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]

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

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Rubi [A]  time = 0.05, antiderivative size = 76, 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} \[ -\frac {2 b^2 \log (x)}{(1-i a)^3}+\frac {2 b^2 \log (a+b x+i)}{(1-i a)^3}+\frac {2 i b}{(a+i)^2 x}-\frac {-a+i}{2 (a+i) x^2} \]

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 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^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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.43, size = 70, normalized size = 0.92 \[ \frac {4 i \, b^{2} x^{2} \log \relax (x) - 4 i \, b^{2} x^{2} \log \left (\frac {b x + a + i}{b}\right ) + a^{3} - 4 \, {\left (-i \, a + 1\right )} b x + i \, a^{2} + a + i}{{\left (2 \, a^{3} + 6 i \, a^{2} - 6 \, a - 2 i\right )} x^{2}} \]

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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giac [A]  time = 0.12, size = 98, normalized size = 1.29 \[ \frac {2 \, b^{3} \log \left (b x + a + i\right )}{a^{3} b i - 3 \, a^{2} b - 3 \, a b i + b} - \frac {2 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{3} i - 3 \, a^{2} - 3 \, a i + 1} + \frac {a^{3} i - a^{2} + a i - 4 \, {\left (a b + b i\right )} x - 1}{2 \, {\left (a + i\right )}^{3} i x^{2}} \]

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)

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maple [B]  time = 0.05, size = 406, normalized size = 5.34 \[ \frac {6 i b^{2} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{2}}{\left (a^{2}+1\right )^{3}}+\frac {a^{2}}{2 \left (a^{2}+1\right ) x^{2}}-\frac {1}{2 \left (a^{2}+1\right ) x^{2}}-\frac {i b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{3}}{\left (a^{2}+1\right )^{3}}+\frac {2 i b \,a^{2}}{\left (a^{2}+1\right )^{2} x}+\frac {4 b a}{\left (a^{2}+1\right )^{2} x}+\frac {2 i b^{2} \ln \relax (x ) a^{3}}{\left (a^{2}+1\right )^{3}}-\frac {2 i b}{\left (a^{2}+1\right )^{2} x}+\frac {6 b^{2} \ln \relax (x ) a^{2}}{\left (a^{2}+1\right )^{3}}-\frac {2 b^{2} \ln \relax (x )}{\left (a^{2}+1\right )^{3}}+\frac {3 i b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{\left (a^{2}+1\right )^{3}}-\frac {i a}{\left (a^{2}+1\right ) x^{2}}-\frac {3 b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{\left (a^{2}+1\right )^{3}}+\frac {b^{2} \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{\left (a^{2}+1\right )^{3}}-\frac {2 i b^{2} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{\left (a^{2}+1\right )^{3}}-\frac {2 b^{2} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a^{3}}{\left (a^{2}+1\right )^{3}}-\frac {6 i b^{2} \ln \relax (x ) a}{\left (a^{2}+1\right )^{3}}+\frac {6 b^{2} \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right ) a}{\left (a^{2}+1\right )^{3}} \]

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]

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

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maxima [B]  time = 0.43, size = 188, normalized size = 2.47 \[ -\frac {{\left (2 \, a^{3} - 6 i \, a^{2} - 6 \, a + 2 i\right )} b^{2} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac {{\left (-2 i \, a^{3} - 6 \, a^{2} + 6 i \, a + 2\right )} b^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )}} + \frac {{\left (2 i \, a^{3} + 6 \, a^{2} - 6 i \, a - 2\right )} b^{2} \log \relax (x)}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac {a^{4} - 2 i \, a^{3} + {\left (4 i \, a^{2} + 8 \, a - 4 i\right )} b x - 2 i \, a - 1}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \]

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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mupad [B]  time = 0.69, size = 154, normalized size = 2.03 \[ \frac {\frac {a-\mathrm {i}}{2\,\left (a+1{}\mathrm {i}\right )}+\frac {b\,x\,2{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^2}}{x^2}+\frac {b^2\,\mathrm {atanh}\left (\frac {-a^3-a^2\,3{}\mathrm {i}+3\,a+1{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^3}+\frac {x\,\left (2\,a^8\,b^2+8\,a^6\,b^2+12\,a^4\,b^2+8\,a^2\,b^2+2\,b^2\right )}{{\left (a+1{}\mathrm {i}\right )}^3\,\left (-b\,a^6+2{}\mathrm {i}\,b\,a^5-b\,a^4+4{}\mathrm {i}\,b\,a^3+b\,a^2+2{}\mathrm {i}\,b\,a+b\right )}\right )\,4{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.88, size = 226, normalized size = 2.97 \[ \frac {2 i b^{2} \log {\left (- \frac {2 a^{4} b^{2}}{\left (a + i\right )^{3}} - \frac {8 i a^{3} b^{2}}{\left (a + i\right )^{3}} + \frac {12 a^{2} b^{2}}{\left (a + i\right )^{3}} + 2 a b^{2} + \frac {8 i a b^{2}}{\left (a + i\right )^{3}} + 4 b^{3} x + 2 i b^{2} - \frac {2 b^{2}}{\left (a + i\right )^{3}} \right )}}{\left (a + i\right )^{3}} - \frac {2 i b^{2} \log {\left (\frac {2 a^{4} b^{2}}{\left (a + i\right )^{3}} + \frac {8 i a^{3} b^{2}}{\left (a + i\right )^{3}} - \frac {12 a^{2} b^{2}}{\left (a + i\right )^{3}} + 2 a b^{2} - \frac {8 i a b^{2}}{\left (a + i\right )^{3}} + 4 b^{3} x + 2 i b^{2} + \frac {2 b^{2}}{\left (a + i\right )^{3}} \right )}}{\left (a + i\right )^{3}} - \frac {a^{2} + 4 i b x + 1}{x^{2} \left (- 2 a^{2} - 4 i a + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*b**2*log(-2*a**4*b**2/(a + I)**3 - 8*I*a**3*b**2/(a + I)**3 + 12*a**2*b**2/(a + I)**3 + 2*a*b**2 + 8*I*a*b
**2/(a + I)**3 + 4*b**3*x + 2*I*b**2 - 2*b**2/(a + I)**3)/(a + I)**3 - 2*I*b**2*log(2*a**4*b**2/(a + I)**3 + 8
*I*a**3*b**2/(a + I)**3 - 12*a**2*b**2/(a + I)**3 + 2*a*b**2 - 8*I*a*b**2/(a + I)**3 + 4*b**3*x + 2*I*b**2 + 2
*b**2/(a + I)**3)/(a + I)**3 - (a**2 + 4*I*b*x + 1)/(x**2*(-2*a**2 - 4*I*a + 2))

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