln(x)*a^3+8*b^3/(a^
2+1)^4*ln(x)*a-2*I*b^3/(a^2+1)^4*ln(x)+2*b/(a^2+1)^2/x^2*a-I*b/(a^2+1)^2/x^2-6*b^2/(a^2+1)^3/x*a^2-2/3*I/(a^2+
1)/x^3*a-4*b^3/(a^2+1)^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a+2*b^3/(a^2+1)^4*arctan(1/2*(2*b^2*x+2*a*b)/b)*a^4-12*b^3/
(a^2+1)^4*arctan(1/2*(2*b^2*x+2*a*b)/b)*a^2+4*b^3/(a^2+1)^4*ln(b^2*x^2+2*a*b*x+a^2+1)*a^3+I*b^3/(a^2+1)^4*ln(b
^2*x^2+2*a*b*x+a^2+1)-1/3/(a^2+1)/x^3

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Maxima [B]  time = 1.532, size = 355, normalized size = 3.82 \begin{align*} \frac{{\left (2 \, a^{4} - 8 i \, a^{3} - 12 \, a^{2} + 8 i \, a + 2\right )} b^{3} \arctan \left (\frac{b^{2} x + a b}{b}\right )}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac{{\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} b^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac{{\left (-2 i \, a^{4} - 8 \, a^{3} + 12 i \, a^{2} + 8 \, a - 2 i\right )} b^{3} \log \left (x\right )}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac{a^{6} - 2 i \, a^{5} -{\left (6 i \, a^{3} + 18 \, a^{2} - 18 i \, a - 6\right )} b^{2} x^{2} + a^{4} - 4 i \, a^{3} -{\left (-3 i \, a^{4} - 6 \, a^{3} - 6 \, a + 3 i\right )} b x - a^{2} - 2 i \, a - 1}{3 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \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^4,x, algorithm="maxima")

[Out]

(2*a^4 - 8*I*a^3 - 12*a^2 + 8*I*a + 2)*b^3*arctan((b^2*x + a*b)/b)/(a^8 + 4*a^6 + 6*a^4 + 4*a^2 + 1) + (I*a^4
+ 4*a^3 - 6*I*a^2 - 4*a + I)*b^3*log(b^2*x^2 + 2*a*b*x + a^2 + 1)/(a^8 + 4*a^6 + 6*a^4 + 4*a^2 + 1) + (-2*I*a^
4 - 8*a^3 + 12*I*a^2 + 8*a - 2*I)*b^3*log(x)/(a^8 + 4*a^6 + 6*a^4 + 4*a^2 + 1) + 1/3*(a^6 - 2*I*a^5 - (6*I*a^3
+ 18*a^2 - 18*I*a - 6)*b^2*x^2 + a^4 - 4*I*a^3 - (-3*I*a^4 - 6*a^3 - 6*a + 3*I)*b*x - a^2 - 2*I*a - 1)/((a^6
+ 3*a^4 + 3*a^2 + 1)*x^3)

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Fricas [A]  time = 1.91116, size = 247, normalized size = 2.66 \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+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+I*(b*x+a))**2/(1+(b*x+a)**2)/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.09622, size = 184, normalized size = 1.98 \begin{align*} -\frac{2 \, b^{4} \log \left (b x + a + i\right )}{a^{4} b i - 4 \, a^{3} b - 6 \, a^{2} b i + 4 \, a b + b i} + \frac{2 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{4} i - 4 \, a^{3} - 6 \, a^{2} i + 4 \, a + i} + \frac{a^{4} i - 2 \, a^{3} + 6 \,{\left (a b^{2} + b^{2} i\right )} x^{2} - 3 \,{\left (a^{2} b + 2 \, a b i - b\right )} x - 2 \, a - i}{3 \,{\left (a + i\right )}^{4} i x^{3}} \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^4,x, algorithm="giac")

[Out]

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