\(\int x^3 (1+x)^{3/2} (1-x+x^2)^{3/2} \, dx\) [496]

Optimal result

Integrand size = 23, antiderivative size = 201 \[ \int x^3 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {54}{935} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {18}{187} x^4 \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{17} x^4 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )-\frac {36\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{935 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \]

[Out]

54/935*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)+18/187*x^4*(1+x)^(1/2)*(x^2-x+1)^(1/2)+2/17*x^4*(x^3+1)*(1+x)^(1/2)*(x^2-
x+1)^(1/2)-36/935*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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {929, 285, 327, 224} \[ \int x^3 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=-\frac {36\ 3^{3/4} \sqrt {2+\sqrt {3}} (x+1)^{3/2} \sqrt {x^2-x+1} \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{935 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \left (x^3+1\right )}+\frac {54}{935} \sqrt {x+1} \sqrt {x^2-x+1} x+\frac {18}{187} \sqrt {x+1} \sqrt {x^2-x+1} x^4+\frac {2}{17} \sqrt {x+1} \sqrt {x^2-x+1} \left (x^3+1\right ) x^4 \]

[In]

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

[Out]

(54*x*Sqrt[1 + x]*Sqrt[1 - x + x^2])/935 + (18*x^4*Sqrt[1 + x]*Sqrt[1 - x + x^2])/187 + (2*x^4*Sqrt[1 + x]*Sqr
t[1 - x + x^2]*(1 + x^3))/17 - (36*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]])/(935*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*} \text {integral}& = \frac {\left (\sqrt {1+x} \sqrt {1-x+x^2}\right ) \int x^3 \left (1+x^3\right )^{3/2} \, dx}{\sqrt {1+x^3}} \\ & = \frac {2}{17} x^4 \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 x^3 \sqrt {1+x^3} \, dx}{17 \sqrt {1+x^3}} \\ & = \frac {18}{187} x^4 \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{17} x^4 \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 {x^3}{\sqrt {1+x^3}} \, dx}{187 \sqrt {1+x^3}} \\ & = \frac {54}{935} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {18}{187} x^4 \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{17} x^4 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )-\frac {\left (54 \sqrt {1+x} \sqrt {1-x+x^2}\right ) \int \frac {1}{\sqrt {1+x^3}} \, dx}{935 \sqrt {1+x^3}} \\ & = \frac {54}{935} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {18}{187} x^4 \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{17} x^4 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )-\frac {36\ 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 )}{935 \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 = 21.56 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.17 \[ \int x^3 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2 \left (x \sqrt {1+x} \left (27-27 x+27 x^2+100 x^3-100 x^4+100 x^5+55 x^6-55 x^7+55 x^8\right )-\frac {9 i \sqrt {6} (1+x) \sqrt {\frac {3 i+\sqrt {3}+\left (-3 i+\sqrt {3}\right ) x}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {\frac {-3 i+\sqrt {3}+\left (3 i+\sqrt {3}\right ) x}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}\right )}{935 \sqrt {1-x+x^2}} \]

[In]

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

[Out]

(2*(x*Sqrt[1 + x]*(27 - 27*x + 27*x^2 + 100*x^3 - 100*x^4 + 100*x^5 + 55*x^6 - 55*x^7 + 55*x^8) - ((9*I)*Sqrt[
6]*(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])])/Sqrt[(-I)/(3*I + Sqrt[3])]))/(935*Sqrt[1 - x + x^2])

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {2 x \left (55 x^{6}+100 x^{3}+27\right ) \sqrt {1+x}\, \sqrt {x^{2}-x +1}}{935}-\frac {108 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{935 \sqrt {x^{3}+1}\, \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\)	\(176\)
elliptic	\(\frac {\sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x^{7} \sqrt {x^{3}+1}}{17}+\frac {40 x^{4} \sqrt {x^{3}+1}}{187}+\frac {54 x \sqrt {x^{3}+1}}{935}-\frac {108 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{935 \sqrt {x^{3}+1}}\right )}{x^{3}+1}\)	\(188\)
default	\(\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (55 x^{10}+155 x^{7}+27 i \sqrt {3}\, \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )-81 \sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{-3+i \sqrt {3}}}\, F\left (\sqrt {-\frac {2 \left (1+x \right )}{-3+i \sqrt {3}}}, \sqrt {-\frac {-3+i \sqrt {3}}{i \sqrt {3}+3}}\right )+127 x^{4}+27 x \right )}{935 \left (x^{3}+1\right )}\)	\(262\)

[In]

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

[Out]

2/935*x*(55*x^6+100*x^3+27)*(1+x)^(1/2)*(x^2-x+1)^(1/2)-108/935*(3/2-1/2*I*3^(1/2))*((1+x)/(3/2-1/2*I*3^(1/2))
)^(1/2)*((x-1/2-1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x-1/2+1/2*I*3^(1/2))/(-3/2+1/2*I*3^(1/2)))^(1/2)/
(x^3+1)^(1/2)*EllipticF(((1+x)/(3/2-1/2*I*3^(1/2)))^(1/2),((-3/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(1/2)))^(1/2))*(
(1+x)*(x^2-x+1))^(1/2)/(1+x)^(1/2)/(x^2-x+1)^(1/2)

Fricas [C] (verification not implemented)

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

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.19 \[ \int x^3 (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2}{935} \, {\left (55 \, x^{7} + 100 \, x^{4} + 27 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} - \frac {108}{935} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \]

[In]

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

[Out]

2/935*(55*x^7 + 100*x^4 + 27*x)*sqrt(x^2 - x + 1)*sqrt(x + 1) - 108/935*weierstrassPInverse(0, -4, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(x^3*(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^3, x)

Giac [F]

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

[In]

integrate(x^3*(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^3, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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