3.896 \(\int \frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{x}-\tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0127445, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {94, 93, 212, 206, 203} \[ -\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{x}-\tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(1/4)/((1 - x)^(1/4)*x^2),x]

[Out]

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

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 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 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]

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 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])

Rubi steps

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

Mathematica [C]  time = 0.0114835, size = 50, normalized size = 0.81 \[ -\frac{(1-x)^{3/4} \left (2 x \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{1-x}{x+1}\right )+3 x+3\right )}{3 x (x+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^(1/4)/((1 - x)^(1/4)*x^2),x]

[Out]

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

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}\sqrt [4]{1+x}{\frac{1}{\sqrt [4]{1-x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^(1/4)/(1-x)^(1/4)/x^2,x)

[Out]

int((1+x)^(1/4)/(1-x)^(1/4)/x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x + 1\right )}^{\frac{1}{4}}}{x^{2}{\left (-x + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate((x + 1)^(1/4)/(x^2*(-x + 1)^(1/4)), x)

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Fricas [A]  time = 1.61204, size = 275, normalized size = 4.44 \begin{align*} \frac{2 \, x \arctan \left (\frac{{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{x - 1}\right ) + x \log \left (\frac{x +{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) - x \log \left (-\frac{x -{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) - 2 \,{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(1/4)/(1-x)**(1/4)/x**2,x)

[Out]

Exception raised: ValueError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x + 1\right )}^{\frac{1}{4}}}{x^{2}{\left (-x + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^2,x, algorithm="giac")

[Out]

integrate((x + 1)^(1/4)/(x^2*(-x + 1)^(1/4)), x)