\(\int \frac {x^3}{(1+x)^{3/2} (1-x+x^2)^{3/2}} \, dx\) [510]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2 x}{3 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]

[Out]

-2/3*x/(1+x)^(1/2)/(x^2-x+1)^(1/2)+4/9*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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {929, 294, 224} \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {4 \sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}}-\frac {2 x}{3 \sqrt {x+1} \sqrt {x^2-x+1}} \]

[In]

Int[x^3/((1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]

[Out]

(-2*x)/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2]) + (4*Sqrt[2 + Sqrt[3]]*Sqrt[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]])/(3*3^(1/4)*Sqrt[(1 + x)/(1 + Sqr
t[3] + x)^2]*Sqrt[1 - x + x^2])

Rule 224

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]*Sq
rt[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]

Rule 294

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] - Dist[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]

Rule 929

Int[((g_.)*(x_))^(n_)*((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d
+ e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]), Int[(g*x)^n*(a*d + c*e*x^3)^p,
x], x] /; FreeQ[{a, b, c, d, e, g, m, n, p}, x] && EqQ[m - p, 0] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+x^3} \int \frac {x^3}{\left (1+x^3\right )^{3/2}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}} \\ & = -\frac {2 x}{3 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {\left (2 \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \\ & = -\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}} F\left (\sin ^{-1}\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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.18 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {-\frac {6 x}{\sqrt {1+x}}+\frac {2 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{9 \sqrt {1-x+x^2}} \]

[In]

Integrate[x^3/((1 + x)^(3/2)*(1 - x + x^2)^(3/2)),x]

[Out]

((-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])])/
Sqrt[(-I)/(3*I + Sqrt[3])])/(9*Sqrt[1 - x + x^2])

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (-\frac {2 x}{3 \sqrt {x^{3}+1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(157\)
risch	\(-\frac {2 x}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}+\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{3 \sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(164\)
default	\(-\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+x \right )}{3 \left (x^{3}+1\right )}\)	\(245\)

[In]

int(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

((1+x)*(x^2-x+1))^(1/2)/(1+x)^(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))*((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)*EllipticF(((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)))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.28 \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} x - 2 \, {\left (x^{3} + 1\right )} {\rm weierstrassPInverse}\left (0, -4, x\right )\right )}}{3 \, {\left (x^{3} + 1\right )}} \]

[In]

integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="fricas")

[Out]

-2/3*(sqrt(x^2 - x + 1)*sqrt(x + 1)*x - 2*(x^3 + 1)*weierstrassPInverse(0, -4, x))/(x^3 + 1)

Sympy [F]

\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {x^{3}}{\left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**3/(1+x)**(3/2)/(x**2-x+1)**(3/2),x)

[Out]

Integral(x**3/((x + 1)**(3/2)*(x**2 - x + 1)**(3/2)), x)

Maxima [F]

\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^3/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)), x)

Giac [F]

\[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {x^{3}}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^3/(1+x)^(3/2)/(x^2-x+1)^(3/2),x, algorithm="giac")

[Out]

integrate(x^3/((x^2 - x + 1)^(3/2)*(x + 1)^(3/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {x^3}{{\left (x+1\right )}^{3/2}\,{\left (x^2-x+1\right )}^{3/2}} \,d x \]

[In]

int(x^3/((x + 1)^(3/2)*(x^2 - x + 1)^(3/2)),x)

[Out]

int(x^3/((x + 1)^(3/2)*(x^2 - x + 1)^(3/2)), x)