Integrand size = 14, antiderivative size = 63 \[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=-\frac {\sqrt {1+a^2 x^2}}{2 x^2}+\frac {i a \sqrt {1+a^2 x^2}}{x}+\frac {1}{2} a^2 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \] Output:
-1/2*(a^2*x^2+1)^(1/2)/x^2+I*a*(a^2*x^2+1)^(1/2)/x+1/2*a^2*arctanh((a^2*x^ 2+1)^(1/2))
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\frac {1}{2} \left (\frac {(-1+2 i a x) \sqrt {1+a^2 x^2}}{x^2}-a^2 \log (x)+a^2 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \] Input:
Integrate[1/(E^(I*ArcTan[a*x])*x^3),x]
Output:
(((-1 + (2*I)*a*x)*Sqrt[1 + a^2*x^2])/x^2 - a^2*Log[x] + a^2*Log[1 + Sqrt[ 1 + a^2*x^2]])/2
Time = 0.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5583, 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.
\(\displaystyle \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {1-i a x}{x^3 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 539 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} \int \frac {a (a x+2 i)}{x^2 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \int \frac {a x+2 i}{x^2 \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 534 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \left (a \int \frac {1}{x \sqrt {a^2 x^2+1}}dx-\frac {2 i \sqrt {a^2 x^2+1}}{x}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \left (\frac {1}{2} a \int \frac {1}{x^2 \sqrt {a^2 x^2+1}}dx^2-\frac {2 i \sqrt {a^2 x^2+1}}{x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \left (\frac {\int \frac {1}{\frac {x^4}{a^2}-\frac {1}{a^2}}d\sqrt {a^2 x^2+1}}{a}-\frac {2 i \sqrt {a^2 x^2+1}}{x}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\sqrt {a^2 x^2+1}}{2 x^2}-\frac {1}{2} a \left (-a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )-\frac {2 i \sqrt {a^2 x^2+1}}{x}\right )\) |
Input:
Int[1/(E^(I*ArcTan[a*x])*x^3),x]
Output:
-1/2*Sqrt[1 + a^2*x^2]/x^2 - (a*(((-2*I)*Sqrt[1 + a^2*x^2])/x - a*ArcTanh[ Sqrt[1 + a^2*x^2]]))/2
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 = 60, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {i \left (2 a^{3} x^{3}+i a^{2} x^{2}+2 a x +i\right )}{2 x^{2} \sqrt {a^{2} x^{2}+1}}+\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )}{2}\) | \(60\) |
default | \(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}}-\frac {a^{2} \left (\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2}+a^{2} \left (\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}+\frac {i a \ln \left (\frac {i a +\left (x -\frac {i}{a}\right ) a^{2}}{\sqrt {a^{2}}}+\sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}\right )}{\sqrt {a^{2}}}\right )-i a \left (-\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+2 a^{2} \left (\frac {\sqrt {a^{2} x^{2}+1}\, x}{2}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{2 \sqrt {a^{2}}}\right )\right )\) | \(217\) |
Input:
int(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
1/2*I*(2*a^3*x^3+I*a^2*x^2+2*a*x+I)/x^2/(a^2*x^2+1)^(1/2)+1/2*a^2*arctanh( 1/(a^2*x^2+1)^(1/2))
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) - a^{2} x^{2} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + 2 i \, a^{2} x^{2} + \sqrt {a^{2} x^{2} + 1} {\left (2 i \, a x - 1\right )}}{2 \, x^{2}} \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^3,x, algorithm="fricas")
Output:
1/2*(a^2*x^2*log(-a*x + sqrt(a^2*x^2 + 1) + 1) - a^2*x^2*log(-a*x + sqrt(a ^2*x^2 + 1) - 1) + 2*I*a^2*x^2 + sqrt(a^2*x^2 + 1)*(2*I*a*x - 1))/x^2
\[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=- i \int \frac {\sqrt {a^{2} x^{2} + 1}}{a x^{4} - i x^{3}}\, dx \] Input:
integrate(1/(1+I*a*x)*(a**2*x**2+1)**(1/2)/x**3,x)
Output:
-I*Integral(sqrt(a**2*x**2 + 1)/(a*x**4 - I*x**3), x)
\[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{3}} \,d x } \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^3,x, algorithm="maxima")
Output:
integrate(sqrt(a^2*x^2 + 1)/((I*a*x + 1)*x^3), x)
\[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\int { \frac {\sqrt {a^{2} x^{2} + 1}}{{\left (i \, a x + 1\right )} x^{3}} \,d x } \] Input:
integrate(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^3,x, algorithm="giac")
Output:
undef
Time = 23.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\frac {a^2\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )}{2}-\frac {\sqrt {a^2\,x^2+1}}{2\,x^2}+\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{x} \] Input:
int((a^2*x^2 + 1)^(1/2)/(x^3*(a*x*1i + 1)),x)
Output:
(a^2*atanh((a^2*x^2 + 1)^(1/2)))/2 - (a^2*x^2 + 1)^(1/2)/(2*x^2) + (a*(a^2 *x^2 + 1)^(1/2)*1i)/x
\[ \int \frac {e^{-i \arctan (a x)}}{x^3} \, dx=\int \frac {\sqrt {a^{2} x^{2}+1}}{a i \,x^{4}+x^{3}}d x \] Input:
int(1/(1+I*a*x)*(a^2*x^2+1)^(1/2)/x^3,x)
Output:
int(sqrt(a**2*x**2 + 1)/(a*i*x**4 + x**3),x)