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

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

[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))

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Rubi [A]  time = 0.0656172, antiderivative size = 170, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [C]  time = 0.926628, size = 221, normalized size = 1.3 \[ \frac{2 \left (x \sqrt{x+1} \left (5 x^5-5 x^4+5 x^3+3 x^2-3 x+3\right )+\sqrt{-\frac{6 i}{\sqrt{3}+3 i}} \left (\sqrt{3}+3 i\right ) (x+1) \sqrt{\frac{\left (\sqrt{3}-3 i\right ) x+\sqrt{3}+3 i}{\left (\sqrt{3}-3 i\right ) (x+1)}} \sqrt{\frac{\left (\sqrt{3}+3 i\right ) x+\sqrt{3}-3 i}{\left (\sqrt{3}+3 i\right ) (x+1)}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{6 i}{\sqrt{3}+3 i}}}{\sqrt{x+1}}\right ),\frac{\sqrt{3}+3 i}{-\sqrt{3}+3 i}\right )\right )}{55 \sqrt{x^2-x+1}} \]

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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Maple [A]  time = 0.589, size = 257, normalized size = 1.5 \begin{align*}{\frac{2}{55\,{x}^{3}+55}\sqrt{1+x}\sqrt{{x}^{2}-x+1} \left ( 5\,{x}^{7}+3\,i\sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{i\sqrt{3}+3}}}\sqrt{{\frac{2\,x-1+i\sqrt{3}}{i\sqrt{3}-3}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}},\sqrt{-{\frac{i\sqrt{3}-3}{i\sqrt{3}+3}}} \right ) \sqrt{3}-9\,\sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}}\sqrt{{\frac{i\sqrt{3}-2\,x+1}{i\sqrt{3}+3}}}\sqrt{{\frac{2\,x-1+i\sqrt{3}}{i\sqrt{3}-3}}}{\it EllipticF} \left ( \sqrt{-2\,{\frac{1+x}{i\sqrt{3}-3}}},\sqrt{-{\frac{i\sqrt{3}-3}{i\sqrt{3}+3}}} \right ) +8\,{x}^{4}+3\,x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{2} - x + 1} \sqrt{x + 1} x^{3}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{x^{2} - x + 1} \sqrt{x + 1} x^{3}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{x + 1} \sqrt{x^{2} - x + 1}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x^{2} - x + 1} \sqrt{x + 1} x^{3}\,{d x} \end{align*}

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)