\(\int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx\) [748]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{-x^2+x^3}}{x}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt {2} \sqrt [4]{-x^2+x^3}}\right )}{\sqrt {2}} \]

[Out]

1/2*arctan(2^(1/2)*(x^3-x^2)^(1/4)/x)*2^(1/2)-1/2*arctanh(1/2*2^(1/2)/(x^3-x^2)^(1/4)*x)*2^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.76, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2081, 107, 504, 1225, 226, 1713, 209, 212} \[ \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx=\frac {\sqrt [4]{x-1} \sqrt {x} \arctan \left (\frac {\sqrt {2} \sqrt [4]{x-1}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{x^3-x^2}}-\frac {\sqrt [4]{x-1} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{x-1}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{x^3-x^2}} \]

[In]

Int[1/((-2 + x)*(-x^2 + x^3)^(1/4)),x]

[Out]

((-1 + x)^(1/4)*Sqrt[x]*ArcTan[(Sqrt[2]*(-1 + x)^(1/4))/Sqrt[x]])/(Sqrt[2]*(-x^2 + x^3)^(1/4)) - ((-1 + x)^(1/
4)*Sqrt[x]*ArcTanh[(Sqrt[2]*(-1 + x)^(1/4))/Sqrt[x]])/(Sqrt[2]*(-x^2 + x^3)^(1/4))

Rule 107

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(1/4)), x_Symbol] :> Dist[-4, Subst[
Int[x^2/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d,
e, f}, x] && GtQ[-f/(d*e - c*f), 0]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1225

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d
^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1713

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \int \frac {1}{(-2+x) \sqrt [4]{-1+x} \sqrt {x}} \, dx}{\sqrt [4]{-x^2+x^3}} \\ & = -\frac {\left (4 \sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1-x^4\right ) \sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-x^2+x^3}} \\ & = -\frac {\left (2 \sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-x^2+x^3}}+\frac {\left (2 \sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-x^2+x^3}}-\frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1+x^2}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\sqrt [4]{-x^2+x^3}} \\ & = -\frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt {x}}\right )}{\sqrt [4]{-x^2+x^3}}+\frac {\left (\sqrt [4]{-1+x} \sqrt {x}\right ) \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt {x}}\right )}{\sqrt [4]{-x^2+x^3}} \\ & = \frac {\sqrt [4]{-1+x} \sqrt {x} \arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^3}}-\frac {\sqrt [4]{-1+x} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{-1+x}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx=\frac {\sqrt [4]{-1+x} \sqrt {x} \left (\arctan \left (\frac {\sqrt {2} \sqrt [4]{-1+x}}{\sqrt {x}}\right )-\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2} \sqrt [4]{-1+x}}\right )\right )}{\sqrt {2} \sqrt [4]{(-1+x) x^2}} \]

[In]

Integrate[1/((-2 + x)*(-x^2 + x^3)^(1/4)),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.19 (sec) , antiderivative size = 200, normalized size of antiderivative = 3.45

method	result	size
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \sqrt {x^{3}-x^{2}}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{3}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}-8 \left (x^{3}-x^{2}\right )^{\frac {3}{4}}+4 \left (x^{3}-x^{2}\right )^{\frac {1}{4}} x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x}{\left (x -2\right )^{2} x}\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \sqrt {x^{3}-x^{2}}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{3}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+8 \left (x^{3}-x^{2}\right )^{\frac {3}{4}}+4 \left (x^{3}-x^{2}\right )^{\frac {1}{4}} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x}{\left (x -2\right )^{2} x}\right )}{4}\)	\(200\)

[In]

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

[Out]

-1/4*RootOf(_Z^2+2)*ln(-(4*RootOf(_Z^2+2)*(x^3-x^2)^(1/2)*x-RootOf(_Z^2+2)*x^3-4*RootOf(_Z^2+2)*x^2-8*(x^3-x^2
)^(3/4)+4*(x^3-x^2)^(1/4)*x^2+4*RootOf(_Z^2+2)*x)/(x-2)^2/x)-1/4*RootOf(_Z^2-2)*ln((4*RootOf(_Z^2-2)*(x^3-x^2)
^(1/2)*x+RootOf(_Z^2-2)*x^3+4*RootOf(_Z^2-2)*x^2+8*(x^3-x^2)^(3/4)+4*(x^3-x^2)^(1/4)*x^2-4*RootOf(_Z^2-2)*x)/(
x-2)^2/x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (48) = 96\).

Time = 1.72 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.33 \[ \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {2} {\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {2} {\left (x^{3} - x^{2}\right )}^{\frac {3}{4}}\right )}}{x^{3} - 4 \, x^{2} + 4 \, x}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {x^{5} + 56 \, x^{4} - 40 \, x^{3} - 8 \, \sqrt {2} {\left (x^{3} - x^{2}\right )}^{\frac {3}{4}} {\left (3 \, x^{2} + 4 \, x - 4\right )} - 32 \, x^{2} - 4 \, \sqrt {2} {\left (x^{4} + 12 \, x^{3} - 12 \, x^{2}\right )} {\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} + 16 \, {\left (x^{3} + 4 \, x^{2} - 4 \, x\right )} \sqrt {x^{3} - x^{2}} + 16 \, x}{x^{5} - 8 \, x^{4} + 24 \, x^{3} - 32 \, x^{2} + 16 \, x}\right ) \]

[In]

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

[Out]

1/4*sqrt(2)*arctan(2*(sqrt(2)*(x^3 - x^2)^(1/4)*x^2 + 2*sqrt(2)*(x^3 - x^2)^(3/4))/(x^3 - 4*x^2 + 4*x)) + 1/8*
sqrt(2)*log(-(x^5 + 56*x^4 - 40*x^3 - 8*sqrt(2)*(x^3 - x^2)^(3/4)*(3*x^2 + 4*x - 4) - 32*x^2 - 4*sqrt(2)*(x^4
+ 12*x^3 - 12*x^2)*(x^3 - x^2)^(1/4) + 16*(x^3 + 4*x^2 - 4*x)*sqrt(x^3 - x^2) + 16*x)/(x^5 - 8*x^4 + 24*x^3 -
32*x^2 + 16*x))

Sympy [F]

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

[In]

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

[Out]

Integral(1/((x**2*(x - 1))**(1/4)*(x - 2)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx=\int \frac {1}{{\left (x^3-x^2\right )}^{1/4}\,\left (x-2\right )} \,d x \]

[In]

int(1/((x^3 - x^2)^(1/4)*(x - 2)),x)

[Out]

int(1/((x^3 - x^2)^(1/4)*(x - 2)), x)