Integrand size = 16, antiderivative size = 83 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=\frac {-i-a}{2 (i-a) x^2}-\frac {2 i b}{(i-a)^2 x}-\frac {2 b^2 \log (x)}{(1+i a)^3}+\frac {2 b^2 \log (i-a-b x)}{(1+i a)^3} \] Output:
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
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=\frac {(-i+a) \left (1+a^2-4 i b x\right )-4 i b^2 x^2 \log (x)+4 i b^2 x^2 \log (i-a-b x)}{2 (-i+a)^3 x^2} \] Input:
Integrate[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]
Output:
((-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)
Time = 0.42 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5618, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 5618 |
\(\displaystyle \int \frac {-i a-i b x+1}{x^3 (i a+i b x+1)}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {2 i b^3}{(a-i)^3 (a+b x-i)}-\frac {2 i b^2}{(a-i)^3 x}+\frac {2 i b}{(a-i)^2 x^2}+\frac {-a-i}{(a-i) x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^2 \log (x)}{(1+i a)^3}+\frac {2 b^2 \log (-a-b x+i)}{(1+i a)^3}-\frac {2 i b}{(-a+i)^2 x}-\frac {a+i}{2 (-a+i) x^2}\) |
Input:
Int[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]
Output:
-1/2*(I + a)/((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
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((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]))))
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]
Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.31
method | result | size |
default | \(-\frac {-a^{4}+2 i a^{3}+2 i a +1}{2 \left (i-a \right )^{4} x^{2}}-\frac {2 b \left (i a^{2}+2 a -i\right )}{\left (i-a \right )^{4} x}-\frac {2 b^{2} \left (i a +1\right ) \ln \left (x \right )}{\left (i-a \right )^{4}}+\frac {2 b^{2} \left (i a +1\right ) \ln \left (-b x -a +i\right )}{\left (i-a \right )^{4}}\) | \(109\) |
risch | \(\frac {-\frac {2 i b x}{a^{2}-2 i a -1}+\frac {i+a}{-2 i+2 a}}{x^{2}}+\frac {2 b^{2} \ln \left (\left (2 a^{4} b +4 a^{2} b +2 b \right ) x \right )}{i a^{3}+3 a^{2}-3 i a -1}-\frac {b^{2} \ln \left (4 a^{8} b^{2} x^{2}+8 a^{9} b x +4 a^{10}+16 a^{6} b^{2} x^{2}+32 a^{7} b x +20 a^{8}+24 a^{4} b^{2} x^{2}+48 a^{5} b x +40 a^{6}+16 a^{2} b^{2} x^{2}+32 a^{3} b x +40 a^{4}+4 b^{2} x^{2}+8 a b x +20 a^{2}+4\right )}{i a^{3}+3 a^{2}-3 i a -1}+\frac {2 i b^{2} \arctan \left (\frac {\left (-2 a^{4} b -4 a^{2} b -2 b \right ) x -2 a^{5}-4 a^{3}-2 a}{2 a^{4}+4 a^{2}+2}\right )}{i a^{3}+3 a^{2}-3 i a -1}\) | \(288\) |
parallelrisch | \(\frac {-b +27 a^{2} b +4 i \ln \left (x \right ) x^{2} a^{7} b^{3}-4 b^{3} x^{2}+8 i a^{9} b -4 \ln \left (x \right ) x^{2} b^{3}-25 a \,b^{2} x -a^{10} b +27 a^{8} b -4 i \ln \left (b x +a -i\right ) x^{2} a^{7} b^{3}-60 i \ln \left (x \right ) x^{3} a^{4} b^{4}+60 i \ln \left (b x +a -i\right ) x^{3} a^{4} b^{4}+60 i \ln \left (x \right ) x^{3} a^{2} b^{4}-60 i \ln \left (b x +a -i\right ) x^{3} a^{2} b^{4}-84 i \ln \left (x \right ) x^{2} a^{5} b^{3}+84 i \ln \left (b x +a -i\right ) x^{2} a^{5} b^{3}+140 i \ln \left (x \right ) x^{2} a^{3} b^{3}-140 i \ln \left (b x +a -i\right ) x^{2} a^{3} b^{3}+24 \ln \left (x \right ) x^{3} a \,b^{4}+84 \ln \left (x \right ) x^{2} a^{2} b^{3}-28 i \ln \left (x \right ) x^{2} a \,b^{3}+140 i x^{2} a^{3} b^{3}-24 \ln \left (b x +a -i\right ) x^{3} a \,b^{4}-84 \ln \left (b x +a -i\right ) x^{2} a^{2} b^{3}+4 i x^{2} a^{7} b^{3}+11 i x \,a^{8} b^{2}-140 i x \,a^{6} b^{2}-4 i \ln \left (x \right ) x^{3} b^{4}+4 i \ln \left (b x +a -i\right ) x^{3} b^{4}-28 i x^{2} a \,b^{3}-24 \ln \left (b x +a -i\right ) x^{3} a^{5} b^{4}-28 \ln \left (b x +a -i\right ) x^{2} a^{6} b^{3}+80 \ln \left (b x +a -i\right ) x^{3} a^{3} b^{4}+140 \ln \left (b x +a -i\right ) x^{2} a^{4} b^{3}+266 i x \,a^{4} b^{2}-92 i x \,a^{2} b^{2}-84 i x^{2} a^{5} b^{3}+24 \ln \left (x \right ) x^{3} a^{5} b^{4}+28 \ln \left (x \right ) x^{2} a^{6} b^{3}-80 \ln \left (x \right ) x^{3} a^{3} b^{4}-140 \ln \left (x \right ) x^{2} a^{4} b^{3}-48 i a^{7} b -8 i a b +4 \ln \left (b x +a -i\right ) x^{2} b^{3}+3 i x \,b^{2}+28 x^{2} a^{6} b^{3}-140 x^{2} a^{4} b^{3}+84 x^{2} a^{2} b^{3}-x \,a^{9} b^{2}+52 x \,a^{7} b^{2}-238 x \,a^{5} b^{2}+196 x \,a^{3} b^{2}-42 a^{6} b +48 i a^{3} b +28 i \ln \left (b x +a -i\right ) x^{2} a \,b^{3}+4 i \ln \left (x \right ) x^{3} a^{6} b^{4}-4 i \ln \left (b x +a -i\right ) x^{3} a^{6} b^{4}-42 a^{4} b}{2 \left (-a^{6}+6 i a^{5}+15 a^{4}-20 i a^{3}-15 a^{2}+6 i a +1\right ) b \left (-a^{3}+3 i a^{2}+3 a -i\right ) \left (-b x -a +i\right ) x^{2}}\) | \(788\) |
Input:
int(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/2/(I-a)^4*(-a^4+2*I*a^3+2*I*a+1)/x^2-2*b*(I*a^2-I+2*a)/(I-a)^4/x-2*b^2* (1+I*a)/(I-a)^4*ln(x)+2*b^2*(1+I*a)/(I-a)^4*ln(I-a-b*x)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=\frac {-4 i \, b^{2} x^{2} \log \left (x\right ) + 4 i \, b^{2} x^{2} \log \left (\frac {b x + a - i}{b}\right ) + a^{3} - 4 \, {\left (i \, a + 1\right )} b x - i \, a^{2} + a - i}{2 \, {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x^{2}} \] Input:
integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="fricas")
Output:
1/2*(-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)/((a^3 - 3*I*a^2 - 3*a + I)*x^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (61) = 122\).
Time = 0.43 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.72 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=- \frac {2 i b^{2} \log {\left (- \frac {2 a^{4} b^{2}}{\left (a - i\right )^{3}} + \frac {8 i a^{3} b^{2}}{\left (a - i\right )^{3}} + \frac {12 a^{2} b^{2}}{\left (a - i\right )^{3}} + 2 a b^{2} - \frac {8 i a b^{2}}{\left (a - i\right )^{3}} + 4 b^{3} x - 2 i b^{2} - \frac {2 b^{2}}{\left (a - i\right )^{3}} \right )}}{\left (a - i\right )^{3}} + \frac {2 i b^{2} \log {\left (\frac {2 a^{4} b^{2}}{\left (a - i\right )^{3}} - \frac {8 i a^{3} b^{2}}{\left (a - i\right )^{3}} - \frac {12 a^{2} b^{2}}{\left (a - i\right )^{3}} + 2 a b^{2} + \frac {8 i a b^{2}}{\left (a - i\right )^{3}} + 4 b^{3} x - 2 i b^{2} + \frac {2 b^{2}}{\left (a - i\right )^{3}} \right )}}{\left (a - i\right )^{3}} - \frac {- a^{2} + 4 i b x - 1}{x^{2} \cdot \left (2 a^{2} - 4 i a - 2\right )} \] Input:
integrate(1/(1+I*(b*x+a))**2*(1+(b*x+a)**2)/x**3,x)
Output:
-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))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (61) = 122\).
Time = 0.04 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.93 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=-\frac {2 \, {\left (-i \, a - 1\right )} b^{2} \log \left (i \, b x + i \, a + 1\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} - \frac {2 \, {\left (i \, a + 1\right )} b^{2} \log \left (x\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} + \frac {4 \, {\left (-i \, a - 1\right )} b^{2} x^{2} + a^{4} - 2 i \, a^{3} + {\left (a^{3} - 5 i \, a^{2} - 7 \, a + 3 i\right )} b x - 2 i \, a - 1}{2 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )}} \] Input:
integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="maxima")
Output:
-2*(-I*a - 1)*b^2*log(I*b*x + I*a + 1)/(a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1) - 2*(I*a + 1)*b^2*log(x)/(a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1) + 1/2*(4*(-I *a - 1)*b^2*x^2 + a^4 - 2*I*a^3 + (a^3 - 5*I*a^2 - 7*a + 3*I)*b*x - 2*I*a - 1)/((a^3 - 3*I*a^2 - 3*a + I)*b*x^3 + (a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1 )*x^2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (61) = 122\).
Time = 0.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=\frac {2 \, b^{3} \log \left (-\frac {i \, a}{i \, b x + i \, a + 1} - \frac {1}{i \, b x + i \, a + 1} + 1\right )}{i \, a^{3} b + 3 \, a^{2} b - 3 i \, a b - b} + \frac {\frac {i \, a b^{2} - 5 \, b^{2}}{-i \, a - 1} + \frac {2 i \, {\left (a b^{3} + 3 i \, b^{3}\right )}}{{\left (i \, b x + i \, a + 1\right )} b}}{2 \, {\left (a - i\right )}^{2} {\left (\frac {i \, a}{i \, b x + i \, a + 1} + \frac {1}{i \, b x + i \, a + 1} - 1\right )}^{2}} \] Input:
integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="giac")
Output:
2*b^3*log(-I*a/(I*b*x + I*a + 1) - 1/(I*b*x + I*a + 1) + 1)/(I*a^3*b + 3*a ^2*b - 3*I*a*b - b) + 1/2*((I*a*b^2 - 5*b^2)/(-I*a - 1) + 2*I*(a*b^3 + 3*I *b^3)/((I*b*x + I*a + 1)*b))/((a - I)^2*(I*a/(I*b*x + I*a + 1) + 1/(I*b*x + I*a + 1) - 1)^2)
Time = 23.35 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=\frac {\frac {a+1{}\mathrm {i}}{2\,\left (a-\mathrm {i}\right )}-\frac {b\,x\,2{}\mathrm {i}}{{\left (a-\mathrm {i}\right )}^2}}{x^2}-\frac {b^2\,\mathrm {atanh}\left (\frac {-a^3+a^2\,3{}\mathrm {i}+3\,a-\mathrm {i}}{{\left (a-\mathrm {i}\right )}^3}-\frac {x\,\left (2\,a^8\,b^2+8\,a^6\,b^2+12\,a^4\,b^2+8\,a^2\,b^2+2\,b^2\right )}{{\left (a-\mathrm {i}\right )}^3\,\left (b\,a^6+2{}\mathrm {i}\,b\,a^5+b\,a^4+4{}\mathrm {i}\,b\,a^3-b\,a^2+2{}\mathrm {i}\,b\,a-b\right )}\right )\,4{}\mathrm {i}}{{\left (a-\mathrm {i}\right )}^3} \] Input:
int(((a + b*x)^2 + 1)/(x^3*(a*1i + b*x*1i + 1)^2),x)
Output:
((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*(a*b*2i - b - a^2*b + a^3*b*4i + a^4*b + a^5*b* 2i + a^6*b)))*4i)/(a - 1i)^3
\[ \int \frac {e^{-2 i \arctan (a+b x)}}{x^3} \, dx=-\left (\int \frac {1}{b^{2} x^{5}+2 a b \,x^{4}-2 b i \,x^{4}+a^{2} x^{3}-2 a i \,x^{3}-x^{3}}d x \right ) a^{2}-\left (\int \frac {1}{b^{2} x^{5}+2 a b \,x^{4}-2 b i \,x^{4}+a^{2} x^{3}-2 a i \,x^{3}-x^{3}}d x \right )-2 \left (\int \frac {1}{b^{2} x^{4}+2 a b \,x^{3}-2 b i \,x^{3}+a^{2} x^{2}-2 a i \,x^{2}-x^{2}}d x \right ) a b -\left (\int \frac {1}{b^{2} x^{3}+2 a b \,x^{2}-2 b i \,x^{2}+a^{2} x -2 a i x -x}d x \right ) b^{2} \] Input:
int(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x)
Output:
- int(1/(a**2*x**3 + 2*a*b*x**4 - 2*a*i*x**3 + b**2*x**5 - 2*b*i*x**4 - x **3),x)*a**2 - int(1/(a**2*x**3 + 2*a*b*x**4 - 2*a*i*x**3 + b**2*x**5 - 2* b*i*x**4 - x**3),x) - 2*int(1/(a**2*x**2 + 2*a*b*x**3 - 2*a*i*x**2 + b**2* x**4 - 2*b*i*x**3 - x**2),x)*a*b - int(1/(a**2*x + 2*a*b*x**2 - 2*a*i*x + b**2*x**3 - 2*b*i*x**2 - x),x)*b**2