\(\int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx\) [913]

Optimal result

Integrand size = 24, antiderivative size = 93 \[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}-\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}+\frac {e^2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{2 \sqrt [4]{1-x^2}} \]

[Out]

-1/3*e*(e*x)^(3/2)*(-x^2+1)^(3/4)-1/2*e^3*(-x^2+1)^(3/4)/(e*x)^(1/2)+1/2*e^2*(1-1/x^2)^(1/4)*(cos(1/2*arccsc(x
))^2)^(1/2)/cos(1/2*arccsc(x))*EllipticE(sin(1/2*arccsc(x)),2^(1/2))*(e*x)^(1/2)/(-x^2+1)^(1/4)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {126, 327, 321, 323, 342, 234} \[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}+\frac {e^2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{2 \sqrt [4]{1-x^2}}-\frac {1}{3} e \left (1-x^2\right )^{3/4} (e x)^{3/2} \]

[In]

Int[(e*x)^(5/2)/((1 - x)^(1/4)*(1 + x)^(1/4)),x]

[Out]

-1/2*(e^3*(1 - x^2)^(3/4))/Sqrt[e*x] - (e*(e*x)^(3/2)*(1 - x^2)^(3/4))/3 + (e^2*(1 - x^(-2))^(1/4)*Sqrt[e*x]*E
llipticE[ArcCsc[x]/2, 2])/(2*(1 - x^2)^(1/4))

Rule 126

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[(a*c + b*d*x^2)
^m*(f*x)^p, x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && GtQ[a, 0] && GtQ[c,
0]

Rule 234

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

Rule 321

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[c*((a + b*x^2)^(3/4)/(b*Sqrt[c*x])), x] + D
ist[a*(c^2/(2*b)), Int[1/((c*x)^(3/2)*(a + b*x^2)^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]

Rule 323

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(c^
2*(a + b*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x^2}} \, dx \\ & = -\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}+\frac {1}{2} e^2 \int \frac {\sqrt {e x}}{\sqrt [4]{1-x^2}} \, dx \\ & = -\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}-\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}-\frac {1}{4} e^4 \int \frac {1}{(e x)^{3/2} \sqrt [4]{1-x^2}} \, dx \\ & = -\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}-\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}-\frac {\left (e^2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x}\right ) \int \frac {1}{\sqrt [4]{1-\frac {1}{x^2}} x^2} \, dx}{4 \sqrt [4]{1-x^2}} \\ & = -\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}-\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}+\frac {\left (e^2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-x^2}} \, dx,x,\frac {1}{x}\right )}{4 \sqrt [4]{1-x^2}} \\ & = -\frac {e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt {e x}}-\frac {1}{3} e (e x)^{3/2} \left (1-x^2\right )^{3/4}+\frac {e^2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{2 \sqrt [4]{1-x^2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.42 \[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=-\frac {1}{3} e (e x)^{3/2} \left (\left (1-x^2\right )^{3/4}-\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {7}{4},x^2\right )\right ) \]

[In]

Integrate[(e*x)^(5/2)/((1 - x)^(1/4)*(1 + x)^(1/4)),x]

[Out]

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

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {5}{2}}}{\left (1-x \right )^{\frac {1}{4}} \left (1+x \right )^{\frac {1}{4}}}d x\]

[In]

int((e*x)^(5/2)/(1-x)^(1/4)/(1+x)^(1/4),x)

[Out]

int((e*x)^(5/2)/(1-x)^(1/4)/(1+x)^(1/4),x)

Fricas [F]

\[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\int { \frac {\left (e x\right )^{\frac {5}{2}}}{{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((e*x)^(5/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

integral(-sqrt(e*x)*e^2*(x + 1)^(3/4)*x^2*(-x + 1)^(3/4)/(x^2 - 1), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 55.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28 \[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\frac {i e^{\frac {5}{2}} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{8}, - \frac {5}{8} & -1, - \frac {3}{4}, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {9}{8}, -1, - \frac {5}{8}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{2}}} \right )} e^{\frac {i \pi }{4}}}{4 \pi \Gamma \left (\frac {1}{4}\right )} - \frac {e^{\frac {5}{2}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {7}{4}, - \frac {13}{8}, - \frac {5}{4}, - \frac {9}{8}, - \frac {3}{4}, 1 & \\- \frac {13}{8}, - \frac {9}{8} & - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi \Gamma \left (\frac {1}{4}\right )} \]

[In]

integrate((e*x)**(5/2)/(1-x)**(1/4)/(1+x)**(1/4),x)

[Out]

I*e**(5/2)*meijerg(((-9/8, -5/8), (-1, -3/4, -1/2, 1)), ((-3/2, -9/8, -1, -5/8, -1/2, 0), ()), exp_polar(-2*I*
pi)/x**2)*exp(I*pi/4)/(4*pi*gamma(1/4)) - e**(5/2)*meijerg(((-7/4, -13/8, -5/4, -9/8, -3/4, 1), ()), ((-13/8,
-9/8), (-7/4, -3/2, -5/4, 0)), x**(-2))/(4*pi*gamma(1/4))

Maxima [F]

\[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\int { \frac {\left (e x\right )^{\frac {5}{2}}}{{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((e*x)^(5/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

integrate((e*x)^(5/2)/((x + 1)^(1/4)*(-x + 1)^(1/4)), x)

Giac [F]

\[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\int { \frac {\left (e x\right )^{\frac {5}{2}}}{{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((e*x)^(5/2)/(1-x)^(1/4)/(1+x)^(1/4),x, algorithm="giac")

[Out]

integrate((e*x)^(5/2)/((x + 1)^(1/4)*(-x + 1)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}}{{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \]

[In]

int((e*x)^(5/2)/((1 - x)^(1/4)*(x + 1)^(1/4)),x)

[Out]

int((e*x)^(5/2)/((1 - x)^(1/4)*(x + 1)^(1/4)), x)