Integrand size = 14, antiderivative size = 118 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}-\frac {4 a^3 \sqrt {1+a^2 x^2}}{i-a x}-\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
-11/2*I*a^3*arctanh((a^2*x^2+1)^(1/2))-1/3*(a^2*x^2+1)^(1/2)/x^3+3/2*I*a*( a^2*x^2+1)^(1/2)/x^2+14/3*a^2*(a^2*x^2+1)^(1/2)/x-4*a^3*(a^2*x^2+1)^(1/2)/ (I-a*x)
Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {1+a^2 x^2} \left (2 i+7 a x-19 i a^2 x^2+52 a^3 x^3\right )}{x^3 (-i+a x)}+33 i a^3 \log (x)-33 i a^3 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \]
((Sqrt[1 + a^2*x^2]*(2*I + 7*a*x - (19*I)*a^2*x^2 + 52*a^3*x^3))/(x^3*(-I + a*x)) + (33*I)*a^3*Log[x] - (33*I)*a^3*Log[1 + Sqrt[1 + a^2*x^2]])/6
Time = 0.48 (sec) , antiderivative size = 118, 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.
| \(\displaystyle \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx\) |
| \(\Big \downarrow \) 5583 |
| \(\displaystyle \int \frac {(1-i a x)^2}{x^4 (1+i a x) \sqrt {a^2 x^2+1}}dx\) |
| \(\Big \downarrow \) 2353 |
| \(\displaystyle \int \left (-\frac {4 a^2}{x^2 \sqrt {a^2 x^2+1}}+\frac {1}{x^4 \sqrt {a^2 x^2+1}}-\frac {3 i a}{x^3 \sqrt {a^2 x^2+1}}-\frac {4 i a^4}{(a x-i) \sqrt {a^2 x^2+1}}+\frac {4 i a^3}{x \sqrt {a^2 x^2+1}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {14 a^2 \sqrt {a^2 x^2+1}}{3 x}+\frac {3 i a \sqrt {a^2 x^2+1}}{2 x^2}-\frac {\sqrt {a^2 x^2+1}}{3 x^3}-\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {4 a^3 \sqrt {a^2 x^2+1}}{-a x+i}\) |
-1/3*Sqrt[1 + a^2*x^2]/x^3 + (((3*I)/2)*a*Sqrt[1 + a^2*x^2])/x^2 + (14*a^2 *Sqrt[1 + a^2*x^2])/(3*x) - (4*a^3*Sqrt[1 + a^2*x^2])/(I - a*x) - ((11*I)/ 2)*a^3*ArcTanh[Sqrt[1 + a^2*x^2]]
3.1.59.3.1 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 = 0.46 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(\frac {28 a^{4} x^{4}+9 i a^{3} x^{3}+26 a^{2} x^{2}+9 i a x -2}{6 x^{3} \sqrt {a^{2} x^{2}+1}}+\frac {i a^{3} \left (-11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )-\frac {8 i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a \left (x -\frac {i}{a}\right )}\right )}{2}\) | \(116\) |
| default | \(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}{3 x^{3}}-\frac {16 a^{2} \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 )}{3}-3 i a \left (-\frac {\left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}{2 x^{2}}+\frac {3 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}\right )+10 i a^{3} \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 )-4 a^{2} \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 )+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 )^{3}}-2 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 )\right )-10 i a^{3} \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 )\) | \(829\) |
1/6*(28*a^4*x^4+9*I*a^3*x^3+26*a^2*x^2+9*I*a*x-2)/x^3/(a^2*x^2+1)^(1/2)+1/ 2*I*a^3*(-11*arctanh(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 = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=\frac {52 \, a^{4} x^{4} - 52 i \, a^{3} x^{3} - 33 \, {\left (i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 33 \, {\left (-i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + {\left (52 \, a^{3} x^{3} - 19 i \, a^{2} x^{2} + 7 \, a x + 2 i\right )} \sqrt {a^{2} x^{2} + 1}}{6 \, {\left (a x^{4} - i \, x^{3}\right )}} \]
1/6*(52*a^4*x^4 - 52*I*a^3*x^3 - 33*(I*a^4*x^4 + a^3*x^3)*log(-a*x + sqrt( a^2*x^2 + 1) + 1) - 33*(-I*a^4*x^4 - a^3*x^3)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + (52*a^3*x^3 - 19*I*a^2*x^2 + 7*a*x + 2*I)*sqrt(a^2*x^2 + 1))/(a*x^ 4 - I*x^3)
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{7} - 3 i a^{2} x^{6} - 3 a x^{5} + i x^{4}}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{7} - 3 i a^{2} x^{6} - 3 a x^{5} + i x^{4}}\, dx\right ) \]
I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**7 - 3*I*a**2*x**6 - 3*a*x**5 + I* x**4), x) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**3*x**7 - 3*I*a**2*x **6 - 3*a*x**5 + I*x**4), x))
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{4}} \,d x } \]
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{4}} \,d x } \]
Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^4} \, dx=\frac {14\,a^2\,\sqrt {a^2\,x^2+1}}{3\,x}-\frac {\sqrt {a^2\,x^2+1}}{3\,x^3}+\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{2\,x^2}-\frac {11\,a^3\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )}{2}-\frac {4\,a^4\,\sqrt {a^2\,x^2+1}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]