3.917 \(\int \frac{1}{\sqrt [4]{1-x} (e x)^{11/2} \sqrt [4]{1+x}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}-\frac{8 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x} E\left (\left .\frac{1}{2} \csc ^{-1}(x)\right |2\right )}{15 e^6 \sqrt [4]{1-x^2}}-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}} \]

[Out]

(-2*(1 - x^2)^(3/4))/(9*e*(e*x)^(9/2)) - (4*(1 - x^2)^(3/4))/(15*e^3*(e*x)^(5/2)) - (8*(1 - x^(-2))^(1/4)*Sqrt
[e*x]*EllipticE[ArcCsc[x]/2, 2])/(15*e^6*(1 - x^2)^(1/4))

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Rubi [A]  time = 0.0324452, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {125, 325, 317, 335, 228} \[ -\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}-\frac{8 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x} E\left (\left .\frac{1}{2} \csc ^{-1}(x)\right |2\right )}{15 e^6 \sqrt [4]{1-x^2}}-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - x)^(1/4)*(e*x)^(11/2)*(1 + x)^(1/4)),x]

[Out]

(-2*(1 - x^2)^(3/4))/(9*e*(e*x)^(9/2)) - (4*(1 - x^2)^(3/4))/(15*e^3*(e*x)^(5/2)) - (8*(1 - x^(-2))^(1/4)*Sqrt
[e*x]*EllipticE[ArcCsc[x]/2, 2])/(15*e^6*(1 - x^2)^(1/4))

Rule 125

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[m - n, 0] && GtQ[a, 0] && GtQ
[c, 0]

Rule 325

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

Rule 317

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 335

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]

Rule 228

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

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [4]{1-x} (e x)^{11/2} \sqrt [4]{1+x}} \, dx &=\int \frac{1}{(e x)^{11/2} \sqrt [4]{1-x^2}} \, dx\\ &=-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}}+\frac{2 \int \frac{1}{(e x)^{7/2} \sqrt [4]{1-x^2}} \, dx}{3 e^2}\\ &=-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}}-\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}+\frac{4 \int \frac{1}{(e x)^{3/2} \sqrt [4]{1-x^2}} \, dx}{15 e^4}\\ &=-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}}-\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}+\frac{\left (4 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x}\right ) \int \frac{1}{\sqrt [4]{1-\frac{1}{x^2}} x^2} \, dx}{15 e^6 \sqrt [4]{1-x^2}}\\ &=-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}}-\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}-\frac{\left (4 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1-x^2}} \, dx,x,\frac{1}{x}\right )}{15 e^6 \sqrt [4]{1-x^2}}\\ &=-\frac{2 \left (1-x^2\right )^{3/4}}{9 e (e x)^{9/2}}-\frac{4 \left (1-x^2\right )^{3/4}}{15 e^3 (e x)^{5/2}}-\frac{8 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x} E\left (\left .\frac{1}{2} \csc ^{-1}(x)\right |2\right )}{15 e^6 \sqrt [4]{1-x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0087668, size = 25, normalized size = 0.26 \[ -\frac{2 x \, _2F_1\left (-\frac{9}{4},\frac{1}{4};-\frac{5}{4};x^2\right )}{9 (e x)^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - x)^(1/4)*(e*x)^(11/2)*(1 + x)^(1/4)),x]

[Out]

(-2*x*Hypergeometric2F1[-9/4, 1/4, -5/4, x^2])/(9*(e*x)^(11/2))

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt [4]{1-x}}} \left ( ex \right ) ^{-{\frac{11}{2}}}{\frac{1}{\sqrt [4]{1+x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-x)^(1/4)/(e*x)^(11/2)/(1+x)^(1/4),x)

[Out]

int(1/(1-x)^(1/4)/(e*x)^(11/2)/(1+x)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e x\right )^{\frac{11}{2}}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(11/2)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((e*x)^(11/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{e x}{\left (x + 1\right )}^{\frac{3}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{e^{6} x^{8} - e^{6} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(11/2)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

integral(-sqrt(e*x)*(x + 1)^(3/4)*(-x + 1)^(3/4)/(e^6*x^8 - e^6*x^6), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)**(1/4)/(e*x)**(11/2)/(1+x)**(1/4),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(11/2)/(1+x)^(1/4),x, algorithm="giac")

[Out]

Timed out