Integrand size = 13, antiderivative size = 57 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\frac {1}{32} \sqrt [4]{-1+x^4} \left (-x^3+4 x^7\right )+\frac {3}{64} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3}{64} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \] Output:
1/32*(x^4-1)^(1/4)*(4*x^7-x^3)+3/64*arctan(x/(x^4-1)^(1/4))-3/64*arctanh(x /(x^4-1)^(1/4))
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\frac {1}{32} x^3 \sqrt [4]{-1+x^4} \left (-1+4 x^4\right )+\frac {3}{64} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {3}{64} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right ) \] Input:
Integrate[x^6*(-1 + x^4)^(1/4),x]
Output:
(x^3*(-1 + x^4)^(1/4)*(-1 + 4*x^4))/32 + (3*ArcTan[x/(-1 + x^4)^(1/4)])/64 - (3*ArcTanh[x/(-1 + x^4)^(1/4)])/64
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.32, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {811, 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.
\(\displaystyle \int x^6 \sqrt [4]{x^4-1} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{8} x^7 \sqrt [4]{x^4-1}-\frac {1}{8} \int \frac {x^6}{\left (x^4-1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \int \frac {x^2}{\left (x^4-1\right )^{3/4}}dx-\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )+\frac {1}{8} \sqrt [4]{x^4-1} x^7\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \int \frac {x^2}{\sqrt {x^4-1} \left (1-\frac {x^4}{x^4-1}\right )}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )+\frac {1}{8} \sqrt [4]{x^4-1} x^7\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{2} \int \frac {1}{\frac {x^2}{\sqrt {x^4-1}}+1}d\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )+\frac {1}{8} \sqrt [4]{x^4-1} x^7\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\frac {x^2}{\sqrt {x^4-1}}}d\frac {x}{\sqrt [4]{x^4-1}}-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )\right )-\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )+\frac {1}{8} \sqrt [4]{x^4-1} x^7\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{8} \left (-\frac {3}{4} \left (\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )\right )-\frac {1}{4} \sqrt [4]{x^4-1} x^3\right )+\frac {1}{8} \sqrt [4]{x^4-1} x^7\) |
Input:
Int[x^6*(-1 + x^4)^(1/4),x]
Output:
(x^7*(-1 + x^4)^(1/4))/8 + (-1/4*(x^3*(-1 + x^4)^(1/4)) - (3*(-1/2*ArcTan[ x/(-1 + x^4)^(1/4)] + ArcTanh[x/(-1 + x^4)^(1/4)]/2))/4)/8
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 = 1.38 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.58
method | result | size |
meijerg | \(\frac {\operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{7} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], x^{4}\right )}{7 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\) | \(33\) |
risch | \(\frac {x^{3} \left (4 x^{4}-1\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{32}-\frac {{\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {3}{4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{4}\right )}{32 \operatorname {signum}\left (x^{4}-1\right )^{\frac {3}{4}}}\) | \(53\) |
pseudoelliptic | \(\frac {16 \left (x^{4}-1\right )^{\frac {1}{4}} x^{7}-4 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-3 \ln \left (\frac {x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )-6 \arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )+3 \ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{128 {\left (x +\left (x^{4}-1\right )^{\frac {1}{4}}\right )}^{2} \left (x^{2}+\sqrt {x^{4}-1}\right )^{2} {\left (\left (x^{4}-1\right )^{\frac {1}{4}}-x \right )}^{2}}\) | \(113\) |
trager | \(\frac {x^{3} \left (4 x^{4}-1\right ) \left (x^{4}-1\right )^{\frac {1}{4}}}{32}-\frac {3 \ln \left (2 \left (x^{4}-1\right )^{\frac {3}{4}} x +2 x^{2} \sqrt {x^{4}-1}+2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}+2 x^{4}-1\right )}{128}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}-1}\, x^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}-1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{128}\) | \(134\) |
Input:
int(x^6*(x^4-1)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/7*signum(x^4-1)^(1/4)/(-signum(x^4-1))^(1/4)*x^7*hypergeom([-1/4,7/4],[1 1/4],x^4)
Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\frac {1}{32} \, {\left (4 \, x^{7} - x^{3}\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + \frac {3}{64} \, \arctan \left (\frac {x}{{\left (x^{4} - 1\right )}^{\frac {1}{4}}}\right ) - \frac {3}{128} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{128} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate(x^6*(x^4-1)^(1/4),x, algorithm="fricas")
Output:
1/32*(4*x^7 - x^3)*(x^4 - 1)^(1/4) + 3/64*arctan(x/(x^4 - 1)^(1/4)) - 3/12 8*log((x + (x^4 - 1)^(1/4))/x) + 3/128*log(-(x - (x^4 - 1)^(1/4))/x)
Result contains complex when optimal does not.
Time = 1.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.56 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\frac {x^{7} e^{\frac {i \pi }{4}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \] Input:
integrate(x**6*(x**4-1)**(1/4),x)
Output:
x**7*exp(I*pi/4)*gamma(7/4)*hyper((-1/4, 7/4), (11/4,), x**4)/(4*gamma(11/ 4))
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (45) = 90\).
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.74 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=-\frac {\frac {3 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + \frac {{\left (x^{4} - 1\right )}^{\frac {5}{4}}}{x^{5}}}{32 \, {\left (\frac {2 \, {\left (x^{4} - 1\right )}}{x^{4}} - \frac {{\left (x^{4} - 1\right )}^{2}}{x^{8}} - 1\right )}} - \frac {3}{64} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{128} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{128} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} - 1\right ) \] Input:
integrate(x^6*(x^4-1)^(1/4),x, algorithm="maxima")
Output:
-1/32*(3*(x^4 - 1)^(1/4)/x + (x^4 - 1)^(5/4)/x^5)/(2*(x^4 - 1)/x^4 - (x^4 - 1)^2/x^8 - 1) - 3/64*arctan((x^4 - 1)^(1/4)/x) - 3/128*log((x^4 - 1)^(1/ 4)/x + 1) + 3/128*log((x^4 - 1)^(1/4)/x - 1)
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.44 \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=-\frac {1}{32} \, x^{8} {\left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}} {\left (\frac {1}{x^{4}} - 1\right )}}{x} - \frac {3 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right )} - \frac {3}{64} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{128} \, \log \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) + \frac {3}{128} \, \log \left (-\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x} + 1\right ) \] Input:
integrate(x^6*(x^4-1)^(1/4),x, algorithm="giac")
Output:
-1/32*x^8*((x^4 - 1)^(1/4)*(1/x^4 - 1)/x - 3*(x^4 - 1)^(1/4)/x) - 3/64*arc tan((x^4 - 1)^(1/4)/x) - 3/128*log((x^4 - 1)^(1/4)/x + 1) + 3/128*log(-(x^ 4 - 1)^(1/4)/x + 1)
Timed out. \[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\int x^6\,{\left (x^4-1\right )}^{1/4} \,d x \] Input:
int(x^6*(x^4 - 1)^(1/4),x)
Output:
int(x^6*(x^4 - 1)^(1/4), x)
\[ \int x^6 \sqrt [4]{-1+x^4} \, dx=\frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{7}}{8}-\frac {\left (x^{4}-1\right )^{\frac {1}{4}} x^{3}}{32}-\frac {3 \left (\int \frac {x^{2}}{\left (x^{4}-1\right )^{\frac {3}{4}}}d x \right )}{32} \] Input:
int(x^6*(x^4-1)^(1/4),x)
Output:
(4*(x**4 - 1)**(1/4)*x**7 - (x**4 - 1)**(1/4)*x**3 - 3*int(((x**4 - 1)**(1 /4)*x**2)/(x**4 - 1),x))/32