3.10.59 \(\int x^2 \sqrt [4]{-x^3+x^4} \, dx\)

Optimal. Leaf size=73 \[ \frac {77 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )}{1024}-\frac {77 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )}{1024}+\frac {\sqrt [4]{x^4-x^3} \left (384 x^3-32 x^2-44 x-77\right )}{1536} \]

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Rubi [B]  time = 0.19, antiderivative size = 162, normalized size of antiderivative = 2.22, number of steps used = 10, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {2021, 2024, 2032, 63, 240, 212, 206, 203} \begin {gather*} \frac {1}{4} \sqrt [4]{x^4-x^3} x^3-\frac {11}{384} \sqrt [4]{x^4-x^3} x-\frac {77 \sqrt [4]{x^4-x^3}}{1536}-\frac {77 (x-1)^{3/4} x^{9/4} \tan ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{1024 \left (x^4-x^3\right )^{3/4}}-\frac {77 (x-1)^{3/4} x^{9/4} \tanh ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{1024 \left (x^4-x^3\right )^{3/4}}-\frac {1}{48} \sqrt [4]{x^4-x^3} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(-x^3 + x^4)^(1/4),x]

[Out]

(-77*(-x^3 + x^4)^(1/4))/1536 - (11*x*(-x^3 + x^4)^(1/4))/384 - (x^2*(-x^3 + x^4)^(1/4))/48 + (x^3*(-x^3 + x^4
)^(1/4))/4 - (77*(-1 + x)^(3/4)*x^(9/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/(1024*(-x^3 + x^4)^(3/4)) - (77*(-1 +
x)^(3/4)*x^(9/4)*ArcTanh[(-1 + x)^(1/4)/x^(1/4)])/(1024*(-x^3 + x^4)^(3/4))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int x^2 \sqrt [4]{-x^3+x^4} \, dx &=\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {1}{16} \int \frac {x^5}{\left (-x^3+x^4\right )^{3/4}} \, dx\\ &=-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {11}{192} \int \frac {x^4}{\left (-x^3+x^4\right )^{3/4}} \, dx\\ &=-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {77 \int \frac {x^3}{\left (-x^3+x^4\right )^{3/4}} \, dx}{1536}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {77 \int \frac {x^2}{\left (-x^3+x^4\right )^{3/4}} \, dx}{2048}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {\left (77 (-1+x)^{3/4} x^{9/4}\right ) \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{2048 \left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {\left (77 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{512 \left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {\left (77 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{512 \left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {\left (77 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{1024 \left (-x^3+x^4\right )^{3/4}}-\frac {\left (77 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{1024 \left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {77 \sqrt [4]{-x^3+x^4}}{1536}-\frac {11}{384} x \sqrt [4]{-x^3+x^4}-\frac {1}{48} x^2 \sqrt [4]{-x^3+x^4}+\frac {1}{4} x^3 \sqrt [4]{-x^3+x^4}-\frac {77 (-1+x)^{3/4} x^{9/4} \tan ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{1024 \left (-x^3+x^4\right )^{3/4}}-\frac {77 (-1+x)^{3/4} x^{9/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{1024 \left (-x^3+x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.48 \begin {gather*} \frac {4 \left ((x-1) x^3\right )^{5/4} \, _2F_1\left (-\frac {11}{4},\frac {5}{4};\frac {9}{4};1-x\right )}{5 x^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(-x^3 + x^4)^(1/4),x]

[Out]

(4*((-1 + x)*x^3)^(5/4)*Hypergeometric2F1[-11/4, 5/4, 9/4, 1 - x])/(5*x^(15/4))

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IntegrateAlgebraic [A]  time = 0.36, size = 73, normalized size = 1.00 \begin {gather*} \frac {\left (-77-44 x-32 x^2+384 x^3\right ) \sqrt [4]{-x^3+x^4}}{1536}+\frac {77 \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )}{1024}-\frac {77 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )}{1024} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-77 - 44*x - 32*x^2 + 384*x^3)*(-x^3 + x^4)^(1/4))/1536 + (77*ArcTan[x/(-x^3 + x^4)^(1/4)])/1024 - (77*ArcTa
nh[x/(-x^3 + x^4)^(1/4)])/1024

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fricas [A]  time = 0.54, size = 90, normalized size = 1.23 \begin {gather*} \frac {1}{1536} \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (384 \, x^{3} - 32 \, x^{2} - 44 \, x - 77\right )} - \frac {77}{1024} \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {77}{2048} \, \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {77}{2048} \, \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^4-x^3)^(1/4),x, algorithm="fricas")

[Out]

1/1536*(x^4 - x^3)^(1/4)*(384*x^3 - 32*x^2 - 44*x - 77) - 77/1024*arctan((x^4 - x^3)^(1/4)/x) - 77/2048*log((x
+ (x^4 - x^3)^(1/4))/x) + 77/2048*log(-(x - (x^4 - x^3)^(1/4))/x)

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giac [A]  time = 0.59, size = 106, normalized size = 1.45 \begin {gather*} \frac {1}{1536} \, {\left (77 \, {\left (\frac {1}{x} - 1\right )}^{3} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 275 \, {\left (\frac {1}{x} - 1\right )}^{2} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 351 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{4}} - 231 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{4} + \frac {77}{1024} \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {77}{2048} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {77}{2048} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^4-x^3)^(1/4),x, algorithm="giac")

[Out]

1/1536*(77*(1/x - 1)^3*(-1/x + 1)^(1/4) + 275*(1/x - 1)^2*(-1/x + 1)^(1/4) - 351*(-1/x + 1)^(5/4) - 231*(-1/x
+ 1)^(1/4))*x^4 + 77/1024*arctan((-1/x + 1)^(1/4)) + 77/2048*log((-1/x + 1)^(1/4) + 1) - 77/2048*log(abs((-1/x
+ 1)^(1/4) - 1))

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maple [C]  time = 0.68, size = 27, normalized size = 0.37

method	result	size
meijerg	\(\frac {4 \mathrm {signum}\left (-1+x \right )^{\frac {1}{4}} x^{\frac {15}{4}} \hypergeom \left (\left [-\frac {1}{4}, \frac {15}{4}\right ], \left [\frac {19}{4}\right ], x\right )}{15 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{4}}}\)	\(27\)
trager	\(\left (\frac {1}{4} x^{3}-\frac {1}{48} x^{2}-\frac {11}{384} x -\frac {77}{1536}\right ) \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\frac {77 \ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-2 x^{3}+x^{2}}{x^{2}}\right )}{2048}+\frac {77 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{2048}\)	\(173\)
risch	\(\frac {\left (384 x^{3}-32 x^{2}-44 x -77\right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{1536}+\frac {\left (-\frac {77 \ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+2 x^{3}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+4 x -1}{\left (-1+x \right )^{2}}\right )}{2048}+\frac {77 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}+5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x -4 \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )}{\left (-1+x \right )^{2}}\right )}{2048}\right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} \left (x \left (-1+x \right )^{3}\right )^{\frac {1}{4}}}{x \left (-1+x \right )}\)	\(407\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^4-x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/15*signum(-1+x)^(1/4)/(-signum(-1+x))^(1/4)*x^(15/4)*hypergeom([-1/4,15/4],[19/4],x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} x^{2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x^4-x^3)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)*x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (x^4-x^3\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^4 - x^3)^(1/4),x)

[Out]

int(x^2*(x^4 - x^3)^(1/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \sqrt [4]{x^{3} \left (x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x**4-x**3)**(1/4),x)

[Out]

Integral(x**2*(x**3*(x - 1))**(1/4), x)

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