Integrand size = 16, antiderivative size = 121 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {10 i a \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}-5 i a \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-5 i a \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:
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))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {-3 \left (1-8 i a x+9 a^2 x^2\right )-10 a x (i+a x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )}{3 x \sqrt [4]{1-i a x} (1+i a x)^{3/4}} \] Input:
Integrate[E^(((5*I)/2)*ArcTan[a*x])/x^2,x]
Output:
(-3*(1 - (8*I)*a*x + 9*a^2*x^2) - 10*a*x*(I + a*x)*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)])/(3*x*(1 - I*a*x)^(1/4)*(1 + I*a*x)^(3/4))
Time = 0.40 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {5585, 105, 105, 104, 756, 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.
\(\displaystyle \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {(1+i a x)^{5/4}}{x^2 (1-i a x)^{5/4}}dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {5}{2} i a \int \frac {\sqrt [4]{i a x+1}}{x (1-i a x)^{5/4}}dx-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {5}{2} i a \left (\int \frac {1}{x \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {5}{2} i a \left (4 \int \frac {1}{\frac {i a x+1}{1-i a x}-1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {5}{2} i a \left (4 \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{2} i a \left (4 \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{2} i a \left (4 \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )+\frac {4 \sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {(1+i a x)^{5/4}}{x \sqrt [4]{1-i a x}}\) |
Input:
Int[E^(((5*I)/2)*ArcTan[a*x])/x^2,x]
Output:
-((1 + I*a*x)^(5/4)/(x*(1 - I*a*x)^(1/4))) + ((5*I)/2)*a*((4*(1 + I*a*x)^( 1/4))/(1 - I*a*x)^(1/4) + 4*(-1/2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/ 4)] - ArcTanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2))
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_)^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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) 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]
\[\int \frac {{\left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}^{\frac {5}{2}}}{x^{2}}d x\]
Input:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
Output:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {-5 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) + 5 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) - 5 \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) + 5 i \, a x \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (-9 i \, a x + 1\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{2 \, x} \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="fricas")
Output:
1/2*(-5*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) + 5*a*x*log(sqr t(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) - 5*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1) /(a*x + I)) - I) + 5*I*a*x*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) - 1) - 2*(-9*I*a*x + 1)*sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)))/x
Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\text {Timed out} \] Input:
integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(5/2)/x**2,x)
Output:
Timed out
\[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int { \frac {\left (\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}{x^{2}} \,d x } \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="maxima")
Output:
integrate(((I*a*x + 1)/sqrt(a^2*x^2 + 1))^(5/2)/x^2, x)
Exception generated. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo r the root of a polynomial with parameters. This might be wrong.The choice was done
Timed out. \[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\int \frac {{\left (\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}\right )}^{5/2}}{x^2} \,d x \] Input:
int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^2,x)
Output:
int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(5/2)/x^2, x)
\[ \int \frac {e^{\frac {5}{2} i \arctan (a x)}}{x^2} \, dx=\frac {8 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}} a i x -2 \sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}+5 \left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}}{a^{4} x^{5}+2 a^{2} x^{3}+x}d x \right ) a^{3} i \,x^{3}+5 \left (\int \frac {\sqrt {a i x +1}\, \left (a^{2} x^{2}+1\right )^{\frac {3}{4}}}{a^{4} x^{5}+2 a^{2} x^{3}+x}d x \right ) a i x}{2 x \left (a^{2} x^{2}+1\right )} \] Input:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(5/2)/x^2,x)
Output:
(8*sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4)*a*i*x - 2*sqrt(a*i*x + 1)*(a**2* x**2 + 1)**(3/4) + 5*int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4))/(a**4*x* *5 + 2*a**2*x**3 + x),x)*a**3*i*x**3 + 5*int((sqrt(a*i*x + 1)*(a**2*x**2 + 1)**(3/4))/(a**4*x**5 + 2*a**2*x**3 + x),x)*a*i*x)/(2*x*(a**2*x**2 + 1))