3.9.37 \(\int \frac {x^2 \sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx\)

Optimal. Leaf size=234 \[ -\frac {1}{3} (1-x)^{3/4} x (x+1)^{5/4}-\frac {1}{12} (1-x)^{3/4} (x+1)^{5/4}-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{x+1}-\frac {3 \log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{16 \sqrt {2}}+\frac {3 \log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{16 \sqrt {2}}+\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )}{8 \sqrt {2}}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{8 \sqrt {2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {90, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {1}{3} (1-x)^{3/4} x (x+1)^{5/4}-\frac {1}{12} (1-x)^{3/4} (x+1)^{5/4}-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{x+1}-\frac {3 \log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{16 \sqrt {2}}+\frac {3 \log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{16 \sqrt {2}}+\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )}{8 \sqrt {2}}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 - x)^(3/4)*(1 + x)^(1/4))/8 - ((1 - x)^(3/4)*(1 + x)^(5/4))/12 - ((1 - x)^(3/4)*x*(1 + x)^(5/4))/3 + (3
*ArcTan[1 - (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)])/(8*Sqrt[2]) - (3*ArcTan[1 + (Sqrt[2]*(1 - x)^(1/4))/(1 + x
)^(1/4)])/(8*Sqrt[2]) - (3*Log[1 + Sqrt[1 - x]/Sqrt[1 + x] - (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)])/(16*Sqrt[
2]) + (3*Log[1 + Sqrt[1 - x]/Sqrt[1 + x] + (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)])/(16*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 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 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 {x^2 \sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx &=-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}-\frac {1}{3} \int \frac {\left (-1-\frac {x}{2}\right ) \sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx\\ &=-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}+\frac {3}{8} \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}+\frac {3}{16} \int \frac {1}{\sqrt [4]{1-x} (1+x)^{3/4}} \, dx\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-x}\right )\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}-\frac {3}{4} \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}-\frac {3}{16} \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 {3}{16} \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 {3 \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 )}{16 \sqrt {2}}-\frac {3 \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 )}{16 \sqrt {2}}\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}-\frac {3 \log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}+\frac {3 \log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}\\ &=-\frac {3}{8} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{12} (1-x)^{3/4} (1+x)^{5/4}-\frac {1}{3} (1-x)^{3/4} x (1+x)^{5/4}+\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}-\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}-\frac {3 \log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}+\frac {3 \log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 58, normalized size = 0.25 \begin {gather*} -\frac {1}{12} (1-x)^{3/4} \left (6 \sqrt [4]{2} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {1-x}{2}\right )+\sqrt [4]{x+1} \left (4 x^2+5 x+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((1 - x)^(3/4)*((1 + x)^(1/4)*(1 + 5*x + 4*x^2) + 6*2^(1/4)*Hypergeometric2F1[-1/4, 3/4, 7/4, (1 - x)/2]
))

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IntegrateAlgebraic [A]  time = 0.66, size = 146, normalized size = 0.62 \begin {gather*} \frac {1}{24} (1-x)^{3/4} \left (-8 (x+1)^{9/4}+6 (x+1)^{5/4}-9 \sqrt [4]{x+1}\right )+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x} \sqrt [4]{x+1}}{\sqrt {1-x}-\sqrt {x+1}}\right )}{8 \sqrt {2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-x} \sqrt [4]{x+1}}{\sqrt {1-x}+\sqrt {x+1}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - x)^(3/4)*(-9*(1 + x)^(1/4) + 6*(1 + x)^(5/4) - 8*(1 + x)^(9/4)))/24 + (3*ArcTan[(Sqrt[2]*(1 - x)^(1/4)*(
1 + x)^(1/4))/(Sqrt[1 - x] - Sqrt[1 + x])])/(8*Sqrt[2]) + (3*ArcTanh[(Sqrt[2]*(1 - x)^(1/4)*(1 + x)^(1/4))/(Sq
rt[1 - x] + Sqrt[1 + x])])/(8*Sqrt[2])

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fricas [A]  time = 1.26, size = 282, normalized size = 1.21 \begin {gather*} -\frac {1}{24} \, {\left (8 \, x^{2} + 10 \, x + 11\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + \frac {3}{8} \, \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 {3}{8} \, \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 {3}{32} \, \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 {3}{32} \, \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*x^2 + 10*x + 11)*(x + 1)^(1/4)*(-x + 1)^(3/4) + 3/8*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)) + 3/8*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)) - 3/32*sqrt(2)*log
(4*(sqrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) + x - sqrt(x + 1)*sqrt(-x + 1) - 1)/(x - 1)) + 3/32*sqrt(2)*log(-4*(s
qrt(2)*(x + 1)^(1/4)*(-x + 1)^(3/4) - x + sqrt(x + 1)*sqrt(-x + 1) + 1)/(x - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.56, size = 451, normalized size = 1.93 \begin {gather*} \frac {\left (8 x^{2}+10 x +11\right ) \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}}}{24 \left (-\left (x -1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}} \left (-x +1\right )^{\frac {1}{4}}}+\frac {\left (\frac {3 \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 )}{16}+\frac {3 \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 )}{16}\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*x^2+10*x+11)*(x+1)^(1/4)*(x-1)/(-(x-1)*(x+1)^3)^(1/4)*((-x+1)*(x+1)^3)^(1/4)/(-x+1)^(1/4)+(3/16*RootOf
(_Z^4+1)^3*ln((RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)*x^2+2*RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)*x-Roo
tOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)*x+RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)-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)^(3/4)+x^3+2*x^2+x)/(x+1)^2)+3/16*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 \begin {gather*} \int \frac {{\left (x + 1\right )}^{\frac {1}{4}} x^{2}}{{\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt [4]{x + 1}}{\sqrt [4]{1 - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(x + 1)**(1/4)/(1 - x)**(1/4), x)

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