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

Optimal. Leaf size=203 \[ -\frac{\log \left (\frac{\sqrt{1-x}}{\sqrt{x+1}}-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1-x}}{\sqrt{x+1}}+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt{2}}-2 \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )+\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )-\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107982, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.65, Rules used = {105, 63, 331, 297, 1162, 617, 204, 1165, 628, 93, 212, 206, 203} \[ -\frac{\log \left (\frac{\sqrt{1-x}}{\sqrt{x+1}}-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt{2}}+\frac{\log \left (\frac{\sqrt{1-x}}{\sqrt{x+1}}+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt{2}}-2 \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )+\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )-\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
 /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 297

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

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 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 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 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]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

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

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx &=\int \frac{1}{\sqrt [4]{1-x} (1+x)^{3/4}} \, dx+\int \frac{1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, dx\\ &=-\left (4 \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-x}\right )\right )+4 \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\right )-2 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-4 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+2 \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-2 \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}-\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{\log \left (1+\frac{\sqrt{1-x}}{\sqrt{1+x}}-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}+\frac{\log \left (1+\frac{\sqrt{1-x}}{\sqrt{1+x}}+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}-\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-2 \tan ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-2 \tanh ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{\log \left (1+\frac{\sqrt{1-x}}{\sqrt{1+x}}-\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}+\frac{\log \left (1+\frac{\sqrt{1-x}}{\sqrt{1+x}}+\frac{\sqrt{2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0235178, size = 74, normalized size = 0.36 \[ -\frac{2 (1-x)^{3/4} \left (\sqrt [4]{2} (x+1)^{3/4} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{1-x}{2}\right )+2 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{1-x}{x+1}\right )\right )}{3 (x+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\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((1+x)^(1/4)/(1-x)^(1/4)/x,x)

[Out]

int((1+x)^(1/4)/(1-x)^(1/4)/x,x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.72465, size = 981, normalized size = 4.83 \begin{align*} 2 \, \sqrt{2} \arctan \left (\frac{\sqrt{2}{\left (x - 1\right )} \sqrt{\frac{\sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} + x - \sqrt{x + 1} \sqrt{-x + 1} - 1}{x - 1}} - \sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - x + 1}{x - 1}\right ) + 2 \, \sqrt{2} \arctan \left (\frac{\sqrt{2}{\left (x - 1\right )} \sqrt{-\frac{\sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - x + \sqrt{x + 1} \sqrt{-x + 1} + 1}{x - 1}} - \sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} + x - 1}{x - 1}\right ) - \frac{1}{2} \, \sqrt{2} \log \left (\frac{4 \,{\left (\sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} + x - \sqrt{x + 1} \sqrt{-x + 1} - 1\right )}}{x - 1}\right ) + \frac{1}{2} \, \sqrt{2} \log \left (-\frac{4 \,{\left (\sqrt{2}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - x + \sqrt{x + 1} \sqrt{-x + 1} + 1\right )}}{x - 1}\right ) + 2 \, \arctan \left (\frac{{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{x - 1}\right ) + \log \left (\frac{x +{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) - \log \left (-\frac{x -{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*arctan((sqrt(2)*(x - 1)*sqrt((sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) + x - sqrt(x + 1)*sqrt(-x + 1) -
1)/(x - 1)) - sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) - x + 1)/(x - 1)) + 2*sqrt(2)*arctan((sqrt(2)*(x - 1)*sqrt(
-(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) - x + sqrt(x + 1)*sqrt(-x + 1) + 1)/(x - 1)) - sqrt(2)*(x + 1)^(1/4)*(-
x + 1)^(3/4) + x - 1)/(x - 1)) - 1/2*sqrt(2)*log(4*(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) + x - sqrt(x + 1)*sqr
t(-x + 1) - 1)/(x - 1)) + 1/2*sqrt(2)*log(-4*(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) - x + sqrt(x + 1)*sqrt(-x +
 1) + 1)/(x - 1)) + 2*arctan((x + 1)^(1/4)*(-x + 1)^(3/4)/(x - 1)) + log((x + (x + 1)^(1/4)*(-x + 1)^(3/4) - 1
)/(x - 1)) - log(-(x - (x + 1)^(1/4)*(-x + 1)^(3/4) - 1)/(x - 1))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [4]{x + 1}}{x \sqrt [4]{1 - x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(1/4)/(1-x)**(1/4)/x,x)

[Out]

Integral((x + 1)**(1/4)/(x*(1 - x)**(1/4)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x,x, algorithm="giac")

[Out]

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