Integrand size = 23, antiderivative size = 137 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2 x}{3 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \] Output:
-2/3*x/(1+x)^(1/2)/(x^2-x+1)^(1/2)+4/9*(1/2*6^(1/2)+1/2*2^(1/2))*(1+x)^(1/ 2)*((x^2-x+1)/(1+x+3^(1/2))^2)^(1/2)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)) ,I*3^(1/2)+2*I)*3^(3/4)/((1+x)/(1+x+3^(1/2))^2)^(1/2)/(x^2-x+1)^(1/2)
Result contains complex when optimal does not.
Time = 21.67 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {-\frac {6 x}{\sqrt {1+x}}+\frac {2 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{9 \sqrt {1-x+x^2}} \] Input:
Integrate[x^3/((1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]
Output:
((-6*x)/Sqrt[1 + x] + ((2*I)*(1 + x)*Sqrt[1 + (6*I)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[6 - (36*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])])/Sq rt[(-I)/(3*I + Sqrt[3])])/(9*Sqrt[1 - x + x^2])
Time = 0.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1210, 817, 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.
\(\displaystyle \int \frac {x^3}{(x+1)^{3/2} \left (x^2-x+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1210 |
\(\displaystyle \frac {\sqrt {x^3+1} \int \frac {x^3}{\left (x^3+1\right )^{3/2}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {2}{3} \int \frac {1}{\sqrt {x^3+1}}dx-\frac {2 x}{3 \sqrt {x^3+1}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {4 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}-\frac {2 x}{3 \sqrt {x^3+1}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
Input:
Int[x^3/((1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]
Output:
(Sqrt[1 + x^3]*((-2*x)/(3*Sqrt[1 + x^3]) + (4*Sqrt[2 + Sqrt[3]]*(1 + x)*Sq rt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/( 1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/(Sqrt[1 + x]*Sqrt[1 - x + x^2])
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^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && 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 = 3.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15
method | result | size |
elliptic | \(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (-\frac {2 x}{3 \sqrt {x^{3}+1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\) | \(157\) |
risch | \(-\frac {2 x}{3 \sqrt {x +1}\, \sqrt {x^{2}-x +1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}}{3 \sqrt {x^{3}+1}\, \sqrt {x +1}\, \sqrt {x^{2}-x +1}}\) | \(164\) |
default | \(-\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )+x \right )}{3 \left (x^{3}+1\right )}\) | \(245\) |
Input:
int(x^3/(x+1)^(3/2)/(x^2-x+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)*(-2/3*x/(x^3+1)^(1/2)+ 4/3*(3/2-1/2*I*3^(1/2))*((x+1)/(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)*EllipticF(((x+1)/(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)))
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.28 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} x - 2 \, {\left (x^{3} + 1\right )} {\rm weierstrassPInverse}\left (0, -4, x\right )\right )}}{3 \, {\left (x^{3} + 1\right )}} \] Input:
integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="fricas")
Output:
-2/3*(sqrt(x^2 - x + 1)*sqrt(x + 1)*x - 2*(x^3 + 1)*weierstrassPInverse(0, -4, x))/(x^3 + 1)
\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**3/(1+x)**(3/2)/(x**2-x+1)**(3/2),x)
Output:
Integral(x**3/((x + 1)**(3/2)*(x**2 - x + 1)**(3/2)), x)
\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="maxima")
Output:
integrate(x^3/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)), x)
\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="giac")
Output:
integrate(x^3/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)), x)
Timed out. \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {x^3}{{\left (x+1\right )}^{3/2}\,{\left (x^2-x+1\right )}^{3/2}} \,d x \] Input:
int(x^3/((x + 1)^(3/2)*(x^2 - x + 1)^(3/2)),x)
Output:
int(x^3/((x + 1)^(3/2)*(x^2 - x + 1)^(3/2)), x)
\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x +2 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{6}+2 x^{3}+1}d x \right ) x^{3}+2 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{6}+2 x^{3}+1}d x \right )}{x^{3}+1} \] Input:
int(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x)
Output:
(2*( - sqrt(x + 1)*sqrt(x**2 - x + 1)*x + int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x**6 + 2*x**3 + 1),x)*x**3 + int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x **6 + 2*x**3 + 1),x)))/(x**3 + 1)