Integrand size = 14, antiderivative size = 93 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}+\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i-a x}+\frac {9}{2} a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
9/2*a^2*arctanh((a^2*x^2+1)^(1/2))-1/2*(a^2*x^2+1)^(1/2)/x^2+3*I*a*(a^2*x^ 2+1)^(1/2)/x-4*I*a^2*(a^2*x^2+1)^(1/2)/(I-a*x)
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=\sqrt {1+a^2 x^2} \left (-\frac {1}{2 x^2}+\frac {3 i a}{x}+\frac {4 i a^2}{-i+a x}\right )-\frac {9}{2} a^2 \log (x)+\frac {9}{2} a^2 \log \left (1+\sqrt {1+a^2 x^2}\right ) \]
Sqrt[1 + a^2*x^2]*(-1/2*1/x^2 + ((3*I)*a)/x + ((4*I)*a^2)/(-I + a*x)) - (9 *a^2*Log[x])/2 + (9*a^2*Log[1 + Sqrt[1 + a^2*x^2]])/2
Time = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5583, 2353, 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.
\(\displaystyle \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {(1-i a x)^2}{x^3 (1+i a x) \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 2353 |
\(\displaystyle \int \left (-\frac {4 a^2}{x \sqrt {a^2 x^2+1}}-\frac {3 i a}{x^2 \sqrt {a^2 x^2+1}}+\frac {1}{x^3 \sqrt {a^2 x^2+1}}+\frac {4 a^3}{(a x-i) \sqrt {a^2 x^2+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {9}{2} a^2 \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {4 i a^2 \sqrt {a^2 x^2+1}}{-a x+i}+\frac {3 i a \sqrt {a^2 x^2+1}}{x}-\frac {\sqrt {a^2 x^2+1}}{2 x^2}\) |
-1/2*Sqrt[1 + a^2*x^2]/x^2 + ((3*I)*a*Sqrt[1 + a^2*x^2])/x - ((4*I)*a^2*Sq rt[1 + a^2*x^2])/(I - a*x) + (9*a^2*ArcTanh[Sqrt[1 + a^2*x^2]])/2
3.1.58.3.1 Defintions of rubi rules used
Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) ^(p_), x_Symbol] :> Int[ExpandIntegrand[Px*(e*x)^m*(c + d*x)^n*(a + b*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && PolyQ[Px, x] && (Integer Q[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0]))
Int[E^(ArcTan[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a* x)^((I*n + 1)/2)/((1 + I*a*x)^((I*n - 1)/2)*Sqrt[1 + a^2*x^2])), x] /; Free Q[{a, m}, x] && IntegerQ[(I*n - 1)/2]
Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {i \left (6 a^{3} x^{3}+i a^{2} x^{2}+6 a x +i\right )}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {a^{2} \left (-9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )-\frac {8 i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a \left (x -\frac {i}{a}\right )}\right )}{2}\) | \(108\) |
default | \(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}{2 x^{2}}-\frac {9 a^{2} \left (\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3}+\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}-3 i a \left (-\frac {\left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}{x}+4 a^{2} \left (\frac {x \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 \sqrt {a^{2} x^{2}+1}\, x}{8}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 \sqrt {a^{2}}}\right )\right )+6 a^{2} \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )\) | \(551\) |
1/2*I*(6*a^3*x^3+I*a^2*x^2+6*a*x+I)/x^2/(a^2*x^2+1)^(1/2)-1/2*a^2*(-9*arct anh(1/(a^2*x^2+1)^(1/2))-8*I/a/(x-I/a)*((x-I/a)^2*a^2+2*I*a*(x-I/a))^(1/2) )
Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=\frac {14 i \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 9 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (14 i \, a^{2} x^{2} + 5 \, a x + i\right )}}{2 \, {\left (a x^{3} - i \, x^{2}\right )}} \]
1/2*(14*I*a^3*x^3 + 14*a^2*x^2 + 9*(a^3*x^3 - I*a^2*x^2)*log(-a*x + sqrt(a ^2*x^2 + 1) + 1) - 9*(a^3*x^3 - I*a^2*x^2)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + sqrt(a^2*x^2 + 1)*(14*I*a^2*x^2 + 5*a*x + I))/(a*x^3 - I*x^2)
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{6} - 3 i a^{2} x^{5} - 3 a x^{4} + i x^{3}}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{6} - 3 i a^{2} x^{5} - 3 a x^{4} + i x^{3}}\, dx\right ) \]
I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**6 - 3*I*a**2*x**5 - 3*a*x**4 + I* x**3), x) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**3*x**6 - 3*I*a**2*x **5 - 3*a*x**4 + I*x**3), x))
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{3}} \,d x } \]
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{3}} \,d x } \]
Time = 0.47 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^3} \, dx=-\frac {a^2\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}+\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{x}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]