\(\int \frac {(1+x)^{3/2} (1-x+x^2)^{3/2}}{x^3} \, dx\) [485]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx=-\frac {\sqrt {1+x} \sqrt {1-x+x^2}}{2 x^2}+\frac {2}{5} x \sqrt {1+x} \sqrt {1-x+x^2}+\frac {9\ 3^{3/4} \sqrt {2+\sqrt {3}} (1+x)^{3/2} \sqrt {1-x+x^2} \sqrt {\frac {1-x+x^2}{\left (1+\sqrt {3}+x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1-\sqrt {3}+x}{1+\sqrt {3}+x}\right ),-7-4 \sqrt {3}\right )}{10 \sqrt {\frac {1+x}{\left (1+\sqrt {3}+x\right )^2}} \left (1+x^3\right )} \] Output:

-1/2*(1+x)^(1/2)*(x^2-x+1)^(1/2)/x^2+2/5*x*(1+x)^(1/2)*(x^2-x+1)^(1/2)+9/1 
0*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 = 10.48 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.13 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {1+x} \left (\frac {2 \left (1-x+x^2\right ) \left (-5+4 x^3\right )}{x^2}-\frac {27 i \sqrt {2} \sqrt {\frac {i+\sqrt {3}-2 i x}{3 i+\sqrt {3}}} \sqrt {\frac {-i+\sqrt {3}+2 i x}{-3 i+\sqrt {3}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {-\frac {i (1+x)}{3 i+\sqrt {3}}}\right ),\frac {3 i+\sqrt {3}}{3 i-\sqrt {3}}\right )}{\sqrt {-\frac {i (1+x)}{3 i+\sqrt {3}}}}\right )}{20 \sqrt {1-x+x^2}} \] Input:

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

Output:

(Sqrt[1 + x]*((2*(1 - x + x^2)*(-5 + 4*x^3))/x^2 - ((27*I)*Sqrt[2]*Sqrt[(I 
 + Sqrt[3] - (2*I)*x)/(3*I + Sqrt[3])]*Sqrt[(-I + Sqrt[3] + (2*I)*x)/(-3*I 
 + Sqrt[3])]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3 
])]], (3*I + Sqrt[3])/(3*I - Sqrt[3])])/Sqrt[((-I)*(1 + x))/(3*I + Sqrt[3] 
)]))/(20*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, 809, 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.

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

\(\Big \downarrow \) 1210

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

\(\Big \downarrow \) 809

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {9}{4} \int \sqrt {x^3+1}dx-\frac {\left (x^3+1\right )^{3/2}}{2 x^2}\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\frac {9}{4} \left (\frac {3}{5} \int \frac {1}{\sqrt {x^3+1}}dx+\frac {2}{5} \sqrt {x^3+1} x\right )-\frac {\left (x^3+1\right )^{3/2}}{2 x^2}\right )}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 759

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

Input:

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

Output:

(Sqrt[1 + x]*Sqrt[1 - x + x^2]*(-1/2*(1 + x^3)^(3/2)/x^2 + (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])))/4))/Sq 
rt[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 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[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 = 2.53 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (4 x^{3}-5\right )}{10 x^{2}}+\frac {27 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right ) \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}}{10 \sqrt {x^{3}+1}\, \sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(173\)
elliptic	\(\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x \sqrt {x^{3}+1}}{5}+\frac {27 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +1}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{10 \sqrt {x^{3}+1}}-\frac {\sqrt {x^{3}+1}}{2 x^{2}}\right )}{x^{3}+1}\)	\(176\)
default	\(-\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (27 i \sqrt {3}\, \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{2}-81 \sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}\, \sqrt {\frac {i \sqrt {3}-2 x +1}{i \sqrt {3}+3}}\, \sqrt {\frac {i \sqrt {3}+2 x -1}{i \sqrt {3}-3}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {2 \left (x +1\right )}{i \sqrt {3}-3}}, \sqrt {-\frac {i \sqrt {3}-3}{i \sqrt {3}+3}}\right ) x^{2}-8 x^{6}+2 x^{3}+10\right )}{20 x^{2} \left (x^{3}+1\right )}\)	\(264\)

Input:

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

Output:

1/10*(x+1)^(1/2)*(x^2-x+1)^(1/2)*(4*x^3-5)/x^2+27/10*(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 = 38, normalized size of antiderivative = 0.22 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx=\frac {27 \, x^{2} {\rm weierstrassPInverse}\left (0, -4, x\right ) + {\left (4 \, x^{3} - 5\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{10 \, x^{2}} \] Input:

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

Output:

1/10*(27*x^2*weierstrassPInverse(0, -4, x) + (4*x^3 - 5)*sqrt(x^2 - x + 1) 
*sqrt(x + 1))/x^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x^3} \, dx=\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, x^{3}-16 \sqrt {x +1}\, \sqrt {x^{2}-x +1}-27 \left (\int \frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}}{x^{6}+x^{3}}d x \right ) x^{2}}{5 x^{2}} \] Input:

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

Output:

(2*sqrt(x + 1)*sqrt(x**2 - x + 1)*x**3 - 16*sqrt(x + 1)*sqrt(x**2 - x + 1) 
 - 27*int((sqrt(x + 1)*sqrt(x**2 - x + 1))/(x**6 + x**3),x)*x**2)/(5*x**2)