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

Optimal. Leaf size=91 \[ -\frac {(1-x)^{3/4} (x+1)^{5/4}}{2 x^2}-\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{4 x}-\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

[Out]

-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))

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Rubi [A]  time = 0.02, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {96, 94, 93, 212, 206, 203} \[ -\frac {(1-x)^{3/4} (x+1)^{5/4}}{2 x^2}-\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{4 x}-\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

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 94

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[(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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] + Dist[(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] || Sum
SimplerQ[m, 1])

Rule 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^3} \, dx &=-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}+\frac {1}{4} \int \frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^2} \, 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}{8} \int \frac {1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, 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}{2} \operatorname {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}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \operatorname {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}}{4 x}-\frac {(1-x)^{3/4} (1+x)^{5/4}}{2 x^2}-\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 57, normalized size = 0.63 \[ -\frac {(1-x)^{3/4} \left (2 x^2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\frac {1-x}{x+1}\right )+9 x^2+15 x+6\right )}{12 x^2 (x+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((1 - x)^(3/4)*(6 + 15*x + 9*x^2 + 2*x^2*Hypergeometric2F1[3/4, 1, 7/4, (1 - x)/(1 + x)]))/(x^2*(1 + x)^
(3/4))

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fricas [A]  time = 0.91, size = 106, normalized size = 1.16 \[ \frac {2 \, x^{2} \arctan \left (\frac {{\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{x - 1}\right ) + x^{2} \log \left (\frac {x + {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - x^{2} \log \left (-\frac {x - {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}} - 1}{x - 1}\right ) - 2 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )}^{\frac {1}{4}} {\left (-x + 1\right )}^{\frac {3}{4}}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{3} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.81, size = 389, normalized size = 4.27 \[ \frac {\left (x +1\right )^{\frac {1}{4}} \left (x -1\right ) \left (3 x +2\right ) \left (\left (-x +1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}}}{4 \left (-\left (x -1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}} \left (-x +1\right )^{\frac {1}{4}} x^{2}}+\frac {\left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-x^{2} \RootOf \left (\textit {\_Z}^{2}+1\right )+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}+\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x \RootOf \left (\textit {\_Z}^{2}+1\right )-2 x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x +\sqrt {-x^{4}-2 x^{3}+2 x +1}\, \RootOf \left (\textit {\_Z}^{2}+1\right )-\RootOf \left (\textit {\_Z}^{2}+1\right )-\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}}{\left (x +1\right )^{2} x}\right )}{8}+\frac {\ln \left (\frac {\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x^{2}-x^{2}-\sqrt {-x^{4}-2 x^{3}+2 x +1}\, x +2 \left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}} x -2 x +\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {3}{4}}-\sqrt {-x^{4}-2 x^{3}+2 x +1}+\left (-x^{4}-2 x^{3}+2 x +1\right )^{\frac {1}{4}}-1}{\left (x +1\right )^{2} x}\right )}{8}\right ) \left (\left (-x +1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}}}{\left (x +1\right )^{\frac {3}{4}} \left (-x +1\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{3} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (x+1\right )}^{1/4}}{x^3\,{\left (1-x\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [4]{x + 1}}{x^{3} \sqrt [4]{1 - x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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