3.899 \(\int \frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x} x^5} \, dx\)

Optimal. Leaf size=137 \[ -\frac{29 (1-x)^{3/4} \sqrt [4]{x+1}}{96 x^2}-\frac{7 (1-x)^{3/4} \sqrt [4]{x+1}}{24 x^3}-\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{4 x^4}-\frac{83 (1-x)^{3/4} \sqrt [4]{x+1}}{192 x}-\frac{11}{64} \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac{11}{64} \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

[Out]

-((1 - x)^(3/4)*(1 + x)^(1/4))/(4*x^4) - (7*(1 - x)^(3/4)*(1 + x)^(1/4))/(24*x^3) - (29*(1 - x)^(3/4)*(1 + x)^
(1/4))/(96*x^2) - (83*(1 - x)^(3/4)*(1 + x)^(1/4))/(192*x) - (11*ArcTan[(1 + x)^(1/4)/(1 - x)^(1/4)])/64 - (11
*ArcTanh[(1 + x)^(1/4)/(1 - x)^(1/4)])/64

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Rubi [A]  time = 0.0376703, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {99, 151, 12, 93, 212, 206, 203} \[ -\frac{29 (1-x)^{3/4} \sqrt [4]{x+1}}{96 x^2}-\frac{7 (1-x)^{3/4} \sqrt [4]{x+1}}{24 x^3}-\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{4 x^4}-\frac{83 (1-x)^{3/4} \sqrt [4]{x+1}}{192 x}-\frac{11}{64} \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac{11}{64} \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(1/4)/((1 - x)^(1/4)*x^5),x]

[Out]

-((1 - x)^(3/4)*(1 + x)^(1/4))/(4*x^4) - (7*(1 - x)^(3/4)*(1 + x)^(1/4))/(24*x^3) - (29*(1 - x)^(3/4)*(1 + x)^
(1/4))/(96*x^2) - (83*(1 - x)^(3/4)*(1 + x)^(1/4))/(192*x) - (11*ArcTan[(1 + x)^(1/4)/(1 - x)^(1/4)])/64 - (11
*ArcTanh[(1 + x)^(1/4)/(1 - x)^(1/4)])/64

Rule 99

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 151

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] && LtQ[m, -1] && IntegerQ[m]

Rule 12

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

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^5} \, dx &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}+\frac{1}{4} \int \frac{\frac{7}{2}+3 x}{\sqrt [4]{1-x} x^4 (1+x)^{3/4}} \, dx\\ &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{1}{12} \int \frac{-\frac{29}{4}-7 x}{\sqrt [4]{1-x} x^3 (1+x)^{3/4}} \, dx\\ &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}+\frac{1}{24} \int \frac{\frac{83}{8}+\frac{29 x}{4}}{\sqrt [4]{1-x} x^2 (1+x)^{3/4}} \, dx\\ &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}-\frac{83 (1-x)^{3/4} \sqrt [4]{1+x}}{192 x}-\frac{1}{24} \int -\frac{33}{16 \sqrt [4]{1-x} x (1+x)^{3/4}} \, dx\\ &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}-\frac{83 (1-x)^{3/4} \sqrt [4]{1+x}}{192 x}+\frac{11}{128} \int \frac{1}{\sqrt [4]{1-x} x (1+x)^{3/4}} \, dx\\ &=-\frac{(1-x)^{3/4} \sqrt [4]{1+x}}{4 x^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}-\frac{83 (1-x)^{3/4} \sqrt [4]{1+x}}{192 x}+\frac{11}{32} \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^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}-\frac{83 (1-x)^{3/4} \sqrt [4]{1+x}}{192 x}-\frac{11}{64} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{11}{64} \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^4}-\frac{7 (1-x)^{3/4} \sqrt [4]{1+x}}{24 x^3}-\frac{29 (1-x)^{3/4} \sqrt [4]{1+x}}{96 x^2}-\frac{83 (1-x)^{3/4} \sqrt [4]{1+x}}{192 x}-\frac{11}{64} \tan ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{11}{64} \tanh ^{-1}\left (\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.018467, size = 67, normalized size = 0.49 \[ -\frac{(1-x)^{3/4} \left (22 x^4 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{1-x}{x+1}\right )+83 x^4+141 x^3+114 x^2+104 x+48\right )}{192 x^4 (x+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^(1/4)/((1 - x)^(1/4)*x^5),x]

[Out]

-((1 - x)^(3/4)*(48 + 104*x + 114*x^2 + 141*x^3 + 83*x^4 + 22*x^4*Hypergeometric2F1[3/4, 1, 7/4, (1 - x)/(1 +
x)]))/(192*x^4*(1 + x)^(3/4))

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5}}\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^5,x)

[Out]

int((1+x)^(1/4)/(1-x)^(1/4)/x^5,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^{5}{\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^5,x, algorithm="maxima")

[Out]

integrate((x + 1)^(1/4)/(x^5*(-x + 1)^(1/4)), x)

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Fricas [A]  time = 1.57685, size = 339, normalized size = 2.47 \begin{align*} \frac{66 \, x^{4} \arctan \left (\frac{{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{x - 1}\right ) + 33 \, x^{4} \log \left (\frac{x +{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) - 33 \, x^{4} \log \left (-\frac{x -{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}} - 1}{x - 1}\right ) - 2 \,{\left (83 \, x^{3} + 58 \, x^{2} + 56 \, x + 48\right )}{\left (x + 1\right )}^{\frac{1}{4}}{\left (-x + 1\right )}^{\frac{3}{4}}}{384 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/4)/(1-x)^(1/4)/x^5,x, algorithm="fricas")

[Out]

1/384*(66*x^4*arctan((x + 1)^(1/4)*(-x + 1)^(3/4)/(x - 1)) + 33*x^4*log((x + (x + 1)^(1/4)*(-x + 1)^(3/4) - 1)
/(x - 1)) - 33*x^4*log(-(x - (x + 1)^(1/4)*(-x + 1)^(3/4) - 1)/(x - 1)) - 2*(83*x^3 + 58*x^2 + 56*x + 48)*(x +
1)^(1/4)*(-x + 1)^(3/4))/x^4

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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**5,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^{5}{\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^5,x, algorithm="giac")

[Out]

integrate((x + 1)^(1/4)/(x^5*(-x + 1)^(1/4)), x)