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

Optimal result

Integrand size = 20, antiderivative size = 173 \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {18}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )+\frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {727, 201, 224} \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} (x+1)^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )}+\frac {18}{55} x \sqrt {x^2-x+1} \sqrt {x+1}+\frac {2}{11} x \sqrt {x^2-x+1} \left (x^3+1\right ) \sqrt {x+1} \]

[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*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 20.48 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02 \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2 x \sqrt {1+x} \left (1-x+x^2\right ) \left (14+5 x^3\right )+\frac {9 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{55 \sqrt {1-x+x^2}} \]

[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])

Maple [F(-1)]

Timed out.

hanged

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2}{55} \, {\left (5 \, x^{4} + 14 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + \frac {54}{55} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \]

[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)

Sympy [F]

\[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\int \left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}\, dx \]

[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)

Maxima [F]

\[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\int { {\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}} \,d x } \]

[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)

Giac [F]

\[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\int { {\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\int {\left (x+1\right )}^{3/2}\,{\left (x^2-x+1\right )}^{3/2} \,d x \]

[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)