Integrand size = 14, antiderivative size = 117 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt {1+a^2 x^2}}{3 x^3}-\frac {3 i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {14 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {4 a^3 \sqrt {1+a^2 x^2}}{i+a x}+\frac {11}{2} i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
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 = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {1+a^2 x^2} \left (-2 i+7 a x+19 i a^2 x^2+52 a^3 x^3\right )}{x^3 (i+a x)}-33 i a^3 \log (x)+33 i a^3 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \]
((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 = 117, 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]]
3.1.26.3.1 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 = 0.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99
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 | \(-\frac {1}{3 x^{3} \sqrt {a^{2} x^{2}+1}}-\frac {13 a^{2} \left (-\frac {1}{x \sqrt {a^{2} x^{2}+1}}-\frac {2 a^{2} x}{\sqrt {a^{2} x^{2}+1}}\right )}{3}-i a^{3} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )+3 i a \left (-\frac {1}{2 x^{2} \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{2} \left (\frac {1}{\sqrt {a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}\right )\) | \(141\) |
meijerg | \(-\frac {-8 a^{4} x^{4}-4 a^{2} x^{2}+1}{3 x^{3} \sqrt {a^{2} x^{2}+1}}+\frac {3 i a^{3} \left (\frac {\sqrt {\pi }\, \left (20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \left (24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{2}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{\sqrt {\pi }}+\frac {3 a^{2} \left (2 a^{2} x^{2}+1\right )}{x \sqrt {a^{2} x^{2}+1}}-\frac {i a^{3} \left (-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}\right )}{\sqrt {\pi }}\) | \(249\) |
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 = 1.19 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {52 \, a^{4} x^{4} + 52 i \, a^{3} x^{3} - 33 \, {\left (-i \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 33 \, {\left (i \, a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + {\left (52 \, a^{3} x^{3} + 19 i \, a^{2} x^{2} + 7 \, a x - 2 i\right )} \sqrt {a^{2} x^{2} + 1}}{6 \, {\left (a x^{4} + i \, x^{3}\right )}} \]
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)
\[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=- i \left (\int \frac {i}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2}}{a^{2} x^{6} \sqrt {a^{2} x^{2} + 1} + x^{4} \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
-I*(Integral(I/(a**2*x**6*sqrt(a**2*x**2 + 1) + x**4*sqrt(a**2*x**2 + 1)), x) + Integral(-3*a*x/(a**2*x**6*sqrt(a**2*x**2 + 1) + x**4*sqrt(a**2*x**2 + 1)), x) + Integral(a**3*x**3/(a**2*x**6*sqrt(a**2*x**2 + 1) + x**4*sqrt (a**2*x**2 + 1)), x) + Integral(-3*I*a**2*x**2/(a**2*x**6*sqrt(a**2*x**2 + 1) + x**4*sqrt(a**2*x**2 + 1)), x))
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {26 \, a^{4} x}{3 \, \sqrt {a^{2} x^{2} + 1}} + \frac {11}{2} i \, a^{3} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) - \frac {11 i \, a^{3}}{2 \, \sqrt {a^{2} x^{2} + 1}} + \frac {13 \, a^{2}}{3 \, \sqrt {a^{2} x^{2} + 1} x} - \frac {3 i \, a}{2 \, \sqrt {a^{2} x^{2} + 1} x^{2}} - \frac {1}{3 \, \sqrt {a^{2} x^{2} + 1} x^{3}} \]
26/3*a^4*x/sqrt(a^2*x^2 + 1) + 11/2*I*a^3*arcsinh(1/(a*abs(x))) - 11/2*I*a ^3/sqrt(a^2*x^2 + 1) + 13/3*a^2/(sqrt(a^2*x^2 + 1)*x) - 3/2*I*a/(sqrt(a^2* x^2 + 1)*x^2) - 1/3/(sqrt(a^2*x^2 + 1)*x^3)
\[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
Time = 0.47 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99 \[ \int \frac {e^{3 i \arctan (a x)}}{x^4} \, dx=\frac {11\,a^3\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{3\,x^3}-\frac {a\,\sqrt {a^2\,x^2+1}\,3{}\mathrm {i}}{2\,x^2}+\frac {14\,a^2\,\sqrt {a^2\,x^2+1}}{3\,x}+\frac {4\,a^4\,\sqrt {a^2\,x^2+1}}{\left (x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \]