Integrand size = 14, antiderivative size = 64 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=-\frac {4 i a (1-i a x)}{\sqrt {1+a^2 x^2}}-\frac {\sqrt {1+a^2 x^2}}{x}+3 i a \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \] Output:
-4*I*a*(1-I*a*x)/(a^2*x^2+1)^(1/2)-(a^2*x^2+1)^(1/2)/x+3*I*a*arctanh((a^2* x^2+1)^(1/2))
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=\sqrt {1+a^2 x^2} \left (-\frac {1}{x}-\frac {4 a}{-i+a x}\right )-3 i a \log (x)+3 i a \log \left (1+\sqrt {1+a^2 x^2}\right ) \] Input:
Integrate[1/(E^((3*I)*ArcTan[a*x])*x^2),x]
Output:
Sqrt[1 + a^2*x^2]*(-x^(-1) - (4*a)/(-I + a*x)) - (3*I)*a*Log[x] + (3*I)*a* Log[1 + Sqrt[1 + a^2*x^2]]
Time = 0.72 (sec) , antiderivative size = 64, 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.
\(\displaystyle \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 5583 |
\(\displaystyle \int \frac {(1-i a x)^2}{x^2 (1+i a x) \sqrt {a^2 x^2+1}}dx\) |
\(\Big \downarrow \) 2353 |
\(\displaystyle \int \left (\frac {4 i a^2}{(a x-i) \sqrt {a^2 x^2+1}}-\frac {3 i a}{x \sqrt {a^2 x^2+1}}+\frac {1}{x^2 \sqrt {a^2 x^2+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 i a \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )+\frac {4 a \sqrt {a^2 x^2+1}}{-a x+i}-\frac {\sqrt {a^2 x^2+1}}{x}\) |
Input:
Int[1/(E^((3*I)*ArcTan[a*x])*x^2),x]
Output:
-(Sqrt[1 + a^2*x^2]/x) + (4*a*Sqrt[1 + a^2*x^2])/(I - a*x) + (3*I)*a*ArcTa nh[Sqrt[1 + a^2*x^2]]
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 = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.28
method | result | size |
risch | \(-\frac {\sqrt {a^{2} x^{2}+1}}{x}+i a \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )+\frac {4 i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a \left (x -\frac {i}{a}\right )}\right )\) | \(82\) |
default | \(-\frac {\left (a^{2} x^{2}+1\right )^{\frac {5}{2}}}{x}+4 a^{2} \left (\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}} x}{4}+\frac {3 \sqrt {a^{2} x^{2}+1}\, x}{8}+\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}+1}\right )}{8 \sqrt {a^{2}}}\right )-3 i a \left (\frac {\left (a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3}+\sqrt {a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )\right )-\frac {i \left (\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{3}}-2 i a \left (-\frac {i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}+3 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )\right )\right )}{a}+9 i a \left (\frac {\left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {3}{2}}}{3}+i a \left (\frac {\left (2 \left (x -\frac {i}{a}\right ) a^{2}+2 i a \right ) \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{4 a^{2}}+\frac {\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 )}{2 \sqrt {a^{2}}}\right )\right )-\frac {2 i \left (\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )\right )^{\frac {5}{2}}}{a \left (x -\frac {i}{a}\right )^{2}}\) | \(580\) |
Input:
int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-(a^2*x^2+1)^(1/2)/x+I*a*(3*arctanh(1/(a^2*x^2+1)^(1/2))+4*I/a/(x-I/a)*((x -I/a)^2*a^2+2*I*a*(x-I/a))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (52) = 104\).
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=-\frac {5 \, a^{2} x^{2} - 5 i \, a x + 3 \, {\left (-i \, a^{2} x^{2} - a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + 3 \, {\left (i \, a^{2} x^{2} + a x\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + \sqrt {a^{2} x^{2} + 1} {\left (5 \, a x - i\right )}}{a x^{2} - i \, x} \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^2,x, algorithm="fricas")
Output:
-(5*a^2*x^2 - 5*I*a*x + 3*(-I*a^2*x^2 - a*x)*log(-a*x + sqrt(a^2*x^2 + 1) + 1) + 3*(I*a^2*x^2 + a*x)*log(-a*x + sqrt(a^2*x^2 + 1) - 1) + sqrt(a^2*x^ 2 + 1)*(5*a*x - I))/(a*x^2 - I*x)
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{5} - 3 i a^{2} x^{4} - 3 a x^{3} + i x^{2}}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{5} - 3 i a^{2} x^{4} - 3 a x^{3} + i x^{2}}\, dx\right ) \] Input:
integrate(1/(1+I*a*x)**3*(a**2*x**2+1)**(3/2)/x**2,x)
Output:
I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**5 - 3*I*a**2*x**4 - 3*a*x**3 + I* x**2), x) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**3*x**5 - 3*I*a**2*x **4 - 3*a*x**3 + I*x**2), x))
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{2}} \,d x } \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^2,x, algorithm="maxima")
Output:
integrate((a^2*x^2 + 1)^(3/2)/((I*a*x + 1)^3*x^2), x)
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{2}} \,d x } \] Input:
integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^2,x, algorithm="giac")
Output:
undef
Time = 22.97 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=a\,\mathrm {atanh}\left (\sqrt {a^2\,x^2+1}\right )\,3{}\mathrm {i}-\frac {\sqrt {a^2\,x^2+1}}{x}+\frac {4\,a^2\,\sqrt {a^2\,x^2+1}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \] Input:
int((a^2*x^2 + 1)^(3/2)/(x^2*(a*x*1i + 1)^3),x)
Output:
a*atanh((a^2*x^2 + 1)^(1/2))*3i - (a^2*x^2 + 1)^(1/2)/x + (4*a^2*(a^2*x^2 + 1)^(1/2))/((((a^2)^(1/2)*1i)/a - x*(a^2)^(1/2))*(a^2)^(1/2))
\[ \int \frac {e^{-3 i \arctan (a x)}}{x^2} \, dx=-\left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{3} i \,x^{5}+3 a^{2} x^{4}-3 a i \,x^{3}-x^{2}}d x \right )-\left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{3} i \,x^{3}+3 a^{2} x^{2}-3 a i x -1}d x \right ) a^{2} \] Input:
int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^2,x)
Output:
- (int(sqrt(a**2*x**2 + 1)/(a**3*i*x**5 + 3*a**2*x**4 - 3*a*i*x**3 - x**2 ),x) + int(sqrt(a**2*x**2 + 1)/(a**3*i*x**3 + 3*a**2*x**2 - 3*a*i*x - 1),x )*a**2)