\(\int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx\) [503]

Optimal result

Integrand size = 23, antiderivative size = 142 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2 x \left (1+x^3\right )}{5 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {4 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \]

[Out]

2/5*x*(x^3+1)/(1+x)^(1/2)/(x^2-x+1)^(1/2)-4/15*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 = 0.03 (sec) , antiderivative size = 142, 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, 327, 224} \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2 x \left (x^3+1\right )}{5 \sqrt {x+1} \sqrt {x^2-x+1}}-\frac {4 \sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}} \]

[In]

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

[Out]

(2*x*(1 + x^3))/(5*Sqrt[1 + x]*Sqrt[1 - x + x^2]) - (4*Sqrt[2 + Sqrt[3]]*Sqrt[1 + x]*Sqrt[(1 - x + x^2)/(1 + S
qrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(5*3^(1/4)*Sqrt[(1 + x)
/(1 + Sqrt[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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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}{\sqrt {1+x^3}} \, dx}{\sqrt {1+x} \sqrt {1-x+x^2}} \\ & = \frac {2 x \left (1+x^3\right )}{5 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {\left (2 \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{5 \sqrt {1+x} \sqrt {1-x+x^2}} \\ & = \frac {2 x \left (1+x^3\right )}{5 \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 )}{5 \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 = 21.43 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.19 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {6 x \sqrt {1+x} \left (1-x+x^2\right )-\frac {2 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{15 \sqrt {1-x+x^2}} \]

[In]

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

[Out]

(6*x*Sqrt[1 + x]*(1 - x + x^2) - ((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])])/(15*Sqrt[1 - x + x^2])

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x \sqrt {x^{3}+1}}{5}-\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{5 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(157\)
risch	\(\frac {2 x \sqrt {1+x}\, \sqrt {x^{2}-x +1}}{5}-\frac {4 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{5 \sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(164\)
default	\(\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+x^{4}+x \right )}{5 \left (x^{3}+1\right )}\)	\(248\)

[In]

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

[Out]

((1+x)*(x^2-x+1))^(1/2)/(1+x)^(1/2)/(x^2-x+1)^(1/2)*(2/5*x*(x^3+1)^(1/2)-4/5*(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 = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.18 \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\frac {2}{5} \, \sqrt {x^{2} - x + 1} \sqrt {x + 1} x - \frac {4}{5} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \]

[In]

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

[Out]

2/5*sqrt(x^2 - x + 1)*sqrt(x + 1)*x - 4/5*weierstrassPInverse(0, -4, x)

Sympy [F]

\[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x^{3}}{\sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {x^{3}}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int { \frac {x^{3}}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx=\int \frac {x^3}{\sqrt {x+1}\,\sqrt {x^2-x+1}} \,d x \]

[In]

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

[Out]

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