\(\int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx\) [507]

Optimal result

Integrand size = 20, antiderivative size = 91 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=-\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \] Output:

-1/4*(1-x)^(3/4)*(1+x)^(1/4)/x-1/2*(1-x)^(3/4)*(1+x)^(5/4)/x^2-1/4*arctan( 
(1+x)^(1/4)/(1-x)^(1/4))-1/4*arctanh((1+x)^(1/4)/(1-x)^(1/4))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\frac {1}{4} \left (-\frac {(1-x)^{3/4} \sqrt [4]{1+x} (2+3 x)}{x^2}-\arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\right ) \] Input:

Integrate[(1 + x)^(1/4)/((1 - x)^(1/4)*x^3),x]
 

Output:

(-(((1 - x)^(3/4)*(1 + x)^(1/4)*(2 + 3*x))/x^2) - ArcTan[(1 + x)^(1/4)/(1 
- x)^(1/4)] - ArcTanh[(1 + x)^(1/4)/(1 - x)^(1/4)])/4
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {107, 105, 104, 756, 216, 219}

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.

\(\displaystyle \int \frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x} x^3} \, dx\)

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[(1 + x)^(1/4)/((1 - x)^(1/4)*x^3),x]
 

Output:

-1/2*((1 - x)^(3/4)*(1 + x)^(5/4))/x^2 + (-(((1 - x)^(3/4)*(1 + x)^(1/4))/ 
x) + 2*(-1/2*ArcTan[(1 + x)^(1/4)/(1 - x)^(1/4)] - ArcTanh[(1 + x)^(1/4)/( 
1 - x)^(1/4)]/2))/4
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

Maple [C] (verified)

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

Time = 0.75 (sec) , antiderivative size = 389, normalized size of antiderivative = 4.27

method	result	size
risch	\(\frac {\left (1+x \right )^{\frac {1}{4}} \left (-1+x \right ) \left (2+3 x \right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{4 x^{2} \left (-\left (-1+x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}}}+\frac {\left (\frac {\ln \left (\frac {\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}-\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-\sqrt {-x^{4}-2 x^{3}+2 x +1}+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -x^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}-2 x -1}{\left (1+x \right )^{2} x}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}}{\left (1+x \right )^{2} x}\right )}{8}\right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{\left (1+x \right )^{\frac {3}{4}} \left (1-x \right )^{\frac {1}{4}}}\)	\(389\)

Input:

int((1+x)^(1/4)/(1-x)^(1/4)/x^3,x,method=_RETURNVERBOSE)
 

Output:

1/4*(1+x)^(1/4)*(-1+x)*(2+3*x)/x^2/(-(-1+x)*(1+x)^3)^(1/4)*((1-x)*(1+x)^3) 
^(1/4)/(1-x)^(1/4)+(1/8*ln(((-x^4-2*x^3+2*x+1)^(3/4)-(-x^4-2*x^3+2*x+1)^(1 
/2)*x+(-x^4-2*x^3+2*x+1)^(1/4)*x^2-(-x^4-2*x^3+2*x+1)^(1/2)+2*(-x^4-2*x^3+ 
2*x+1)^(1/4)*x-x^2+(-x^4-2*x^3+2*x+1)^(1/4)-2*x-1)/(1+x)^2/x)-1/8*RootOf(_ 
Z^2+1)*ln(-(RootOf(_Z^2+1)*(-x^4-2*x^3+2*x+1)^(1/2)*x+RootOf(_Z^2+1)*(-x^4 
-2*x^3+2*x+1)^(1/2)-RootOf(_Z^2+1)*x^2-(-x^4-2*x^3+2*x+1)^(3/4)+(-x^4-2*x^ 
3+2*x+1)^(1/4)*x^2-2*RootOf(_Z^2+1)*x+2*(-x^4-2*x^3+2*x+1)^(1/4)*x-RootOf( 
_Z^2+1)+(-x^4-2*x^3+2*x+1)^(1/4))/(1+x)^2/x))/(1+x)^(3/4)*((1-x)*(1+x)^3)^ 
(1/4)/(1-x)^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\frac {2 \, x^{2} \arctan \left (\frac {{\left (-x + 1\right )}^{\frac {1}{4}}}{{\left (x + 1\right )}^{\frac {1}{4}}}\right ) + x^{2} \log \left (\frac {x + {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - x^{2} \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - 2 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{8 \, x^{2}} \] Input:

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^3,x, algorithm="fricas")
 

Output:

1/8*(2*x^2*arctan((-x + 1)^(1/4)/(x + 1)^(1/4)) + x^2*log((x + (x + 1)^(1/ 
4)*(-x + 1)^(3/4) - 1)/(x - 1)) - x^2*log(-(x - (x + 1)^(1/4)*(-x + 1)^(3/ 
4) - 1)/(x - 1)) - 2*(3*x + 2)*(x + 1)^(1/4)*(-x + 1)^(3/4))/x^2
 

Sympy [F]

\[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int \frac {\sqrt [4]{x + 1}}{x^{3} \sqrt [4]{1 - x}}\, dx \] Input:

integrate((1+x)**(1/4)/(1-x)**(1/4)/x**3,x)
 

Output:

Integral((x + 1)**(1/4)/(x**3*(1 - x)**(1/4)), x)
 

Maxima [F]

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

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx=\int \frac {{\left (x+1\right )}^{1/4}}{x^3\,{\left (1-x\right )}^{1/4}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int((1+x)^(1/4)/(1-x)^(1/4)/x^3,x)
 

Output:

int((x + 1)**(1/4)/(( - x + 1)**(1/4)*x**3),x)