3.5.99 \(\int (1+x)^{3/2} (1-x+x^2)^{3/2} \, dx\) [499]

Optimal. Leaf size=173 \[ \frac {18}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )+\frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \]

[Out]

18/55*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)+2/11*x*(x^3+1)*(1+x)^(1/2)*(x^2-x+1)^(1/2)+18/55*3^(3/4)*(1+x)^(3/2)*Ellip
ticF((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 = 0.05, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {727, 201, 224} \begin {gather*} \frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} (x+1)^{3/2} 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 {18}{55} x \sqrt {x^2-x+1} \sqrt {x+1}+\frac {2}{11} x \sqrt {x^2-x+1} \left (x^3+1\right ) \sqrt {x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(18*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/55 + (2*x*Sqrt[1 + x]*Sqrt[1 - x + x^2]*(1 + x^3))/11 + (18*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 - Sqr
t[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(55*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*(1 + x^3))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 727

Int[((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[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && IGtQ[m - p + 1, 0] &&  !Intege
rQ[p]

Rubi steps

\begin {align*} \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx &=\frac {\left (\sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \left (1+x^3\right )^{3/2} \, dx}{\sqrt {1+x^3}}\\ &=\frac {2}{11} x \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )+\frac {\left (9 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \sqrt {1+x^3} \, dx}{11 \sqrt {1+x^3}}\\ &=\frac {18}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )+\frac {\left (27 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{55 \sqrt {1+x^3}}\\ &=\frac {18}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )+\frac {18\ 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 = 20.44, size = 176, normalized size = 1.02 \begin {gather*} \frac {2 x \sqrt {1+x} \left (1-x+x^2\right ) \left (14+5 x^3\right )+\frac {9 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)}} 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 )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{55 \sqrt {1-x+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[1 + x]*(1 - x + x^2)*(14 + 5*x^3) + ((9*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 + Sqr
t[3])/(3*I - Sqrt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(55*Sqrt[1 - x + x^2])

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Maple [A]
time = 0.11, size = 257, normalized size = 1.49

method	result	size
risch	\(\frac {2 x \left (5 x^{3}+14\right ) \sqrt {1+x}\, \sqrt {x^{2}-x +1}}{55}+\frac {54 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{55 \sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(171\)
elliptic	\(\frac {\sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x^{4} \sqrt {x^{3}+1}}{11}+\frac {28 x \sqrt {x^{3}+1}}{55}+\frac {54 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \EllipticF \left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{55 \sqrt {x^{3}+1}}\right )}{x^{3}+1}\)	\(176\)
default	\(-\frac {\sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (-10 x^{7}+27 i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-81 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, \EllipticF \left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-38 x^{4}-28 x \right )}{55 \left (x^{3}+1\right )}\)	\(257\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/55*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(-10*x^7+27*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))-81*(-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))-38*x^4-28*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((1+x)^(3/2)*(x^2-x+1)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.44, size = 33, normalized size = 0.19 \begin {gather*} \frac {2}{55} \, {\left (5 \, x^{4} + 14 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + \frac {54}{55} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/55*(5*x^4 + 14*x)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 54/55*weierstrassPInverse(0, -4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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