\(\int (1+x)^{3/2} (1-x+x^2)^{3/2} \, dx\) [482]

Optimal result

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

18/55*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)+2/11*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(x^ 
3+1)+18/55*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 = 20.83 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02 \[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2 x \sqrt {1+x} \left (1-x+x^2\right ) \left (14+5 x^3\right )+\frac {9 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{55 \sqrt {1-x+x^2}} \] Input:

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

Output:

(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))]*E 
llipticF[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])])/(55*Sqrt[1 - x + x^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1151, 748, 748, 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.

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

\(\Big \downarrow \) 1151

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

\(\Big \downarrow \) 748

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {9}{11} \int \sqrt {x^3+1}dx+\frac {2}{11} x \left (x^3+1\right )^{3/2}\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {9}{11} \left (\frac {3}{5} \int \frac {1}{\sqrt {x^3+1}}dx+\frac {2}{5} \sqrt {x^3+1} x\right )+\frac {2}{11} x \left (x^3+1\right )^{3/2}\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {9}{11} \left (\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} (x+1) \sqrt {\frac {x^2-x+1}{\left (x+\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {x-\sqrt {3}+1}{x+\sqrt {3}+1}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt {\frac {x+1}{\left (x+\sqrt {3}+1\right )^2}} \sqrt {x^3+1}}+\frac {2}{5} \sqrt {x^3+1} x\right )+\frac {2}{11} x \left (x^3+1\right )^{3/2}\right )}{\sqrt {x^3+1}}\)

Input:

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

Output:

(Sqrt[1 + x]*Sqrt[1 - x + x^2]*((2*x*(1 + x^3)^(3/2))/11 + (9*((2*x*Sqrt[1 
 + x^3])/5 + (2*3^(3/4)*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*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/11))/S 
qrt[1 + x^3]
 

Defintions of rubi rules used

rule 748

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

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 1151

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[(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] /; F 
reeQ[{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] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99

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

Input:

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

Output:

2/55*x*(5*x^3+14)*(x+1)^(1/2)*(x^2-x+1)^(1/2)+54/55*(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))*((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)
 

Fricas [A] (verification not implemented)

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (1+x)^{3/2} \left (1-x+x^2\right )^{3/2} \, dx=\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x^{4}}{11}+\frac {28 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x}{55}+\frac {27 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{3}+1}d x \right )}{55} \] Input:

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

Output:

(10*sqrt(x + 1)*sqrt(x**2 - x + 1)*x**4 + 28*sqrt(x + 1)*sqrt(x**2 - x + 1 
)*x + 27*int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x**3 + 1),x))/55