Integrand size = 16, antiderivative size = 170 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}-\frac {5 i a (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{12 x^2}+\frac {11 a^2 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{24 x}+\frac {3}{8} i a^3 \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )+\frac {3}{8} i a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right ) \] Output:
-1/3*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/x^3-5/12*I*a*(1-I*a*x)^(3/4)*(1+I*a*x )^(1/4)/x^2+11/24*a^2*(1-I*a*x)^(3/4)*(1+I*a*x)^(1/4)/x+3/8*I*a^3*arctan(( 1+I*a*x)^(1/4)/(1-I*a*x)^(1/4))+3/8*I*a^3*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 = 93, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\frac {(1-i a x)^{3/4} \left (-8-18 i a x+21 a^2 x^2+11 i a^3 x^3+6 i a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {i+a x}{i-a x}\right )\right )}{24 x^3 (1+i a x)^{3/4}} \] Input:
Integrate[E^((I/2)*ArcTan[a*x])/x^4,x]
Output:
((1 - I*a*x)^(3/4)*(-8 - (18*I)*a*x + 21*a^2*x^2 + (11*I)*a^3*x^3 + (6*I)* a^3*x^3*Hypergeometric2F1[3/4, 1, 7/4, (I + a*x)/(I - a*x)]))/(24*x^3*(1 + I*a*x)^(3/4))
Time = 0.47 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {5585, 110, 27, 168, 27, 168, 27, 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.
\(\displaystyle \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx\) |
\(\Big \downarrow \) 5585 |
\(\displaystyle \int \frac {\sqrt [4]{1+i a x}}{x^4 \sqrt [4]{1-i a x}}dx\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {1}{3} \int \frac {a (5 i-4 a x)}{2 x^3 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \int \frac {5 i-4 a x}{x^3 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{2} \int \frac {a (10 i a x+11)}{2 x^2 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \int \frac {10 i a x+11}{x^2 \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (-\int -\frac {9 i a}{2 x \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (\frac {9}{2} i a \int \frac {1}{x \sqrt [4]{1-i a x} (i a x+1)^{3/4}}dx-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (18 i a \int \frac {1}{\frac {i a x+1}{1-i a x}-1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (18 i a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \int \frac {1}{\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}+1}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}\right )-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (18 i a \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}}d\frac {\sqrt [4]{i a x+1}}{\sqrt [4]{1-i a x}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} a \left (-\frac {1}{4} a \left (18 i a \left (-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{1+i a x}}{\sqrt [4]{1-i a x}}\right )\right )-\frac {11 (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{x}\right )-\frac {5 i (1-i a x)^{3/4} \sqrt [4]{1+i a x}}{2 x^2}\right )-\frac {(1-i a x)^{3/4} \sqrt [4]{1+i a x}}{3 x^3}\) |
Input:
Int[E^((I/2)*ArcTan[a*x])/x^4,x]
Output:
-1/3*((1 - I*a*x)^(3/4)*(1 + I*a*x)^(1/4))/x^3 + (a*((((-5*I)/2)*(1 - I*a* x)^(3/4)*(1 + I*a*x)^(1/4))/x^2 - (a*((-11*(1 - I*a*x)^(3/4)*(1 + I*a*x)^( 1/4))/x + (18*I)*a*(-1/2*ArcTan[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)] - Arc Tanh[(1 + I*a*x)^(1/4)/(1 - I*a*x)^(1/4)]/2)))/4))/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[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]
\[\int \frac {\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}}{x^{4}}d x\]
Input:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
Output:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\frac {9 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + 1\right ) - 9 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} + i\right ) + 9 \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - i\right ) - 9 i \, a^{3} x^{3} \log \left (\sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}} - 1\right ) - 2 \, {\left (11 i \, a^{3} x^{3} - a^{2} x^{2} + 2 i \, a x + 8\right )} \sqrt {\frac {i \, \sqrt {a^{2} x^{2} + 1}}{a x + i}}}{48 \, x^{3}} \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x, algorithm="fricas")
Output:
1/48*(9*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + 1) - 9*a^3*x^3 *log(sqrt(I*sqrt(a^2*x^2 + 1)/(a*x + I)) + I) + 9*a^3*x^3*log(sqrt(I*sqrt( a^2*x^2 + 1)/(a*x + I)) - I) - 9*I*a^3*x^3*log(sqrt(I*sqrt(a^2*x^2 + 1)/(a *x + I)) - 1) - 2*(11*I*a^3*x^3 - a^2*x^2 + 2*I*a*x + 8)*sqrt(I*sqrt(a^2*x ^2 + 1)/(a*x + I)))/x^3
\[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\int \frac {\sqrt {\frac {i \left (a x - i\right )}{\sqrt {a^{2} x^{2} + 1}}}}{x^{4}}\, dx \] Input:
integrate(((1+I*a*x)/(a**2*x**2+1)**(1/2))**(1/2)/x**4,x)
Output:
Integral(sqrt(I*(a*x - I)/sqrt(a**2*x**2 + 1))/x**4, x)
\[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\int { \frac {\sqrt {\frac {i \, a x + 1}{\sqrt {a^{2} x^{2} + 1}}}}{x^{4}} \,d x } \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x, algorithm="maxima")
Output:
integrate(sqrt((I*a*x + 1)/sqrt(a^2*x^2 + 1))/x^4, x)
Exception generated. \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,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. \[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\int \frac {\sqrt {\frac {1+a\,x\,1{}\mathrm {i}}{\sqrt {a^2\,x^2+1}}}}{x^4} \,d x \] Input:
int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/2)/x^4,x)
Output:
int(((a*x*1i + 1)/(a^2*x^2 + 1)^(1/2))^(1/2)/x^4, x)
\[ \int \frac {e^{\frac {1}{2} i \arctan (a x)}}{x^4} \, dx=\int \frac {\sqrt {\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}}}{x^{4}}d x \] Input:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)
Output:
int(((1+I*a*x)/(a^2*x^2+1)^(1/2))^(1/2)/x^4,x)