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

Optimal. Leaf size=114 \[ -\frac{5 (1-x)^{3/4} \sqrt [4]{x+1}}{12 x^2}-\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{3 x^3}-\frac{11 (1-x)^{3/4} \sqrt [4]{x+1}}{24 x}-\frac{3}{8} \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac{3}{8} \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

[Out]

-((1 - x)^(3/4)*(1 + x)^(1/4))/(3*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

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Rubi [A]  time = 0.0289435, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{5 (1-x)^{3/4} \sqrt [4]{x+1}}{12 x^2}-\frac{(1-x)^{3/4} \sqrt [4]{x+1}}{3 x^3}-\frac{11 (1-x)^{3/4} \sqrt [4]{x+1}}{24 x}-\frac{3}{8} \tan ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right )-\frac{3}{8} \tanh ^{-1}\left (\frac{\sqrt [4]{x+1}}{\sqrt [4]{1-x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - x)^(3/4)*(1 + x)^(1/4))/(3*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 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^4} \, dx &=-\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} \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}}{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} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+x}}{\sqrt [4]{1-x}}\right )-\frac{3}{8} \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}}{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 [C]  time = 0.016317, size = 62, normalized size = 0.54 \[ -\frac{(1-x)^{3/4} \left (6 x^3 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{1-x}{x+1}\right )+11 x^3+21 x^2+18 x+8\right )}{24 x^3 (x+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - x)^(3/4)*(8 + 18*x + 21*x^2 + 11*x^3 + 6*x^3*Hypergeometric2F1[3/4, 1, 7/4, (1 - x)/(1 + x)]))/(24*x^3*
(1 + x)^(3/4))

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Maple [F]  time = 0.021, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\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^4,x)

[Out]

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

[Out]

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

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Fricas [A]  time = 1.56994, size = 321, normalized size = 2.82 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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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**4,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^{4}{\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^4,x, algorithm="giac")

[Out]

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