\(\int \frac {(1-x)^{5/3}}{(1+x)^{2/3}} \, dx\) [565]

Optimal result

Integrand size = 17, antiderivative size = 112 \[ \int \frac {(1-x)^{5/3}}{(1+x)^{2/3}} \, dx=\frac {5}{3} (1-x)^{2/3} \sqrt [3]{1+x}+\frac {1}{2} (1-x)^{5/3} \sqrt [3]{1+x}+\frac {20 \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{1+x}}\right )}{3 \sqrt {3}}+\frac {10}{9} \log (1+x)+\frac {10}{3} \log \left (1+\frac {\sqrt [3]{1-x}}{\sqrt [3]{1+x}}\right ) \] Output:

5/3*(1-x)^(2/3)*(1+x)^(1/3)+1/2*(1-x)^(5/3)*(1+x)^(1/3)-20/9*arctan(-1/3*3 
^(1/2)+2/3*(1-x)^(1/3)*3^(1/2)/(1+x)^(1/3))*3^(1/2)+10/9*ln(1+x)+10/3*ln(1 
+(1-x)^(1/3)/(1+x)^(1/3))
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10 \[ \int \frac {(1-x)^{5/3}}{(1+x)^{2/3}} \, dx=\frac {1}{18} \left (3 (13-3 x) (1-x)^{2/3} \sqrt [3]{1+x}-40 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{-2 \sqrt [3]{1-x}+\sqrt [3]{1+x}}\right )+40 \log \left (\sqrt [3]{1-x}+\sqrt [3]{1+x}\right )-20 \log \left ((1-x)^{2/3}+(1+x)^{2/3}-\sqrt [3]{1-x^2}\right )\right ) \] Input:

Integrate[(1 - x)^(5/3)/(1 + x)^(2/3),x]
 

Output:

(3*(13 - 3*x)*(1 - x)^(2/3)*(1 + x)^(1/3) - 40*Sqrt[3]*ArcTan[(Sqrt[3]*(1 
+ x)^(1/3))/(-2*(1 - x)^(1/3) + (1 + x)^(1/3))] + 40*Log[(1 - x)^(1/3) + ( 
1 + x)^(1/3)] - 20*Log[(1 - x)^(2/3) + (1 + x)^(2/3) - (1 - x^2)^(1/3)])/1 
8
 

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {60, 60, 72}

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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 72

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

Input:

Int[(1 - x)^(5/3)/(1 + x)^(2/3),x]
 

Output:

((1 - x)^(5/3)*(1 + x)^(1/3))/2 + (5*((1 - x)^(2/3)*(1 + x)^(1/3) + (4*(Sq 
rt[3]*ArcTan[1/Sqrt[3] - (2*(1 - x)^(1/3))/(Sqrt[3]*(1 + x)^(1/3))] + Log[ 
1 + x]/2 + (3*Log[1 + (1 - x)^(1/3)/(1 + x)^(1/3)])/2))/3))/3
 

Defintions of rubi rules used

rule 60

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

rule 72

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> 
 With[{q = Rt[-d/b, 3]}, Simp[Sqrt[3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b* 
x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a + b* 
x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; F 
reeQ[{a, b, c, d}, x] && NegQ[d/b]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.41 (sec) , antiderivative size = 417, normalized size of antiderivative = 3.72

method	result	size
risch	\(\frac {\left (-13+3 x \right ) \left (1+x \right )^{\frac {1}{3}} \left (-1+x \right ) \left (\left (1-x \right ) \left (1+x \right )^{2}\right )^{\frac {1}{3}}}{6 \left (-\left (-1+x \right ) \left (1+x \right )^{2}\right )^{\frac {1}{3}} \left (1-x \right )^{\frac {1}{3}}}+\frac {\left (\frac {20 \ln \left (\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}-x^{2}+x +1\right )^{\frac {2}{3}}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}}+3 \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -x^{2}+3 \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1}{1+x}\right )}{9}+\frac {20 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (-x^{3}-x^{2}+x +1\right )^{\frac {2}{3}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+3 \left (-x^{3}-x^{2}+x +1\right )^{\frac {2}{3}}+3 \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}} x -6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -2 x^{2}+3 \left (-x^{3}-x^{2}+x +1\right )^{\frac {1}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-4 x -2}{1+x}\right )}{9}\right ) \left (\left (1-x \right ) \left (1+x \right )^{2}\right )^{\frac {1}{3}}}{\left (1+x \right )^{\frac {2}{3}} \left (1-x \right )^{\frac {1}{3}}}\)	\(417\)

Input:

int((1-x)^(5/3)/(1+x)^(2/3),x,method=_RETURNVERBOSE)
 

Output:

1/6*(-13+3*x)*(1+x)^(1/3)*(-1+x)/(-(-1+x)*(1+x)^2)^(1/3)*((1-x)*(1+x)^2)^( 
1/3)/(1-x)^(1/3)+(20/9*ln((-4*RootOf(_Z^2+_Z+1)^2*x^2+3*RootOf(_Z^2+_Z+1)* 
(-x^3-x^2+x+1)^(2/3)+3*RootOf(_Z^2+_Z+1)*(-x^3-x^2+x+1)^(1/3)*x-4*RootOf(_ 
Z^2+_Z+1)^2*x-4*RootOf(_Z^2+_Z+1)*x^2+3*RootOf(_Z^2+_Z+1)*(-x^3-x^2+x+1)^( 
1/3)+3*(-x^3-x^2+x+1)^(1/3)*x-2*RootOf(_Z^2+_Z+1)*x-x^2+3*(-x^3-x^2+x+1)^( 
1/3)+2*RootOf(_Z^2+_Z+1)+1)/(1+x))+20/9*RootOf(_Z^2+_Z+1)*ln(-(-2*RootOf(_ 
Z^2+_Z+1)^2*x^2+3*RootOf(_Z^2+_Z+1)*(-x^3-x^2+x+1)^(2/3)-2*RootOf(_Z^2+_Z+ 
1)^2*x-5*RootOf(_Z^2+_Z+1)*x^2+3*(-x^3-x^2+x+1)^(2/3)+3*(-x^3-x^2+x+1)^(1/ 
3)*x-6*RootOf(_Z^2+_Z+1)*x-2*x^2+3*(-x^3-x^2+x+1)^(1/3)-RootOf(_Z^2+_Z+1)- 
4*x-2)/(1+x)))/(1+x)^(2/3)*((1-x)*(1+x)^2)^(1/3)/(1-x)^(1/3)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.11 \[ \int \frac {(1-x)^{5/3}}{(1+x)^{2/3}} \, dx=-\frac {1}{6} \, {\left (3 \, x - 13\right )} {\left (x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {2}{3}} - \frac {20}{9} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {2}{3}}}{3 \, {\left (x - 1\right )}}\right ) - \frac {10}{9} \, \log \left (\frac {x - {\left (x + 1\right )}^{\frac {2}{3}} {\left (-x + 1\right )}^{\frac {1}{3}} + {\left (x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {2}{3}} - 1}{x - 1}\right ) + \frac {20}{9} \, \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {2}{3}} - 1}{x - 1}\right ) \] Input:

integrate((1-x)^(5/3)/(1+x)^(2/3),x, algorithm="fricas")
 

Output:

-1/6*(3*x - 13)*(x + 1)^(1/3)*(-x + 1)^(2/3) - 20/9*sqrt(3)*arctan(1/3*(sq 
rt(3)*(x - 1) + 2*sqrt(3)*(x + 1)^(1/3)*(-x + 1)^(2/3))/(x - 1)) - 10/9*lo 
g((x - (x + 1)^(2/3)*(-x + 1)^(1/3) + (x + 1)^(1/3)*(-x + 1)^(2/3) - 1)/(x 
 - 1)) + 20/9*log(-(x - (x + 1)^(1/3)*(-x + 1)^(2/3) - 1)/(x - 1))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.38 \[ \int \frac {(1-x)^{5/3}}{(1+x)^{2/3}} \, dx=\frac {2 \cdot 2^{\frac {2}{3}} \sqrt [3]{x + 1} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {\left (x + 1\right ) e^{2 i \pi }}{2}} \right )}}{\Gamma \left (\frac {4}{3}\right )} \] Input:

integrate((1-x)**(5/3)/(1+x)**(2/3),x)
 

Output:

2*2**(2/3)*(x + 1)**(1/3)*gamma(1/3)*hyper((-5/3, 1/3), (4/3,), (x + 1)*ex 
p_polar(2*I*pi)/2)/gamma(4/3)
 

Maxima [F]

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

integrate((1-x)^(5/3)/(1+x)^(2/3),x, algorithm="maxima")
 

Output:

integrate((-x + 1)^(5/3)/(x + 1)^(2/3), x)
 

Giac [F]

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

integrate((1-x)^(5/3)/(1+x)^(2/3),x, algorithm="giac")
 

Output:

integrate((-x + 1)^(5/3)/(x + 1)^(2/3), x)
 

Mupad [F(-1)]

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

int((1 - x)^(5/3)/(x + 1)^(2/3),x)
 

Output:

int((1 - x)^(5/3)/(x + 1)^(2/3), x)
 

Reduce [F]

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

int((1-x)^(5/3)/(1+x)^(2/3),x)
 

Output:

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