3.9.54 \(\int \frac {1}{\sqrt [4]{1-x} \sqrt {e x} \sqrt [4]{1+x}} \, dx\)

Optimal. Leaf size=216 \[ -\frac {\log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {\log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}+1\right )}{\sqrt {2} \sqrt {e}} \]

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Rubi [A]  time = 0.15, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {125, 329, 240, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {\log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {\log \left (\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}+\sqrt {e}\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}+1\right )}{\sqrt {2} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[e*x])/(Sqrt[e]*(1 - x^2)^(1/4))]/(Sqrt[2]*Sqrt[e])) + ArcTan[1 + (Sqrt[2]*Sqrt[e*x]
)/(Sqrt[e]*(1 - x^2)^(1/4))]/(Sqrt[2]*Sqrt[e]) - Log[Sqrt[e] + (Sqrt[e]*x)/Sqrt[1 - x^2] - (Sqrt[2]*Sqrt[e*x])
/(1 - x^2)^(1/4)]/(2*Sqrt[2]*Sqrt[e]) + Log[Sqrt[e] + (Sqrt[e]*x)/Sqrt[1 - x^2] + (Sqrt[2]*Sqrt[e*x])/(1 - x^2
)^(1/4)]/(2*Sqrt[2]*Sqrt[e])

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [4]{1-x} \sqrt {e x} \sqrt [4]{1+x}} \, dx &=\int \frac {1}{\sqrt {e x} \sqrt [4]{1-x^2}} \, dx\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{e}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e-x^2}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{e^2}+\frac {\operatorname {Subst}\left (\int \frac {e+x^2}{1+\frac {x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{e^2}\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}\\ &=-\frac {\log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {\log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e x}}{\sqrt {e} \sqrt [4]{1-x^2}}\right )}{\sqrt {2} \sqrt {e}}-\frac {\log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}+\frac {\log \left (\sqrt {e}+\frac {\sqrt {e} x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {e x}}{\sqrt [4]{1-x^2}}\right )}{2 \sqrt {2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 156, normalized size = 0.72 \begin {gather*} \frac {\sqrt {x} \left (-\log \left (\frac {x}{\sqrt {1-x^2}}-\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )+\log \left (\frac {x}{\sqrt {1-x^2}}+\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt [4]{1-x^2}}+1\right )\right )}{2 \sqrt {2} \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)] -
Log[1 + x/Sqrt[1 - x^2] - (Sqrt[2]*Sqrt[x])/(1 - x^2)^(1/4)] + Log[1 + x/Sqrt[1 - x^2] + (Sqrt[2]*Sqrt[x])/(1
- x^2)^(1/4)]))/(2*Sqrt[2]*Sqrt[e*x])

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IntegrateAlgebraic [F]  time = 8.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{1-x} \sqrt {e x} \sqrt [4]{1+x}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.59, size = 455, normalized size = 2.11 \begin {gather*} \sqrt {2} \frac {1}{e^{2}}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {3}{4}} - \sqrt {2} {\left (e x^{2} - e\right )} \sqrt {-\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {1}{4}} + e \sqrt {x + 1} x \sqrt {-x + 1} - {\left (e^{2} x^{2} - e^{2}\right )} \sqrt {\frac {1}{e^{2}}}}{x^{2} - 1}} \frac {1}{e^{2}}^{\frac {3}{4}} - x^{2} + 1}{x^{2} - 1}\right ) + \sqrt {2} \frac {1}{e^{2}}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {3}{4}} - \sqrt {2} {\left (e x^{2} - e\right )} \sqrt {\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {1}{4}} - e \sqrt {x + 1} x \sqrt {-x + 1} + {\left (e^{2} x^{2} - e^{2}\right )} \sqrt {\frac {1}{e^{2}}}}{x^{2} - 1}} \frac {1}{e^{2}}^{\frac {3}{4}} + x^{2} - 1}{x^{2} - 1}\right ) + \frac {1}{4} \, \sqrt {2} \frac {1}{e^{2}}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {1}{4}} + e \sqrt {x + 1} x \sqrt {-x + 1} - {\left (e^{2} x^{2} - e^{2}\right )} \sqrt {\frac {1}{e^{2}}}}{x^{2} - 1}\right ) - \frac {1}{4} \, \sqrt {2} \frac {1}{e^{2}}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} \sqrt {e x} e {\left (x + 1\right )}^{\frac {3}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \frac {1}{e^{2}}^{\frac {1}{4}} - e \sqrt {x + 1} x \sqrt {-x + 1} + {\left (e^{2} x^{2} - e^{2}\right )} \sqrt {\frac {1}{e^{2}}}}{x^{2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(e^(-2))^(1/4)*arctan(-(sqrt(2)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(3/4) - sqrt(2)*(e*x
^2 - e)*sqrt(-(sqrt(2)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(1/4) + e*sqrt(x + 1)*x*sqrt(-x + 1)
- (e^2*x^2 - e^2)*sqrt(e^(-2)))/(x^2 - 1))*(e^(-2))^(3/4) - x^2 + 1)/(x^2 - 1)) + sqrt(2)*(e^(-2))^(1/4)*arcta
n(-(sqrt(2)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(3/4) - sqrt(2)*(e*x^2 - e)*sqrt((sqrt(2)*sqrt(e
*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(1/4) - e*sqrt(x + 1)*x*sqrt(-x + 1) + (e^2*x^2 - e^2)*sqrt(e^(-2)
))/(x^2 - 1))*(e^(-2))^(3/4) + x^2 - 1)/(x^2 - 1)) + 1/4*sqrt(2)*(e^(-2))^(1/4)*log(-(sqrt(2)*sqrt(e*x)*e*(x +
1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(1/4) + e*sqrt(x + 1)*x*sqrt(-x + 1) - (e^2*x^2 - e^2)*sqrt(e^(-2)))/(x^2 -
1)) - 1/4*sqrt(2)*(e^(-2))^(1/4)*log((sqrt(2)*sqrt(e*x)*e*(x + 1)^(3/4)*(-x + 1)^(3/4)*(e^(-2))^(1/4) - e*sqrt
(x + 1)*x*sqrt(-x + 1) + (e^2*x^2 - e^2)*sqrt(e^(-2)))/(x^2 - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {e x} {\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)^(1/2)/(1+x)^(1/4),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (-x +1\right )^{\frac {1}{4}} \sqrt {e x}\, \left (x +1\right )^{\frac {1}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {e x} {\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)^(1/2)/(1+x)^(1/4),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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sympy [C]  time = 7.97, size = 90, normalized size = 0.42 \begin {gather*} - \frac {i {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {3}{8}, \frac {7}{8} & \frac {1}{2}, \frac {3}{4}, 1, 1 \\0, \frac {3}{8}, \frac {1}{2}, \frac {7}{8}, 1, 0 & \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{2}}} \right )} e^{\frac {i \pi }{4}}}{4 \pi \sqrt {e} \Gamma \left (\frac {1}{4}\right )} - \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{8}, \frac {1}{4}, \frac {3}{8}, \frac {3}{4}, 1 & \\- \frac {1}{8}, \frac {3}{8} & - \frac {1}{4}, 0, \frac {1}{4}, 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi \sqrt {e} \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)**(1/2)/(1+x)**(1/4),x)

[Out]

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

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