\(\int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx\) [895]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 203 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx=-2 \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\sqrt {2} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}} \]

[Out]

-2*arctan((1+x)^(1/4)/(1-x)^(1/4))-2*arctanh((1+x)^(1/4)/(1-x)^(1/4))-1/2*ln(1-(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4)
+(1-x)^(1/2)/(1+x)^(1/2))*2^(1/2)+1/2*ln(1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4)+(1-x)^(1/2)/(1+x)^(1/2))*2^(1/2)-ar
ctan(-1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4))*2^(1/2)-arctan(1+(1-x)^(1/4)*2^(1/2)/(1+x)^(1/4))*2^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {132, 65, 338, 303, 1176, 631, 210, 1179, 642, 95, 218, 212, 209} \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx=-2 \arctan \left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )+\sqrt {2} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt {2}}+\frac {\log \left (\frac {\sqrt {1-x}}{\sqrt {x+1}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{x+1}}+1\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[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] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 132

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

Rule 209

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

Rule 210

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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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

Rule 303

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

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt [4]{1-x} (1+x)^{3/4}} \, dx+\int \frac {1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, dx \\ & = -\left (4 \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-x}\right )\right )+4 \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ & = -\left (2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\right )-2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-4 \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right ) \\ & = -2 \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+2 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-2 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right ) \\ & = -2 \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}-\text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right ) \\ & = -2 \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}-\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )+\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right ) \\ & = -2 \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}-\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}}+\frac {\log \left (1+\frac {\sqrt {1-x}}{\sqrt {1+x}}+\frac {\sqrt {2} \sqrt [4]{1-x}}{\sqrt [4]{1+x}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx=-2 \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-x^2}}{\sqrt {1-x}-\sqrt {1+x}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1-x^2}}{\sqrt {1-x}+\sqrt {1+x}}\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (1+x \right )^{\frac {1}{4}}}{\left (1-x \right )^{\frac {1}{4}} x}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x} \, dx=-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x - i - 1\right )} + 2 \, {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x + i - 1\right )} + 2 \, {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) - \left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x - i + 1\right )} + 2 \, {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + \left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x + i + 1\right )} + 2 \, {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + 2 \, \arctan \left (\frac {{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + \log \left (\frac {x + {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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