3.9.13 \(\int x^{10} \sqrt [4]{-1+x^4} \, dx\)

Optimal. Leaf size=62 \[ \frac {7}{256} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {7}{256} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{384} \sqrt [4]{x^4-1} \left (32 x^{11}-4 x^7-7 x^3\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 81, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {279, 321, 331, 298, 203, 206} \begin {gather*} \frac {7}{256} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {7}{256} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}-\frac {1}{96} \sqrt [4]{x^4-1} x^7-\frac {7}{384} \sqrt [4]{x^4-1} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^10*(-1 + x^4)^(1/4),x]

[Out]

(-7*x^3*(-1 + x^4)^(1/4))/384 - (x^7*(-1 + x^4)^(1/4))/96 + (x^11*(-1 + x^4)^(1/4))/12 + (7*ArcTan[x/(-1 + x^4
)^(1/4)])/256 - (7*ArcTanh[x/(-1 + x^4)^(1/4)])/256

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 279

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 321

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

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rubi steps

\begin {align*} \int x^{10} \sqrt [4]{-1+x^4} \, dx &=\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}-\frac {1}{12} \int \frac {x^{10}}{\left (-1+x^4\right )^{3/4}} \, dx\\ &=-\frac {1}{96} x^7 \sqrt [4]{-1+x^4}+\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}-\frac {7}{96} \int \frac {x^6}{\left (-1+x^4\right )^{3/4}} \, dx\\ &=-\frac {7}{384} x^3 \sqrt [4]{-1+x^4}-\frac {1}{96} x^7 \sqrt [4]{-1+x^4}+\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}-\frac {7}{128} \int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx\\ &=-\frac {7}{384} x^3 \sqrt [4]{-1+x^4}-\frac {1}{96} x^7 \sqrt [4]{-1+x^4}+\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}-\frac {7}{128} \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {7}{384} x^3 \sqrt [4]{-1+x^4}-\frac {1}{96} x^7 \sqrt [4]{-1+x^4}+\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}-\frac {7}{256} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {7}{256} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ &=-\frac {7}{384} x^3 \sqrt [4]{-1+x^4}-\frac {1}{96} x^7 \sqrt [4]{-1+x^4}+\frac {1}{12} x^{11} \sqrt [4]{-1+x^4}+\frac {7}{256} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {7}{256} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 67, normalized size = 1.08 \begin {gather*} \frac {x^3 \sqrt [4]{x^4-1} \left (7 \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};x^4\right )+\sqrt [4]{1-x^4} \left (8 x^8-x^4-7\right )\right )}{96 \sqrt [4]{1-x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^10*(-1 + x^4)^(1/4),x]

[Out]

(x^3*(-1 + x^4)^(1/4)*((1 - x^4)^(1/4)*(-7 - x^4 + 8*x^8) + 7*Hypergeometric2F1[-1/4, 3/4, 7/4, x^4]))/(96*(1
- x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.19, size = 62, normalized size = 1.00 \begin {gather*} \frac {1}{384} \sqrt [4]{-1+x^4} \left (-7 x^3-4 x^7+32 x^{11}\right )+\frac {7}{256} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {7}{256} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^10*(-1 + x^4)^(1/4),x]

[Out]

((-1 + x^4)^(1/4)*(-7*x^3 - 4*x^7 + 32*x^11))/384 + (7*ArcTan[x/(-1 + x^4)^(1/4)])/256 - (7*ArcTanh[x/(-1 + x^
4)^(1/4)])/256

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fricas [A]  time = 0.47, size = 75, normalized size = 1.21 \begin {gather*} \frac {1}{384} \, {\left (32 \, x^{11} - 4 \, x^{7} - 7 \, x^{3}\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}} - \frac {7}{256} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{512} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{512} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(x^4-1)^(1/4),x, algorithm="fricas")

[Out]

1/384*(32*x^11 - 4*x^7 - 7*x^3)*(x^4 - 1)^(1/4) - 7/256*arctan((x^4 - 1)^(1/4)/x) - 7/512*log((x + (x^4 - 1)^(
1/4))/x) + 7/512*log(-(x - (x^4 - 1)^(1/4))/x)

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giac [B]  time = 0.29, size = 105, normalized size = 1.69 \begin {gather*} \frac {1}{384} \, x^{12} {\left (\frac {18 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 1\right )}}{x} - \frac {21 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {7 \, {\left (x^{8} - 2 \, x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{9}}\right )} + \frac {7}{256} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) - \frac {7}{512} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(x^4-1)^(1/4),x, algorithm="giac")

[Out]

1/384*x^12*(18*(x^4 - 1)^(1/4)*(1/x^4 - 1)/x - 21*(x^4 - 1)^(1/4)/x + 7*(x^8 - 2*x^4 + 1)*(x^4 - 1)^(1/4)/x^9)
+ 7/256*arctan((x^4 - 1)^(1/4)/x) + 7/512*log((x^4 - 1)^(1/4)/x + 1) - 7/512*log(-(x^4 - 1)^(1/4)/x + 1)

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maple [C]  time = 1.88, size = 33, normalized size = 0.53

method	result	size
meijerg	\(\frac {\mathrm {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{11} \hypergeom \left (\left [-\frac {1}{4}, \frac {11}{4}\right ], \left [\frac {15}{4}\right ], x^{4}\right )}{11 \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {1}{4}}}\)	\(33\)
risch	\(\frac {x^{3} \left (32 x^{8}-4 x^{4}-7\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{384}-\frac {7 \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {3}{4}} x^{3} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{384 \mathrm {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\)	\(58\)
trager	\(\frac {x^{3} \left (32 x^{8}-4 x^{4}-7\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{384}+\frac {7 \ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-2 x^{4}+1\right )}{512}+\frac {7 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{512}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(x^4-1)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/11*signum(x^4-1)^(1/4)/(-signum(x^4-1))^(1/4)*x^11*hypergeom([-1/4,11/4],[15/4],x^4)

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maxima [B]  time = 0.40, size = 123, normalized size = 1.98 \begin {gather*} -\frac {\frac {21 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {18 \, {\left (x^{4} - 1\right )}^{\frac {5}{4}}}{x^{5}} - \frac {7 \, {\left (x^{4} - 1\right )}^{\frac {9}{4}}}{x^{9}}}{384 \, {\left (\frac {3 \, {\left (x^{4} - 1\right )}}{x^{4}} - \frac {3 \, {\left (x^{4} - 1\right )}^{2}}{x^{8}} + \frac {{\left (x^{4} - 1\right )}^{3}}{x^{12}} - 1\right )}} - \frac {7}{256} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(x^4-1)^(1/4),x, algorithm="maxima")

[Out]

-1/384*(21*(x^4 - 1)^(1/4)/x + 18*(x^4 - 1)^(5/4)/x^5 - 7*(x^4 - 1)^(9/4)/x^9)/(3*(x^4 - 1)/x^4 - 3*(x^4 - 1)^
2/x^8 + (x^4 - 1)^3/x^12 - 1) - 7/256*arctan((x^4 - 1)^(1/4)/x) - 7/512*log((x^4 - 1)^(1/4)/x + 1) + 7/512*log
((x^4 - 1)^(1/4)/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^{10}\,{\left (x^4-1\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(x^4 - 1)^(1/4),x)

[Out]

int(x^10*(x^4 - 1)^(1/4), x)

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sympy [C]  time = 1.40, size = 36, normalized size = 0.58 \begin {gather*} - \frac {x^{11} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10*(x**4-1)**(1/4),x)

[Out]

-x**11*exp(-3*I*pi/4)*gamma(11/4)*hyper((-1/4, 11/4), (15/4,), x**4)/(4*gamma(15/4))

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