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

Optimal. Leaf size=58 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^3-x^2}}{x}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^3-x^2}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 102, normalized size of antiderivative = 1.76, number of steps used = 11, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2056, 106, 490, 1211, 220, 1699, 203, 206} \begin {gather*} \frac {\sqrt [4]{x-1} \sqrt {x} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x-1}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{x^3-x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 106

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 203

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

Rule 206

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 490

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), In
t[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 1211

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 1699

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 2056

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*} \int \frac {1}{(-2+x) \sqrt [4]{-x^2+x^3}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-1+x}}{\sqrt {x}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 40, normalized size = 0.69 \begin {gather*} -\frac {\sqrt [4]{1-x} x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x,\frac {x}{2}\right )}{\sqrt [4]{(x-1) x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((1 - x)^(1/4)*x*AppellF1[1/2, 1/4, 1, 3/2, x, x/2])/((-1 + x)*x^2)^(1/4))

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IntegrateAlgebraic [A]  time = 0.21, size = 58, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{-x^2+x^3}}{x}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{-x^2+x^3}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.10, size = 193, normalized size = 3.33 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 2.28, size = 200, normalized size = 3.45

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{4}} {\left (x - 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (x^3-x^2\right )}^{1/4}\,\left (x-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^3 - x^2)^(1/4)*(x - 2)),x)

[Out]

int(1/((x^3 - x^2)^(1/4)*(x - 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{x^{2} \left (x - 1\right )} \left (x - 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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