Integrand size = 13, antiderivative size = 59 \[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{2} x \sqrt [4]{-x^2+x^4}+\frac {1}{4} \arctan \left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\frac {1}{4} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right ) \] Output:
1/2*x*(x^4-x^2)^(1/4)+1/4*arctan(x/(x^4-x^2)^(1/4))-1/4*arctanh(x/(x^4-x^2 )^(1/4))
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.39 \[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {x^{3/2} \left (-1+x^2\right )^{3/4} \left (2 x^{3/2} \sqrt [4]{-1+x^2}+\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )-\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )\right )}{4 \left (x^2 \left (-1+x^2\right )\right )^{3/4}} \] Input:
Integrate[(-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) + ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] - ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]))/(4*(x^2*(-1 + x^2))^( 3/4))
Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.56, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {1402, 248, 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.
\(\displaystyle \int \sqrt [4]{x^4-x^2} \, dx\) |
\(\Big \downarrow \) 1402 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \int \sqrt {x} \sqrt [4]{x^2-1}dx}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} x^{3/2} \sqrt [4]{x^2-1}-\frac {1}{4} \int \frac {\sqrt {x}}{\left (x^2-1\right )^{3/4}}dx\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} x^{3/2} \sqrt [4]{x^2-1}-\frac {1}{2} \int \frac {x}{\left (x^2-1\right )^{3/4}}d\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} x^{3/2} \sqrt [4]{x^2-1}-\frac {1}{2} \int \frac {x}{1-x^2}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x+1}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}-\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )+\frac {1}{2} \sqrt [4]{x^2-1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} \left (\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-\frac {1}{2} \int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )+\frac {1}{2} \sqrt [4]{x^2-1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^2} \left (\frac {1}{2} \left (\frac {1}{2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )+\frac {1}{2} \sqrt [4]{x^2-1} x^{3/2}\right )}{\sqrt {x} \sqrt [4]{x^2-1}}\) |
Input:
Int[(-x^2 + x^4)^(1/4),x]
Output:
((-x^2 + x^4)^(1/4)*((x^(3/2)*(-1 + x^2)^(1/4))/2 + (ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)]/2 - ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)]/2)/2))/(Sqrt[x]*(-1 + x^ 2)^(1/4))
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] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^p /(x^(2*p)*(b + c*x^2)^p) Int[x^(2*p)*(b + c*x^2)^p, x], x] /; FreeQ[{b, c , p}, x] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.56
\[\frac {2 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x^{2}\right )}{3 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{4}}}\]
Input:
int((x^4-x^2)^(1/4),x)
Output:
2/3*signum(x^2-1)^(1/4)/(-signum(x^2-1))^(1/4)*x^(3/2)*hypergeom([-1/4,3/4 ],[7/4],x^2)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (47) = 94\).
Time = 0.62 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.03 \[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{2} \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x + \frac {1}{8} \, \arctan \left (-\frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} - 1\right )} - {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{2 \, {\left (x^{3} - x\right )}}\right ) + \frac {1}{8} \, \log \left (-\frac {2 \, x^{3} - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} - x^{2}} x - x - 2 \, {\left (x^{4} - x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \] Input:
integrate((x^4-x^2)^(1/4),x, algorithm="fricas")
Output:
1/2*(x^4 - x^2)^(1/4)*x + 1/8*arctan(-1/2*((x^4 - x^2)^(1/4)*(x^2 - 1) - ( x^4 - x^2)^(3/4))/(x^3 - x)) + 1/8*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)
\[ \int \sqrt [4]{-x^2+x^4} \, dx=\int \sqrt [4]{x^{4} - x^{2}}\, dx \] Input:
integrate((x**4-x**2)**(1/4),x)
Output:
Integral((x**4 - x**2)**(1/4), x)
\[ \int \sqrt [4]{-x^2+x^4} \, dx=\int { {\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((x^4-x^2)^(1/4),x, algorithm="maxima")
Output:
integrate((x^4 - x^2)^(1/4), x)
Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {1}{2} \, x^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - \frac {1}{4} \, \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{8} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{8} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \] Input:
integrate((x^4-x^2)^(1/4),x, algorithm="giac")
Output:
1/2*x^2*(-1/x^2 + 1)^(1/4) - 1/4*arctan((-1/x^2 + 1)^(1/4)) - 1/8*log((-1/ x^2 + 1)^(1/4) + 1) + 1/8*log(-(-1/x^2 + 1)^(1/4) + 1)
Time = 8.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.53 \[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {2\,x\,{\left (x^4-x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ x^2\right )}{3\,{\left (1-x^2\right )}^{1/4}} \] Input:
int((x^4 - x^2)^(1/4),x)
Output:
(2*x*(x^4 - x^2)^(1/4)*hypergeom([-1/4, 3/4], 7/4, x^2))/(3*(1 - x^2)^(1/4 ))
\[ \int \sqrt [4]{-x^2+x^4} \, dx=\frac {\sqrt {x}\, \left (x^{2}-1\right )^{\frac {1}{4}} x}{2}-\frac {\left (\int \frac {\sqrt {x}}{\left (x^{2}-1\right )^{\frac {3}{4}}}d x \right )}{4} \] Input:
int((x^4-x^2)^(1/4),x)
Output:
(2*sqrt(x)*(x**2 - 1)**(1/4)*x - int((sqrt(x)*(x**2 - 1)**(1/4))/(x**2 - 1 ),x))/4