Integrand size = 14, antiderivative size = 113 \[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=-\frac {\sqrt {1+a^2 x^2}}{4 x^4}+\frac {i a \sqrt {1+a^2 x^2}}{3 x^3}+\frac {3 a^2 \sqrt {1+a^2 x^2}}{8 x^2}-\frac {2 i a^3 \sqrt {1+a^2 x^2}}{3 x}-\frac {3}{8} a^4 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \] Output:
-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-3/8*a^4*arctanh((a^2*x^2+1)^(1/ 2))
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=\frac {1}{24} \left (\frac {\sqrt {1+a^2 x^2} \left (-6+8 i a x+9 a^2 x^2-16 i a^3 x^3\right )}{x^4}+9 a^4 \log (x)-9 a^4 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \] Input:
Integrate[1/(E^(I*ArcTan[a*x])*x^5),x]
Output:
((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 = 0.48 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {5583, 539, 27, 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.
| \(\displaystyle \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx\) |
| \(\Big \downarrow \) 5583 |
| \(\displaystyle \int \frac {1-i a x}{x^5 \sqrt {a^2 x^2+1}}dx\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} \int \frac {a (3 a x+4 i)}{x^4 \sqrt {a^2 x^2+1}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \int \frac {3 a x+4 i}{x^4 \sqrt {a^2 x^2+1}}dx\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (-\frac {1}{3} \int -\frac {a (9-8 i a x)}{x^3 \sqrt {a^2 x^2+1}}dx-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (\frac {1}{3} \int \frac {a (9-8 i a x)}{x^3 \sqrt {a^2 x^2+1}}dx-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (\frac {1}{3} a \int \frac {9-8 i a x}{x^3 \sqrt {a^2 x^2+1}}dx-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} \int \frac {a (9 a x+16 i)}{x^2 \sqrt {a^2 x^2+1}}dx\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \int \frac {9 a x+16 i}{x^2 \sqrt {a^2 x^2+1}}dx\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
| \(\Big \downarrow \) 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 |
| \(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{4 x^4}-\frac {1}{4} a \left (\frac {1}{3} a \left (-\frac {9 \sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \left (-9 a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {16 i \sqrt {a^2 x^2+1}}{x}\right )\right )-\frac {4 i \sqrt {a^2 x^2+1}}{3 x^3}\right )\) |
Input:
Int[1/(E^(I*ArcTan[a*x])*x^5),x]
Output:
-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
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 = 0.19 (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 | \(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 x^{4}}-\frac {5 a^{2} \left (-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}}+\frac {a^{2} \left (\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}\right )}{4}+a^{4} \left (\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )+\frac {i a \left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 x^{3}}+i a^{3} \left (-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+2 a^{2} \left (\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}\right )\right )-a^{4} \left (\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}\right )\) | \(290\) |
Input:
int(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-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 = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=-\frac {9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - 9 \, a^{4} x^{4} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 16 i \, a^{4} x^{4} - {\left (-16 i \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 8 i \, a x - 6\right )} \sqrt {a^{2} x^{2} + 1}}{24 \, x^{4}} \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^5,x, algorithm="fricas")
Output:
-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
\[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=- i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{6} - i x^{5}}\, dx \] Input:
integrate(1/(1+I*a*x)*(a**2*x**2+1)**(1/2)/x**5,x)
Output:
-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**6 - I*x**5), x)
\[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{5}} \,d x } \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^5,x, algorithm="maxima")
Output:
integrate(sqrt(a^2*x^2 + 1)/((I*a*x + 1)*x^5), x)
\[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{5}} \,d x } \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^5,x, algorithm="giac")
Output:
undef
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=\frac {a^4\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8}-\frac {\sqrt {a^2\,x^2+1}}{4\,x^4}+\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{3\,x^3}+\frac {3\,a^2\,\sqrt {a^2\,x^2+1}}{8\,x^2}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,2{}\mathrm {i}}{3\,x} \] Input:
int((a^2*x^2 + 1)^(1/2)/(x^5*(a*x*1i + 1)),x)
Output:
(a^4*atan((a^2*x^2 + 1)^(1/2)*1i)*3i)/8 - (a^2*x^2 + 1)^(1/2)/(4*x^4) + (a *(a^2*x^2 + 1)^(1/2)*1i)/(3*x^3) + (3*a^2*(a^2*x^2 + 1)^(1/2))/(8*x^2) - ( a^3*(a^2*x^2 + 1)^(1/2)*2i)/(3*x)
\[ \int \frac {e^{-i \arctan (a x)}}{x^5} \, dx=\int \frac {\sqrt {a^{2} x^{2}+1}}{a i \,x^{6}+x^{5}}d x \] Input:
int(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^5,x)
Output:
int(sqrt(a**2*x**2 + 1)/(a*i*x**6 + x**5),x)