3.260 \(\int \frac{x^2 (2-\sqrt{1+x^2})}{\sqrt{1+x^2} (1-x^3+(1+x^2)^{3/2})} \, dx\)

Optimal. Leaf size=136 \[ -\frac{x^2}{6}-\frac{1}{6} \sqrt{x^2+1} x+\frac{8 \sqrt{x^2+1}}{9}-\frac{7}{54} \log \left (3 x^2+2 x+3\right )+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{x+1}{\sqrt{2} \sqrt{x^2+1}}\right )+\frac{7}{27} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+1}}\right )+\frac{8 x}{9}+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{3 x+1}{2 \sqrt{2}}\right )-\frac{41}{54} \sinh ^{-1}(x) \]

[Out]

(8*x)/9 - x^2/6 + (8*Sqrt[1 + x^2])/9 - (x*Sqrt[1 + x^2])/6 - (41*ArcSinh[x])/54 + (4*Sqrt[2]*ArcTan[(1 + 3*x)
/(2*Sqrt[2])])/27 + (4*Sqrt[2]*ArcTan[(1 + x)/(Sqrt[2]*Sqrt[1 + x^2])])/27 + (7*ArcTanh[(1 - x)/(2*Sqrt[1 + x^
2])])/27 - (7*Log[3 + 2*x + 3*x^2])/54

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Rubi [A]  time = 1.49609, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 14, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6742, 195, 215, 634, 618, 204, 628, 1020, 12, 1081, 1037, 1031, 206, 261} \[ -\frac{x^2}{6}-\frac{1}{6} \sqrt{x^2+1} x+\frac{8 \sqrt{x^2+1}}{9}-\frac{7}{54} \log \left (3 x^2+2 x+3\right )+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{x+1}{\sqrt{2} \sqrt{x^2+1}}\right )+\frac{7}{27} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{x^2+1}}\right )+\frac{8 x}{9}+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{3 x+1}{2 \sqrt{2}}\right )-\frac{41}{54} \sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(2 - Sqrt[1 + x^2]))/(Sqrt[1 + x^2]*(1 - x^3 + (1 + x^2)^(3/2))),x]

[Out]

(8*x)/9 - x^2/6 + (8*Sqrt[1 + x^2])/9 - (x*Sqrt[1 + x^2])/6 - (41*ArcSinh[x])/54 + (4*Sqrt[2]*ArcTan[(1 + 3*x)
/(2*Sqrt[2])])/27 + (4*Sqrt[2]*ArcTan[(1 + x)/(Sqrt[2]*Sqrt[1 + x^2])])/27 + (7*ArcTanh[(1 - x)/(2*Sqrt[1 + x^
2])])/27 - (7*Log[3 + 2*x + 3*x^2])/54

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 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 1020

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(h*(a + c*x^2)^p*(d + e*x + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] + Dist[1/(2*f*(p + q + 1)), Int[(a + c*x^2)
^(p - 1)*(d + e*x + f*x^2)^q*Simp[a*h*e*p - a*(h*e - 2*g*f)*(p + q + 1) - 2*h*p*(c*d - a*f)*x - (h*c*e*p + c*(
h*e - 2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] && GtQ[
p, 0] && NeQ[p + q + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1081

Int[((A_.) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Symbol] :> Dist[
C/c, Int[1/Sqrt[d + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C - b*C*x)/((a + b*x + c*x^2)*Sqrt[d + f*x^2]), x]
, x] /; FreeQ[{a, b, c, d, f, A, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1037

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(c*d - a*f)^2 + b^2*d*f, 2]}, Dist[1/(2*q), Int[Simp[h*b*d - g*(c*d - a*f - q) + (h*(c*d - a*f + q) + g*
b*f)*x, x]/((a + b*x + c*x^2)*Sqrt[d + f*x^2]), x], x] - Dist[1/(2*q), Int[Simp[h*b*d - g*(c*d - a*f + q) + (h
*(c*d - a*f - q) + g*b*f)*x, x]/((a + b*x + c*x^2)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x
] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 1031

Int[((g_) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*
g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - b*d*x^2, x], x], x, Simp[g*b - 2*a*h - (b*h
- 2*g*c)*x, x]/Sqrt[d + f*x^2]], x] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[b*h^2*d -
2*g*h*(c*d - a*f) - g^2*b*f, 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{x^2 \left (2-\sqrt{1+x^2}\right )}{\sqrt{1+x^2} \left (1-x^3+\left (1+x^2\right )^{3/2}\right )} \, dx &=\int \left (-\frac{x^2}{1-x^3+\sqrt{1+x^2}+x^2 \sqrt{1+x^2}}-\frac{2 x^2}{\sqrt{1+x^2} \left (-1+x^3-\left (1+x^2\right )^{3/2}\right )}\right ) \, dx\\ &=-\left (2 \int \frac{x^2}{\sqrt{1+x^2} \left (-1+x^3-\left (1+x^2\right )^{3/2}\right )} \, dx\right )-\int \frac{x^2}{1-x^3+\sqrt{1+x^2}+x^2 \sqrt{1+x^2}} \, dx\\ &=-\left (2 \int \left (-\frac{1}{3}+\frac{2}{9 \sqrt{1+x^2}}-\frac{x}{3 \sqrt{1+x^2}}+\frac{2 x}{3 \left (3+2 x+3 x^2\right )}+\frac{3+5 x}{9 \sqrt{1+x^2} \left (3+2 x+3 x^2\right )}\right ) \, dx\right )-\int \left (-\frac{2}{9}+\frac{x}{3}+\frac{\sqrt{1+x^2}}{3}+\frac{-3-5 x}{9 \left (3+2 x+3 x^2\right )}-\frac{2 x \sqrt{1+x^2}}{3 \left (3+2 x+3 x^2\right )}\right ) \, dx\\ &=\frac{8 x}{9}-\frac{x^2}{6}-\frac{1}{9} \int \frac{-3-5 x}{3+2 x+3 x^2} \, dx-\frac{2}{9} \int \frac{3+5 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx-\frac{1}{3} \int \sqrt{1+x^2} \, dx-\frac{4}{9} \int \frac{1}{\sqrt{1+x^2}} \, dx+\frac{2}{3} \int \frac{x}{\sqrt{1+x^2}} \, dx+\frac{2}{3} \int \frac{x \sqrt{1+x^2}}{3+2 x+3 x^2} \, dx-\frac{4}{3} \int \frac{x}{3+2 x+3 x^2} \, dx\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{4}{9} \sinh ^{-1}(x)+\frac{1}{18} \int \frac{4-4 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx-\frac{1}{18} \int \frac{16+16 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx+\frac{5}{54} \int \frac{2+6 x}{3+2 x+3 x^2} \, dx+\frac{4}{27} \int \frac{1}{3+2 x+3 x^2} \, dx-\frac{1}{6} \int \frac{1}{\sqrt{1+x^2}} \, dx-\frac{2}{9} \int \frac{2+6 x}{3+2 x+3 x^2} \, dx+\frac{2}{9} \int -\frac{2 x^2}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx+\frac{4}{9} \int \frac{1}{3+2 x+3 x^2} \, dx\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{11}{18} \sinh ^{-1}(x)-\frac{7}{54} \log \left (3+2 x+3 x^2\right )-\frac{8}{27} \operatorname{Subst}\left (\int \frac{1}{-32-x^2} \, dx,x,2+6 x\right )-\frac{4}{9} \int \frac{x^2}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx-\frac{8}{9} \operatorname{Subst}\left (\int \frac{1}{-32-x^2} \, dx,x,2+6 x\right )-\frac{128}{9} \operatorname{Subst}\left (\int \frac{1}{-4096-2 x^2} \, dx,x,\frac{32+32 x}{\sqrt{1+x^2}}\right )-\frac{1024}{9} \operatorname{Subst}\left (\int \frac{1}{32768-2 x^2} \, dx,x,\frac{-64+64 x}{\sqrt{1+x^2}}\right )\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{11}{18} \sinh ^{-1}(x)+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{1+3 x}{2 \sqrt{2}}\right )+\frac{1}{9} \sqrt{2} \tan ^{-1}\left (\frac{1+x}{\sqrt{2} \sqrt{1+x^2}}\right )+\frac{4}{9} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x^2}}\right )-\frac{7}{54} \log \left (3+2 x+3 x^2\right )-\frac{4}{27} \int \frac{1}{\sqrt{1+x^2}} \, dx-\frac{4}{27} \int \frac{-3-2 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{41}{54} \sinh ^{-1}(x)+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{1+3 x}{2 \sqrt{2}}\right )+\frac{1}{9} \sqrt{2} \tan ^{-1}\left (\frac{1+x}{\sqrt{2} \sqrt{1+x^2}}\right )+\frac{4}{9} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x^2}}\right )-\frac{7}{54} \log \left (3+2 x+3 x^2\right )-\frac{1}{27} \int \frac{-10-10 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx+\frac{1}{27} \int \frac{2-2 x}{\sqrt{1+x^2} \left (3+2 x+3 x^2\right )} \, dx\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{41}{54} \sinh ^{-1}(x)+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{1+3 x}{2 \sqrt{2}}\right )+\frac{1}{9} \sqrt{2} \tan ^{-1}\left (\frac{1+x}{\sqrt{2} \sqrt{1+x^2}}\right )+\frac{4}{9} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x^2}}\right )-\frac{7}{54} \log \left (3+2 x+3 x^2\right )-\frac{64}{27} \operatorname{Subst}\left (\int \frac{1}{-1024-2 x^2} \, dx,x,\frac{16+16 x}{\sqrt{1+x^2}}\right )-\frac{800}{27} \operatorname{Subst}\left (\int \frac{1}{12800-2 x^2} \, dx,x,\frac{40-40 x}{\sqrt{1+x^2}}\right )\\ &=\frac{8 x}{9}-\frac{x^2}{6}+\frac{8 \sqrt{1+x^2}}{9}-\frac{1}{6} x \sqrt{1+x^2}-\frac{41}{54} \sinh ^{-1}(x)+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{1+3 x}{2 \sqrt{2}}\right )+\frac{4}{27} \sqrt{2} \tan ^{-1}\left (\frac{1+x}{\sqrt{2} \sqrt{1+x^2}}\right )+\frac{7}{27} \tanh ^{-1}\left (\frac{1-x}{2 \sqrt{1+x^2}}\right )-\frac{7}{54} \log \left (3+2 x+3 x^2\right )\\ \end{align*}

Mathematica [C]  time = 0.862711, size = 261, normalized size = 1.92 \[ \frac{1}{162} \left (-27 x^2-27 \sqrt{x^2+1} x+144 \sqrt{x^2+1}-21 \log \left (3 x^2+2 x+3\right )+\sqrt{1-2 i \sqrt{2}} \left (11 \sqrt{2}-i\right ) \tanh ^{-1}\left (\frac{-2 i \sqrt{2} x-x+3}{\sqrt{2+4 i \sqrt{2}} \sqrt{x^2+1}}\right )+11 \sqrt{2+4 i \sqrt{2}} \tanh ^{-1}\left (\frac{2 i \sqrt{2} x-x+3}{\sqrt{2-4 i \sqrt{2}} \sqrt{x^2+1}}\right )+i \sqrt{1+2 i \sqrt{2}} \tanh ^{-1}\left (\frac{2 i \sqrt{2} x-x+3}{\sqrt{2-4 i \sqrt{2}} \sqrt{x^2+1}}\right )+144 x+24 \sqrt{2} \tan ^{-1}\left (\frac{3 x+1}{2 \sqrt{2}}\right )-123 \sinh ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(2 - Sqrt[1 + x^2]))/(Sqrt[1 + x^2]*(1 - x^3 + (1 + x^2)^(3/2))),x]

[Out]

(144*x - 27*x^2 + 144*Sqrt[1 + x^2] - 27*x*Sqrt[1 + x^2] - 123*ArcSinh[x] + 24*Sqrt[2]*ArcTan[(1 + 3*x)/(2*Sqr
t[2])] + Sqrt[1 - (2*I)*Sqrt[2]]*(-I + 11*Sqrt[2])*ArcTanh[(3 - x - (2*I)*Sqrt[2]*x)/(Sqrt[2 + (4*I)*Sqrt[2]]*
Sqrt[1 + x^2])] + I*Sqrt[1 + (2*I)*Sqrt[2]]*ArcTanh[(3 - x + (2*I)*Sqrt[2]*x)/(Sqrt[2 - (4*I)*Sqrt[2]]*Sqrt[1
+ x^2])] + 11*Sqrt[2 + (4*I)*Sqrt[2]]*ArcTanh[(3 - x + (2*I)*Sqrt[2]*x)/(Sqrt[2 - (4*I)*Sqrt[2]]*Sqrt[1 + x^2]
)] - 21*Log[3 + 2*x + 3*x^2])/162

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Maple [B]  time = 0.044, size = 654, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(2-(x^2+1)^(1/2))/(1-x^3+(x^2+1)^(3/2))/(x^2+1)^(1/2),x)

[Out]

8/9*x-1/6*x^2-7/54*ln(3*x^2+2*x+3)+4/27*2^(1/2)*arctan(1/8*(6*x+2)*2^(1/2))-41/54*arcsinh(x)-1/12*2^(1/2)*(2*(
1+x)^2/(1-x)^2+2)^(1/2)*(-2^(1/2)*arctan(1/2*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)/((1+x)^2/(1-x)^2+1)*(1+x)/(1-
x))+5*arctanh((2*(1+x)^2/(1-x)^2+2)^(1/2)))/(((1+x)^2/(1-x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))+3/8*
2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)*(-2^(1/2)*arctan(1/2*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)/((1+x)^2/(1-x)^2+
1)*(1+x)/(1-x))+arctanh((2*(1+x)^2/(1-x)^2+2)^(1/2)))/(((1+x)^2/(1-x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(
1-x))-1/6*x*(x^2+1)^(1/2)+8/9*(x^2+1)^(1/2)+1/216*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)*(13*2^(1/2)*arctan(1/2*2
^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)/((1+x)^2/(1-x)^2+1)*(1+x)/(1-x))+43*arctanh((2*(1+x)^2/(1-x)^2+2)^(1/2)))/(
((1+x)^2/(1-x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))-1/36*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)*(-11*2^(
1/2)*arctan(1/2*2^(1/2)*(2*(1+x)^2/(1-x)^2+2)^(1/2)/((1+x)^2/(1-x)^2+1)*(1+x)/(1-x))+arctanh((2*(1+x)^2/(1-x)^
2+2)^(1/2)))/(((1+x)^2/(1-x)^2+1)/(1+(1+x)/(1-x))^2)^(1/2)/(1+(1+x)/(1-x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{x}{2 \,{\left (x^{2} + 1\right )}} + \frac{1}{2} \, \arctan \left (x\right ) + \int -\frac{3 \, x^{10} - 4 \, x^{9} + 5 \, x^{8} - 2 \, x^{7} + 15 \, x^{6} + 6 \, x^{5} + 9 \, x^{4}}{2 \,{\left (2 \, x^{13} + 7 \, x^{11} - 4 \, x^{10} + 11 \, x^{9} - 11 \, x^{8} + 13 \, x^{7} - 13 \, x^{6} + 11 \, x^{5} - 11 \, x^{4} + 4 \, x^{3} - 7 \, x^{2} - 2 \,{\left (x^{12} + 3 \, x^{10} - 2 \, x^{9} + 3 \, x^{8} - 6 \, x^{7} + 2 \, x^{6} - 6 \, x^{5} + 3 \, x^{4} - 2 \, x^{3} + 3 \, x^{2} + 1\right )} \sqrt{x^{2} + 1} - 2\right )}}\,{d x} + \frac{1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{6} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2-(x^2+1)^(1/2))/(1-x^3+(x^2+1)^(3/2))/(x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-1/2*x/(x^2 + 1) + 1/2*arctan(x) + integrate(-1/2*(3*x^10 - 4*x^9 + 5*x^8 - 2*x^7 + 15*x^6 + 6*x^5 + 9*x^4)/(2
*x^13 + 7*x^11 - 4*x^10 + 11*x^9 - 11*x^8 + 13*x^7 - 13*x^6 + 11*x^5 - 11*x^4 + 4*x^3 - 7*x^2 - 2*(x^12 + 3*x^
10 - 2*x^9 + 3*x^8 - 6*x^7 + 2*x^6 - 6*x^5 + 3*x^4 - 2*x^3 + 3*x^2 + 1)*sqrt(x^2 + 1) - 2), x) + 1/6*log(x^2 +
x + 1) + 1/6*log(x - 1)

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Fricas [A]  time = 2.00707, size = 541, normalized size = 3.98 \begin{align*} -\frac{1}{6} \, x^{2} - \frac{1}{18} \, \sqrt{x^{2} + 1}{\left (3 \, x - 16\right )} + \frac{4}{27} \, \sqrt{2} \arctan \left (\frac{1}{4} \, \sqrt{2}{\left (3 \, x + 1\right )}\right ) + \frac{4}{27} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (3 \, x - 1\right )} + \frac{3}{2} \, \sqrt{2} \sqrt{x^{2} + 1}\right ) - \frac{4}{27} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (x + 1\right )} + \frac{1}{2} \, \sqrt{2} \sqrt{x^{2} + 1}\right ) + \frac{8}{9} \, x + \frac{7}{54} \, \log \left (3 \, x^{2} - \sqrt{x^{2} + 1}{\left (3 \, x - 1\right )} - x + 2\right ) - \frac{7}{54} \, \log \left (3 \, x^{2} + 2 \, x + 3\right ) - \frac{7}{54} \, \log \left (x^{2} - \sqrt{x^{2} + 1}{\left (x + 1\right )} + x + 2\right ) + \frac{41}{54} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2-(x^2+1)^(1/2))/(1-x^3+(x^2+1)^(3/2))/(x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/6*x^2 - 1/18*sqrt(x^2 + 1)*(3*x - 16) + 4/27*sqrt(2)*arctan(1/4*sqrt(2)*(3*x + 1)) + 4/27*sqrt(2)*arctan(-1
/2*sqrt(2)*(3*x - 1) + 3/2*sqrt(2)*sqrt(x^2 + 1)) - 4/27*sqrt(2)*arctan(-1/2*sqrt(2)*(x + 1) + 1/2*sqrt(2)*sqr
t(x^2 + 1)) + 8/9*x + 7/54*log(3*x^2 - sqrt(x^2 + 1)*(3*x - 1) - x + 2) - 7/54*log(3*x^2 + 2*x + 3) - 7/54*log
(x^2 - sqrt(x^2 + 1)*(x + 1) + x + 2) + 41/54*log(-x + sqrt(x^2 + 1))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(2-(x**2+1)**(1/2))/(1-x**3+(x**2+1)**(3/2))/(x**2+1)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.10647, size = 238, normalized size = 1.75 \begin{align*} -\frac{1}{6} \, x^{2} - \frac{1}{18} \, \sqrt{x^{2} + 1}{\left (3 \, x - 16\right )} + \frac{4}{27} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (3 \, x - 3 \, \sqrt{x^{2} + 1} - 1\right )}\right ) + \frac{4}{27} \, \sqrt{2} \arctan \left (\frac{1}{4} \, \sqrt{2}{\left (3 \, x + 1\right )}\right ) - \frac{4}{27} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (x - \sqrt{x^{2} + 1} + 1\right )}\right ) + \frac{8}{9} \, x + \frac{7}{54} \, \log \left (3 \,{\left (x - \sqrt{x^{2} + 1}\right )}^{2} - 2 \, x + 2 \, \sqrt{x^{2} + 1} + 1\right ) - \frac{7}{54} \, \log \left ({\left (x - \sqrt{x^{2} + 1}\right )}^{2} + 2 \, x - 2 \, \sqrt{x^{2} + 1} + 3\right ) - \frac{7}{54} \, \log \left (3 \, x^{2} + 2 \, x + 3\right ) + \frac{41}{54} \, \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2-(x^2+1)^(1/2))/(1-x^3+(x^2+1)^(3/2))/(x^2+1)^(1/2),x, algorithm="giac")

[Out]

-1/6*x^2 - 1/18*sqrt(x^2 + 1)*(3*x - 16) + 4/27*sqrt(2)*arctan(-1/2*sqrt(2)*(3*x - 3*sqrt(x^2 + 1) - 1)) + 4/2
7*sqrt(2)*arctan(1/4*sqrt(2)*(3*x + 1)) - 4/27*sqrt(2)*arctan(-1/2*sqrt(2)*(x - sqrt(x^2 + 1) + 1)) + 8/9*x +
7/54*log(3*(x - sqrt(x^2 + 1))^2 - 2*x + 2*sqrt(x^2 + 1) + 1) - 7/54*log((x - sqrt(x^2 + 1))^2 + 2*x - 2*sqrt(
x^2 + 1) + 3) - 7/54*log(3*x^2 + 2*x + 3) + 41/54*log(-x + sqrt(x^2 + 1))