Integrand size = 23, antiderivative size = 203 \[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {26}{27 x^2 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2}{9 x^2 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {91 \left (1+x^3\right )}{54 x^2 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {91 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{54 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]
26/27/x^2/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9/x^2/(x^3+1)/(1+x)^(1/2)/(x^2-x+1 )^(1/2)-91/54*(x^3+1)/x^2/(1+x)^(1/2)/(x^2-x+1)^(1/2)-91/162*EllipticF((1+ x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(1+x)^(1/2)*(1/2*6^(1/2)+1/2*2^(1/ 2))*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*3^(3/4)/(x^2-x+1)^(1/2)/((1+x)/(1+x+ 3^(1/2))^2)^(1/2)
Result contains complex when optimal does not.
Time = 10.43 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {-\frac {6 \left (27+130 x^3+91 x^6\right )}{x^2 (1+x)^{3/2}}-\frac {91 i (1+x) \left (1-x+x^2\right ) \sqrt {6+\frac {36 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {1-\frac {6 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{324 \left (1-x+x^2\right )^{3/2}} \]
((-6*(27 + 130*x^3 + 91*x^6))/(x^2*(1 + x)^(3/2)) - ((91*I)*(1 + x)*(1 - x + x^2)*Sqrt[6 + (36*I)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[1 - (6*I)/((3*I + Sqrt[3])*(1 + x))]*EllipticF[I*ArcSinh[Sqrt[(-6*I)/(3*I + Sqrt[3])]/Sqrt[ 1 + x]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(32 4*(1 - x + x^2)^(3/2))
Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1210, 819, 819, 847, 759}
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.
| \(\displaystyle \int \frac {1}{x^3 (x+1)^{5/2} \left (x^2-x+1\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1210 |
| \(\displaystyle \frac {\sqrt {x^3+1} \int \frac {1}{x^3 \left (x^3+1\right )^{5/2}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {13}{9} \int \frac {1}{x^3 \left (x^3+1\right )^{3/2}}dx+\frac {2}{9 x^2 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 819 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {13}{9} \left (\frac {7}{3} \int \frac {1}{x^3 \sqrt {x^3+1}}dx+\frac {2}{3 x^2 \sqrt {x^3+1}}\right )+\frac {2}{9 x^2 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {13}{9} \left (\frac {7}{3} \left (-\frac {1}{4} \int \frac {1}{\sqrt {x^3+1}}dx-\frac {\sqrt {x^3+1}}{2 x^2}\right )+\frac {2}{3 x^2 \sqrt {x^3+1}}\right )+\frac {2}{9 x^2 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {13}{9} \left (\frac {7}{3} \left (-\frac {\sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {\sqrt {x^3+1}}{2 x^2}\right )+\frac {2}{3 x^2 \sqrt {x^3+1}}\right )+\frac {2}{9 x^2 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
(Sqrt[1 + x^3]*(2/(9*x^2*(1 + x^3)^(3/2)) + (13*(2/(3*x^2*Sqrt[1 + x^3]) + (7*(-1/2*Sqrt[1 + x^3]/x^2 - (Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2 )/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x )], -7 - 4*Sqrt[3]])/(2*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/3))/9))/(Sqrt[1 + x]*Sqrt[1 - x + x^2])
3.6.23.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]) Int[(f + g*x)^n*(a*d + c* e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
Time = 0.68 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.88
| method | result | size |
| elliptic | \(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (-\frac {\sqrt {x^{3}+1}}{2 x^{2}}-\frac {2 x}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}-\frac {32 x}{27 \sqrt {x^{3}+1}}-\frac {91 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{54 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\) | \(179\) |
| default | \(\frac {91 i \sqrt {3}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{5} \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}-273 F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{5} \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}+91 i \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) \sqrt {3}\, x^{2}-273 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right ) x^{2}-182 x^{6}-260 x^{3}-54}{108 x^{2} \left (x^{2}-x +1\right )^{\frac {3}{2}} \left (1+x \right )^{\frac {3}{2}}}\) | \(481\) |
((1+x)*(x^2-x+1))^(1/2)/(1+x)^(1/2)/(x^2-x+1)^(1/2)*(-1/2/x^2*(x^3+1)^(1/2 )-2/9*x/(x^3+1)^(3/2)-32/27*x/(x^3+1)^(1/2)-91/54*(3/2-1/2*I*3^(1/2))*((1+ x)/(3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2))) ^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/(x^3+1)^(1/2)*El lipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2* I*3^(1/2)))^(1/2)))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=-\frac {{\left (91 \, x^{6} + 130 \, x^{3} + 27\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + 91 \, {\left (x^{8} + 2 \, x^{5} + x^{2}\right )} {\rm weierstrassPInverse}\left (0, -4, x\right )}{54 \, {\left (x^{8} + 2 \, x^{5} + x^{2}\right )}} \]
-1/54*((91*x^6 + 130*x^3 + 27)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 91*(x^8 + 2 *x^5 + x^2)*weierstrassPInverse(0, -4, x))/(x^8 + 2*x^5 + x^2)
\[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{x^{3} \left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}} x^{3}} \,d x } \]
\[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{x^3\,{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \]