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

Optimal. Leaf size=93 \[ -\frac{e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt{e x}}+\frac{e^2 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x} E\left (\left .\frac{1}{2} \csc ^{-1}(x)\right |2\right )}{2 \sqrt [4]{1-x^2}}-\frac{1}{3} e \left (1-x^2\right )^{3/4} (e x)^{3/2} \]

[Out]

-(e^3*(1 - x^2)^(3/4))/(2*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))

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Rubi [A]  time = 0.0387342, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {125, 321, 315, 317, 335, 228} \[ -\frac{e^3 \left (1-x^2\right )^{3/4}}{2 \sqrt{e x}}+\frac{e^2 \sqrt [4]{1-\frac{1}{x^2}} \sqrt{e x} E\left (\left .\frac{1}{2} \csc ^{-1}(x)\right |2\right )}{2 \sqrt [4]{1-x^2}}-\frac{1}{3} e \left (1-x^2\right )^{3/4} (e x)^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*(1 - x^2)^(3/4))/(2*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 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 321

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 315

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 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{(e x)^{5/2}}{\sqrt [4]{1-x} \sqrt [4]{1+x}} \, dx &=\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 ) \operatorname{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]  time = 0.0145962, size = 39, normalized size = 0.42 \[ -\frac{1}{3} e (e x)^{3/2} \left (\left (1-x^2\right )^{3/4}-\, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{7}{4};x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt [4]{1-x}}}{\frac{1}{\sqrt [4]{1+x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{e x} e^{2}{\left (x + 1\right )}^{\frac{3}{4}} x^{2}{\left (-x + 1\right )}^{\frac{3}{4}}}{x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out