3.5.89 \(\int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx\) [489]

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

[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 = 0.04, 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 = {929, 285, 327, 224} \begin {gather*} -\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 (\text {ArcSin}\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 {6}{55} \sqrt {x+1} \sqrt {x^2-x+1} x+\frac {2}{11} \sqrt {x+1} \sqrt {x^2-x+1} x^4 \end {gather*}

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 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 285

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 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*} \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] Result contains complex when optimal does not.
time = 40.63, size = 221, normalized size = 1.30 \begin {gather*} \frac {2 \left (x \sqrt {1+x} \left (3-3 x+3 x^2+5 x^3-5 x^4+5 x^5\right )+\sqrt {-\frac {6 i}{3 i+\sqrt {3}}} \left (3 i+\sqrt {3}\right ) (1+x) \sqrt {\frac {3 i+\sqrt {3}+\left (-3 i+\sqrt {3}\right ) x}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {\frac {-3 i+\sqrt {3}+\left (3 i+\sqrt {3}\right ) x}{\left (3 i+\sqrt {3}\right ) (1+x)}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right )|\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )\right )}{55 \sqrt {1-x+x^2}} \end {gather*}

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 = 0.24, size = 257, normalized size = 1.51

method	result	size
elliptic	\(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x^{4} \sqrt {x^{3}+1}}{11}+\frac {6 x \sqrt {x^{3}+1}}{55}-\frac {12 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{55 \sqrt {x^{3}+1}}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(169\)
risch	\(\frac {2 x \left (5 x^{3}+3\right ) \sqrt {1+x}\, \sqrt {x^{2}-x +1}}{55}-\frac {12 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{55 \sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(171\)
default	\(\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (5 x^{7}+3 i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-9 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+8 x^{4}+3 x \right )}{55 \left (x^{3}+1\right )}\)	\(257\)

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,method=_RETURNVERBOSE)

[Out]

2/55*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(5*x^7+3*I*3^(1/2)*(-2*(1+x)/(-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*(1+x)/(-3+I*3^(1/2)))^(1/2),(-(-3+I*3^(1/
2))/(I*3^(1/2)+3))^(1/2))-9*(-2*(1+x)/(-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*(1+x)/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.55, size = 33, normalized size = 0.19 \begin {gather*} \frac {2}{55} \, {\left (5 \, x^{4} + 3 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} - \frac {12}{55} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \end {gather*}

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]

2/55*(5*x^4 + 3*x)*sqrt(x^2 - x + 1)*sqrt(x + 1) - 12/55*weierstrassPInverse(0, -4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sqrt {x + 1} \sqrt {x^{2} - x + 1}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\sqrt {x+1}\,\sqrt {x^2-x+1} \,d x \end {gather*}

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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