\(\int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx\) [1209]

Optimal result

Integrand size = 25, antiderivative size = 46 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

2*(x^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*EllipticE(sin(1/2*arctan(x)),2^(1/2))/a^2/(a-I
*a*x)^(1/4)/(a+I*a*x)^(1/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {42, 203, 202} \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\frac {2 \sqrt [4]{x^2+1} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[In]

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

[Out]

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

Rule 42

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[(a + b*x)^FracPart[m]*((c + d*x)^Frac
Part[m]/(a*c + b*d*x^2)^FracPart[m]), Int[(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 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{a^2+a^2 x^2} \int \frac {1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = \frac {\sqrt [4]{1+x^2} \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = \frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=-\frac {i 2^{3/4} \sqrt [4]{1+i x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {3}{4},\frac {1}{2}-\frac {i x}{2}\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[In]

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

[Out]

((-I)*2^(3/4)*(1 + I*x)^(1/4)*Hypergeometric2F1[-1/4, 5/4, 3/4, 1/2 - (I/2)*x])/(a^2*(a - I*a*x)^(1/4)*(a + I*
a*x)^(1/4))

Maple [C] (verified)

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

Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.98

method	result	size
risch	\(\frac {2 x}{a^{2} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (a^{2}\right )^{\frac {1}{4}} a^{2} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(91\)

[In]

int(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {5}{4}} {\left (-i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.11 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=- \frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{8}, \frac {9}{8}, 1 & \frac {1}{2}, \frac {5}{4}, \frac {7}{4} \\\frac {5}{8}, \frac {3}{4}, \frac {9}{8}, \frac {5}{4}, \frac {7}{4} & 0 \end {matrix} \middle | {\frac {e^{- 3 i \pi }}{x^{2}}} \right )} e^{- \frac {3 i \pi }{4}}}{4 \pi a^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, 0, \frac {1}{8}, \frac {1}{2}, \frac {5}{8}, 1 & \\\frac {1}{8}, \frac {5}{8} & - \frac {1}{2}, 0, \frac {3}{4}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{x^{2}}} \right )}}{4 \pi a^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(1/(a-I*a*x)**(5/4)/(a+I*a*x)**(5/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {5}{4}} {\left (-i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(5/4)*(-I*a*x + a)^(5/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(5/4),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]ext_reduce Error: Bad Argument TypeDone

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{5/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \]

[In]

int(1/((a - a*x*1i)^(5/4)*(a + a*x*1i)^(5/4)),x)

[Out]

int(1/((a - a*x*1i)^(5/4)*(a + a*x*1i)^(5/4)), x)