Integrand size = 33, antiderivative size = 337 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=-\frac {3}{2} \sqrt {x+\sqrt {1+x}}+\sqrt {1+x} \sqrt {x+\sqrt {1+x}}-\frac {7}{4} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right )+8 \text {RootSum}\left [625+1000 \text {$\#$1}+300 \text {$\#$1}^2+120 \text {$\#$1}^3+470 \text {$\#$1}^4+24 \text {$\#$1}^5+12 \text {$\#$1}^6+8 \text {$\#$1}^7+\text {$\#$1}^8\&,\frac {25 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}+20 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^2+14 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^3+4 \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^4+\log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}+\text {$\#$1}\right ) \text {$\#$1}^5}{125+75 \text {$\#$1}+45 \text {$\#$1}^2+235 \text {$\#$1}^3+15 \text {$\#$1}^4+9 \text {$\#$1}^5+7 \text {$\#$1}^6+\text {$\#$1}^7}\&\right ] \] Output:
Unintegrable
Time = 0.27 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (-3+2 \sqrt {1+x}\right )-\frac {7}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right )+\text {RootSum}\left [1-8 \text {$\#$1}+40 \text {$\#$1}^2-48 \text {$\#$1}^3+20 \text {$\#$1}^4+8 \text {$\#$1}^5-4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {-\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right )+2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}-2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^2+4 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^3-\log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^4+2 \log \left (-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}-\text {$\#$1}\right ) \text {$\#$1}^5}{-1+10 \text {$\#$1}-18 \text {$\#$1}^2+10 \text {$\#$1}^3+5 \text {$\#$1}^4-3 \text {$\#$1}^5+\text {$\#$1}^7}\&\right ] \] Input:
Integrate[(Sqrt[1 + x]*(-1 + x^2))/((1 + x^2)*Sqrt[x + Sqrt[1 + x]]),x]
Output:
(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 + RootSum[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[-S qrt[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) & ]
Result contains complex when optimal does not.
Time = 2.13 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2003, 7267, 25, 7292, 7279, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {x+1} \left (x^2-1\right )}{\left (x^2+1\right ) \sqrt {x+\sqrt {x+1}}} \, dx\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle \int \frac {(x-1) (x+1)^{3/2}}{\left (x^2+1\right ) \sqrt {x+\sqrt {x+1}}}dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 \int -\frac {(1-x) (x+1)^2}{\left (x^2+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {(1-x) (x+1)^2}{\left (x^2+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -2 \int \frac {(1-x) (x+1)^2}{\sqrt {x+\sqrt {x+1}} \left ((x+1)^2-2 (x+1)+2\right )}d\sqrt {x+1}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -2 \int \left (\frac {2 (x+1)}{\sqrt {x+\sqrt {x+1}} \left ((x+1)^2-2 (x+1)+2\right )}-\frac {x+1}{\sqrt {x+\sqrt {x+1}}}\right )d\sqrt {x+1}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \arctan \left (\frac {-2 \left ((-2+2 i)+\sqrt {1-i}\right ) \sqrt {x+1}+4 \sqrt {1-i}+(2-2 i)}{4 \sqrt {(1+i)+(1-i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1+i)+(1-i)^{3/2}}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \arctan \left (\frac {2 \left ((2-2 i)+\sqrt {1-i}\right ) \sqrt {x+1}-4 \sqrt {1-i}+(2-2 i)}{4 \sqrt {(1+i)-(1-i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1+i)-(1-i)^{3/2}}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \arctan \left (\frac {-2 \left ((-2-2 i)+\sqrt {1+i}\right ) \sqrt {x+1}+4 \sqrt {1+i}+(2+2 i)}{4 \sqrt {(1-i)+(1+i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1-i)+(1+i)^{3/2}}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \arctan \left (\frac {2 \left ((2+2 i)+\sqrt {1+i}\right ) \sqrt {x+1}-4 \sqrt {1+i}+(2+2 i)}{4 \sqrt {(1-i)-(1+i)^{3/2}} \sqrt {x+\sqrt {x+1}}}\right )}{\sqrt {(1-i)-(1+i)^{3/2}}}+\frac {7}{8} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )+\frac {1}{2} \sqrt {x+1} \sqrt {x+\sqrt {x+1}}-\frac {3}{4} \sqrt {x+\sqrt {x+1}}\right )\) |
Input:
Int[(Sqrt[1 + x]*(-1 + x^2))/((1 + x^2)*Sqrt[x + Sqrt[1 + x]]),x]
Output:
2*((-3*Sqrt[x + Sqrt[1 + x]])/4 + (Sqrt[1 + x]*Sqrt[x + Sqrt[1 + x]])/2 - ((1/2 + I/2)*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)] - ((1/2 + I/2)*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)] - ((1/2 - I /2)*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)] - ((1/2 - I/2)*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)]*S qrt[x + Sqrt[1 + x]])])/Sqrt[(1 - I) - (1 + I)^(3/2)] + (7*ArcTanh[(1 + 2* Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/8)
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
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] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
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 | \(\sqrt {1+x}\, \sqrt {x +\sqrt {1+x}}-\frac {3 \sqrt {x +\sqrt {1+x}}}{2}+\frac {7 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+8 \textit {\_Z}^{5}+20 \textit {\_Z}^{4}-48 \textit {\_Z}^{3}+40 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )}{\sum }\frac {\left (2 \textit {\_R}^{5}-\textit {\_R}^{4}+4 \textit {\_R}^{3}-2 \textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (\sqrt {x +\sqrt {1+x}}-\sqrt {1+x}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5}+5 \textit {\_R}^{4}+10 \textit {\_R}^{3}-18 \textit {\_R}^{2}+10 \textit {\_R} -1}\right )\) | \(161\) |
Input:
int((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x,method=_RETURNVERB OSE)
Output:
(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))
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 9.61 (sec) , antiderivative size = 5376, normalized size of antiderivative = 15.95 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\text {Too large to display} \] Input:
integrate((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x, algorithm=" fricas")
Output:
Too large to include
Not integrable
Time = 11.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.08 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {x + \sqrt {x + 1}} \left (x^{2} + 1\right )}\, dx \] Input:
integrate((1+x)**(1/2)*(x**2-1)/(x**2+1)/(x+(1+x)**(1/2))**(1/2),x)
Output:
Integral((x - 1)*(x + 1)**(3/2)/(sqrt(x + sqrt(x + 1))*(x**2 + 1)), x)
Not integrable
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int { \frac {{\left (x^{2} - 1\right )} \sqrt {x + 1}}{{\left (x^{2} + 1\right )} \sqrt {x + \sqrt {x + 1}}} \,d x } \] Input:
integrate((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x, algorithm=" maxima")
Output:
integrate((x^2 - 1)*sqrt(x + 1)/((x^2 + 1)*sqrt(x + sqrt(x + 1))), x)
Exception generated. \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x, algorithm=" giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Invalid _EXT in replace_ext Error: Bad Argument ValueInvalid _EXT in replace_ext Error: Bad Argument ValueInv alid _EXT
Not integrable
Time = 10.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=\int \frac {\left (x^2-1\right )\,\sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}\,\left (x^2+1\right )} \,d x \] Input:
int(((x^2 - 1)*(x + 1)^(1/2))/((x + (x + 1)^(1/2))^(1/2)*(x^2 + 1)),x)
Output:
int(((x^2 - 1)*(x + 1)^(1/2))/((x + (x + 1)^(1/2))^(1/2)*(x^2 + 1)), x)
Not integrable
Time = 0.48 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt {1+x} \left (-1+x^2\right )}{\left (1+x^2\right ) \sqrt {x+\sqrt {1+x}}} \, dx=-\left (\int \frac {\sqrt {x +1}}{\sqrt {\sqrt {x +1}+x}\, x^{2}+\sqrt {\sqrt {x +1}+x}}d x \right )+\int \frac {\sqrt {x +1}\, x^{2}}{\sqrt {\sqrt {x +1}+x}\, x^{2}+\sqrt {\sqrt {x +1}+x}}d x \] Input:
int((1+x)^(1/2)*(x^2-1)/(x^2+1)/(x+(1+x)^(1/2))^(1/2),x)
Output:
- int(sqrt(x + 1)/(sqrt(sqrt(x + 1) + x)*x**2 + sqrt(sqrt(x + 1) + x)),x) + int((sqrt(x + 1)*x**2)/(sqrt(sqrt(x + 1) + x)*x**2 + sqrt(sqrt(x + 1) + x)),x)