\(\int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} (\sqrt {1-x}-\sqrt {1+x})} \, dx\) [221]

Optimal result

Integrand size = 52, antiderivative size = 304 \[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=\frac {5}{16} (1-x)^{3/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {7 \left (1-x^2\right )^{5/4}}{24 \sqrt {1-x}}+\frac {x \left (1-x^2\right )^{5/4}}{6 \sqrt {1-x}}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}-\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}} \]

[Out]

5/16*(1-x)^(3/4)*(1+x)^(1/4)-1/16*(1-x)^(1/4)*(1+x)^(3/4)+1/24*(1-x)^(5/4)*(1+x)^(3/4)+3/16*arctan(-1+(1-x)^(1
/4)*2^(1/2)/(1+x)^(1/4))*2^(1/2)+3/16*arctan(1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4))*2^(1/2)+1/16*ln(1-(1-x)^(1/4)*
2^(1/2)/(1+x)^(1/4)+(1-x)^(1/2)/(1+x)^(1/2))*2^(1/2)-1/16*ln(1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4)+(1-x)^(1/2)/(1+
x)^(1/2))*2^(1/2)+7/24*(-x^2+1)^(5/4)/(1-x)^(1/2)+1/6*x*(-x^2+1)^(5/4)/(1-x)^(1/2)+1/6*(-x^2+1)^(5/4)*(1+x)^(1
/2)

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.05, number of steps used = 33, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2128, 809, 689, 52, 65, 246, 217, 1179, 642, 1176, 631, 210, 1647, 807, 338, 303} \[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )}{8 \sqrt {2}}+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{8 \sqrt {2}}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}+\frac {1}{6} \sqrt {x+1} \left (1-x^2\right )^{5/4}+\frac {1}{6} (1-x)^{7/4} (x+1)^{5/4}+\frac {1}{24} (1-x)^{5/4} (x+1)^{3/4}-\frac {1}{16} \sqrt [4]{1-x} (x+1)^{3/4}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{x+1}-\frac {5}{48} (1-x)^{3/4} \sqrt [4]{x+1}+\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{8 \sqrt {2}}-\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{8 \sqrt {2}} \]

[In]

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

[Out]

(-5*(1 - x)^(3/4)*(1 + x)^(1/4))/48 + (5*(1 - x)^(7/4)*(1 + x)^(1/4))/24 - ((1 - x)^(1/4)*(1 + x)^(3/4))/16 +
((1 - x)^(5/4)*(1 + x)^(3/4))/24 + ((1 - x)^(7/4)*(1 + x)^(5/4))/6 + (Sqrt[1 + x]*(1 - x^2)^(5/4))/6 + (1 - x^
2)^(9/4)/(3*(1 - x)^(3/2)) - (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]) + Log[1 + Sqrt[1 - x]/Sqrt[1 + x] - (Sqrt[2]*(1 - x)^(1/4
))/(1 + x)^(1/4)]/(8*Sqrt[2]) - Log[1 + Sqrt[1 - x]/Sqrt[1 + x] + (Sqrt[2]*(1 - x)^(1/4))/(1 + x)^(1/4)]/(8*Sq
rt[2])

Rule 52

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 65

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 210

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

Rule 217

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 246

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 303

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 338

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 631

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 642

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 689

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p,
x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && GtQ[a, 0] && GtQ[d, 0] &&  !I
GtQ[m, 0]

Rule 807

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d
+ e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 0]

Rule 809

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*
((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 1176

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 1179

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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 2128

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[c/(e*(b*c - a*d)
), Int[(u*Sqrt[a + b*x])/x, x], x] - Dist[a/(f*(b*c - a*d)), Int[(u*Sqrt[c + d*x])/x, x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a*e^2 - c*f^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int x \sqrt {1+x} \sqrt [4]{1-x^2} \, dx\right )-\frac {1}{2} \int \frac {x (1+x) \sqrt [4]{1-x^2}}{\sqrt {1-x}} \, dx \\ & = \frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}-\frac {1}{12} \int \sqrt {1+x} \sqrt [4]{1-x^2} \, dx-\frac {1}{2} \int \frac {x \left (1-x^2\right )^{5/4}}{(1-x)^{3/2}} \, dx \\ & = \frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {1}{12} \int \sqrt [4]{1-x} (1+x)^{3/4} \, dx-\frac {1}{2} \int \frac {\left (1-x^2\right )^{5/4}}{\sqrt {1-x}} \, dx \\ & = \frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {1}{16} \int \frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}} \, dx-\frac {1}{2} \int (1-x)^{3/4} (1+x)^{5/4} \, dx \\ & = -\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {1}{32} \int \frac {1}{(1-x)^{3/4} \sqrt [4]{1+x}} \, dx-\frac {5}{12} \int (1-x)^{3/4} \sqrt [4]{1+x} \, dx \\ & = \frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {5}{48} \int \frac {(1-x)^{3/4}}{(1+x)^{3/4}} \, dx+\frac {1}{8} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-x}\right ) \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\frac {5}{32} \int \frac {1}{\sqrt [4]{1-x} (1+x)^{3/4}} \, dx \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}+\frac {1}{16} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {1}{16} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {5}{8} \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-x}\right ) \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}+\frac {1}{32} \text {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}{32} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {5}{8} \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\frac {\text {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 )}{32 \sqrt {2}}-\frac {\text {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 )}{32 \sqrt {2}} \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{32 \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 )}{32 \sqrt {2}}-\frac {5}{16} \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {5}{16} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}} \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \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 )}{32 \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 )}{32 \sqrt {2}}+\frac {5}{32} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {5}{32} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\frac {5 \text {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 )}{32 \sqrt {2}}+\frac {5 \text {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 )}{32 \sqrt {2}} \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \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 )}{8 \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 )}{8 \sqrt {2}}+\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}}-\frac {5 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{16 \sqrt {2}} \\ & = -\frac {5}{48} (1-x)^{3/4} \sqrt [4]{1+x}+\frac {5}{24} (1-x)^{7/4} \sqrt [4]{1+x}-\frac {1}{16} \sqrt [4]{1-x} (1+x)^{3/4}+\frac {1}{24} (1-x)^{5/4} (1+x)^{3/4}+\frac {1}{6} (1-x)^{7/4} (1+x)^{5/4}+\frac {1}{6} \sqrt {1+x} \left (1-x^2\right )^{5/4}+\frac {\left (1-x^2\right )^{9/4}}{3 (1-x)^{3/2}}-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \sqrt {2}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{8 \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 )}{8 \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 )}{8 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.24 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.52 \[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=-\frac {1}{48} \sqrt {1+x} \sqrt [4]{1-x^2} \left (-7+2 x+8 x^2-\frac {\sqrt {1-x^2} \left (29+22 x+8 x^2\right )}{1+x}\right )+\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt {1+x} \sqrt [4]{1-x^2}}{1+x-\sqrt {1-x^2}}\right )-2 \text {arctanh}\left (\frac {1+x+\sqrt {1-x^2}}{\sqrt {2} \sqrt {1+x} \sqrt [4]{1-x^2}}\right )}{8 \sqrt {2}} \]

[In]

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

[Out]

-1/48*(Sqrt[1 + x]*(1 - x^2)^(1/4)*(-7 + 2*x + 8*x^2 - (Sqrt[1 - x^2]*(29 + 22*x + 8*x^2))/(1 + x))) + (3*ArcT
an[(Sqrt[2]*Sqrt[1 + x]*(1 - x^2)^(1/4))/(1 + x - Sqrt[1 - x^2])] - 2*ArcTanh[(1 + x + Sqrt[1 - x^2])/(Sqrt[2]
*Sqrt[1 + x]*(1 - x^2)^(1/4))])/(8*Sqrt[2])

Maple [F]

\[\int \frac {x^{2} \left (-x^{2}+1\right )^{\frac {1}{4}} \sqrt {1+x}}{\sqrt {1-x}\, \left (\sqrt {1-x}-\sqrt {1+x}\right )}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.21 \[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=-\frac {1}{48} \, {\left (8 \, x^{2} + 2 \, x - 7\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1} + \frac {1}{48} \, {\left (8 \, x^{2} + 22 \, x + 29\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {-x + 1} + \left (\frac {1}{64} i + \frac {1}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x + i + 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1}}{x + 1}\right ) - \left (\frac {1}{64} i - \frac {1}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x - i + 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1}}{x + 1}\right ) + \left (\frac {1}{64} i - \frac {1}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x + i - 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1}}{x + 1}\right ) - \left (\frac {1}{64} i + \frac {1}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x - i - 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1}}{x + 1}\right ) + \left (\frac {5}{64} i + \frac {5}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x - i - 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {-x + 1}}{x - 1}\right ) - \left (\frac {5}{64} i - \frac {5}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x + i - 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {-x + 1}}{x - 1}\right ) + \left (\frac {5}{64} i - \frac {5}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x - i + 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {-x + 1}}{x - 1}\right ) - \left (\frac {5}{64} i + \frac {5}{64}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x + i + 1\right )} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {-x + 1}}{x - 1}\right ) \]

[In]

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

[Out]

-1/48*(8*x^2 + 2*x - 7)*(-x^2 + 1)^(1/4)*sqrt(x + 1) + 1/48*(8*x^2 + 22*x + 29)*(-x^2 + 1)^(1/4)*sqrt(-x + 1)
+ (1/64*I + 1/64)*sqrt(2)*log((sqrt(2)*((I + 1)*x + I + 1) + 2*(-x^2 + 1)^(1/4)*sqrt(x + 1))/(x + 1)) - (1/64*
I - 1/64)*sqrt(2)*log((sqrt(2)*(-(I - 1)*x - I + 1) + 2*(-x^2 + 1)^(1/4)*sqrt(x + 1))/(x + 1)) + (1/64*I - 1/6
4)*sqrt(2)*log((sqrt(2)*((I - 1)*x + I - 1) + 2*(-x^2 + 1)^(1/4)*sqrt(x + 1))/(x + 1)) - (1/64*I + 1/64)*sqrt(
2)*log((sqrt(2)*(-(I + 1)*x - I - 1) + 2*(-x^2 + 1)^(1/4)*sqrt(x + 1))/(x + 1)) + (5/64*I + 5/64)*sqrt(2)*log(
(sqrt(2)*((I + 1)*x - I - 1) + 2*(-x^2 + 1)^(1/4)*sqrt(-x + 1))/(x - 1)) - (5/64*I - 5/64)*sqrt(2)*log((sqrt(2
)*(-(I - 1)*x + I - 1) + 2*(-x^2 + 1)^(1/4)*sqrt(-x + 1))/(x - 1)) + (5/64*I - 5/64)*sqrt(2)*log((sqrt(2)*((I
- 1)*x - I + 1) + 2*(-x^2 + 1)^(1/4)*sqrt(-x + 1))/(x - 1)) - (5/64*I + 5/64)*sqrt(2)*log((sqrt(2)*(-(I + 1)*x
+ I + 1) + 2*(-x^2 + 1)^(1/4)*sqrt(-x + 1))/(x - 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=\int { -\frac {{\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1} x^{2}}{\sqrt {-x + 1} {\left (\sqrt {x + 1} - \sqrt {-x + 1}\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=\int { -\frac {{\left (-x^{2} + 1\right )}^{\frac {1}{4}} \sqrt {x + 1} x^{2}}{\sqrt {-x + 1} {\left (\sqrt {x + 1} - \sqrt {-x + 1}\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \sqrt {1+x} \sqrt [4]{1-x^2}}{\sqrt {1-x} \left (\sqrt {1-x}-\sqrt {1+x}\right )} \, dx=-\int \frac {x^2\,{\left (1-x^2\right )}^{1/4}\,\sqrt {x+1}}{\left (\sqrt {x+1}-\sqrt {1-x}\right )\,\sqrt {1-x}} \,d x \]

[In]

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

[Out]

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