\(\int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx\) [472]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\frac {6}{55} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{11} x^4 \sqrt {1+x} \sqrt {1-x+x^2}-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{55 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \] Output:

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)*(1/2*6^(1/2)+1/2*2^(1/2))*(1+x)^(3/2)*(x^2-x+1)^(1/2)*((x^2-x+1 
)/(1+x+3^(1/2))^2)^(1/2)*EllipticF((1+x-3^(1/2))/(1+x+3^(1/2)),I*3^(1/2)+2 
*I)/((1+x)/(1+x+3^(1/2))^2)^(1/2)/(x^3+1)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 33.11 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.30 \[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\frac {2 \left (x \sqrt {1+x} \left (3-3 x+3 x^2+5 x^3-5 x^4+5 x^5\right )+\sqrt {-\frac {6 i}{3 i+\sqrt {3}}} \left (3 i+\sqrt {3}\right ) (1+x) \sqrt {\frac {3 i+\sqrt {3}+\left (-3 i+\sqrt {3}\right ) x}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {\frac {-3 i+\sqrt {3}+\left (3 i+\sqrt {3}\right ) x}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )\right )}{55 \sqrt {1-x+x^2}} \] Input:

Integrate[x^3*Sqrt[1 + x]*Sqrt[1 - x + x^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1210, 811, 843, 759}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1210

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \int x^3 \sqrt {x^3+1}dx}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {3}{11} \int \frac {x^3}{\sqrt {x^3+1}}dx+\frac {2}{11} \sqrt {x^3+1} x^4\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {3}{11} \left (\frac {2}{5} x \sqrt {x^3+1}-\frac {2}{5} \int \frac {1}{\sqrt {x^3+1}}dx\right )+\frac {2}{11} \sqrt {x^3+1} x^4\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {3}{11} \left (\frac {2}{5} x \sqrt {x^3+1}-\frac {4 \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}\right )+\frac {2}{11} \sqrt {x^3+1} x^4\right )}{\sqrt {x^3+1}}\)

Input:

Int[x^3*Sqrt[1 + x]*Sqrt[1 - x + x^2],x]
 

Output:

(Sqrt[1 + x]*Sqrt[1 - x + x^2]*((2*x^4*Sqrt[1 + x^3])/11 + (3*((2*x*Sqrt[1 
 + x^3])/5 - (4*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] 
+ x)^2]*EllipticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt 
[3]])/(5*3^(1/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/11))/S 
qrt[1 + x^3]
 

Defintions of rubi rules used

rule 759

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]*Sqrt[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 811

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] + Simp[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] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 843

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] - Simp[ 
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 1210

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + 
c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p])   Int[(f + g*x)^n*(a*d + c* 
e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 
 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
 

Maple [A] (verified)

Time = 3.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.99

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

Input:

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

Output:

((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)*(2/11*x^4*(x^3+1)^(1/2 
)+6/55*x*(x^3+1)^(1/2)-12/55*(3/2-1/2*I*3^(1/2))*((x+1)/(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(((x+1)/(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)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.19 \[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\frac {2}{55} \, {\left (5 \, x^{4} + 3 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} - \frac {12}{55} \, {\rm weierstrassPInverse}\left (0, -4, x\right ) \] Input:

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

Output:

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

Sympy [F]

\[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\int x^{3} \sqrt {x + 1} \sqrt {x^{2} - x + 1}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\int { \sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{3} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\int { \sqrt {x^{2} - x + 1} \sqrt {x + 1} x^{3} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\int x^3\,\sqrt {x+1}\,\sqrt {x^2-x+1} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^3 \sqrt {1+x} \sqrt {1-x+x^2} \, dx=\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x^{4}}{11}+\frac {6 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x}{55}-\frac {6 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{3}+1}d x \right )}{55} \] Input:

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

Output:

(2*(5*sqrt(x + 1)*sqrt(x**2 - x + 1)*x**4 + 3*sqrt(x + 1)*sqrt(x**2 - x + 
1)*x - 3*int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x**3 + 1),x)))/55