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

Optimal. Leaf size=186 \[ -(1-x)^{3/4} \sqrt [4]{x+1}-\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt {2}} \]

[Out]

-(1-x)^(3/4)*(1+x)^(1/4)-1/2*arctan(-1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4))*2^(1/2)-1/2*arctan(1+(1-x)^(1/4)*2^(1/
2)/(1+x)^(1/4))*2^(1/2)-1/4*ln(1-(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4)+(1-x)^(1/2)/(1+x)^(1/2))*2^(1/2)+1/4*ln(1+(1-
x)^(1/4)*2^(1/2)/(1+x)^(1/4)+(1-x)^(1/2)/(1+x)^(1/2))*2^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -(1-x)^{3/4} \sqrt [4]{x+1}-\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(1/4)/(1 - x)^(1/4),x]

[Out]

-((1 - x)^(3/4)*(1 + x)^(1/4)) + ArcTan[1 - (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)]/Sqrt[2] - ArcTan[1 + (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)]/(2*Sqrt[2]) + Log[1 + Sqrt[1 - x]/Sqrt[1 + x] + (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)]/(2*Sqrt[2])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [C]  time = 0.01, size = 37, normalized size = 0.20 \[ -\frac {4}{3} \sqrt [4]{2} (1-x)^{3/4} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {1-x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 270, normalized size = 1.45 \[ \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x - 1\right )} \sqrt {\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + x - \sqrt {x + 1} \sqrt {-x + 1} - 1}{x - 1}} - \sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - x + 1}{x - 1}\right ) + \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x - 1\right )} \sqrt {-\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - x + \sqrt {x + 1} \sqrt {-x + 1} + 1}{x - 1}} - \sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + x - 1}{x - 1}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\frac {4 \, {\left (\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + x - \sqrt {x + 1} \sqrt {-x + 1} - 1\right )}}{x - 1}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\frac {4 \, {\left (\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - x + \sqrt {x + 1} \sqrt {-x + 1} + 1\right )}}{x - 1}\right ) - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)) + sqrt(2)*arctan((sqrt(2)*(x - 1)*sqrt(-(sq
rt(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)) - 1/4*sqrt(2)*log(4*(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) + x - sqrt(x + 1)*sqrt(-x
+ 1) - 1)/(x - 1)) + 1/4*sqrt(2)*log(-4*(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) - x + sqrt(x + 1)*sqrt(-x + 1)
+ 1)/(x - 1)) - (x + 1)^(1/4)*(-x + 1)^(3/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.47, size = 443, normalized size = 2.38 \[ \frac {\left (x +1\right )^{\frac {1}{4}} \left (x -1\right ) \left (\left (-x +1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}}}{\left (-\left (x -1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}} \left (-x +1\right )^{\frac {1}{4}}}+\frac {\left (-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-x^{3}+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 x^{2}+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}-x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )}{\left (x +1\right )^{2}}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {x^{3}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2} \RootOf \left (\textit {\_Z}^{4}+1\right )+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}+2 x^{2}+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x \RootOf \left (\textit {\_Z}^{4}+1\right )+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{2}+x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )}{\left (x +1\right )^{2}}\right )}{2}\right ) \left (\left (-x +1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}}}{\left (x +1\right )^{\frac {3}{4}} \left (-x +1\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x+1)^(1/4)*(x-1)/(-(x-1)*(x+1)^3)^(1/4)*((-x+1)*(x+1)^3)^(1/4)/(-x+1)^(1/4)+(-1/2*RootOf(_Z^4+1)^3*ln(-(RootO
f(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)*x^2+RootOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)*x+2*RootOf(_Z^4+1)^3*(-x^4-
2*x^3+2*x+1)^(1/4)*x+RootOf(_Z^4+1)*(-x^4-2*x^3+2*x+1)^(3/4)+RootOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)+RootOf(
_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)-x^3-2*x^2-x)/(x+1)^2)+1/2*RootOf(_Z^4+1)*ln((RootOf(_Z^4+1)^3*(-x^4-2*x^3+2
*x+1)^(3/4)+RootOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)*x+RootOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)+RootOf(_Z^4+
1)*(-x^4-2*x^3+2*x+1)^(1/4)*x^2+2*RootOf(_Z^4+1)*(-x^4-2*x^3+2*x+1)^(1/4)*x+x^3+RootOf(_Z^4+1)*(-x^4-2*x^3+2*x
+1)^(1/4)+2*x^2+x)/(x+1)^2))/(x+1)^(3/4)*((-x+1)*(x+1)^3)^(1/4)/(-x+1)^(1/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^(1/4)/(1 - x)^(1/4),x)

[Out]

int((x + 1)^(1/4)/(1 - x)^(1/4), x)

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sympy [C]  time = 2.17, size = 41, normalized size = 0.22 \[ \frac {2^{\frac {3}{4}} \left (x + 1\right )^{\frac {5}{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {\left (x + 1\right ) e^{2 i \pi }}{2}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(3/4)*(x + 1)**(5/4)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), (x + 1)*exp_polar(2*I*pi)/2)/(2*gamma(9/4))

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