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

Optimal. Leaf size=142 \[ \frac {2 x \left (x^3+1\right )}{5 \sqrt {x+1} \sqrt {x^2-x+1}}-\frac {4 \sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}} \]

[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]  time = 0.05, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {915, 321, 218} \[ \frac {2 x \left (x^3+1\right )}{5 \sqrt {x+1} \sqrt {x^2-x+1}}-\frac {4 \sqrt {2+\sqrt {3}} \sqrt {x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[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 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((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]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 321

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 915

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*} \int \frac {x^3}{\sqrt {1+x} \sqrt {1-x+x^2}} \, dx &=\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]  time = 0.61, size = 169, normalized size = 1.19 \[ \frac {6 x \sqrt {x+1} \left (x^2-x+1\right )-\frac {2 i (x+1) \sqrt {1+\frac {6 i}{\left (\sqrt {3}-3 i\right ) (x+1)}} \sqrt {6-\frac {36 i}{\left (\sqrt {3}+3 i\right ) (x+1)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {x+1}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{\sqrt {3}+3 i}}}}{15 \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[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])

________________________________________________________________________________________

fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{3}}{x^{3} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 0.04, size = 248, normalized size = 1.75 \[ \frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (x^{4}+x +i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-3 \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )\right )}{5 \left (x^{3}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(x+1)^(1/2)*(x^2-x+1)^(1/2)*(I*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((-2*x+I*3^(1/2)+1)/(I*3^(1/2)+3))^(1/2)*((
2*x+I*3^(1/2)-1)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3
))^(1/2))*3^(1/2)-3*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((-2*x+I*3^(1/2)+1)/(I*3^(1/2)+3))^(1/2)*((2*x+I*3^(1/2)-1
)/(-3+I*3^(1/2)))^(1/2)*EllipticF((-2*(x+1)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/2))/(I*3^(1/2)+3))^(1/2))+x^4+x
)/(x^3+1)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x^{2} - x + 1} \sqrt {x + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{\sqrt {x+1}\,\sqrt {x^2-x+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {x + 1} \sqrt {x^{2} - x + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________