3.1209 \(\int \frac{1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx\)

Optimal. Leaf size=46 \[ \frac{2 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[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))

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Rubi [A]  time = 0.0085974, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {42, 197, 196} \[ \frac{2 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[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 197

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]

Rule 196

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

Rubi steps

\begin{align*} \int \frac{1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx &=\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]  time = 0.0229696, size = 68, normalized size = 1.48 \[ -\frac{i 2^{3/4} \sqrt [4]{1+i x} \, _2F_1\left (-\frac{1}{4},\frac{5}{4};\frac{3}{4};\frac{1}{2}-\frac{i x}{2}\right )}{a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[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))

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Maple [C]  time = 0.038, size = 91, normalized size = 2. \begin{align*} 2\,{\frac{x}{{a}^{2}\sqrt [4]{-a \left ( -1+ix \right ) }\sqrt [4]{a \left ( 1+ix \right ) }}}-{\frac{x}{{a}^{2}}{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,-{x}^{2})}\sqrt [4]{-{a}^{2} \left ( -1+ix \right ) \left ( 1+ix \right ) }{\frac{1}{\sqrt [4]{{a}^{2}}}}{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a x + a\right )}^{\frac{5}{4}}{\left (-i \, a x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}} x +{\left (a^{4} x^{2} + a^{4}\right )}{\rm integral}\left (-\frac{{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{a^{4} x^{2} + a^{4}}, x\right )}{a^{4} x^{2} + a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 54.0124, size = 97, normalized size = 2.11 \begin{align*} - \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError