\(\int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx\) [504]

Optimal result

Integrand size = 17, antiderivative size = 133 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx=-(1-x)^{3/4} \sqrt [4]{1+x}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+x}}{\sqrt [4]{1-x} \left (1+\frac {\sqrt {1+x}}{\sqrt {1-x}}\right )}\right )}{\sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx=-(1-x)^{3/4} \sqrt [4]{1+x}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-x^2}}{\sqrt {1-x}-\sqrt {1+x}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-x^2}}{\sqrt {1-x}+\sqrt {1+x}}\right )}{\sqrt {2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.38, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {60, 73, 854, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 854

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

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

rule 826

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

rule 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 
 1)/n)   Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n 
)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 
2^(-1)] && IntegersQ[m, p + (m + 1)/n]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

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

Time = 0.54 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.31

method	result	size
risch	\(\frac {\left (1+x \right )^{\frac {1}{4}} \left (-1+x \right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{\left (-\left (-1+x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}}}+\frac {\left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {-x^{4}-2 x^{3}+2 x +1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x +x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}+2 x^{2}+x}{\left (1+x \right )^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {-x^{4}-2 x^{3}+2 x +1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+x^{3}+2 x^{2}+x}{\left (1+x \right )^{2}}\right )}{2}\right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{\left (1+x \right )^{\frac {3}{4}} \left (1-x \right )^{\frac {1}{4}}}\)	\(440\)

Input:

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

Output:

(1+x)^(1/4)*(-1+x)/(-(-1+x)*(1+x)^3)^(1/4)*((1-x)*(1+x)^3)^(1/4)/(1-x)^(1/ 
4)+(1/2*RootOf(_Z^4+1)*ln((RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(3/4)+RootO 
f(_Z^4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)*x+RootOf(_Z^4+1)^2*(-x^4-2*x^3+2*x+1) 
^(1/2)+RootOf(_Z^4+1)*(-x^4-2*x^3+2*x+1)^(1/4)*x^2+2*RootOf(_Z^4+1)*(-x^4- 
2*x^3+2*x+1)^(1/4)*x+x^3+RootOf(_Z^4+1)*(-x^4-2*x^3+2*x+1)^(1/4)+2*x^2+x)/ 
(1+x)^2)+1/2*RootOf(_Z^4+1)^3*ln((RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4 
)*x^2+2*RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)*x-RootOf(_Z^4+1)^2*(-x^4 
-2*x^3+2*x+1)^(1/2)*x+RootOf(_Z^4+1)^3*(-x^4-2*x^3+2*x+1)^(1/4)-RootOf(_Z^ 
4+1)^2*(-x^4-2*x^3+2*x+1)^(1/2)+RootOf(_Z^4+1)*(-x^4-2*x^3+2*x+1)^(3/4)+x^ 
3+2*x^2+x)/(1+x)^2))/(1+x)^(3/4)*((1-x)*(1+x)^3)^(1/4)/(1-x)^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + x - 1}{x - 1}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - x + 1}{x - 1}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} + x - \sqrt {x + 1} \sqrt {-x + 1} - 1}{x - 1}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\frac {\sqrt {2} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - x + \sqrt {x + 1} \sqrt {-x + 1} + 1}{x - 1}\right ) - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.31 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}} \, dx=\frac {2^{\frac {3}{4}} \left (x + 1\right )^{\frac {5}{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {\left (x + 1\right ) e^{2 i \pi }}{2}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \] Input:

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

Output:

2**(3/4)*(x + 1)**(5/4)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), (x + 1)*exp_p 
olar(2*I*pi)/2)/(2*gamma(9/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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