Integrand size = 15, antiderivative size = 45 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \sqrt {-x+x^4} \left (-x+2 x^4\right )-\frac {1}{12} \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right ) \]
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.36 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {\sqrt {x \left (-1+x^3\right )} \left (x^{3/2} \left (-1+2 x^3\right )-\frac {\log \left (x^{3/2}+\sqrt {-1+x^3}\right )}{\sqrt {-1+x^3}}\right )}{12 \sqrt {x}} \]
(Sqrt[x*(-1 + x^3)]*(x^(3/2)*(-1 + 2*x^3) - Log[x^(3/2) + Sqrt[-1 + x^3]]/ Sqrt[-1 + x^3]))/(12*Sqrt[x])
Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1927, 1930, 1935, 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.
\(\displaystyle \int x^3 \sqrt {x^4-x} \, dx\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {1}{6} x^4 \sqrt {x^4-x}-\frac {1}{4} \int \frac {x^4}{\sqrt {x^4-x}}dx\) |
\(\Big \downarrow \) 1930 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {x}{\sqrt {x^4-x}}dx-\frac {1}{3} \sqrt {x^4-x} x\right )+\frac {1}{6} \sqrt {x^4-x} x^4\) |
\(\Big \downarrow \) 1935 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{3} \int \frac {1}{1-\frac {x^4}{x^4-x}}d\frac {x^2}{\sqrt {x^4-x}}-\frac {1}{3} \sqrt {x^4-x} x\right )+\frac {1}{6} \sqrt {x^4-x} x^4\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{3} \text {arctanh}\left (\frac {x^2}{\sqrt {x^4-x}}\right )-\frac {1}{3} \sqrt {x^4-x} x\right )+\frac {1}{6} \sqrt {x^4-x} x^4\) |
3.6.80.3.1 Defintions of rubi rules used
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_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))) I nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp [-2/(n - j) Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
Time = 3.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96
method | result | size |
trager | \(\frac {x \left (2 x^{3}-1\right ) \sqrt {x^{4}-x}}{12}+\frac {\ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{24}\) | \(43\) |
risch | \(\frac {x^{2} \left (2 x^{3}-1\right ) \left (x^{3}-1\right )}{12 \sqrt {x \left (x^{3}-1\right )}}+\frac {\ln \left (2 x^{3}-2 x \sqrt {x^{4}-x}-1\right )}{24}\) | \(50\) |
meijerg | \(-\frac {i \sqrt {\operatorname {signum}\left (x^{3}-1\right )}\, \left (-\frac {i \sqrt {\pi }\, x^{\frac {3}{2}} \left (-6 x^{3}+3\right ) \sqrt {-x^{3}+1}}{6}+\frac {i \sqrt {\pi }\, \arcsin \left (x^{\frac {3}{2}}\right )}{2}\right )}{6 \sqrt {\pi }\, \sqrt {-\operatorname {signum}\left (x^{3}-1\right )}}\) | \(61\) |
default | \(\frac {x^{2} \left (\left (4 x^{4}-2 x \right ) \sqrt {x^{4}-x}+\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right )}{24 \left (x^{2}-\sqrt {x^{4}-x}\right )^{2} \left (x^{2}+\sqrt {x^{4}-x}\right )^{2}}\) | \(98\) |
pseudoelliptic | \(\frac {x^{2} \left (\left (4 x^{4}-2 x \right ) \sqrt {x^{4}-x}+\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right )}{24 \left (x^{2}-\sqrt {x^{4}-x}\right )^{2} \left (x^{2}+\sqrt {x^{4}-x}\right )^{2}}\) | \(98\) |
elliptic | \(\frac {x^{4} \sqrt {x^{4}-x}}{6}-\frac {x \sqrt {x^{4}-x}}{12}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (x -1\right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (\operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )-\operatorname {EllipticPi}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \frac {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{4 \left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (x -1\right ) \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\) | \(315\) |
Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \, {\left (2 \, x^{4} - x\right )} \sqrt {x^{4} - x} + \frac {1}{24} \, \log \left (2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x - 1\right ) \]
\[ \int x^3 \sqrt {-x+x^4} \, dx=\int x^{3} \sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )}\, dx \]
\[ \int x^3 \sqrt {-x+x^4} \, dx=\int { \sqrt {x^{4} - x} x^{3} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \, \sqrt {x^{4} - x} {\left (2 \, x^{3} - 1\right )} x - \frac {1}{24} \, \log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right ) + \frac {1}{24} \, \log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right ) \]
1/12*sqrt(x^4 - x)*(2*x^3 - 1)*x - 1/24*log(sqrt(-1/x^3 + 1) + 1) + 1/24*l og(abs(sqrt(-1/x^3 + 1) - 1))
Timed out. \[ \int x^3 \sqrt {-x+x^4} \, dx=\int x^3\,\sqrt {x^4-x} \,d x \]