Integrand size = 30, antiderivative size = 39 \[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\frac {\sqrt {x+x^3}}{1+x+x^2}-\arctan \left (\frac {\sqrt {x+x^3}}{1+x^2}\right ) \]
Time = 0.80 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.31 \[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\sqrt {x+x^3} \left (\frac {1}{1+x+x^2}-\frac {\arctan \left (\frac {\sqrt {x}}{\sqrt {1+x^2}}\right )}{\sqrt {x} \sqrt {1+x^2}}\right ) \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 8.42 (sec) , antiderivative size = 1709, normalized size of antiderivative = 43.82, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2467, 25, 2035, 7293, 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.
\(\displaystyle \int \frac {\left (x^2-1\right ) \sqrt {x^3+x}}{\left (x^2+1\right ) \left (x^2+x+1\right )^2} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x^3+x} \int -\frac {\sqrt {x} \left (1-x^2\right )}{\sqrt {x^2+1} \left (x^2+x+1\right )^2}dx}{\sqrt {x} \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x^3+x} \int \frac {\sqrt {x} \left (1-x^2\right )}{\sqrt {x^2+1} \left (x^2+x+1\right )^2}dx}{\sqrt {x} \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x^3+x} \int \frac {x \left (1-x^2\right )}{\sqrt {x^2+1} \left (x^2+x+1\right )^2}d\sqrt {x}}{\sqrt {x} \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2 \sqrt {x^3+x} \int \left (\frac {-\sqrt {x}-1}{4 \left (x-\sqrt {x}+1\right ) \sqrt {x^2+1}}+\frac {\sqrt {x}-1}{4 \left (x+\sqrt {x}+1\right ) \sqrt {x^2+1}}+\frac {\sqrt {x}+1}{4 \left (x-\sqrt {x}+1\right )^2 \sqrt {x^2+1}}+\frac {1-\sqrt {x}}{4 \left (x+\sqrt {x}+1\right )^2 \sqrt {x^2+1}}\right )d\sqrt {x}}{\sqrt {x} \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x^3+x} \left (\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )}{4 \left (i-\sqrt {3}\right )}+\frac {\left (i-\sqrt {3}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )}{4 \left (i+\sqrt {3}\right )}+\frac {\arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )}{2 \sqrt {3} \left (i+\sqrt {3}\right )}+\frac {1}{12} \left (1+i \sqrt {3}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )+\frac {1}{12} \left (1-i \sqrt {3}\right ) \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )-\frac {\arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )}{2 \sqrt {3} \left (i-\sqrt {3}\right )}+\frac {1}{3} \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+1}}\right )+\frac {\text {arctanh}\left (\frac {2-\left (1-i \sqrt {3}\right ) x}{\sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {x^2+1}}\right )}{4 \left (i+\sqrt {3}\right )}-\frac {\text {arctanh}\left (\frac {\left (1+i \sqrt {3}\right )^2 x+4}{2 \sqrt {2 \left (1-i \sqrt {3}\right )} \sqrt {x^2+1}}\right )}{4 \left (i+\sqrt {3}\right )}+\frac {\left (i+\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \arctan \left (\sqrt {x}\right )|\frac {1}{2}\right )}{6 \left (i-\sqrt {3}\right ) \sqrt {x^2+1}}+\frac {\left (i-\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \arctan \left (\sqrt {x}\right )|\frac {1}{2}\right )}{6 \left (i+\sqrt {3}\right ) \sqrt {x^2+1}}+\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \arctan \left (\sqrt {x}\right )|\frac {1}{2}\right )}{3 \left (1+i \sqrt {3}\right ) \sqrt {x^2+1}}+\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} E\left (2 \arctan \left (\sqrt {x}\right )|\frac {1}{2}\right )}{3 \left (1-i \sqrt {3}\right ) \sqrt {x^2+1}}+\frac {\left (1+i \sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{8 \sqrt {x^2+1}}-\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{6 \left (1+i \sqrt {3}\right ) \sqrt {x^2+1}}+\frac {\left (1-i \sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{8 \sqrt {x^2+1}}-\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{6 \left (1-i \sqrt {3}\right ) \sqrt {x^2+1}}-\frac {(x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{6 \sqrt {x^2+1}}-\frac {\left (i+\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{8 \left (i-\sqrt {3}\right ) \sqrt {x^2+1}}-\frac {\left (i-\sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{8 \left (i+\sqrt {3}\right ) \sqrt {x^2+1}}-\frac {\left (3+i \sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{48 \sqrt {x^2+1}}-\frac {\left (3-i \sqrt {3}\right ) (x+1) \sqrt {\frac {x^2+1}{(x+1)^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan \left (\sqrt {x}\right ),\frac {1}{2}\right )}{48 \sqrt {x^2+1}}-\frac {\left (i+\sqrt {3}\right ) \sqrt {x^2+1}}{6 \left (i-\sqrt {3}\right ) \left (-2 \sqrt {x}-i \sqrt {3}+1\right )}-\frac {\sqrt {x^2+1}}{3 \left (1+i \sqrt {3}\right ) \left (-2 \sqrt {x}-i \sqrt {3}+1\right )}-\frac {\left (i-\sqrt {3}\right ) \sqrt {x^2+1}}{6 \left (i+\sqrt {3}\right ) \left (-2 \sqrt {x}+i \sqrt {3}+1\right )}-\frac {\sqrt {x^2+1}}{3 \left (1-i \sqrt {3}\right ) \left (-2 \sqrt {x}+i \sqrt {3}+1\right )}+\frac {\left (i+\sqrt {3}\right ) \sqrt {x^2+1}}{6 \left (i-\sqrt {3}\right ) \left (2 \sqrt {x}-i \sqrt {3}+1\right )}+\frac {\sqrt {x^2+1}}{3 \left (1+i \sqrt {3}\right ) \left (2 \sqrt {x}-i \sqrt {3}+1\right )}+\frac {\left (i-\sqrt {3}\right ) \sqrt {x^2+1}}{6 \left (i+\sqrt {3}\right ) \left (2 \sqrt {x}+i \sqrt {3}+1\right )}+\frac {\sqrt {x^2+1}}{3 \left (1-i \sqrt {3}\right ) \left (2 \sqrt {x}+i \sqrt {3}+1\right )}-\frac {\left (i+\sqrt {3}\right ) \sqrt {x} \sqrt {x^2+1}}{6 \left (i-\sqrt {3}\right ) (x+1)}-\frac {\left (i-\sqrt {3}\right ) \sqrt {x} \sqrt {x^2+1}}{6 \left (i+\sqrt {3}\right ) (x+1)}-\frac {\sqrt {x} \sqrt {x^2+1}}{3 \left (1+i \sqrt {3}\right ) (x+1)}-\frac {\sqrt {x} \sqrt {x^2+1}}{3 \left (1-i \sqrt {3}\right ) (x+1)}\right )}{\sqrt {x} \sqrt {x^2+1}}\) |
(-2*Sqrt[x + x^3]*(-1/3*Sqrt[1 + x^2]/((1 + I*Sqrt[3])*(1 - I*Sqrt[3] - 2* Sqrt[x])) - ((I + Sqrt[3])*Sqrt[1 + x^2])/(6*(I - Sqrt[3])*(1 - I*Sqrt[3] - 2*Sqrt[x])) - Sqrt[1 + x^2]/(3*(1 - I*Sqrt[3])*(1 + I*Sqrt[3] - 2*Sqrt[x ])) - ((I - Sqrt[3])*Sqrt[1 + x^2])/(6*(I + Sqrt[3])*(1 + I*Sqrt[3] - 2*Sq rt[x])) + Sqrt[1 + x^2]/(3*(1 + I*Sqrt[3])*(1 - I*Sqrt[3] + 2*Sqrt[x])) + ((I + Sqrt[3])*Sqrt[1 + x^2])/(6*(I - Sqrt[3])*(1 - I*Sqrt[3] + 2*Sqrt[x]) ) + Sqrt[1 + x^2]/(3*(1 - I*Sqrt[3])*(1 + I*Sqrt[3] + 2*Sqrt[x])) + ((I - Sqrt[3])*Sqrt[1 + x^2])/(6*(I + Sqrt[3])*(1 + I*Sqrt[3] + 2*Sqrt[x])) - (S qrt[x]*Sqrt[1 + x^2])/(3*(1 - I*Sqrt[3])*(1 + x)) - (Sqrt[x]*Sqrt[1 + x^2] )/(3*(1 + I*Sqrt[3])*(1 + x)) - ((I - Sqrt[3])*Sqrt[x]*Sqrt[1 + x^2])/(6*( I + Sqrt[3])*(1 + x)) - ((I + Sqrt[3])*Sqrt[x]*Sqrt[1 + x^2])/(6*(I - Sqrt [3])*(1 + x)) + ArcTan[Sqrt[x]/Sqrt[1 + x^2]]/3 - ArcTan[Sqrt[x]/Sqrt[1 + x^2]]/(2*Sqrt[3]*(I - Sqrt[3])) + ((1 - I*Sqrt[3])*ArcTan[Sqrt[x]/Sqrt[1 + x^2]])/12 + ((1 + I*Sqrt[3])*ArcTan[Sqrt[x]/Sqrt[1 + x^2]])/12 + ArcTan[S qrt[x]/Sqrt[1 + x^2]]/(2*Sqrt[3]*(I + Sqrt[3])) + ((I - Sqrt[3])*ArcTan[Sq rt[x]/Sqrt[1 + x^2]])/(4*(I + Sqrt[3])) + ((I + Sqrt[3])*ArcTan[Sqrt[x]/Sq rt[1 + x^2]])/(4*(I - Sqrt[3])) + ArcTanh[(2 - (1 - I*Sqrt[3])*x)/(Sqrt[2* (1 - I*Sqrt[3])]*Sqrt[1 + x^2])]/(4*(I + Sqrt[3])) - ArcTanh[(4 + (1 + I*S qrt[3])^2*x)/(2*Sqrt[2*(1 - I*Sqrt[3])]*Sqrt[1 + x^2])]/(4*(I + Sqrt[3])) + ((1 + x)*Sqrt[(1 + x^2)/(1 + x)^2]*EllipticE[2*ArcTan[Sqrt[x]], 1/2])...
3.6.2.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 7.37 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {\left (x^{2}+1\right ) x}{\left (x^{2}+x +1\right ) \sqrt {\left (x^{2}+1\right ) x}}+\arctan \left (\frac {\sqrt {\left (x^{2}+1\right ) x}}{x}\right )\) | \(40\) |
default | \(\frac {\left (x^{2}+x +1\right ) \arctan \left (\frac {\sqrt {\left (x^{2}+1\right ) x}}{x}\right )+\sqrt {\left (x^{2}+1\right ) x}}{x^{2}+x +1}\) | \(41\) |
pseudoelliptic | \(\frac {\left (x^{2}+x +1\right ) \arctan \left (\frac {\sqrt {\left (x^{2}+1\right ) x}}{x}\right )+\sqrt {\left (x^{2}+1\right ) x}}{x^{2}+x +1}\) | \(41\) |
trager | \(\frac {\sqrt {x^{3}+x}}{x^{2}+x +1}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {x^{3}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{2}+x +1}\right )}{2}\) | \(71\) |
elliptic | \(\frac {\sqrt {x^{3}+x}}{x^{2}+x +1}+\frac {i \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \operatorname {EllipticF}\left (\sqrt {-i \left (i+x \right )}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {x^{3}+x}}+\frac {i \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \left (i \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )+1+i\right ) \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-i+\frac {i \sqrt {3}}{2}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {x^{3}+x}}+\frac {i \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {-i \left (i+x \right )}\, \sqrt {2}\, \sqrt {i \left (-i+x \right )}\, \sqrt {i x}\, \left (i \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )+1+i\right ) \operatorname {EllipticPi}\left (\sqrt {-i \left (i+x \right )}, \frac {1}{2}-i-\frac {i \sqrt {3}}{2}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {x^{3}+x}}\) | \(240\) |
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\frac {{\left (x^{2} + x + 1\right )} \arctan \left (\frac {x^{2} - x + 1}{2 \, \sqrt {x^{3} + x}}\right ) + 2 \, \sqrt {x^{3} + x}}{2 \, {\left (x^{2} + x + 1\right )}} \]
\[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\int \frac {\sqrt {x \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right )}{\left (x^{2} + 1\right ) \left (x^{2} + x + 1\right )^{2}}\, dx \]
\[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\int { \frac {\sqrt {x^{3} + x} {\left (x^{2} - 1\right )}}{{\left (x^{2} + x + 1\right )}^{2} {\left (x^{2} + 1\right )}} \,d x } \]
\[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\int { \frac {\sqrt {x^{3} + x} {\left (x^{2} - 1\right )}}{{\left (x^{2} + x + 1\right )}^{2} {\left (x^{2} + 1\right )}} \,d x } \]
Time = 5.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.26 \[ \int \frac {\left (-1+x^2\right ) \sqrt {x+x^3}}{\left (1+x^2\right ) \left (1+x+x^2\right )^2} \, dx=\frac {\sqrt {x^3+x}}{x^2+x+1}-\frac {\ln \left (x^2+x+1\right )\,1{}\mathrm {i}}{2}+\frac {\ln \left (x^2-x+1+\sqrt {x^3+x}\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]