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

Optimal. Leaf size=170 \[ \frac {6}{55} \sqrt {x+1} \sqrt {x^2-x+1} x+\frac {2}{11} \sqrt {x+1} \sqrt {x^2-x+1} x^4-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} (x+1)^{3/2} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )} \]

[Out]

6/55*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)+2/11*x^4*(1+x)^(1/2)*(x^2-x+1)^(1/2)-4/55*3^(3/4)*(1+x)^(3/2)*EllipticF((1+
x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2*I)*(x^2-x+1)^(1/2)*(1/2*6^(1/2)+1/2*2^(1/2))*((x^2-x+1)/(1+x+3^(1/2))^2)^
(1/2)/(x^3+1)/((1+x)/(1+x+3^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {915, 279, 321, 218} \[ \frac {2}{11} \sqrt {x+1} \sqrt {x^2-x+1} x^4+\frac {6}{55} \sqrt {x+1} \sqrt {x^2-x+1} x-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} (x+1)^{3/2} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/55 + (2*x^4*Sqrt[1 + x]*Sqrt[1 - x + x^2])/11 - (4*3^(3/4)*Sqrt[2 + Sqrt[3
]]*(1 + x)^(3/2)*Sqrt[1 - x + x^2]*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/
(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(55*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx &=\frac {\left (\sqrt {1+x} \sqrt {1-x+x^2}\right ) \int x^3 \sqrt {1+x^3} \, dx}{\sqrt {1+x^3}}\\ &=\frac {2}{11} x^4 \sqrt {1+x} \sqrt {1-x+x^2}+\frac {\left (3 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {x^3}{\sqrt {1+x^3}} \, dx}{11 \sqrt {1+x^3}}\\ &=\frac {6}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x^4 \sqrt {1+x} \sqrt {1-x+x^2}-\frac {\left (6 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{55 \sqrt {1+x^3}}\\ &=\frac {6}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x^4 \sqrt {1+x} \sqrt {1-x+x^2}-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \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 )}{55 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )}\\ \end {align*}

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Mathematica [C]  time = 0.92, size = 221, normalized size = 1.30 \[ \frac {2 \left (x \sqrt {x+1} \left (5 x^5-5 x^4+5 x^3+3 x^2-3 x+3\right )+\sqrt {-\frac {6 i}{\sqrt {3}+3 i}} \left (\sqrt {3}+3 i\right ) (x+1) \sqrt {\frac {\left (\sqrt {3}-3 i\right ) x+\sqrt {3}+3 i}{\left (\sqrt {3}-3 i\right ) (x+1)}} \sqrt {\frac {\left (\sqrt {3}+3 i\right ) x+\sqrt {3}-3 i}{\left (\sqrt {3}+3 i\right ) (x+1)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {x+1}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )\right )}{55 \sqrt {x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*Sqrt[1 + x]*(3 - 3*x + 3*x^2 + 5*x^3 - 5*x^4 + 5*x^5) + Sqrt[(-6*I)/(3*I + Sqrt[3])]*(3*I + Sqrt[3])*(1
+ x)*Sqrt[(3*I + Sqrt[3] + (-3*I + Sqrt[3])*x)/((-3*I + Sqrt[3])*(1 + x))]*Sqrt[(-3*I + Sqrt[3] + (3*I + Sqrt[
3])*x)/((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])]))/(55*Sqrt[1 - x + x^2])

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{3}, x\right ) \]

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)

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

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

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maple [A]  time = 0.14, size = 257, normalized size = 1.51 \[ \frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (5 x^{7}+8 x^{4}+3 x +3 i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-9 \sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {-2 x +i \sqrt {3}+1}{i \sqrt {3}+3}}\, \sqrt {\frac {2 x +i \sqrt {3}-1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (x +1\right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )\right )}{55 \left (x^{3}+1\right )} \]

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]

2/55*(x+1)^(1/2)*(x^2-x+1)^(1/2)*(5*x^7+3*I*3^(1/2)*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((I*3^(1/2)-2*x+1)/(I*3^(1
/2)+3))^(1/2)*((I*3^(1/2)+2*x-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))-9*(-2*(x+1)/(-3+I*3^(1/2)))^(1/2)*((I*3^(1/2)-2*x+1)/(I*3^(1/2)+3))^(1/2)*((I*3^(1/2
)+2*x-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)
)+8*x^4+3*x)/(x^3+1)

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

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

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

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)

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

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)

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