Integrand size = 14, antiderivative size = 90 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=-\frac {\sqrt {1+a^2 x^2}}{3 x^3}-\frac {i a \sqrt {1+a^2 x^2}}{2 x^2}+\frac {2 a^2 \sqrt {1+a^2 x^2}}{3 x}+\frac {1}{2} i a^3 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \] Output:
-1/3*(a^2*x^2+1)^(1/2)/x^3-1/2*I*a*(a^2*x^2+1)^(1/2)/x^2+2/3*a^2*(a^2*x^2+ 1)^(1/2)/x+1/2*I*a^3*arctanh((a^2*x^2+1)^(1/2))
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.78 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {1+a^2 x^2} \left (-2-3 i a x+4 a^2 x^2\right )}{x^3}-3 i a^3 \log (x)+3 i a^3 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \] Input:
Integrate[E^(I*ArcTan[a*x])/x^4,x]
Output:
((Sqrt[1 + a^2*x^2]*(-2 - (3*I)*a*x + 4*a^2*x^2))/x^3 - (3*I)*a^3*Log[x] + (3*I)*a^3*Log[1 + Sqrt[1 + a^2*x^2]])/6
Time = 0.44 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5583, 539, 25, 27, 539, 27, 534, 243, 73, 221}
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.
\(\displaystyle \int \frac {e^{i \arctan (a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {1+i a x}{x^4 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}-\frac {1}{3} \int -\frac {a (3 i-2 a x)}{x^3 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} \int \frac {a (3 i-2 a x)}{x^3 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \int \frac {3 i-2 a x}{x^3 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} \int \frac {a (3 i a x+4)}{x^2 \sqrt {a^2 x^2+1}}dx-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} a \int \frac {3 i a x+4}{x^2 \sqrt {a^2 x^2+1}}dx-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} a \left (-\frac {4 \sqrt {a^2 x^2+1}}{x}+3 i a \int \frac {1}{x \sqrt {a^2 x^2+1}}dx\right )-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} a \left (-\frac {4 \sqrt {a^2 x^2+1}}{x}+\frac {3}{2} i a \int \frac {1}{x^2 \sqrt {a^2 x^2+1}}dx^2\right )-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} a \left (-\frac {4 \sqrt {a^2 x^2+1}}{x}+\frac {3 i \int \frac {1}{\frac {x^4}{a^2}-\frac {1}{a^2}}d\sqrt {a^2 x^2+1}}{a}\right )-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{3 x^3}+\frac {1}{3} a \left (-\frac {1}{2} a \left (-\frac {4 \sqrt {a^2 x^2+1}}{x}-3 i a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )\right )-\frac {3 i \sqrt {a^2 x^2+1}}{2 x^2}\right )\) |
Input:
Int[E^(I*ArcTan[a*x])/x^4,x]
Output:
-1/3*Sqrt[1 + a^2*x^2]/x^3 + (a*((((-3*I)/2)*Sqrt[1 + a^2*x^2])/x^2 - (a*( (-4*Sqrt[1 + a^2*x^2])/x - (3*I)*a*ArcTanh[Sqrt[1 + a^2*x^2]]))/2))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
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 = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {4 a^{4} x^{4}-3 i a^{3} x^{3}+2 a^{2} x^{2}-3 i a x -2}{6 x^{3} \sqrt {a^{2} x^{2}+1}}+\frac {i a^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}\) | \(68\) |
default | \(-\frac {\sqrt {a^{2} x^{2}+1}}{3 x^{3}}+\frac {2 a^{2} \sqrt {a^{2} x^{2}+1}}{3 x}+i a \left (-\frac {\sqrt {a^{2} x^{2}+1}}{2 x^{2}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}\right )\) | \(75\) |
meijerg | \(-\frac {\left (-2 a^{2} x^{2}+1\right ) \sqrt {a^{2} x^{2}+1}}{3 x^{3}}+\frac {i a^{3} \left (-\frac {\sqrt {\pi }}{x^{2} a^{2}}-\frac {\left (1-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{2}+\frac {\sqrt {\pi }\, \left (4 a^{2} x^{2}+8\right )}{8 a^{2} x^{2}}-\frac {\sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}}{a^{2} x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )\right )}{2 \sqrt {\pi }}\) | \(131\) |
Input:
int((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
1/6*(4*a^4*x^4-3*I*a^3*x^3+2*a^2*x^2-3*I*a*x-2)/x^3/(a^2*x^2+1)^(1/2)+1/2* I*a^3*arctanh(1/(a^2*x^2+1)^(1/2))
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {3 i \, a^{3} x^{3} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 3 i \, a^{3} x^{3} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 4 \, a^{3} x^{3} + {\left (4 \, a^{2} x^{2} - 3 i \, a x - 2\right )} \sqrt {a^{2} x^{2} + 1}}{6 \, x^{3}} \] Input:
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^4,x, algorithm="fricas")
Output:
1/6*(3*I*a^3*x^3*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 3*I*a^3*x^3*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + 4*a^3*x^3 + (4*a^2*x^2 - 3*I*a*x - 2)*sqrt(a^2* x^2 + 1))/x^3
Time = 1.73 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {2 a^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3} + \frac {i a^{3} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a^{2} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{2 x} - \frac {a \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3 x^{2}} \] Input:
integrate((1+I*a*x)/(a**2*x**2+1)**(1/2)/x**4,x)
Output:
2*a**3*sqrt(1 + 1/(a**2*x**2))/3 + I*a**3*asinh(1/(a*x))/2 - I*a**2*sqrt(1 + 1/(a**2*x**2))/(2*x) - a*sqrt(1 + 1/(a**2*x**2))/(3*x**2)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {1}{2} i \, a^{3} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {2 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{3 \, x} - \frac {i \, \sqrt {a^{2} x^{2} + 1} a}{2 \, x^{2}} - \frac {\sqrt {a^{2} x^{2} + 1}}{3 \, x^{3}} \] Input:
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^4,x, algorithm="maxima")
Output:
1/2*I*a^3*arcsinh(1/(a*abs(x))) + 2/3*sqrt(a^2*x^2 + 1)*a^2/x - 1/2*I*sqrt (a^2*x^2 + 1)*a/x^2 - 1/3*sqrt(a^2*x^2 + 1)/x^3
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (70) = 140\).
Time = 0.14 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.79 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {1}{2} i \, a^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) - \frac {1}{2} i \, a^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) - \frac {-3 i \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{5} a^{3} - 12 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} a^{2} {\left | a \right |} + 3 \, {\left (i \, x {\left | a \right |} - i \, \sqrt {a^{2} x^{2} + 1}\right )} a^{3} + 4 \, a^{2} {\left | a \right |}}{3 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} \] Input:
integrate((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^4,x, algorithm="giac")
Output:
1/2*I*a^3*log(abs(-x*abs(a) + sqrt(a^2*x^2 + 1) + 1)) - 1/2*I*a^3*log(abs( -x*abs(a) + sqrt(a^2*x^2 + 1) - 1)) - 1/3*(-3*I*(x*abs(a) - sqrt(a^2*x^2 + 1))^5*a^3 - 12*(x*abs(a) - sqrt(a^2*x^2 + 1))^2*a^2*abs(a) + 3*(I*x*abs(a ) - I*sqrt(a^2*x^2 + 1))*a^3 + 4*a^2*abs(a))/((x*abs(a) - sqrt(a^2*x^2 + 1 ))^2 - 1)^3
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {a^3\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{3\,x^3}+\frac {2\,a^2\,\sqrt {a^2\,x^2+1}}{3\,x}-\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{2\,x^2} \] Input:
int((a*x*1i + 1)/(x^4*(a^2*x^2 + 1)^(1/2)),x)
Output:
(a^3*atan((a^2*x^2 + 1)^(1/2)*1i))/2 - (a^2*x^2 + 1)^(1/2)/(3*x^3) - (a*(a ^2*x^2 + 1)^(1/2)*1i)/(2*x^2) + (2*a^2*(a^2*x^2 + 1)^(1/2))/(3*x)
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.21 \[ \int \frac {e^{i \arctan (a x)}}{x^4} \, dx=\frac {4 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}-3 \sqrt {a^{2} x^{2}+1}\, a i x -2 \sqrt {a^{2} x^{2}+1}-3 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x -1\right ) a^{3} i \,x^{3}+3 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x +1\right ) a^{3} i \,x^{3}-4 a^{3} x^{3}}{6 x^{3}} \] Input:
int((1+I*a*x)/(a^2*x^2+1)^(1/2)/x^4,x)
Output:
(4*sqrt(a**2*x**2 + 1)*a**2*x**2 - 3*sqrt(a**2*x**2 + 1)*a*i*x - 2*sqrt(a* *2*x**2 + 1) - 3*log(sqrt(a**2*x**2 + 1) + a*x - 1)*a**3*i*x**3 + 3*log(sq rt(a**2*x**2 + 1) + a*x + 1)*a**3*i*x**3 - 4*a**3*x**3)/(6*x**3)