Integrand size = 13, antiderivative size = 62 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=\frac {1}{384} \sqrt [4]{-1+x^4} \left (-7 x^3-4 x^7+32 x^{11}\right )+\frac {7}{256} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {7}{256} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \] Output:
1/384*(x^4-1)^(1/4)*(32*x^11-4*x^7-7*x^3)+7/256*arctan(x/(x^4-1)^(1/4))-7/ 256*arctanh(x/(x^4-1)^(1/4))
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=\frac {1}{384} x^3 \sqrt [4]{-1+x^4} \left (-7-4 x^4+32 x^8\right )+\frac {7}{256} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {7}{256} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \] Input:
Integrate[x^10*(-1 + x^4)^(1/4),x]
Output:
(x^3*(-1 + x^4)^(1/4)*(-7 - 4*x^4 + 32*x^8))/384 + (7*ArcTan[x/(-1 + x^4)^ (1/4)])/256 - (7*ArcTanh[x/(-1 + x^4)^(1/4)])/256
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.55, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {811, 843, 843, 854, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^{10} \sqrt [4]{x^4-1} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{12} x^{11} \sqrt [4]{x^4-1}-\frac {1}{12} \int \frac {x^{10}}{\left (x^4-1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \int \frac {x^6}{\left (x^4-1\right )^{3/4}}dx-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \int \frac {x^2}{\left (x^4-1\right )^{3/4}}dx+\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \int \frac {x^2}{\sqrt {x^4-1} \left (1-\frac {x^4}{x^4-1}\right )}d\frac {x}{\sqrt [4]{x^4-1}}+\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4-1}}+1}d\frac {x}{\sqrt [4]{x^4-1}}\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \left (\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )\right )+\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )-\frac {1}{8} \sqrt [4]{x^4-1} x^7\right )+\frac {1}{12} \sqrt [4]{x^4-1} x^{11}\) |
Input:
Int[x^10*(-1 + x^4)^(1/4),x]
Output:
(x^11*(-1 + x^4)^(1/4))/12 + (-1/8*(x^7*(-1 + x^4)^(1/4)) - (7*((x^3*(-1 + x^4)^(1/4))/4 + (3*(-1/2*ArcTan[x/(-1 + x^4)^(1/4)] + ArcTanh[x/(-1 + x^4 )^(1/4)]/2))/4))/8)/12
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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] + Simp[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] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
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] - Simp[ 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.41 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.53
method | result | size |
meijerg | \(\frac {\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{11} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {11}{4}\right ], \left [\frac {15}{4}\right ], x^{4}\right )}{11 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\) | \(33\) |
risch | \(\frac {x^{3} \left (32 x^{8}-4 x^{4}-7\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{384}-\frac {7 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {3}{4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{384 \operatorname {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\) | \(58\) |
pseudoelliptic | \(\frac {\left (128 x^{11}-16 x^{7}-28 x^{3}\right ) \left (x^{4}-1\right )^{\frac {1}{4}}-21 \ln \left (\frac {x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )-42 \arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )+21 \ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{1536 {\left (x +\left (x^{4}-1\right )^{\frac {1}{4}}\right )}^{3} \left (x^{2}+\sqrt {x^{4}-1}\right )^{3} {\left (-\left (x^{4}-1\right )^{\frac {1}{4}}+x \right )}^{3}}\) | \(113\) |
trager | \(\frac {x^{3} \left (32 x^{8}-4 x^{4}-7\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{384}-\frac {7 \ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{512}+\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{512}\) | \(139\) |
Input:
int(x^10*(x^4-1)^(1/4),x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=\frac {1}{384} \, {\left (32 \, x^{11} - 4 \, x^{7} - 7 \, x^{3}\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \frac {7}{256} \, \arctan \left (\frac {x}{{\left (x^{4} - 1\right )}^{\frac {1}{4}}}\right ) - \frac {7}{512} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{512} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate(x^10*(x^4-1)^(1/4),x, algorithm="fricas")
Output:
1/384*(32*x^11 - 4*x^7 - 7*x^3)*(x^4 - 1)^(1/4) + 7/256*arctan(x/(x^4 - 1) ^(1/4)) - 7/512*log((x + (x^4 - 1)^(1/4))/x) + 7/512*log(-(x - (x^4 - 1)^( 1/4))/x)
Result contains complex when optimal does not.
Time = 3.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.58 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=- \frac {x^{11} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \] Input:
integrate(x**10*(x**4-1)**(1/4),x)
Output:
-x**11*exp(-3*I*pi/4)*gamma(11/4)*hyper((-1/4, 11/4), (15/4,), x**4)/(4*ga mma(15/4))
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (50) = 100\).
Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.98 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=-\frac {\frac {21 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {18 \, {\left (x^{4} - 1\right )}^{\frac {5}{4}}}{x^{5}} - \frac {7 \, {\left (x^{4} - 1\right )}^{\frac {9}{4}}}{x^{9}}}{384 \, {\left (\frac {3 \, {\left (x^{4} - 1\right )}}{x^{4}} - \frac {3 \, {\left (x^{4} - 1\right )}^{2}}{x^{8}} + \frac {{\left (x^{4} - 1\right )}^{3}}{x^{12}} - 1\right )}} - \frac {7}{256} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \] Input:
integrate(x^10*(x^4-1)^(1/4),x, algorithm="maxima")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (50) = 100\).
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.69 \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=-\frac {1}{384} \, x^{12} {\left (\frac {18 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 1\right )}}{x} - \frac {21 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {7 \, {\left (x^{8} - 2 \, x^{4} + 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x^{9}}\right )} - \frac {7}{256} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{512} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {7}{512} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \] Input:
integrate(x^10*(x^4-1)^(1/4),x, algorithm="giac")
Output:
-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)
Timed out. \[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=\int x^{10}\,{\left (x^4-1\right )}^{1/4} \,d x \] Input:
int(x^10*(x^4 - 1)^(1/4),x)
Output:
int(x^10*(x^4 - 1)^(1/4), x)
\[ \int x^{10} \sqrt [4]{-1+x^4} \, dx=\frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{11}}{12}-\frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{7}}{96}-\frac {7 \left (x^{4}-1\right )^{\frac {1}{4}} x^{3}}{384}-\frac {7 \left (\int \frac {x^{2}}{\left (x^{4}-1\right )^{\frac {3}{4}}}d x \right )}{128} \] Input:
int(x^10*(x^4-1)^(1/4),x)
Output:
(32*(x**4 - 1)**(1/4)*x**11 - 4*(x**4 - 1)**(1/4)*x**7 - 7*(x**4 - 1)**(1/ 4)*x**3 - 21*int(((x**4 - 1)**(1/4)*x**2)/(x**4 - 1),x))/384