\(\int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx\) [111]

Optimal result

Integrand size = 16, antiderivative size = 121 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-\frac {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]

[Out]

-10*I*a*(1-I*a*x)^(1/4)/(1+I*a*x)^(1/4)-(1-I*a*x)^(5/4)/x/(1+I*a*x)^(1/4)-5*I*a*arctan((1+I*a*x)^(1/4)/(1-I*a*
x)^(1/4))+5*I*a*arctanh((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5170, 96, 95, 304, 209, 212} \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-\frac {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}} \]

[In]

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

[Out]

((-10*I)*a*(1 - I*a*x)^(1/4))/(1 + I*a*x)^(1/4) - (1 - I*a*x)^(5/4)/(x*(1 + I*a*x)^(1/4)) - (5*I)*a*ArcTan[(1
+ I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] + (5*I)*a*ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]

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 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^2 (1+i a x)^{5/4}} \, dx \\ & = -\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-\frac {1}{2} (5 i a) \int \frac {\sqrt [4]{1-i a x}}{x (1+i a x)^{5/4}} \, dx \\ & = -\frac {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-\frac {1}{2} (5 i a) \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{1+i a x}} \, dx \\ & = -\frac {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-(10 i a) \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 {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}+(5 i a) \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 )-(5 i a) \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 {10 i a \sqrt [4]{1-i a x}}{\sqrt [4]{1+i a x}}-\frac {(1-i a x)^{5/4}}{x \sqrt [4]{1+i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+5 i a \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 = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {i \sqrt [4]{1-i a x} \left (i-9 a x+10 a x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{x \sqrt [4]{1+i a x}} \]

[In]

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

[Out]

(I*(1 - I*a*x)^(1/4)*(I - 9*a*x + 10*a*x*Hypergeometric2F1[1/4, 1, 5/4, (I + a*x)/(I - a*x)]))/(x*(1 + I*a*x)^
(1/4))

Maple [F]

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

[In]

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

[Out]

int(1/((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,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. 212 vs. \(2 (83) = 166\).

Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.75 \[ \int \frac {e^{-\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=-\frac {2 \, \sqrt {a^{2} x^{2} + 1} {\left (9 \, a x - i\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 5 \, {\left (-i \, a^{2} x^{2} - a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 5 \, {\left (a^{2} x^{2} - i \, a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 5 \, {\left (a^{2} x^{2} - i \, a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 5 \, {\left (i \, a^{2} x^{2} + a x\right )} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right )}{2 \, {\left (a x^{2} - i \, x\right )}} \]

[In]

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

[Out]

-1/2*(2*sqrt(a^2*x^2 + 1)*(9*a*x - I)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 5*(-I*a^2*x^2 - a*x)*log(sqrt(I*sq
rt(a^2*x^2 + 1)/(a*x + I)) + 1) - 5*(a^2*x^2 - I*a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + 5*(a^2*x^
2 - I*a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - I) + 5*(I*a^2*x^2 + a*x)*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*
x + I)) - 1))/(a*x^2 - I*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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