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

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

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Rubi [A]  time = 0.04, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {99, 151, 12, 93, 212, 206, 203} \begin {gather*} -\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 ) \end {gather*}

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 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 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 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^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*}

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Mathematica [C]  time = 0.02, size = 67, normalized size = 0.49 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*((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)]))/(x^4*(1 + x)^(3/4))

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IntegrateAlgebraic [A]  time = 0.43, size = 108, normalized size = 0.79 \begin {gather*} \frac {(1-x)^{3/4} \left (-83 (x+1)^{13/4}+191 (x+1)^{9/4}-189 (x+1)^{5/4}+33 \sqrt [4]{x+1}\right )}{192 x^4}+\frac {11}{64} \tan ^{-1}\left (\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{x-1}\right )+\frac {11}{64} \tanh ^{-1}\left (\frac {(1-x)^{3/4} \sqrt [4]{x+1}}{x-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - x)^(3/4)*(33*(1 + x)^(1/4) - 189*(1 + x)^(5/4) + 191*(1 + x)^(9/4) - 83*(1 + x)^(13/4)))/(192*x^4) + (11
*ArcTan[((1 - x)^(3/4)*(1 + x)^(1/4))/(-1 + x)])/64 + (11*ArcTanh[((1 - x)^(3/4)*(1 + x)^(1/4))/(-1 + x)])/64

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fricas [A]  time = 1.44, size = 117, normalized size = 0.85 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{5} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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)

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maple [C]  time = 0.65, size = 398, normalized size = 2.91 \begin {gather*} \frac {\left (x +1\right )^{\frac {1}{4}} \left (x -1\right ) \left (83 x^{3}+58 x^{2}+56 x +48\right ) \left (\left (-x +1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}}}{192 \left (-\left (x -1\right ) \left (x +1\right )^{3}\right )^{\frac {1}{4}} \left (-x +1\right )^{\frac {1}{4}} x^{4}}+\frac {\left (-\frac {11 \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 )}{128}+\frac {11 \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 )}{128}\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)^(1/4)/(-x+1)^(1/4)/x^5,x)

[Out]

1/192*(x+1)^(1/4)*(x-1)*(83*x^3+58*x^2+56*x+48)/x^4/(-(x-1)*(x+1)^3)^(1/4)*((-x+1)*(x+1)^3)^(1/4)/(-x+1)^(1/4)
+(-11/128*RootOf(_Z^2+1)*ln((-RootOf(_Z^2+1)*(-x^4-2*x^3+2*x+1)^(1/2)*x+(-x^4-2*x^3+2*x+1)^(3/4)-RootOf(_Z^2+1
)*(-x^4-2*x^3+2*x+1)^(1/2)-(-x^4-2*x^3+2*x+1)^(1/4)*x^2+RootOf(_Z^2+1)*x^2-2*(-x^4-2*x^3+2*x+1)^(1/4)*x+2*Root
Of(_Z^2+1)*x-(-x^4-2*x^3+2*x+1)^(1/4)+RootOf(_Z^2+1))/x/(x+1)^2)+11/128*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 \begin {gather*} \int \frac {{\left (x + 1\right )}^{\frac {1}{4}}}{x^{5} {\left (-x + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x+1\right )}^{1/4}}{x^5\,{\left (1-x\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x + 1}}{x^{5} \sqrt [4]{1 - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 1)**(1/4)/(x**5*(1 - x)**(1/4)), x)

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