3.10.15 \(\int \frac {1}{\sqrt [4]{1-x} (e x)^{3/2} \sqrt [4]{1+x}} \, dx\) [915]

Optimal. Leaf size=42 \[ -\frac {2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{e^2 \sqrt [4]{1-x^2}} \]

[Out]

-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)/e^2/(-x^2+1)^(1/4)

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Rubi [A]
time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {126, 323, 342, 234} \begin {gather*} -\frac {2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{e^2 \sqrt [4]{1-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - x)^(1/4)*(e*x)^(3/2)*(1 + x)^(1/4)),x]

[Out]

(-2*(1 - x^(-2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCsc[x]/2, 2])/(e^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 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 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*} \int \frac {1}{\sqrt [4]{1-x} (e x)^{3/2} \sqrt [4]{1+x}} \, dx &=\int \frac {1}{(e x)^{3/2} \sqrt [4]{1-x^2}} \, dx\\ &=\frac {\left (\sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x}\right ) \int \frac {1}{\sqrt [4]{1-\frac {1}{x^2}} x^2} \, dx}{e^2 \sqrt [4]{1-x^2}}\\ &=-\frac {\left (\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 )}{e^2 \sqrt [4]{1-x^2}}\\ &=-\frac {2 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{e^2 \sqrt [4]{1-x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 23, normalized size = 0.55 \begin {gather*} -\frac {2 x \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};x^2\right )}{(e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - x)^(1/4)*(e*x)^(3/2)*(1 + x)^(1/4)),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (1-x \right )^{\frac {1}{4}} \left (e x \right )^{\frac {3}{2}} \left (1+x \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(3/2)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(3/2)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 14.87, size = 87, normalized size = 2.07 \begin {gather*} - \frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {7}{8}, \frac {11}{8}, 1 & 1, \frac {5}{4}, \frac {3}{2} \\\frac {1}{2}, \frac {7}{8}, 1, \frac {11}{8}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{2}}} \right )} e^{\frac {i \pi }{4}}}{4 \pi e^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} - \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{4}, \frac {3}{8}, \frac {3}{4}, \frac {7}{8}, \frac {5}{4}, 1 & \\\frac {3}{8}, \frac {7}{8} & \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi e^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)**(1/4)/(e*x)**(3/2)/(1+x)**(1/4),x)

[Out]

-I*meijerg(((7/8, 11/8, 1), (1, 5/4, 3/2)), ((1/2, 7/8, 1, 11/8, 3/2), (0,)), exp_polar(-2*I*pi)/x**2)*exp(I*p
i/4)/(4*pi*e**(3/2)*gamma(1/4)) - meijerg(((1/4, 3/8, 3/4, 7/8, 5/4, 1), ()), ((3/8, 7/8), (1/4, 1/2, 3/4, 0))
, x**(-2))/(4*pi*e**(3/2)*gamma(1/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(3/2)/(1+x)^(1/4),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (e\,x\right )}^{3/2}\,{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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