Integrand size = 25, antiderivative size = 59 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=-\arctan \left (\frac {\sqrt {x+x^2+x^3}}{1+x+x^2}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} \sqrt {x+x^2+x^3}}{1+x+x^2}\right )}{\sqrt {3}} \]
-arctan((x^3+x^2+x)^(1/2)/(x^2+x+1))-1/3*arctanh(3^(1/2)*(x^3+x^2+x)^(1/2) /(x^2+x+1))*3^(1/2)
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.34 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=-\frac {\sqrt {x} \sqrt {1+x+x^2} \left (3 \arctan \left (\frac {\sqrt {x}}{\sqrt {1+x+x^2}}\right )+\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {1+x+x^2}}\right )\right )}{3 \sqrt {x \left (1+x+x^2\right )}} \]
-1/3*(Sqrt[x]*Sqrt[1 + x + x^2]*(3*ArcTan[Sqrt[x]/Sqrt[1 + x + x^2]] + Sqr t[3]*ArcTanh[(Sqrt[3]*Sqrt[x])/Sqrt[1 + x + x^2]]))/Sqrt[x*(1 + x + x^2)]
Time = 0.68 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2467, 25, 2035, 7276, 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.
\(\displaystyle \int \frac {x^2+1}{\left (x^2-1\right ) \sqrt {x^3+x^2+x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt {x^2+x+1} \int -\frac {x^2+1}{\sqrt {x} \left (1-x^2\right ) \sqrt {x^2+x+1}}dx}{\sqrt {x^3+x^2+x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt {x^2+x+1} \int \frac {x^2+1}{\sqrt {x} \left (1-x^2\right ) \sqrt {x^2+x+1}}dx}{\sqrt {x^3+x^2+x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2+x+1} \int \frac {x^2+1}{\left (1-x^2\right ) \sqrt {x^2+x+1}}d\sqrt {x}}{\sqrt {x^3+x^2+x}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2+x+1} \int \left (\frac {2}{\left (1-x^2\right ) \sqrt {x^2+x+1}}-\frac {1}{\sqrt {x^2+x+1}}\right )d\sqrt {x}}{\sqrt {x^3+x^2+x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt {x^2+x+1} \left (\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+x+1}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {3} \sqrt {x}}{\sqrt {x^2+x+1}}\right )}{2 \sqrt {3}}\right )}{\sqrt {x^3+x^2+x}}\) |
(-2*Sqrt[x]*Sqrt[1 + x + x^2]*(ArcTan[Sqrt[x]/Sqrt[1 + x + x^2]]/2 + ArcTa nh[(Sqrt[3]*Sqrt[x])/Sqrt[1 + x + x^2]]/(2*Sqrt[3])))/Sqrt[x + x^2 + x^3]
3.8.52.3.1 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]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 2.55 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69
method | result | size |
default | \(\arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{2}+x +1\right )}\, \sqrt {3}}{3 x}\right )}{3}\) | \(41\) |
pseudoelliptic | \(\arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (x^{2}+x +1\right )}\, \sqrt {3}}{3 x}\right )}{3}\) | \(41\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \sqrt {x^{3}+x^{2}+x}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (1+x \right )^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )-6 \sqrt {x^{3}+x^{2}+x}}{\left (x -1\right )^{2}}\right )}{6}\) | \(103\) |
elliptic | \(\text {Expression too large to display}\) | \(910\) |
Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.51 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=\frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{4} + 20 \, x^{3} - 4 \, \sqrt {3} \sqrt {x^{3} + x^{2} + x} {\left (x^{2} + 4 \, x + 1\right )} + 30 \, x^{2} + 20 \, x + 1}{x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1}\right ) + \frac {1}{2} \, \arctan \left (\frac {x^{2} + 1}{2 \, \sqrt {x^{3} + x^{2} + x}}\right ) \]
1/12*sqrt(3)*log((x^4 + 20*x^3 - 4*sqrt(3)*sqrt(x^3 + x^2 + x)*(x^2 + 4*x + 1) + 30*x^2 + 20*x + 1)/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)) + 1/2*arctan(1/ 2*(x^2 + 1)/sqrt(x^3 + x^2 + x))
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=\int \frac {x^{2} + 1}{\sqrt {x \left (x^{2} + x + 1\right )} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=\int { \frac {x^{2} + 1}{\sqrt {x^{3} + x^{2} + x} {\left (x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=\int { \frac {x^{2} + 1}{\sqrt {x^{3} + x^{2} + x} {\left (x^{2} - 1\right )}} \,d x } \]
Time = 5.62 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.78 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt {x+x^2+x^3}} \, dx=-\frac {\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (\sqrt {3}+1{}\mathrm {i}\right )\,\left (-\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+\Pi \left (\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+\Pi \left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,1{}\mathrm {i}}{\sqrt {x^3+x^2-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,x}} \]
-((x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)* 1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 + 1/2))^(1 /2)*(3^(1/2) + 1i)*(ellipticPi(1/2 - (3^(1/2)*1i)/2, asin((x/((3^(1/2)*1i) /2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)) - ellip ticF(asin((x/((3^(1/2)*1i)/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^( 1/2)*1i)/2 + 1/2)) + ellipticPi((3^(1/2)*1i)/2 - 1/2, asin((x/((3^(1/2)*1i )/2 - 1/2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))*1i)/( x^2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)