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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 93, 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, 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 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]))))

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^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.51, size = 94, normalized size = 1.01 \[ \frac {-6 i \, b^{3} x^{3} \log \relax (x) + 6 i \, b^{3} x^{3} \log \left (\frac {b x + a + i}{b}\right ) - 6 \, {\left (i \, a - 1\right )} b^{2} x^{2} + a^{4} + 2 i \, a^{3} + {\left (3 i \, a^{2} - 6 \, a - 3 i\right )} b x + 2 i \, a - 1}{{\left (3 \, a^{4} + 12 i \, a^{3} - 18 \, a^{2} - 12 i \, a + 3\right )} x^{3}} \]

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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giac [A]  time = 0.12, size = 136, normalized size = 1.46 \[ -\frac {2 \, b^{4} \log \left (b x + a + i\right )}{a^{4} b i - 4 \, a^{3} b - 6 \, a^{2} b i + 4 \, a b + b i} + \frac {2 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{4} i - 4 \, a^{3} - 6 \, a^{2} i + 4 \, a + i} + \frac {a^{4} i - 2 \, a^{3} + 6 \, {\left (a b^{2} + b^{2} i\right )} x^{2} - 3 \, {\left (a^{2} b + 2 \, a b i - b\right )} x - 2 \, a - i}{3 \, {\left (a + i\right )}^{4} i x^{3}} \]

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)

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maple [B]  time = 0.06, size = 560, normalized size = 6.02 \[ \frac {2 b^{3} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{\left (a^{2}+1\right )^{4}}+\frac {a^{2}}{3 \left (a^{2}+1\right ) x^{3}}+\frac {2 b^{2}}{\left (a^{2}+1\right )^{3} x}-\frac {8 b^{3} \ln \relax (x ) a^{3}}{\left (a^{2}+1\right )^{4}}+\frac {2 b a}{\left (a^{2}+1\right )^{2} x^{2}}-\frac {6 b^{2} a^{2}}{\left (a^{2}+1\right )^{3} x}-\frac {4 b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{\left (a^{2}+1\right )^{4}}+\frac {2 b^{3} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{4}}{\left (a^{2}+1\right )^{4}}-\frac {12 b^{3} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{\left (a^{2}+1\right )^{4}}+\frac {4 b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{\left (a^{2}+1\right )^{4}}+\frac {8 b^{3} \ln \relax (x ) a}{\left (a^{2}+1\right )^{4}}-\frac {1}{3 \left (a^{2}+1\right ) x^{3}}-\frac {2 i b^{3} \ln \relax (x ) a^{4}}{\left (a^{2}+1\right )^{4}}+\frac {12 i b^{3} \ln \relax (x ) a^{2}}{\left (a^{2}+1\right )^{4}}+\frac {i b \,a^{2}}{\left (a^{2}+1\right )^{2} x^{2}}-\frac {2 i b^{2} a^{3}}{\left (a^{2}+1\right )^{3} x}-\frac {6 i b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{\left (a^{2}+1\right )^{4}}-\frac {8 i b^{3} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{3}}{\left (a^{2}+1\right )^{4}}+\frac {8 i b^{3} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{\left (a^{2}+1\right )^{4}}+\frac {i b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{4}}{\left (a^{2}+1\right )^{4}}+\frac {6 i b^{2} a}{\left (a^{2}+1\right )^{3} x}-\frac {2 i b^{3} \ln \relax (x )}{\left (a^{2}+1\right )^{4}}-\frac {2 i a}{3 \left (a^{2}+1\right ) x^{3}}-\frac {i b}{\left (a^{2}+1\right )^{2} x^{2}}+\frac {i b^{3} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{\left (a^{2}+1\right )^{4}} \]

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

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maxima [B]  time = 0.44, size = 263, normalized size = 2.83 \[ \frac {{\left (2 \, a^{4} - 8 i \, a^{3} - 12 \, a^{2} + 8 i \, a + 2\right )} b^{3} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac {{\left (i \, a^{4} + 4 \, a^{3} - 6 i \, a^{2} - 4 \, a + i\right )} b^{3} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac {{\left (-2 i \, a^{4} - 8 \, a^{3} + 12 i \, a^{2} + 8 \, a - 2 i\right )} b^{3} \log \relax (x)}{a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1} + \frac {a^{6} - 2 i \, a^{5} - {\left (6 i \, a^{3} + 18 \, a^{2} - 18 i \, a - 6\right )} b^{2} x^{2} + a^{4} - 4 i \, a^{3} - {\left (-3 i \, a^{4} - 6 \, a^{3} - 6 \, a + 3 i\right )} b x - a^{2} - 2 i \, a - 1}{3 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \]

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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mupad [B]  time = 0.72, size = 199, normalized size = 2.14 \[ \frac {\frac {a-\mathrm {i}}{3\,\left (a+1{}\mathrm {i}\right )}-\frac {b^2\,x^2\,2{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^3}+\frac {b\,x\,1{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^2}}{x^3}+\frac {b^3\,\mathrm {atanh}\left (\frac {a^4+a^3\,4{}\mathrm {i}-6\,a^2-a\,4{}\mathrm {i}+1}{{\left (a+1{}\mathrm {i}\right )}^4}-\frac {x\,\left (2\,a^{12}\,b^2+12\,a^{10}\,b^2+30\,a^8\,b^2+40\,a^6\,b^2+30\,a^4\,b^2+12\,a^2\,b^2+2\,b^2\right )}{{\left (a+1{}\mathrm {i}\right )}^4\,\left (-b\,a^9+3{}\mathrm {i}\,b\,a^8+8{}\mathrm {i}\,b\,a^6+6\,b\,a^5+6{}\mathrm {i}\,b\,a^4+8\,b\,a^3+3\,b\,a-b\,1{}\mathrm {i}\right )}\right )\,4{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*1i + b*x*1i + 1)^2/(x^4*((a + b*x)^2 + 1)),x)

[Out]

((a - 1i)/(3*(a + 1i)) - (b^2*x^2*2i)/(a + 1i)^3 + (b*x*1i)/(a + 1i)^2)/x^3 + (b^3*atanh((a^3*4i - 6*a^2 - a*4
i + a^4 + 1)/(a + 1i)^4 - (x*(2*b^2 + 12*a^2*b^2 + 30*a^4*b^2 + 40*a^6*b^2 + 30*a^8*b^2 + 12*a^10*b^2 + 2*a^12
*b^2))/((a + 1i)^4*(3*a*b - b*1i + 8*a^3*b + a^4*b*6i + 6*a^5*b + a^6*b*8i + a^8*b*3i - a^9*b)))*4i)/(a + 1i)^
4

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sympy [B]  time = 1.15, size = 286, normalized size = 3.08 \[ - \frac {2 i b^{3} \log {\left (- \frac {2 a^{5} b^{3}}{\left (a + i\right )^{4}} - \frac {10 i a^{4} b^{3}}{\left (a + i\right )^{4}} + \frac {20 a^{3} b^{3}}{\left (a + i\right )^{4}} + \frac {20 i a^{2} b^{3}}{\left (a + i\right )^{4}} + 2 a b^{3} - \frac {10 a b^{3}}{\left (a + i\right )^{4}} + 4 b^{4} x + 2 i b^{3} - \frac {2 i b^{3}}{\left (a + i\right )^{4}} \right )}}{\left (a + i\right )^{4}} + \frac {2 i b^{3} \log {\left (\frac {2 a^{5} b^{3}}{\left (a + i\right )^{4}} + \frac {10 i a^{4} b^{3}}{\left (a + i\right )^{4}} - \frac {20 a^{3} b^{3}}{\left (a + i\right )^{4}} - \frac {20 i a^{2} b^{3}}{\left (a + i\right )^{4}} + 2 a b^{3} + \frac {10 a b^{3}}{\left (a + i\right )^{4}} + 4 b^{4} x + 2 i b^{3} + \frac {2 i b^{3}}{\left (a + i\right )^{4}} \right )}}{\left (a + i\right )^{4}} - \frac {- i a^{3} + a^{2} - i a - 6 b^{2} x^{2} + x \left (3 a b + 3 i b\right ) + 1}{x^{3} \left (3 i a^{3} - 9 a^{2} - 9 i a + 3\right )} \]

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]

-2*I*b**3*log(-2*a**5*b**3/(a + I)**4 - 10*I*a**4*b**3/(a + I)**4 + 20*a**3*b**3/(a + I)**4 + 20*I*a**2*b**3/(
a + I)**4 + 2*a*b**3 - 10*a*b**3/(a + I)**4 + 4*b**4*x + 2*I*b**3 - 2*I*b**3/(a + I)**4)/(a + I)**4 + 2*I*b**3
*log(2*a**5*b**3/(a + I)**4 + 10*I*a**4*b**3/(a + I)**4 - 20*a**3*b**3/(a + I)**4 - 20*I*a**2*b**3/(a + I)**4
+ 2*a*b**3 + 10*a*b**3/(a + I)**4 + 4*b**4*x + 2*I*b**3 + 2*I*b**3/(a + I)**4)/(a + I)**4 - (-I*a**3 + a**2 -
I*a - 6*b**2*x**2 + x*(3*a*b + 3*I*b) + 1)/(x**3*(3*I*a**3 - 9*a**2 - 9*I*a + 3))

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