\(\int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx\) [112]

Optimal result

Integrand size = 16, antiderivative size = 163 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

-25/2*a^2*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)+5/4*I*a*(1-I*a*x)^(5/4)/x/(1+I*a*x)^(1/4)-1/2*(1-I*a*x)^(9/4)/x^2/(1
+I*a*x)^(1/4)-25/4*a^2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+25/4*a^2*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4
))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5170, 98, 96, 95, 304, 209, 212} \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}} \]

[In]

Int[1/(E^(((5*I)/2)*ArcTan[a*x])*x^3),x]

[Out]

(-25*a^2*(1 - I*a*x)^(1/4))/(2*(1 + I*a*x)^(1/4)) + (((5*I)/4)*a*(1 - I*a*x)^(5/4))/(x*(1 + I*a*x)^(1/4)) - (1
- I*a*x)^(9/4)/(2*x^2*(1 + I*a*x)^(1/4)) - (25*a^2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4 + (25*a^2*A
rcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)])/4

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 5170

Int[E^(ArcTan[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 - I*a*x)^(I*(n/2))/(1 + I*a*x)^(I*(n/2))
), x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[(I*n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(1-i a x)^{5/4}}{x^3 (1+i a x)^{5/4}} \, dx \\ & = -\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {1}{4} (5 i a) \int \frac {(1-i a x)^{5/4}}{x^2 (1+i a x)^{5/4}} \, dx \\ & = \frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {\sqrt [4]{1-i a x}}{x (1+i a x)^{5/4}} \, dx \\ & = -\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {1}{8} \left (25 a^2\right ) \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {1}{2} \left (25 a^2\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ & = -\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}+\frac {1}{4} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{4} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ & = -\frac {25 a^2 \sqrt [4]{1-i a x}}{2 \sqrt [4]{1+i a x}}+\frac {5 i a (1-i a x)^{5/4}}{4 x \sqrt [4]{1+i a x}}-\frac {(1-i a x)^{9/4}}{2 x^2 \sqrt [4]{1+i a x}}-\frac {25}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {25}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.50 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\frac {\sqrt [4]{1-i a x} \left (-2+9 i a x-43 a^2 x^2+50 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{4 x^2 \sqrt [4]{1+i a x}} \]

[In]

Integrate[1/(E^(((5*I)/2)*ArcTan[a*x])*x^3),x]

[Out]

((1 - I*a*x)^(1/4)*(-2 + (9*I)*a*x - 43*a^2*x^2 + 50*a^2*x^2*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x
)]))/(4*x^2*(1 + I*a*x)^(1/4))

Maple [F]

\[\int \frac {1}{{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{3}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (111) = 222\).

Time = 0.27 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=-\frac {2 \, \sqrt {a^{2} x^{2} + 1} {\left (-43 i \, a^{2} x^{2} - 9 \, a x - 2 i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 25 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 25 \, {\left (i \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 25 \, {\left (-i \, a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 25 \, {\left (a^{3} x^{3} - i \, a^{2} x^{2}\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right )}{8 \, {\left (a x^{3} - i \, x^{2}\right )}} \]

[In]

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

[Out]

-1/8*(2*sqrt(a^2*x^2 + 1)*(-43*I*a^2*x^2 - 9*a*x - 2*I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 25*(a^3*x^3 - I*
a^2*x^2)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 25*(I*a^3*x^3 + a^2*x^2)*log(sqrt(I*sqrt(a^2*x^2 + 1)/
(a*x + I)) + I) + 25*(-I*a^3*x^3 - a^2*x^2)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 25*(a^3*x^3 - I*a^2
*x^2)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1))/(a*x^3 - I*x^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]Warning, replacing 0 by 81, a substitution variable should perhaps be p
urged.Warni

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {1}{x^3\,{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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