\(\int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx\) [75]

Optimal result

Integrand size = 14, antiderivative size = 135 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\frac {4 a^4 (1-i a x)}{\sqrt {1+a^2 x^2}}-\frac {\sqrt {1+a^2 x^2}}{4 x^4}+\frac {i a \sqrt {1+a^2 x^2}}{x^3}+\frac {19 a^2 \sqrt {1+a^2 x^2}}{8 x^2}-\frac {6 i a^3 \sqrt {1+a^2 x^2}}{x}-\frac {51}{8} a^4 \text {arctanh}\left (\sqrt {1+a^2 x^2}\right ) \] Output:

4*a^4*(1-I*a*x)/(a^2*x^2+1)^(1/2)-1/4*(a^2*x^2+1)^(1/2)/x^4+I*a*(a^2*x^2+1 
)^(1/2)/x^3+19/8*a^2*(a^2*x^2+1)^(1/2)/x^2-6*I*a^3*(a^2*x^2+1)^(1/2)/x-51/ 
8*a^4*arctanh((a^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\frac {1}{8} \left (\frac {\sqrt {1+a^2 x^2} \left (2 i+6 a x-11 i a^2 x^2-29 a^3 x^3-80 i a^4 x^4\right )}{x^4 (-i+a x)}+51 a^4 \log (x)-51 a^4 \log \left (1+\sqrt {1+a^2 x^2}\right )\right ) \] Input:

Integrate[1/(E^((3*I)*ArcTan[a*x])*x^5),x]
 

Output:

((Sqrt[1 + a^2*x^2]*(2*I + 6*a*x - (11*I)*a^2*x^2 - 29*a^3*x^3 - (80*I)*a^ 
4*x^4))/(x^4*(-I + a*x)) + 51*a^4*Log[x] - 51*a^4*Log[1 + Sqrt[1 + a^2*x^2 
]])/8
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03, 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.

\(\displaystyle \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx\)

\(\Big \downarrow \) 5583

\(\displaystyle \int \frac {(1-i a x)^2}{x^5 (1+i a x) \sqrt {a^2 x^2+1}}dx\)

\(\Big \downarrow \) 2353

\(\displaystyle \int \left (\frac {1}{x^5 \sqrt {a^2 x^2+1}}-\frac {3 i a}{x^4 \sqrt {a^2 x^2+1}}-\frac {4 a^2}{x^3 \sqrt {a^2 x^2+1}}-\frac {4 a^5}{(a x-i) \sqrt {a^2 x^2+1}}+\frac {4 a^4}{x \sqrt {a^2 x^2+1}}+\frac {4 i a^3}{x^2 \sqrt {a^2 x^2+1}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {19 a^2 \sqrt {a^2 x^2+1}}{8 x^2}-\frac {\sqrt {a^2 x^2+1}}{4 x^4}+\frac {i a \sqrt {a^2 x^2+1}}{x^3}-\frac {51}{8} a^4 \text {arctanh}\left (\sqrt {a^2 x^2+1}\right )+\frac {4 i a^4 \sqrt {a^2 x^2+1}}{-a x+i}-\frac {6 i a^3 \sqrt {a^2 x^2+1}}{x}\)

Input:

Int[1/(E^((3*I)*ArcTan[a*x])*x^5),x]
 

Output:

-1/4*Sqrt[1 + a^2*x^2]/x^4 + (I*a*Sqrt[1 + a^2*x^2])/x^3 + (19*a^2*Sqrt[1 
+ a^2*x^2])/(8*x^2) - ((6*I)*a^3*Sqrt[1 + a^2*x^2])/x + ((4*I)*a^4*Sqrt[1 
+ a^2*x^2])/(I - a*x) - (51*a^4*ArcTanh[Sqrt[1 + a^2*x^2]])/8
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2353

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]))
 

rule 5583

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]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {i \left (48 a^{5} x^{5}+19 i a^{4} x^{4}+40 a^{3} x^{3}+17 i a^{2} x^{2}-8 a x -2 i\right )}{8 x^{4} \sqrt {a^{2} x^{2}+1}}+\frac {a^{4} \left (-51 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {a^{2} x^{2}+1}}\right )-\frac {32 i \sqrt {\left (x -\frac {i}{a}\right )^{2} a^{2}+2 i a \left (x -\frac {i}{a}\right )}}{a \left (x -\frac {i}{a}\right )}\right )}{8}\)	\(125\)
default	\(\text {Expression too large to display}\)	\(937\)

Input:

int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/8*I*(48*a^5*x^5+19*I*a^4*x^4+40*a^3*x^3+17*I*a^2*x^2-8*a*x-2*I)/x^4/(a^ 
2*x^2+1)^(1/2)+1/8*a^4*(-51*arctanh(1/(a^2*x^2+1)^(1/2))-32*I/a/(x-I/a)*(( 
x-I/a)^2*a^2+2*I*a*(x-I/a))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\frac {-80 i \, a^{5} x^{5} - 80 \, a^{4} x^{4} - 51 \, {\left (a^{5} x^{5} - i \, a^{4} x^{4}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} + 1\right ) + 51 \, {\left (a^{5} x^{5} - i \, a^{4} x^{4}\right )} \log \left (-a x + \sqrt {a^{2} x^{2} + 1} - 1\right ) + {\left (-80 i \, a^{4} x^{4} - 29 \, a^{3} x^{3} - 11 i \, a^{2} x^{2} + 6 \, a x + 2 i\right )} \sqrt {a^{2} x^{2} + 1}}{8 \, {\left (a x^{5} - i \, x^{4}\right )}} \] Input:

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^5,x, algorithm="fricas")
 

Output:

1/8*(-80*I*a^5*x^5 - 80*a^4*x^4 - 51*(a^5*x^5 - I*a^4*x^4)*log(-a*x + sqrt 
(a^2*x^2 + 1) + 1) + 51*(a^5*x^5 - I*a^4*x^4)*log(-a*x + sqrt(a^2*x^2 + 1) 
 - 1) + (-80*I*a^4*x^4 - 29*a^3*x^3 - 11*I*a^2*x^2 + 6*a*x + 2*I)*sqrt(a^2 
*x^2 + 1))/(a*x^5 - I*x^4)
 

Sympy [F]

\[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=i \left (\int \frac {\sqrt {a^{2} x^{2} + 1}}{a^{3} x^{8} - 3 i a^{2} x^{7} - 3 a x^{6} + i x^{5}}\, dx + \int \frac {a^{2} x^{2} \sqrt {a^{2} x^{2} + 1}}{a^{3} x^{8} - 3 i a^{2} x^{7} - 3 a x^{6} + i x^{5}}\, dx\right ) \] Input:

integrate(1/(1+I*a*x)**3*(a**2*x**2+1)**(3/2)/x**5,x)
 

Output:

I*(Integral(sqrt(a**2*x**2 + 1)/(a**3*x**8 - 3*I*a**2*x**7 - 3*a*x**6 + I* 
x**5), x) + Integral(a**2*x**2*sqrt(a**2*x**2 + 1)/(a**3*x**8 - 3*I*a**2*x 
**7 - 3*a*x**6 + I*x**5), x))
 

Maxima [F]

\[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{5}} \,d x } \] Input:

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^5,x, algorithm="maxima")
 

Output:

integrate((a^2*x^2 + 1)^(3/2)/((I*a*x + 1)^3*x^5), x)
 

Giac [F]

\[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\int { \frac {{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (i \, a x + 1\right )}^{3} x^{5}} \,d x } \] Input:

integrate(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^5,x, algorithm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 23.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\frac {a^4\,\mathrm {atan}\left (\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}\right )\,51{}\mathrm {i}}{8}-\frac {\sqrt {a^2\,x^2+1}}{4\,x^4}+\frac {a\,\sqrt {a^2\,x^2+1}\,1{}\mathrm {i}}{x^3}+\frac {19\,a^2\,\sqrt {a^2\,x^2+1}}{8\,x^2}-\frac {a^3\,\sqrt {a^2\,x^2+1}\,6{}\mathrm {i}}{x}+\frac {a^5\,\sqrt {a^2\,x^2+1}\,4{}\mathrm {i}}{\left (-x\,\sqrt {a^2}+\frac {\sqrt {a^2}\,1{}\mathrm {i}}{a}\right )\,\sqrt {a^2}} \] Input:

int((a^2*x^2 + 1)^(3/2)/(x^5*(a*x*1i + 1)^3),x)
 

Output:

(a^4*atan((a^2*x^2 + 1)^(1/2)*1i)*51i)/8 - (a^2*x^2 + 1)^(1/2)/(4*x^4) + ( 
a*(a^2*x^2 + 1)^(1/2)*1i)/x^3 + (19*a^2*(a^2*x^2 + 1)^(1/2))/(8*x^2) - (a^ 
3*(a^2*x^2 + 1)^(1/2)*6i)/x + (a^5*(a^2*x^2 + 1)^(1/2)*4i)/((((a^2)^(1/2)* 
1i)/a - x*(a^2)^(1/2))*(a^2)^(1/2))
 

Reduce [F]

\[ \int \frac {e^{-3 i \arctan (a x)}}{x^5} \, dx=\frac {448 \sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}-4512 \sqrt {a^{2} x^{2}+1}\, a^{3} i \,x^{3}-1038 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+368 \sqrt {a^{2} x^{2}+1}\, a i x -12 \sqrt {a^{2} x^{2}+1}+2496 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} x^{9}-3 a^{4} i \,x^{8}-2 a^{3} x^{7}-2 a^{2} i \,x^{6}-3 a \,x^{5}+i \,x^{4}}d x \right ) a \,x^{4}-960 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} x^{7}-3 a^{4} i \,x^{6}-2 a^{3} x^{5}-2 a^{2} i \,x^{4}-3 a \,x^{3}+i \,x^{2}}d x \right ) a^{3} x^{4}-3456 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} i \,x^{9}+3 a^{4} x^{8}-2 a^{3} i \,x^{7}+2 a^{2} x^{6}-3 a i \,x^{5}-x^{4}}d x \right ) a i \,x^{4}+5184 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} i \,x^{8}+3 a^{4} x^{7}-2 a^{3} i \,x^{6}+2 a^{2} x^{5}-3 a i \,x^{4}-x^{3}}d x \right ) a^{2} x^{4}+14336 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} i \,x^{7}+3 a^{4} x^{6}-2 a^{3} i \,x^{5}+2 a^{2} x^{4}-3 a i \,x^{3}-x^{2}}d x \right ) a^{3} i \,x^{4}-17472 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} i \,x^{6}+3 a^{4} x^{5}-2 a^{3} i \,x^{4}+2 a^{2} x^{3}-3 a i \,x^{2}-x}d x \right ) a^{4} x^{4}-9408 \left (\int \frac {\sqrt {a^{2} x^{2}+1}}{a^{5} i \,x^{5}+3 a^{4} x^{4}-2 a^{3} i \,x^{3}+2 a^{2} x^{2}-3 a i x -1}d x \right ) a^{5} i \,x^{4}-448 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x^{4}}{a^{5} i \,x^{5}+3 a^{4} x^{4}-2 a^{3} i \,x^{3}+2 a^{2} x^{2}-3 a i x -1}d x \right ) a^{9} i \,x^{4}-1344 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x^{3}}{a^{5} i \,x^{5}+3 a^{4} x^{4}-2 a^{3} i \,x^{3}+2 a^{2} x^{2}-3 a i x -1}d x \right ) a^{8} x^{4}+192 \left (\int \frac {\sqrt {a^{2} x^{2}+1}\, x}{a^{5} i \,x^{5}+3 a^{4} x^{4}-2 a^{3} i \,x^{3}+2 a^{2} x^{2}-3 a i x -1}d x \right ) a^{6} x^{4}+153 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}-1\right ) a^{4} x^{4}-153 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+1\right ) a^{4} x^{4}}{48 x^{4}} \] Input:

int(1/(1+I*a*x)^3*(a^2*x^2+1)^(3/2)/x^5,x)
 

Output:

(448*sqrt(a**2*x**2 + 1)*a**4*x**4 - 4512*sqrt(a**2*x**2 + 1)*a**3*i*x**3 
- 1038*sqrt(a**2*x**2 + 1)*a**2*x**2 + 368*sqrt(a**2*x**2 + 1)*a*i*x - 12* 
sqrt(a**2*x**2 + 1) + 2496*int(sqrt(a**2*x**2 + 1)/(a**5*x**9 - 3*a**4*i*x 
**8 - 2*a**3*x**7 - 2*a**2*i*x**6 - 3*a*x**5 + i*x**4),x)*a*x**4 - 960*int 
(sqrt(a**2*x**2 + 1)/(a**5*x**7 - 3*a**4*i*x**6 - 2*a**3*x**5 - 2*a**2*i*x 
**4 - 3*a*x**3 + i*x**2),x)*a**3*x**4 - 3456*int(sqrt(a**2*x**2 + 1)/(a**5 
*i*x**9 + 3*a**4*x**8 - 2*a**3*i*x**7 + 2*a**2*x**6 - 3*a*i*x**5 - x**4),x 
)*a*i*x**4 + 5184*int(sqrt(a**2*x**2 + 1)/(a**5*i*x**8 + 3*a**4*x**7 - 2*a 
**3*i*x**6 + 2*a**2*x**5 - 3*a*i*x**4 - x**3),x)*a**2*x**4 + 14336*int(sqr 
t(a**2*x**2 + 1)/(a**5*i*x**7 + 3*a**4*x**6 - 2*a**3*i*x**5 + 2*a**2*x**4 
- 3*a*i*x**3 - x**2),x)*a**3*i*x**4 - 17472*int(sqrt(a**2*x**2 + 1)/(a**5* 
i*x**6 + 3*a**4*x**5 - 2*a**3*i*x**4 + 2*a**2*x**3 - 3*a*i*x**2 - x),x)*a* 
*4*x**4 - 9408*int(sqrt(a**2*x**2 + 1)/(a**5*i*x**5 + 3*a**4*x**4 - 2*a**3 
*i*x**3 + 2*a**2*x**2 - 3*a*i*x - 1),x)*a**5*i*x**4 - 448*int((sqrt(a**2*x 
**2 + 1)*x**4)/(a**5*i*x**5 + 3*a**4*x**4 - 2*a**3*i*x**3 + 2*a**2*x**2 - 
3*a*i*x - 1),x)*a**9*i*x**4 - 1344*int((sqrt(a**2*x**2 + 1)*x**3)/(a**5*i* 
x**5 + 3*a**4*x**4 - 2*a**3*i*x**3 + 2*a**2*x**2 - 3*a*i*x - 1),x)*a**8*x* 
*4 + 192*int((sqrt(a**2*x**2 + 1)*x)/(a**5*i*x**5 + 3*a**4*x**4 - 2*a**3*i 
*x**3 + 2*a**2*x**2 - 3*a*i*x - 1),x)*a**6*x**4 + 153*log(sqrt(a**2*x**2 + 
 1) - 1)*a**4*x**4 - 153*log(sqrt(a**2*x**2 + 1) + 1)*a**4*x**4)/(48*x*...