\(\int \frac {\sqrt {1+x} (-1+x^2)}{(1+x^2) \sqrt {x+\sqrt {1+x}}} \, dx\) [2924]

Optimal result

Integrand size = 33, antiderivative size = 337 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=-\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-\frac {7}{4} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right )+8 \text {RootSum}\left [625+1000 \text {$\#$1}+300 \text {$\#$1}^2+120 \text {$\#$1}^3+470 \text {$\#$1}^4+24 \text {$\#$1}^5+12 \text {$\#$1}^6+8 \text {$\#$1}^7+\text {$\#$1}^8\&,\frac {25 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}+20 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^2+14 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^3+4 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^4+\log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^5}{125+75 \text {$\#$1}+45 \text {$\#$1}^2+235 \text {$\#$1}^3+15 \text {$\#$1}^4+9 \text {$\#$1}^5+7 \text {$\#$1}^6+\text {$\#$1}^7}\&\right ] \]

[Out]

Unintegrable

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.89 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.25, number of steps used = 19, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {6860, 756, 654, 635, 212, 998, 738, 210} \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=-\frac {(1+i) \arctan \left (\frac {-2 \left ((-2+2 i)+\sqrt {1-i}\right ) \sqrt {x+1}+4 \sqrt {1-i}+(2-2 i)}{4 \sqrt {(1+i)+(1-i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1+i)+(1-i)^{3/2}}}-\frac {(1+i) \arctan \left (\frac {2 \left ((2-2 i)+\sqrt {1-i}\right ) \sqrt {x+1}-4 \sqrt {1-i}+(2-2 i)}{4 \sqrt {(1+i)-(1-i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1+i)-(1-i)^{3/2}}}-\frac {(1-i) \arctan \left (\frac {-2 \left ((-2-2 i)+\sqrt {1+i}\right ) \sqrt {x+1}+4 \sqrt {1+i}+(2+2 i)}{4 \sqrt {(1-i)+(1+i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1-i)+(1+i)^{3/2}}}-\frac {(1-i) \arctan \left (\frac {2 \left ((2+2 i)+\sqrt {1+i}\right ) \sqrt {x+1}-4 \sqrt {1+i}+(2+2 i)}{4 \sqrt {(1-i)-(1+i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1-i)-(1+i)^{3/2}}}+\frac {7}{4} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )+\sqrt {x+1} \sqrt {x+\sqrt {x+1}}-\frac {3}{2} \sqrt {x+\sqrt {x+1}} \]

[In]

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

[Out]

(-3*Sqrt[x + Sqrt[1 + x]])/2 + Sqrt[1 + x]*Sqrt[x + Sqrt[1 + x]] - ((1 + I)*ArcTan[((2 - 2*I) + 4*Sqrt[1 - I]
- 2*((-2 + 2*I) + Sqrt[1 - I])*Sqrt[1 + x])/(4*Sqrt[(1 + I) + (1 - I)^(3/2)]*Sqrt[x + Sqrt[1 + x]])])/Sqrt[(1
+ I) + (1 - I)^(3/2)] - ((1 + I)*ArcTan[((2 - 2*I) - 4*Sqrt[1 - I] + 2*((2 - 2*I) + Sqrt[1 - I])*Sqrt[1 + x])/
(4*Sqrt[(1 + I) - (1 - I)^(3/2)]*Sqrt[x + Sqrt[1 + x]])])/Sqrt[(1 + I) - (1 - I)^(3/2)] - ((1 - I)*ArcTan[((2
+ 2*I) + 4*Sqrt[1 + I] - 2*((-2 - 2*I) + Sqrt[1 + I])*Sqrt[1 + x])/(4*Sqrt[(1 - I) + (1 + I)^(3/2)]*Sqrt[x + S
qrt[1 + x]])])/Sqrt[(1 - I) + (1 + I)^(3/2)] - ((1 - I)*ArcTan[((2 + 2*I) - 4*Sqrt[1 + I] + 2*((2 + 2*I) + Sqr
t[1 + I])*Sqrt[1 + x])/(4*Sqrt[(1 - I) - (1 + I)^(3/2)]*Sqrt[x + Sqrt[1 + x]])])/Sqrt[(1 - I) - (1 + I)^(3/2)]
+ (7*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/4

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 212

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

Rule 635

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

Rule 654

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

Rule 738

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

Rule 756

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 998

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

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^4 \left (-2+x^2\right )}{\sqrt {-1+x+x^2} \left (2-2 x^2+x^4\right )} \, dx,x,\sqrt {1+x}\right ) \\ & = 2 \text {Subst}\left (\int \left (\frac {x^2}{\sqrt {-1+x+x^2}}-\frac {2 x^2}{\sqrt {-1+x+x^2} \left (2-2 x^2+x^4\right )}\right ) \, dx,x,\sqrt {1+x}\right ) \\ & = 2 \text {Subst}\left (\int \frac {x^2}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )-4 \text {Subst}\left (\int \frac {x^2}{\sqrt {-1+x+x^2} \left (2-2 x^2+x^4\right )} \, dx,x,\sqrt {1+x}\right ) \\ & = \sqrt {1+x} \sqrt {x+\sqrt {1+x}}-4 \text {Subst}\left (\int \left (\frac {1-i}{\sqrt {-1+x+x^2} \left ((-2-2 i)+2 x^2\right )}+\frac {1+i}{\sqrt {-1+x+x^2} \left ((-2+2 i)+2 x^2\right )}\right ) \, dx,x,\sqrt {1+x}\right )+\text {Subst}\left (\int \frac {1-\frac {3 x}{2}}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right ) \\ & = -\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}+\frac {7}{4} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )-(4-4 i) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2} \left ((-2-2 i)+2 x^2\right )} \, dx,x,\sqrt {1+x}\right )-(4+4 i) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2} \left ((-2+2 i)+2 x^2\right )} \, dx,x,\sqrt {1+x}\right ) \\ & = -\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-(2-2 i) \text {Subst}\left (\int \frac {1}{\left ((-2-2 i)-2 \sqrt {1+i} x\right ) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )-(2-2 i) \text {Subst}\left (\int \frac {1}{\left ((-2-2 i)+2 \sqrt {1+i} x\right ) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )-(2+2 i) \text {Subst}\left (\int \frac {1}{\left ((-2+2 i)-2 \sqrt {1-i} x\right ) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )-(2+2 i) \text {Subst}\left (\int \frac {1}{\left ((-2+2 i)+2 \sqrt {1-i} x\right ) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )+\frac {7}{2} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right ) \\ & = -\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}+\frac {7}{4} \text {arctanh}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )-(-4-4 i) \text {Subst}\left (\int \frac {1}{(-16-16 i)-16 (1-i)^{3/2}-x^2} \, dx,x,\frac {(2-2 i)+4 \sqrt {1-i}-\left ((-4+4 i)+2 \sqrt {1-i}\right ) \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )-(-4-4 i) \text {Subst}\left (\int \frac {1}{(-16-16 i)+16 (1-i)^{3/2}-x^2} \, dx,x,\frac {(2-2 i)-4 \sqrt {1-i}-\left ((-4+4 i)-2 \sqrt {1-i}\right ) \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )-(-4+4 i) \text {Subst}\left (\int \frac {1}{(-16+16 i)-16 (1+i)^{3/2}-x^2} \, dx,x,\frac {(2+2 i)+4 \sqrt {1+i}-\left ((-4-4 i)+2 \sqrt {1+i}\right ) \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )-(-4+4 i) \text {Subst}\left (\int \frac {1}{(-16+16 i)+16 (1+i)^{3/2}-x^2} \, dx,x,\frac {(2+2 i)-4 \sqrt {1+i}-\left ((-4-4 i)-2 \sqrt {1+i}\right ) \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right ) \\ & = -\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-\frac {(1+i) \arctan \left (\frac {(2-2 i)+4 \sqrt {1-i}-2 \left ((-2+2 i)+\sqrt {1-i}\right ) \sqrt {1+x}}{4 \sqrt {(1+i)+(1-i)^{3/2}} \sqrt {x+\sqrt {1+x}}}\right )}{\sqrt {(1+i)+(1-i)^{3/2}}}-\frac {(1+i) \arctan \left (\frac {(2-2 i)-4 \sqrt {1-i}+2 \left ((2-2 i)+\sqrt {1-i}\right ) \sqrt {1+x}}{4 \sqrt {(1+i)-(1-i)^{3/2}} \sqrt {x+\sqrt {1+x}}}\right )}{\sqrt {(1+i)-(1-i)^{3/2}}}-\frac {(1-i) \arctan \left (\frac {(2+2 i)+4 \sqrt {1+i}-2 \left ((-2-2 i)+\sqrt {1+i}\right ) \sqrt {1+x}}{4 \sqrt {(1-i)+(1+i)^{3/2}} \sqrt {x+\sqrt {1+x}}}\right )}{\sqrt {(1-i)+(1+i)^{3/2}}}-\frac {(1-i) \arctan \left (\frac {(2+2 i)-4 \sqrt {1+i}+2 \left ((2+2 i)+\sqrt {1+i}\right ) \sqrt {1+x}}{4 \sqrt {(1-i)-(1+i)^{3/2}} \sqrt {x+\sqrt {1+x}}}\right )}{\sqrt {(1-i)-(1+i)^{3/2}}}+\frac {7}{4} \text {arctanh}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (-3+2 \sqrt {1+x}\right )-\frac {7}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right )+\text {RootSum}\left [1-8 \text {$\#$1}+40 \text {$\#$1}^2-48 \text {$\#$1}^3+20 \text {$\#$1}^4+8 \text {$\#$1}^5-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {-\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right )+2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+4 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3-\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4+2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^5}{-1+10 \text {$\#$1}-18 \text {$\#$1}^2+10 \text {$\#$1}^3+5 \text {$\#$1}^4-3 \text {$\#$1}^5+\text {$\#$1}^7}\&\right ] \]

[In]

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

[Out]

(Sqrt[x + Sqrt[1 + x]]*(-3 + 2*Sqrt[1 + x]))/2 - (7*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]])/4 + Roo
tSum[1 - 8*#1 + 40*#1^2 - 48*#1^3 + 20*#1^4 + 8*#1^5 - 4*#1^6 + #1^8 & , (-Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1
+ x]] - #1] + 2*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]] - #1]*#1 - 2*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]
- #1]*#1^2 + 4*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]] - #1]*#1^3 - Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]
- #1]*#1^4 + 2*Log[-Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]] - #1]*#1^5)/(-1 + 10*#1 - 18*#1^2 + 10*#1^3 + 5*#1^4
- 3*#1^5 + #1^7) & ]

Maple [N/A] (verified)

Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.48

method	result	size
derivativedivides	\(\sqrt {1+x}\, \sqrt {x +\sqrt {1+x}}-\frac {3 \sqrt {x +\sqrt {1+x}}}{2}+\frac {7 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+8 \textit {\_Z}^{5}+20 \textit {\_Z}^{4}-48 \textit {\_Z}^{3}+40 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )}{\sum }\frac {\left (2 \textit {\_R}^{5}-\textit {\_R}^{4}+4 \textit {\_R}^{3}-2 \textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5}+5 \textit {\_R}^{4}+10 \textit {\_R}^{3}-18 \textit {\_R}^{2}+10 \textit {\_R} -1}\right )\)	\(161\)
default	\(\sqrt {1+x}\, \sqrt {x +\sqrt {1+x}}-\frac {3 \sqrt {x +\sqrt {1+x}}}{2}+\frac {7 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+8 \textit {\_Z}^{5}+20 \textit {\_Z}^{4}-48 \textit {\_Z}^{3}+40 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )}{\sum }\frac {\left (2 \textit {\_R}^{5}-\textit {\_R}^{4}+4 \textit {\_R}^{3}-2 \textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5}+5 \textit {\_R}^{4}+10 \textit {\_R}^{3}-18 \textit {\_R}^{2}+10 \textit {\_R} -1}\right )\)	\(161\)

[In]

int((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2)-3/2*(x+(1+x)^(1/2))^(1/2)+7/4*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(1/2))+sum(
(2*_R^5-_R^4+4*_R^3-2*_R^2+2*_R-1)/(_R^7-3*_R^5+5*_R^4+10*_R^3-18*_R^2+10*_R-1)*ln((x+(1+x)^(1/2))^(1/2)-(1+x)
^(1/2)-_R),_R=RootOf(_Z^8-4*_Z^6+8*_Z^5+20*_Z^4-48*_Z^3+40*_Z^2-8*_Z+1))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 9.95 (sec) , antiderivative size = 5812, normalized size of antiderivative = 17.25 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [N/A]

Not integrable

Time = 11.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.08 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {x + \sqrt {x + 1}} \left (x^{2} + 1\right )}\, dx \]

[In]

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

[Out]

Integral((x - 1)*(x + 1)**(3/2)/(sqrt(x + sqrt(x + 1))*(x**2 + 1)), x)

Maxima [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {{\left (x^{2} - 1\right )} \sqrt {x + 1}}{{\left (x^{2} + 1\right )} \sqrt {x + \sqrt {x + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Invalid _EXT in replace_ext Error: Bad Argument ValueInvalid _EXT in replace_ext Error: Bad Argument ValueI
nvalid _EXT

Mupad [N/A]

Not integrable

Time = 7.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {\left (x^2-1\right )\,\sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}\,\left (x^2+1\right )} \,d x \]

[In]

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

[Out]

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