Integrand size = 16, antiderivative size = 132 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {i a \sqrt [4]{1-i a x} (1+i a x)^{3/4}}{4 x}-\frac {(1-i a x)^{5/4} (1+i a x)^{3/4}}{2 x^2}-\frac {1}{4} a^2 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {1}{4} a^2 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \]
1/4*I*a*(1-I*a*x)^(1/4)*(1+I*a*x)^(3/4)/x-1/2*(1-I*a*x)^(5/4)*(1+I*a*x)^(3 /4)/x^2-1/4*a^2*arctan((1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+1/4*a^2*arctanh((1 +I*a*x)^(1/4)/(1-I*a*x)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {\sqrt [4]{1-i a x} \left (-2+i a x-3 a^2 x^2+2 a^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {i+a x}{i-a x}\right )\right )}{4 x^2 \sqrt [4]{1+i a x}} \]
((1 - I*a*x)^(1/4)*(-2 + I*a*x - 3*a^2*x^2 + 2*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))
Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5585, 107, 105, 104, 25, 827, 216, 219}
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.
\(\displaystyle \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {\sqrt [4]{1-i a x}}{x^3 \sqrt [4]{1+i a x}}dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle -\frac {1}{4} i a \int \frac {\sqrt [4]{1-i a x}}{x^2 \sqrt [4]{i a x+1}}dx-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -\frac {1}{4} i a \left (-\frac {1}{2} i a \int \frac {1}{x (1-i a x)^{3/4} \sqrt [4]{i a x+1}}dx-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {1}{4} i a \left (-2 i a \int -\frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} i a \left (2 i a \int \frac {\sqrt {i a x+1}}{\sqrt {1-i a x} \left (1-\frac {i a x+1}{1-i a x}\right )}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{4} i a \left (-2 i a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {\sqrt [4]{1-i a x} (1+i a x)^{3/4}}{x}\right )-\frac {(1+i a x)^{3/4} (1-i a x)^{5/4}}{2 x^2}\) |
-1/2*((1 - I*a*x)^(5/4)*(1 + I*a*x)^(3/4))/x^2 - (I/4)*a*(-(((1 - I*a*x)^( 1/4)*(1 + I*a*x)^(3/4))/x) - (2*I)*a*(ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x) ^(1/4)]/2 - ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2))
3.1.94.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[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] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[(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] || SumSimplerQ[m, 1])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
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] && !Intege rQ[(I*n - 1)/2]
\[\int \frac {1}{\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}\, x^{3}}d x\]
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\frac {a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + i \, a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - a^{2} x^{2} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} {\left (-3 i \, a x + 2\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{8 \, x^{2}} \]
1/8*(a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - I*a^2*x^2*log( sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + I*a^2*x^2*log(sqrt(I*sqrt(a^2*x ^2 + 1)/(a*x + I)) - I) - a^2*x^2*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1) - 2*sqrt(a^2*x^2 + 1)*(-3*I*a*x + 2)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x^2
\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {1}{x^{3} \sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}\, dx \]
\[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int { \frac {1}{x^{3} \sqrt {\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}}} \,d x } \]
Exception generated. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ]Warning, replacing 0 by -46, a substitution variable should perhaps be pu rged.Warn
Timed out. \[ \int \frac {e^{-\frac {1}{2} i \arctan (a x)}}{x^3} \, dx=\int \frac {1}{x^3\,\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}} \,d x \]