Integrand size = 17, antiderivative size = 72 \[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{192} \sqrt [4]{-x^2+x^4} \left (-7 x-4 x^3+32 x^5\right )+\frac {7}{128} \arctan \left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {7}{128} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \] Output:
1/192*(x^4-x^2)^(1/4)*(32*x^5-4*x^3-7*x)+7/128*arctan(x/(x^4-x^2)^(1/4))-7 /128*arctanh(x/(x^4-x^2)^(1/4))
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33 \[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\frac {x^{3/2} \left (-1+x^2\right )^{3/4} \left (2 x^{3/2} \sqrt [4]{-1+x^2} \left (-7-4 x^2+32 x^4\right )+21 \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )-21 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{384 \left (x^2 \left (-1+x^2\right )\right )^{3/4}} \] Input:
Integrate[x^4*(-x^2 + x^4)^(1/4),x]
Output:
(x^(3/2)*(-1 + x^2)^(3/4)*(2*x^(3/2)*(-1 + x^2)^(1/4)*(-7 - 4*x^2 + 32*x^4 ) + 21*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] - 21*ArcTanh[Sqrt[x]/(-1 + x^2)^(1 /4)]))/(384*(x^2*(-1 + x^2))^(3/4))
Time = 0.30 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.96, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {1426, 1429, 1429, 1431, 266, 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.
| \(\displaystyle \int x^4 \sqrt [4]{x^4-x^2} \, dx\) |
| \(\Big \downarrow \) 1426 |
| \(\displaystyle \frac {1}{6} x^5 \sqrt [4]{x^4-x^2}-\frac {1}{12} \int \frac {x^6}{\left (x^4-x^2\right )^{3/4}}dx\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \int \frac {x^4}{\left (x^4-x^2\right )^{3/4}}dx-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 1429 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3}{4} \int \frac {x^2}{\left (x^4-x^2\right )^{3/4}}dx+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 1431 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {\sqrt {x}}{\left (x^2-1\right )^{3/4}}dx}{4 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {x}{\left (x^2-1\right )^{3/4}}d\sqrt {x}}{2 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \int \frac {x}{1-x^2}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}}{2 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}-\frac {1}{2} \int \frac {1}{x+1}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{2 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{2 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{12} \left (-\frac {7}{8} \left (\frac {3 \left (x^2-1\right )^{3/4} x^{3/2} \left (\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{2 \left (x^4-x^2\right )^{3/4}}+\frac {1}{2} \sqrt [4]{x^4-x^2} x\right )-\frac {1}{4} \sqrt [4]{x^4-x^2} x^3\right )+\frac {1}{6} \sqrt [4]{x^4-x^2} x^5\) |
Input:
Int[x^4*(-x^2 + x^4)^(1/4),x]
Output:
(x^5*(-x^2 + x^4)^(1/4))/6 + (-1/4*(x^3*(-x^2 + x^4)^(1/4)) - (7*((x*(-x^2 + x^4)^(1/4))/2 + (3*x^(3/2)*(-1 + x^2)^(3/4)*(-1/2*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] + ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]/2))/(2*(-x^2 + x^4)^(3/4)) ))/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_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, 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[(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]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[m + 2*p - 1, 0] && NeQ[m + 4*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Timed out.
\[\int x^{4} \left (x^{4}-x^{2}\right )^{\frac {1}{4}}d x\]
Input:
int(x^4*(x^4-x^2)^(1/4),x)
Output:
int(x^4*(x^4-x^2)^(1/4),x)
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (60) = 120\).
Time = 0.79 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.85 \[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{192} \, {\left (32 \, x^{5} - 4 \, x^{3} - 7 \, x\right )} {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} + \frac {7}{256} \, \arctan \left (-\frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )} - {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{3} - x\right )}}\right ) + \frac {7}{256} \, \log \left (-\frac {2 \, x^{3} - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} - x^{2}} x - x - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \] Input:
integrate(x^4*(x^4-x^2)^(1/4),x, algorithm="fricas")
Output:
1/192*(32*x^5 - 4*x^3 - 7*x)*(x^4 - x^2)^(1/4) + 7/256*arctan(-1/2*((x^4 - x^2)^(1/4)*(x^2 - 1) - (x^4 - x^2)^(3/4))/(x^3 - x)) + 7/256*log(-(2*x^3 - 2*(x^4 - x^2)^(1/4)*x^2 + 2*sqrt(x^4 - x^2)*x - x - 2*(x^4 - x^2)^(3/4)) /x)
\[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\int x^{4} \sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right )}\, dx \] Input:
integrate(x**4*(x**4-x**2)**(1/4),x)
Output:
Integral(x**4*(x**2*(x - 1)*(x + 1))**(1/4), x)
\[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\int { {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{4} \,d x } \] Input:
integrate(x^4*(x^4-x^2)^(1/4),x, algorithm="maxima")
Output:
integrate((x^4 - x^2)^(1/4)*x^4, x)
Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.24 \[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=-\frac {1}{192} \, {\left (7 \, {\left (\frac {1}{x^{2}} - 1\right )}^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 18 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} - 21 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{6} - \frac {7}{128} \, \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {7}{256} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {7}{256} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \] Input:
integrate(x^4*(x^4-x^2)^(1/4),x, algorithm="giac")
Output:
-1/192*(7*(1/x^2 - 1)^2*(-1/x^2 + 1)^(1/4) - 18*(-1/x^2 + 1)^(5/4) - 21*(- 1/x^2 + 1)^(1/4))*x^6 - 7/128*arctan((-1/x^2 + 1)^(1/4)) - 7/256*log((-1/x ^2 + 1)^(1/4) + 1) + 7/256*log(-(-1/x^2 + 1)^(1/4) + 1)
Timed out. \[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\int x^4\,{\left (x^4-x^2\right )}^{1/4} \,d x \] Input:
int(x^4*(x^4 - x^2)^(1/4),x)
Output:
int(x^4*(x^4 - x^2)^(1/4), x)
\[ \int x^4 \sqrt [4]{-x^2+x^4} \, dx=\frac {\sqrt {x}\, \left (x^{2}-1\right )^{\frac {1}{4}} x^{5}}{6}-\frac {\sqrt {x}\, \left (x^{2}-1\right )^{\frac {1}{4}} x^{3}}{48}-\frac {7 \sqrt {x}\, \left (x^{2}-1\right )^{\frac {1}{4}} x}{192}-\frac {7 \left (\int \frac {\sqrt {x}}{\left (x^{2}-1\right )^{\frac {3}{4}}}d x \right )}{128} \] Input:
int(x^4*(x^4-x^2)^(1/4),x)
Output:
(64*sqrt(x)*(x**2 - 1)**(1/4)*x**5 - 8*sqrt(x)*(x**2 - 1)**(1/4)*x**3 - 14 *sqrt(x)*(x**2 - 1)**(1/4)*x - 21*int((sqrt(x)*(x**2 - 1)**(1/4))/(x**2 - 1),x))/384