\(\int x^3 \sqrt {-x+x^4} \, dx\) [580]

Optimal result

Integrand size = 15, antiderivative size = 45 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \sqrt {-x+x^4} \left (-x+2 x^4\right )-\frac {1}{12} \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right ) \]

[Out]

1/12*(x^4-x)^(1/2)*(2*x^4-x)-1/12*arctanh(x^2/(x^4-x)^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2046, 2049, 2054, 212} \[ \int x^3 \sqrt {-x+x^4} \, dx=-\frac {1}{12} \text {arctanh}\left (\frac {x^2}{\sqrt {x^4-x}}\right )+\frac {1}{6} \sqrt {x^4-x} x^4-\frac {1}{12} \sqrt {x^4-x} x \]

[In]

Int[x^3*Sqrt[-x + x^4],x]

[Out]

-1/12*(x*Sqrt[-x + x^4]) + (x^4*Sqrt[-x + x^4])/6 - ArcTanh[x^2/Sqrt[-x + x^4]]/12

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2046

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] + Dist[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] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2049

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] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{6} x^4 \sqrt {-x+x^4}-\frac {1}{4} \int \frac {x^4}{\sqrt {-x+x^4}} \, dx \\ & = -\frac {1}{12} x \sqrt {-x+x^4}+\frac {1}{6} x^4 \sqrt {-x+x^4}-\frac {1}{8} \int \frac {x}{\sqrt {-x+x^4}} \, dx \\ & = -\frac {1}{12} x \sqrt {-x+x^4}+\frac {1}{6} x^4 \sqrt {-x+x^4}-\frac {1}{12} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {-x+x^4}}\right ) \\ & = -\frac {1}{12} x \sqrt {-x+x^4}+\frac {1}{6} x^4 \sqrt {-x+x^4}-\frac {1}{12} \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.36 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {\sqrt {x \left (-1+x^3\right )} \left (x^{3/2} \left (-1+2 x^3\right )-\frac {\log \left (x^{3/2}+\sqrt {-1+x^3}\right )}{\sqrt {-1+x^3}}\right )}{12 \sqrt {x}} \]

[In]

Integrate[x^3*Sqrt[-x + x^4],x]

[Out]

(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])

Maple [A] (verified)

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	\(\frac {x^{4} \sqrt {x^{4}-x}}{6}-\frac {x \sqrt {x^{4}-x}}{12}-\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (x -1\right )^{2} \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}\, \left (\operatorname {EllipticF}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )-\operatorname {EllipticPi}\left (\sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -1\right )}}, \frac {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{4 \left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (x -1\right ) \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\)	\(315\)

[In]

int(x^3*(x^4-x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/12*x*(2*x^3-1)*(x^4-x)^(1/2)+1/24*ln(2*x^3-2*x*(x^4-x)^(1/2)-1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \, {\left (2 \, x^{4} - x\right )} \sqrt {x^{4} - x} + \frac {1}{24} \, \log \left (2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x - 1\right ) \]

[In]

integrate(x^3*(x^4-x)^(1/2),x, algorithm="fricas")

[Out]

1/12*(2*x^4 - x)*sqrt(x^4 - x) + 1/24*log(2*x^3 - 2*sqrt(x^4 - x)*x - 1)

Sympy [F]

\[ \int x^3 \sqrt {-x+x^4} \, dx=\int x^{3} \sqrt {x \left (x - 1\right ) \left (x^{2} + x + 1\right )}\, dx \]

[In]

integrate(x**3*(x**4-x)**(1/2),x)

[Out]

Integral(x**3*sqrt(x*(x - 1)*(x**2 + x + 1)), x)

Maxima [F]

\[ \int x^3 \sqrt {-x+x^4} \, dx=\int { \sqrt {x^{4} - x} x^{3} \,d x } \]

[In]

integrate(x^3*(x^4-x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 - x)*x^3, x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int x^3 \sqrt {-x+x^4} \, dx=\frac {1}{12} \, \sqrt {x^{4} - x} {\left (2 \, x^{3} - 1\right )} x - \frac {1}{24} \, \log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right ) + \frac {1}{24} \, \log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right ) \]

[In]

integrate(x^3*(x^4-x)^(1/2),x, algorithm="giac")

[Out]

1/12*sqrt(x^4 - x)*(2*x^3 - 1)*x - 1/24*log(sqrt(-1/x^3 + 1) + 1) + 1/24*log(abs(sqrt(-1/x^3 + 1) - 1))

Mupad [F(-1)]

Timed out. \[ \int x^3 \sqrt {-x+x^4} \, dx=\int x^3\,\sqrt {x^4-x} \,d x \]

[In]

int(x^3*(x^4 - x)^(1/2),x)

[Out]

int(x^3*(x^4 - x)^(1/2), x)