\(\int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^4} \, dx\) [898]

Optimal result

Integrand size = 20, antiderivative size = 114 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^4} \, dx=-\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}-\frac {3}{8} \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {3}{8} \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \]

[Out]

-1/3*(1-x)^(3/4)*(1+x)^(1/4)/x^3-5/12*(1-x)^(3/4)*(1+x)^(1/4)/x^2-11/24*(1-x)^(3/4)*(1+x)^(1/4)/x-3/8*arctan((
1+x)^(1/4)/(1-x)^(1/4))-3/8*arctanh((1+x)^(1/4)/(1-x)^(1/4))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {101, 156, 12, 95, 218, 212, 209} \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^4} \, dx=-\frac {3}{8} \arctan \left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {3}{8} \text {arctanh}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{x+1}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{x+1}}{24 x} \]

[In]

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

[Out]

-1/3*((1 - x)^(3/4)*(1 + x)^(1/4))/x^3 - (5*(1 - x)^(3/4)*(1 + x)^(1/4))/(12*x^2) - (11*(1 - x)^(3/4)*(1 + x)^
(1/4))/(24*x) - (3*ArcTan[(1 + x)^(1/4)/(1 - x)^(1/4)])/8 - (3*ArcTanh[(1 + x)^(1/4)/(1 - x)^(1/4)])/8

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 101

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

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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] + Dist[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -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 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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}+\frac {1}{3} \int \frac {\frac {5}{2}+2 x}{\sqrt [4]{1-x} x^3 (1+x)^{3/4}} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {1}{6} \int \frac {-\frac {11}{4}-\frac {5 x}{2}}{\sqrt [4]{1-x} x^2 (1+x)^{3/4}} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}+\frac {1}{6} \int \frac {9}{8 \sqrt [4]{1-x} x (1+x)^{3/4}} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}+\frac {3}{16} \int \frac {1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, dx \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}-\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {3}{8} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ & = -\frac {(1-x)^{3/4} \sqrt [4]{1+x}}{3 x^3}-\frac {5 (1-x)^{3/4} \sqrt [4]{1+x}}{12 x^2}-\frac {11 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x}-\frac {3}{8} \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {3}{8} \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^4} \, dx=\frac {1}{24} \left (-\frac {(1-x)^{3/4} \sqrt [4]{1+x} \left (8+10 x+11 x^2\right )}{x^3}-9 \arctan \left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-9 \text {arctanh}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\right ) \]

[In]

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

[Out]

(-(((1 - x)^(3/4)*(1 + x)^(1/4)*(8 + 10*x + 11*x^2))/x^3) - 9*ArcTan[(1 + x)^(1/4)/(1 - x)^(1/4)] - 9*ArcTanh[
(1 + x)^(1/4)/(1 - x)^(1/4)])/24

Maple [C] (verified)

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

Time = 1.11 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.46

method	result	size
risch	\(\frac {\left (-1+x \right ) \left (1+x \right )^{\frac {1}{4}} \left (11 x^{2}+10 x +8\right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{24 x^{3} \left (-\left (-1+x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}}}+\frac {\left (\frac {3 \ln \left (\frac {\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}-\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-\sqrt {-x^{4}-2 x^{3}+2 x +1}+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -x^{2}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}-2 x -1}{x \left (1+x \right )^{2}}\right )}{16}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{4}-2 x^{3}+2 x +1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}}{\left (1+x \right )^{2} x}\right )}{16}\right ) \left (\left (1-x \right ) \left (1+x \right )^{3}\right )^{\frac {1}{4}}}{\left (1+x \right )^{\frac {3}{4}} \left (1-x \right )^{\frac {1}{4}}}\)	\(394\)

[In]

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

[Out]

1/24*(-1+x)*(1+x)^(1/4)*(11*x^2+10*x+8)/x^3/(-(-1+x)*(1+x)^3)^(1/4)*((1-x)*(1+x)^3)^(1/4)/(1-x)^(1/4)+(3/16*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)/x/(1+x)^2)-3/16*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)

none

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^4} \, dx=\frac {18 \, x^{3} \arctan \left (\frac {{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + 9 \, x^{3} \log \left (\frac {x + {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - 9 \, x^{3} \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - 2 \, {\left (11 \, x^{2} + 10 \, x + 8\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{48 \, x^{3}} \]

[In]

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

[Out]

1/48*(18*x^3*arctan((x + 1)^(1/4)*(-x + 1)^(3/4)/(x - 1)) + 9*x^3*log((x + (x + 1)^(1/4)*(-x + 1)^(3/4) - 1)/(
x - 1)) - 9*x^3*log(-(x - (x + 1)^(1/4)*(-x + 1)^(3/4) - 1)/(x - 1)) - 2*(11*x^2 + 10*x + 8)*(x + 1)^(1/4)*(-x
+ 1)^(3/4))/x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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