Integrand size = 14, antiderivative size = 92 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}-\frac {3 i a \sqrt {1+a^2 x^2}}{x}-\frac {4 i a^2 \sqrt {1+a^2 x^2}}{i+a x}+\frac {9}{2} a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
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 = 0.86 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\sqrt {1+a^2 x^2} \left (-\frac {1}{2 x^2}-\frac {3 i a}{x}-\frac {4 i a^2}{i+a x}\right )-\frac {9}{2} a^2 \log (x)+\frac {9}{2} a^2 \log \left (1+\sqrt {1+a^2 x^2}\right ) \]
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 = 92, 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
3.1.25.3.1 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 = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14
method | result | size |
default | \(-\frac {1}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {9 a^{2} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}+3 i a \left (-\frac {1}{x \sqrt {a^{2} x^{2}+1}}-\frac {2 a^{2} x}{\sqrt {a^{2} x^{2}+1}}\right )\) | \(105\) |
risch | \(-\frac {i \left (6 a^{3} x^{3}-i a^{2} x^{2}+6 a x -i\right )}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {a^{2} \left (-9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {8 i \sqrt {\left (x +\frac {i}{a}\right )^{2} a^{2}-2 i a \left (x +\frac {i}{a}\right )}}{a \left (x +\frac {i}{a}\right )}\right )}{2}\) | \(108\) |
meijerg | \(\frac {a^{2} \left (\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \left (24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{\sqrt {\pi }}-\frac {3 i a \left (2 a^{2} x^{2}+1\right )}{x \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{2} \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }}-\frac {i a^{3} x}{\sqrt {a^{2} x^{2}+1}}\) | \(229\) |
-1/2/x^2/(a^2*x^2+1)^(1/2)-9/2*a^2*(1/(a^2*x^2+1)^(1/2)-arctanh(1/(a^2*x^2 +1)^(1/2)))-I*a^3*x/(a^2*x^2+1)^(1/2)+3*I*a*(-1/x/(a^2*x^2+1)^(1/2)-2/(a^2 *x^2+1)^(1/2)*a^2*x)
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.41 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\frac {-14 i \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (-14 i \, a^{2} x^{2} + 5 \, a x - i\right )}}{2 \, {\left (a x^{3} + i \, x^{2}\right )}} \]
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)
\[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=- i \left (\int \frac {i}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{5} \sqrt {a^{2} x^{2} + 1} + x^{3} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
-I*(Integral(I/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**3/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt (a**2*x**2 + 1)), x) + Integral(-3*I*a**2*x**2/(a**2*x**5*sqrt(a**2*x**2 + 1) + x**3*sqrt(a**2*x**2 + 1)), x))
Time = 0.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {7 i \, a^{3} x}{\sqrt {a^{2} x^{2} + 1}} + \frac {9}{2} \, a^{2} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {9 \, a^{2}}{2 \, \sqrt {a^{2} x^{2} + 1}} - \frac {3 i \, a}{\sqrt {a^{2} x^{2} + 1} x} - \frac {1}{2 \, \sqrt {a^{2} x^{2} + 1} x^{2}} \]
-7*I*a^3*x/sqrt(a^2*x^2 + 1) + 9/2*a^2*arcsinh(1/(a*abs(x))) - 9/2*a^2/sqr t(a^2*x^2 + 1) - 3*I*a/(sqrt(a^2*x^2 + 1)*x) - 1/2/(sqrt(a^2*x^2 + 1)*x^2)
\[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08 \[ \int \frac {e^{3 i \arctan (a x)}}{x^3} \, dx=-\frac {a^2\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}-\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{x}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]