\(\int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx\) [265]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\frac {2 x}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac {2 \left (1+x^2\right )^{3/4} \operatorname {EllipticF}\left (\frac {\arctan (x)}{2},2\right )}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}} \] Output:

2/3*x/a^2/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)+2/3*(x^2+1)^(3/4)*InverseJacobiA 
M(1/2*arctan(x),2^(1/2))/a^2/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=-\frac {i \sqrt [4]{2} (1+i x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {7}{4},\frac {1}{4},\frac {1}{2}-\frac {i x}{2}\right )}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}} \] Input:

Integrate[1/((a - I*a*x)^(7/4)*(a + I*a*x)^(7/4)),x]
 

Output:

((-1/3*I)*2^(1/4)*(1 + I*x)^(3/4)*Hypergeometric2F1[-3/4, 7/4, 1/4, 1/2 - 
(I/2)*x])/(a^2*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4))
 

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {46, 215, 231, 229}

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.

\(\displaystyle \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx\)

\(\Big \downarrow \) 46

\(\displaystyle \frac {\left (a^2 x^2+a^2\right )^{3/4} \int \frac {1}{\left (x^2 a^2+a^2\right )^{7/4}}dx}{(a-i a x)^{3/4} (a+i a x)^{3/4}}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\left (a^2 x^2+a^2\right )^{3/4} \left (\frac {\int \frac {1}{\left (x^2 a^2+a^2\right )^{3/4}}dx}{3 a^2}+\frac {2 x}{3 a^2 \left (a^2 x^2+a^2\right )^{3/4}}\right )}{(a-i a x)^{3/4} (a+i a x)^{3/4}}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\left (a^2 x^2+a^2\right )^{3/4} \left (\frac {\left (x^2+1\right )^{3/4} \int \frac {1}{\left (x^2+1\right )^{3/4}}dx}{3 a^2 \left (a^2 x^2+a^2\right )^{3/4}}+\frac {2 x}{3 a^2 \left (a^2 x^2+a^2\right )^{3/4}}\right )}{(a-i a x)^{3/4} (a+i a x)^{3/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\left (a^2 x^2+a^2\right )^{3/4} \left (\frac {2 \left (x^2+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {\arctan (x)}{2},2\right )}{3 a^2 \left (a^2 x^2+a^2\right )^{3/4}}+\frac {2 x}{3 a^2 \left (a^2 x^2+a^2\right )^{3/4}}\right )}{(a-i a x)^{3/4} (a+i a x)^{3/4}}\)

Input:

Int[1/((a - I*a*x)^(7/4)*(a + I*a*x)^(7/4)),x]
 

Output:

((a^2 + a^2*x^2)^(3/4)*((2*x)/(3*a^2*(a^2 + a^2*x^2)^(3/4)) + (2*(1 + x^2) 
^(3/4)*EllipticF[ArcTan[x]/2, 2])/(3*a^2*(a^2 + a^2*x^2)^(3/4))))/((a - I* 
a*x)^(3/4)*(a + I*a*x)^(3/4))
 

Defintions of rubi rules used

rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(a 
 + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^FracPart[m])   I 
nt[(a*c + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 
 0] &&  !IntegerQ[2*m]
 

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

Maple [F]

\[\int \frac {1}{\left (-i a x +a \right )^{\frac {7}{4}} \left (i a x +a \right )^{\frac {7}{4}}}d x\]

Input:

int(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),x)
 

Output:

int(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),x)
 

Fricas [F]

\[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {7}{4}} {\left (-i \, a x + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),x, algorithm="fricas")
 

Output:

1/3*(3*(a^4*x^2 + a^4)*integral(1/3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)/( 
a^4*x^2 + a^4), x) + 2*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)*x)/(a^4*x^2 + 
a^4)
 

Sympy [A] (verification not implemented)

Time = 15.68 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=- \frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {7}{8}, \frac {11}{8}, 1 & \frac {1}{2}, \frac {7}{4}, \frac {9}{4} \\\frac {7}{8}, \frac {5}{4}, \frac {11}{8}, \frac {7}{4}, \frac {9}{4} & 0 \end {matrix} \middle | {\frac {e^{- 3 i \pi }}{x^{2}}} \right )} e^{- \frac {i \pi }{4}}}{4 \pi a^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, 0, \frac {3}{8}, \frac {1}{2}, \frac {7}{8}, 1 & \\\frac {3}{8}, \frac {7}{8} & - \frac {1}{2}, 0, \frac {5}{4}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{x^{2}}} \right )}}{4 \pi a^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(1/(a-I*a*x)**(7/4)/(a+I*a*x)**(7/4),x)
 

Output:

-I*meijerg(((7/8, 11/8, 1), (1/2, 7/4, 9/4)), ((7/8, 5/4, 11/8, 7/4, 9/4), 
 (0,)), exp_polar(-3*I*pi)/x**2)*exp(-I*pi/4)/(4*pi*a**(7/2)*gamma(7/4)) + 
 I*meijerg(((-1/2, 0, 3/8, 1/2, 7/8, 1), ()), ((3/8, 7/8), (-1/2, 0, 5/4, 
0)), exp_polar(-I*pi)/x**2)/(4*pi*a**(7/2)*gamma(7/4))
 

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {7}{4}} {\left (-i \, a x + a\right )}^{\frac {7}{4}}} \,d x } \] Input:

integrate(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),x, algorithm="maxima")
 

Output:

integrate(1/((I*a*x + a)^(7/4)*(-I*a*x + a)^(7/4)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),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
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{7/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{7/4}} \,d x \] Input:

int(1/((a - a*x*1i)^(7/4)*(a + a*x*1i)^(7/4)),x)
 

Output:

int(1/((a - a*x*1i)^(7/4)*(a + a*x*1i)^(7/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx=\frac {2 \sqrt {a}\, \left (i x +1\right )^{\frac {3}{4}} \left (2 i x -1\right )}{3 \sqrt {i x +1}\, \left (-i x +1\right )^{\frac {3}{4}} a^{4} \left (i -x \right )} \] Input:

int(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(7/4),x)
 

Output:

(2*sqrt(a)*(i*x + 1)**(3/4)*(2*i*x - 1))/(3*sqrt(i*x + 1)*( - i*x + 1)**(3 
/4)*a**4*(i - x))