\(\int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x} x^2} \, dx\) [491]

Optimal result

Integrand size = 22, antiderivative size = 150 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x} x^2} \, dx=-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {\sqrt [3]{2} (2-x)^{2/3}}{\sqrt {3} \sqrt [3]{1-x}}\right )}{4 \sqrt [3]{2}}-\frac {1}{4} (1-x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-1+x\right )+\frac {3 \log \left (-\sqrt [3]{1-x}+\frac {(2-x)^{2/3}}{2^{2/3}}\right )}{8 \sqrt [3]{2}}-\frac {\log (x)}{4 \sqrt [3]{2}} \] Output:

-1/2*(1-x)^(2/3)*(2-x)^(2/3)/x-1/8*3^(1/2)*arctan(1/3*3^(1/2)+1/3*2^(1/3)* 
(2-x)^(2/3)*3^(1/2)/(1-x)^(1/3))*2^(2/3)-1/4*(1-x)^(2/3)*hypergeom([1/3, 2 
/3],[5/3],-1+x)+3/16*ln(-(1-x)^(1/3)+1/2*(2-x)^(2/3)*2^(1/3))*2^(2/3)-1/8* 
ln(x)*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 21.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x} x^2} \, dx=-\frac {(1-x)^{2/3} \left (5 (2-x)^{2/3}+10 x \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},-1+x,1-x\right )+(-1+x) x \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{3},1,\frac {8}{3},-1+x,1-x\right )\right )}{10 x} \] Input:

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

Output:

-1/10*((1 - x)^(2/3)*(5*(2 - x)^(2/3) + 10*x*AppellF1[2/3, 1/3, 1, 5/3, -1 
 + x, 1 - x] + (-1 + x)*x*AppellF1[5/3, 1/3, 1, 8/3, -1 + x, 1 - x]))/x
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {134, 27, 175, 79, 133}

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.

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

\(\Big \downarrow \) 134

\(\displaystyle \frac {1}{12} \int \frac {2 (x+3)}{\sqrt [3]{1-x} \sqrt [3]{2-x} x}dx-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \frac {x+3}{\sqrt [3]{1-x} \sqrt [3]{2-x} x}dx-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{6} \left (\int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x}}dx+3 \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x} x}dx\right )-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {1}{6} \left (3 \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x} x}dx-\frac {3}{2} (1-x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x-1\right )\right )-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}\)

\(\Big \downarrow \) 133

\(\displaystyle \frac {1}{6} \left (3 \left (-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{2} (2-x)^{2/3}}{\sqrt {3} \sqrt [3]{1-x}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {3 \log \left (\frac {(2-x)^{2/3}}{2^{2/3}}-\sqrt [3]{1-x}\right )}{4 \sqrt [3]{2}}-\frac {\log (x)}{2 \sqrt [3]{2}}\right )-\frac {3}{2} (1-x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x-1\right )\right )-\frac {(1-x)^{2/3} (2-x)^{2/3}}{2 x}\)

Input:

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

Output:

-1/2*((1 - x)^(2/3)*(2 - x)^(2/3))/x + ((-3*(1 - x)^(2/3)*Hypergeometric2F 
1[1/3, 2/3, 5/3, -1 + x])/2 + 3*(-1/2*(Sqrt[3]*ArcTan[1/Sqrt[3] + (2^(1/3) 
*(2 - x)^(2/3))/(Sqrt[3]*(1 - x)^(1/3))])/2^(1/3) + (3*Log[-(1 - x)^(1/3) 
+ (2 - x)^(2/3)/2^(2/3)])/(4*2^(1/3)) - Log[x]/(2*2^(1/3))))/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 133

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x_)) 
^(1/3)), x_] :> With[{q = Rt[b*((b*e - a*f)/(b*c - a*d)^2), 3]}, Simp[-Log[ 
a + b*x]/(2*q*(b*c - a*d)), x] + (-Simp[Sqrt[3]*(ArcTan[1/Sqrt[3] + 2*q*((c 
 + d*x)^(2/3)/(Sqrt[3]*(e + f*x)^(1/3)))]/(2*q*(b*c - a*d))), x] + Simp[3*( 
Log[q*(c + d*x)^(2/3) - (e + f*x)^(1/3)]/(4*q*(b*c - a*d))), x])] /; FreeQ[ 
{a, b, c, d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0]
 

rule 134

Int[((a_.) + (b_.)*(x_))^(m_)/(((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x 
_))^(1/3)), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(2/3)*((e + f*x)^(2/3 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[f/(6*(m + 1)*(b*c - a*d)*(b 
*e - a*f))   Int[(a + b*x)^(m + 1)*((a*d*(3*m + 1) - 3*b*c*(3*m + 5) - 2*b* 
d*(3*m + 7)*x)/((c + d*x)^(1/3)*(e + f*x)^(1/3))), x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0] && ILtQ[m, -1]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((-x + 2)^(2/3)*(-x + 1)^(2/3)/(x^4 - 3*x^3 + 2*x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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