3.916 \(\int \frac {1}{\sqrt [4]{1-x} (e x)^{7/2} \sqrt [4]{1+x}} \, dx\)

Optimal. Leaf size=70 \[ -\frac {4 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{5 e^4 \sqrt [4]{1-x^2}}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}} \]

[Out]

-2/5*(-x^2+1)^(3/4)/e/(e*x)^(5/2)-4/5*(1-1/x^2)^(1/4)*(cos(1/2*arccsc(x))^2)^(1/2)/cos(1/2*arccsc(x))*Elliptic
E(sin(1/2*arccsc(x)),2^(1/2))*(e*x)^(1/2)/e^4/(-x^2+1)^(1/4)

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {4 \sqrt [4]{1-\frac {1}{x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \csc ^{-1}(x)\right |2\right )}{5 e^4 \sqrt [4]{1-x^2}}-\frac {2 \left (1-x^2\right )^{3/4}}{5 e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - x)^(1/4)*(e*x)^(7/2)*(1 + x)^(1/4)),x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 25, normalized size = 0.36 \[ -\frac {2 x \, _2F_1\left (-\frac {5}{4},\frac {1}{4};-\frac {1}{4};x^2\right )}{5 (e x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - x)^(1/4)*(e*x)^(7/2)*(1 + x)^(1/4)),x]

[Out]

(-2*x*Hypergeometric2F1[-5/4, 1/4, -1/4, x^2])/(5*(e*x)^(7/2))

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e x} {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{e^{4} x^{6} - e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x\right )^{\frac {7}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((e*x)^(7/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-x +1\right )^{\frac {1}{4}} \left (e x \right )^{\frac {7}{2}} \left (x +1\right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x+1)^(1/4)/(e*x)^(7/2)/(x+1)^(1/4),x)

[Out]

int(1/(-x+1)^(1/4)/(e*x)^(7/2)/(x+1)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x\right )^{\frac {7}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(7/2)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((e*x)^(7/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (e\,x\right )}^{7/2}\,{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*x)^(7/2)*(1 - x)^(1/4)*(x + 1)^(1/4)),x)

[Out]

int(1/((e*x)^(7/2)*(1 - x)^(1/4)*(x + 1)^(1/4)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)**(1/4)/(e*x)**(7/2)/(1+x)**(1/4),x)

[Out]

Timed out

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