\(\int \frac {1}{(1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\) [503]

Optimal result

Integrand size = 20, antiderivative size = 168 \[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {14 x}{27 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2 x}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}+\frac {14 \sqrt {2+\sqrt {3}} \sqrt {1+x} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \sqrt {1-x+x^2}} \] Output:

14/27*x/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9*x/(1+x)^(1/2)/(x^2-x+1)^(1/2)/(x^3 
+1)+14/81*(1/2*6^(1/2)+1/2*2^(1/2))*(1+x)^(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)*3^(3/4)/((1+x 
)/(1+x+3^(1/2))^2)^(1/2)/(x^2-x+1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 20.94 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {\frac {6 x \left (10+7 x^3\right )}{(1+x)^{3/2} \left (1-x+x^2\right )}+\frac {7 i (1+x) \sqrt {1+\frac {6 i}{\left (-3 i+\sqrt {3}\right ) (1+x)}} \sqrt {6-\frac {36 i}{\left (3 i+\sqrt {3}\right ) (1+x)}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {6 i}{3 i+\sqrt {3}}}}{\sqrt {1+x}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i}{3 i+\sqrt {3}}}}}{81 \sqrt {1-x+x^2}} \] Input:

Integrate[1/((1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
 

Output:

((6*x*(10 + 7*x^3))/((1 + x)^(3/2)*(1 - x + x^2)) + ((7*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 + 
 Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[(-I)/(3*I + Sqrt[3])])/(81*Sqrt[1 - x + x 
^2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1151, 749, 749, 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.

\(\displaystyle \int \frac {1}{(x+1)^{5/2} \left (x^2-x+1\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1151

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

\(\Big \downarrow \) 749

\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {7}{9} \int \frac {1}{\left (x^3+1\right )^{3/2}}dx+\frac {2 x}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 749

\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {7}{9} \left (\frac {1}{3} \int \frac {1}{\sqrt {x^3+1}}dx+\frac {2 x}{3 \sqrt {x^3+1}}\right )+\frac {2 x}{9 \left (x^3+1\right )^{3/2}}\right )}{\sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 759

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

Input:

Int[1/((1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
 

Output:

(Sqrt[1 + x^3]*((2*x)/(9*(1 + x^3)^(3/2)) + (7*((2*x)/(3*Sqrt[1 + x^3]) + 
(2*Sqrt[2 + Sqrt[3]]*(1 + x)*Sqrt[(1 - x + x^2)/(1 + Sqrt[3] + x)^2]*Ellip 
ticF[ArcSin[(1 - Sqrt[3] + x)/(1 + Sqrt[3] + x)], -7 - 4*Sqrt[3]])/(3*3^(1 
/4)*Sqrt[(1 + x)/(1 + Sqrt[3] + x)^2]*Sqrt[1 + x^3])))/9))/(Sqrt[1 + x]*Sq 
rt[1 - x + x^2])
 

Defintions of rubi rules used

rule 749

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 
 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1))   Int[(a + b*x^ 
n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte 
gerQ[2*p] || Denominator[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 = 2.79 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.99

method	result	size
elliptic	\(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}+\frac {14 x}{27 \sqrt {x^{3}+1}}+\frac {14 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{27 \sqrt {x^{3}+1}}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(167\)
default	\(-\frac {7 i \sqrt {3}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{3} \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}-21 \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{3} \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}+7 i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )-21 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right )-14 x^{4}-20 x}{27 \left (x^{2}-x +1\right )^{\frac {3}{2}} \left (x +1\right )^{\frac {3}{2}}}\)	\(469\)

Input:

int(1/(x+1)^(5/2)/(x^2-x+1)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)*(2/9*x/(x^3+1)^(3/2)+1 
4/27*x/(x^3+1)^(1/2)+14/27*(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 = 56, normalized size of antiderivative = 0.33 \[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left (7 \, x^{4} + 10 \, x\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} + 7 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} {\rm weierstrassPInverse}\left (0, -4, x\right )\right )}}{27 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \] Input:

integrate(1/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="fricas")
 

Output:

2/27*((7*x^4 + 10*x)*sqrt(x^2 - x + 1)*sqrt(x + 1) + 7*(x^6 + 2*x^3 + 1)*w 
eierstrassPInverse(0, -4, x))/(x^6 + 2*x^3 + 1)
 

Sympy [F]

\[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(1+x)**(5/2)/(x**2-x+1)**(5/2),x)
 

Output:

Integral(1/((x + 1)**(5/2)*(x**2 - x + 1)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)), x)
 

Giac [F]

\[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="giac")
 

Output:

integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \] Input:

int(1/((x + 1)^(5/2)*(x^2 - x + 1)^(5/2)),x)
 

Output:

int(1/((x + 1)^(5/2)*(x^2 - x + 1)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{9}+3 x^{6}+3 x^{3}+1}d x \] Input:

int(1/(1+x)^(5/2)/(x^2-x+1)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x**9 + 3*x**6 + 3*x**3 + 1),x)