Integrand size = 14, antiderivative size = 113 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}-\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}+\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}-\frac {3}{8} a^4 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \]
-3/8*a^4*arctanh((a^2*x^2+1)^(1/2))-1/4*(a^2*x^2+1)^(1/2)/x^4-1/3*I*a*(a^2 *x^2+1)^(1/2)/x^3+3/8*a^2*(a^2*x^2+1)^(1/2)/x^2+2/3*I*a^3*(a^2*x^2+1)^(1/2 )/x
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=\frac {1}{24} \left (\frac {\sqrt {1+a^2 x^2} \left (-6-8 i a x+9 a^2 x^2+16 i a^3 x^3\right )}{x^4}+9 a^4 \log (x)-9 a^4 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \]
((Sqrt[1 + a^2*x^2]*(-6 - (8*I)*a*x + 9*a^2*x^2 + (16*I)*a^3*x^3))/x^4 + 9 *a^4*Log[x] - 9*a^4*Log[1 + Sqrt[1 + a^2*x^2]])/24
Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {5583, 539, 25, 27, 539, 27, 539, 25, 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.
\(\displaystyle \int \frac {e^{i \arctan (a x)}}{x^5} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {1+i a x}{x^5 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} \int -\frac {a (4 i-3 a x)}{x^4 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} \int \frac {a (4 i-3 a x)}{x^4 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \int \frac {4 i-3 a x}{x^4 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} \int \frac {a (8 i a x+9)}{x^3 \sqrt {a^2 x^2+1}}dx-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \int \frac {8 i a x+9}{x^3 \sqrt {a^2 x^2+1}}dx-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} \int -\frac {a (16 i-9 a x)}{x^2 \sqrt {a^2 x^2+1}}dx\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} \int \frac {a (16 i-9 a x)}{x^2 \sqrt {a^2 x^2+1}}dx\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a \int \frac {16 i-9 a x}{x^2 \sqrt {a^2 x^2+1}}dx\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a \left (-9 a \int \frac {1}{x \sqrt {a^2 x^2+1}}dx-\frac {16 i \sqrt {a^2 x^2+1}}{x}\right )\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a \left (-\frac {9}{2} a \int \frac {1}{x^2 \sqrt {a^2 x^2+1}}dx^2-\frac {16 i \sqrt {a^2 x^2+1}}{x}\right )\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a \left (-\frac {9 \int \frac {1}{\frac {x^4}{a^2}-\frac {1}{a^2}}d\sqrt {a^2 x^2+1}}{a}-\frac {16 i \sqrt {a^2 x^2+1}}{x}\right )\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}+\frac {1}{2} a \left (9 a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {16 i \sqrt {a^2 x^2+1}}{x}\right )\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
-1/4*Sqrt[1 + a^2*x^2]/x^4 + (a*((((-4*I)/3)*Sqrt[1 + a^2*x^2])/x^3 - (a*( (-9*Sqrt[1 + a^2*x^2])/(2*x^2) + (a*(((-16*I)*Sqrt[1 + a^2*x^2])/x + 9*a*A rcTanh[Sqrt[1 + a^2*x^2]]))/2))/3))/4
3.1.10.3.1 Defintions of rubi rules used
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 = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68
method | result | size |
risch | \(\frac {i \left (16 a^{5} x^{5}-9 i a^{4} x^{4}+8 a^{3} x^{3}-3 i a^{2} x^{2}-8 a x +6 i\right )}{24 x^{4} \sqrt {a^{2} x^{2}+1}}-\frac {3 a^{4} \operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{8}\) | \(77\) |
default | \(-\frac {\sqrt {a^{2} x^{2}+1}}{4 x^{4}}-\frac {3 a^{2} \left (-\frac {\sqrt {a^{2} x^{2}+1}}{2 x^{2}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}\right )}{4}+i a \left (-\frac {\sqrt {a^{2} x^{2}+1}}{3 x^{3}}+\frac {2 a^{2} \sqrt {a^{2} x^{2}+1}}{3 x}\right )\) | \(97\) |
meijerg | \(\frac {a^{4} \left (\frac {\sqrt {\pi }\, \left (-7 a^{4} x^{4}-8 a^{2} x^{2}+8\right )}{16 a^{4} x^{4}}-\frac {\sqrt {\pi }\, \left (-12 a^{2} x^{2}+8\right ) \sqrt {a^{2} x^{2}+1}}{16 a^{4} x^{4}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {a^{2} x^{2}+1}}{2}\right )}{4}+\frac {3 \left (\frac {7}{6}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right ) \sqrt {\pi }}{8}-\frac {\sqrt {\pi }}{2 x^{4} a^{4}}+\frac {\sqrt {\pi }}{2 x^{2} a^{2}}\right )}{2 \sqrt {\pi }}-\frac {i a \left (-2 a^{2} x^{2}+1\right ) \sqrt {a^{2} x^{2}+1}}{3 x^{3}}\) | \(162\) |
1/24*I*(16*a^5*x^5-9*I*a^4*x^4+8*a^3*x^3-3*I*a^2*x^2-8*a*x+6*I)/x^4/(a^2*x ^2+1)^(1/2)-3/8*a^4*arctanh(1/(a^2*x^2+1)^(1/2))
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=-\frac {9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) - 16 i \, a^{4} x^{4} - {\left (16 i \, a^{3} x^{3} + 9 \, a^{2} x^{2} - 8 i \, a x - 6\right )} \sqrt {a^{2} x^{2} + 1}}{24 \, x^{4}} \]
-1/24*(9*a^4*x^4*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - 9*a^4*x^4*log(-a*x + sqrt(a^2*x^2 + 1) - 1) - 16*I*a^4*x^4 - (16*I*a^3*x^3 + 9*a^2*x^2 - 8*I*a* x - 6)*sqrt(a^2*x^2 + 1))/x^4
Time = 2.96 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.08 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=\frac {2 i a^{4} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3} - \frac {3 a^{4} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {i a^{2} \sqrt {1 + \frac {1}{a^{2} x^{2}}}}{3 x^{2}} + \frac {a}{8 x^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} \]
2*I*a**4*sqrt(1 + 1/(a**2*x**2))/3 - 3*a**4*asinh(1/(a*x))/8 + 3*a**3/(8*x *sqrt(1 + 1/(a**2*x**2))) - I*a**2*sqrt(1 + 1/(a**2*x**2))/(3*x**2) + a/(8 *x**3*sqrt(1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sqrt(1 + 1/(a**2*x**2)))
Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=-\frac {3}{8} \, a^{4} \operatorname {arsinh}\left (\frac {1}{a {\left | x \right |}}\right ) + \frac {2 i \, \sqrt {a^{2} x^{2} + 1} a^{3}}{3 \, x} + \frac {3 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{8 \, x^{2}} - \frac {i \, \sqrt {a^{2} x^{2} + 1} a}{3 \, x^{3}} - \frac {\sqrt {a^{2} x^{2} + 1}}{4 \, x^{4}} \]
-3/8*a^4*arcsinh(1/(a*abs(x))) + 2/3*I*sqrt(a^2*x^2 + 1)*a^3/x + 3/8*sqrt( a^2*x^2 + 1)*a^2/x^2 - 1/3*I*sqrt(a^2*x^2 + 1)*a/x^3 - 1/4*sqrt(a^2*x^2 + 1)/x^4
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (89) = 178\).
Time = 0.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.10 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=-\frac {3}{8} \, a^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} + 1 \right |}\right ) + \frac {3}{8} \, a^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1} - 1 \right |}\right ) - \frac {9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{7} a^{4} - 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{5} a^{4} - 48 i \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{4} a^{3} {\left | a \right |} - 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{3} a^{4} + 64 i \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} a^{3} {\left | a \right |} + 9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )} a^{4} - 16 i \, a^{3} {\left | a \right |}}{12 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{4}} \]
-3/8*a^4*log(abs(-x*abs(a) + sqrt(a^2*x^2 + 1) + 1)) + 3/8*a^4*log(abs(-x* abs(a) + sqrt(a^2*x^2 + 1) - 1)) - 1/12*(9*(x*abs(a) - sqrt(a^2*x^2 + 1))^ 7*a^4 - 33*(x*abs(a) - sqrt(a^2*x^2 + 1))^5*a^4 - 48*I*(x*abs(a) - sqrt(a^ 2*x^2 + 1))^4*a^3*abs(a) - 33*(x*abs(a) - sqrt(a^2*x^2 + 1))^3*a^4 + 64*I* (x*abs(a) - sqrt(a^2*x^2 + 1))^2*a^3*abs(a) + 9*(x*abs(a) - sqrt(a^2*x^2 + 1))*a^4 - 16*I*a^3*abs(a))/((x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)^4
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {e^{i \arctan (a x)}}{x^5} \, dx=\frac {a^4\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8}-\frac {\sqrt {a^2\,x^2+1}}{4\,x^4}-\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{3\,x^3}+\frac {3\,a^2\,\sqrt {a^2\,x^2+1}}{8\,x^2}+\frac {a^3\,\sqrt {a^2\,x^2+1}\,2{}\mathrm {i}}{3\,x} \]