3.9.57 \(\int \frac {1}{\sqrt [4]{1-x} (e x)^{13/2} \sqrt [4]{1+x}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {64 \left (1-x^2\right )^{11/4}}{231 e (e x)^{11/2}}+\frac {16 \left (1-x^2\right )^{7/4}}{21 e (e x)^{11/2}}-\frac {2 \left (1-x^2\right )^{3/4}}{3 e (e x)^{11/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {125, 273, 264} \begin {gather*} -\frac {64 \left (1-x^2\right )^{11/4}}{231 e (e x)^{11/2}}+\frac {16 \left (1-x^2\right )^{7/4}}{21 e (e x)^{11/2}}-\frac {2 \left (1-x^2\right )^{3/4}}{3 e (e x)^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - x)^(1/4)*(e*x)^(13/2)*(1 + x)^(1/4)),x]

[Out]

(-2*(1 - x^2)^(3/4))/(3*e*(e*x)^(11/2)) + (16*(1 - x^2)^(7/4))/(21*e*(e*x)^(11/2)) - (64*(1 - x^2)^(11/4))/(23
1*e*(e*x)^(11/2))

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 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 273

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{1-x} (e x)^{13/2} \sqrt [4]{1+x}} \, dx &=\int \frac {1}{(e x)^{13/2} \sqrt [4]{1-x^2}} \, dx\\ &=-\frac {2 \left (1-x^2\right )^{3/4}}{3 e (e x)^{11/2}}-\frac {8}{3} \int \frac {\left (1-x^2\right )^{3/4}}{(e x)^{13/2}} \, dx\\ &=-\frac {2 \left (1-x^2\right )^{3/4}}{3 e (e x)^{11/2}}+\frac {16 \left (1-x^2\right )^{7/4}}{21 e (e x)^{11/2}}+\frac {32}{21} \int \frac {\left (1-x^2\right )^{7/4}}{(e x)^{13/2}} \, dx\\ &=-\frac {2 \left (1-x^2\right )^{3/4}}{3 e (e x)^{11/2}}+\frac {16 \left (1-x^2\right )^{7/4}}{21 e (e x)^{11/2}}-\frac {64 \left (1-x^2\right )^{11/4}}{231 e (e x)^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 0.53 \begin {gather*} -\frac {2 \left (1-x^2\right )^{3/4} \left (32 x^4+24 x^2+21\right ) \sqrt {e x}}{231 e^7 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - x)^(1/4)*(e*x)^(13/2)*(1 + x)^(1/4)),x]

[Out]

(-2*Sqrt[e*x]*(1 - x^2)^(3/4)*(21 + 24*x^2 + 32*x^4))/(231*e^7*x^6)

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IntegrateAlgebraic [F]  time = 14.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{1-x} (e x)^{13/2} \sqrt [4]{1+x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((1 - x)^(1/4)*(e*x)^(13/2)*(1 + x)^(1/4)),x]

[Out]

Defer[IntegrateAlgebraic][1/((1 - x)^(1/4)*(e*x)^(13/2)*(1 + x)^(1/4)), x]

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fricas [A]  time = 1.10, size = 37, normalized size = 0.49 \begin {gather*} -\frac {2 \, {\left (32 \, x^{4} + 24 \, x^{2} + 21\right )} \sqrt {e x} {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{231 \, e^{7} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(13/2)/(1+x)^(1/4),x, algorithm="fricas")

[Out]

-2/231*(32*x^4 + 24*x^2 + 21)*sqrt(e*x)*(x + 1)^(3/4)*(-x + 1)^(3/4)/(e^7*x^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (e x\right )^{\frac {13}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(13/2)/(1+x)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((e*x)^(13/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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maple [A]  time = 0.00, size = 33, normalized size = 0.43 \begin {gather*} -\frac {2 \left (x +1\right )^{\frac {3}{4}} \left (32 x^{4}+24 x^{2}+21\right ) \left (-x +1\right )^{\frac {3}{4}} x}{231 \left (e x \right )^{\frac {13}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x+1)^(1/4)/(e*x)^(13/2)/(x+1)^(1/4),x)

[Out]

-2/231*x*(x+1)^(3/4)*(32*x^4+24*x^2+21)*(-x+1)^(3/4)/(e*x)^(13/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (e x\right )^{\frac {13}{2}} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)^(1/4)/(e*x)^(13/2)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((e*x)^(13/2)*(x + 1)^(1/4)*(-x + 1)^(1/4)), x)

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mupad [B]  time = 1.26, size = 52, normalized size = 0.68 \begin {gather*} -\frac {\sqrt {e\,x}\,\left (\frac {2}{11\,e^7}+\frac {2\,x^2}{77\,e^7}+\frac {16\,x^4}{231\,e^7}-\frac {64\,x^6}{231\,e^7}\right )}{x^6\,{\left (1-x\right )}^{1/4}\,{\left (x+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*x)^(13/2)*(1 - x)^(1/4)*(x + 1)^(1/4)),x)

[Out]

-((e*x)^(1/2)*(2/(11*e^7) + (2*x^2)/(77*e^7) + (16*x^4)/(231*e^7) - (64*x^6)/(231*e^7)))/(x^6*(1 - x)^(1/4)*(x
+ 1)^(1/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-x)**(1/4)/(e*x)**(13/2)/(1+x)**(1/4),x)

[Out]

Timed out

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