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

Optimal. Leaf size=33 \[ \frac {4 \sqrt [4]{(x-1)^3} \left (693 x^4+154 x^3-1029 x^2-549 x+731\right )}{4389} \]

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Rubi [B]  time = 0.19, antiderivative size = 72, normalized size of antiderivative = 2.18, number of steps used = 17, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6742, 2067, 15, 30, 2081, 43} \begin {gather*} \frac {12}{19} \sqrt [4]{(x-1)^3} (1-x)^4+\frac {36}{11} \sqrt [4]{(x-1)^3} (1-x)^2-\frac {4}{7} \sqrt [4]{(x-1)^3} (1-x)+\frac {8}{3} \left ((x-1)^3\right )^{5/4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - x)*((-1 + x)^3)^(1/4))/7 + (36*(1 - x)^2*((-1 + x)^3)^(1/4))/11 + (12*(1 - x)^4*((-1 + x)^3)^(1/4))/1
9 + (8*((-1 + x)^3)^(5/4))/3

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rule 2081

Int[(P3_)^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = C
oeff[P3, x, 2], d = Coeff[P3, x, 3]}, Subst[Int[((3*d*e - c*f)/(3*d) + f*x)^m*Simp[(2*c^3 - 9*b*c*d + 27*a*d^2
)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x, x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[{e, f, m, p}, x
] && PolyQ[P3, x, 3]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \sqrt [4]{-1+3 x-3 x^2+x^3} \left (-1-2 x+x^2+3 x^3\right ) \, dx &=\int \left (-\sqrt [4]{-1+3 x-3 x^2+x^3}-2 x \sqrt [4]{-1+3 x-3 x^2+x^3}+x^2 \sqrt [4]{-1+3 x-3 x^2+x^3}+3 x^3 \sqrt [4]{-1+3 x-3 x^2+x^3}\right ) \, dx\\ &=-\left (2 \int x \sqrt [4]{-1+3 x-3 x^2+x^3} \, dx\right )+3 \int x^3 \sqrt [4]{-1+3 x-3 x^2+x^3} \, dx-\int \sqrt [4]{-1+3 x-3 x^2+x^3} \, dx+\int x^2 \sqrt [4]{-1+3 x-3 x^2+x^3} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \sqrt [4]{x^3} (1+x) \, dx,x,-1+x\right )\right )+3 \operatorname {Subst}\left (\int \sqrt [4]{x^3} (1+x)^3 \, dx,x,-1+x\right )-\operatorname {Subst}\left (\int \sqrt [4]{x^3} \, dx,x,-1+x\right )+\operatorname {Subst}\left (\int \sqrt [4]{x^3} (1+x)^2 \, dx,x,-1+x\right )\\ &=-\frac {\sqrt [4]{(-1+x)^3} \operatorname {Subst}\left (\int x^{3/4} \, dx,x,-1+x\right )}{(-1+x)^{3/4}}+\frac {\sqrt [4]{(-1+x)^3} \operatorname {Subst}\left (\int x^{3/4} (1+x)^2 \, dx,x,-1+x\right )}{(-1+x)^{3/4}}-\frac {\left (2 \sqrt [4]{(-1+x)^3}\right ) \operatorname {Subst}\left (\int x^{3/4} (1+x) \, dx,x,-1+x\right )}{(-1+x)^{3/4}}+\frac {\left (3 \sqrt [4]{(-1+x)^3}\right ) \operatorname {Subst}\left (\int x^{3/4} (1+x)^3 \, dx,x,-1+x\right )}{(-1+x)^{3/4}}\\ &=\frac {4}{7} (1-x) \sqrt [4]{(-1+x)^3}+\frac {\sqrt [4]{(-1+x)^3} \operatorname {Subst}\left (\int \left (x^{3/4}+2 x^{7/4}+x^{11/4}\right ) \, dx,x,-1+x\right )}{(-1+x)^{3/4}}-\frac {\left (2 \sqrt [4]{(-1+x)^3}\right ) \operatorname {Subst}\left (\int \left (x^{3/4}+x^{7/4}\right ) \, dx,x,-1+x\right )}{(-1+x)^{3/4}}+\frac {\left (3 \sqrt [4]{(-1+x)^3}\right ) \operatorname {Subst}\left (\int \left (x^{3/4}+3 x^{7/4}+3 x^{11/4}+x^{15/4}\right ) \, dx,x,-1+x\right )}{(-1+x)^{3/4}}\\ &=-\frac {4}{7} (1-x) \sqrt [4]{(-1+x)^3}+\frac {36}{11} (1-x)^2 \sqrt [4]{(-1+x)^3}+\frac {12}{19} (1-x)^4 \sqrt [4]{(-1+x)^3}+\frac {8}{3} \left ((-1+x)^3\right )^{5/4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.94 \begin {gather*} \frac {4 (x-1) \sqrt [4]{(x-1)^3} \left (693 x^3+847 x^2-182 x-731\right )}{4389} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1 + x)*((-1 + x)^3)^(1/4)*(-731 - 182*x + 847*x^2 + 693*x^3))/4389

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IntegrateAlgebraic [A]  time = 4.95, size = 57, normalized size = 1.73 \begin {gather*} \frac {4 \left (627 (-1+x)^{7/4}+3591 (-1+x)^{11/4}+2926 (-1+x)^{15/4}+693 (-1+x)^{19/4}\right ) \sqrt [4]{(-1+x)^3}}{4389 (-1+x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(627*(-1 + x)^(7/4) + 3591*(-1 + x)^(11/4) + 2926*(-1 + x)^(15/4) + 693*(-1 + x)^(19/4))*((-1 + x)^3)^(1/4)
)/(4389*(-1 + x)^(3/4))

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fricas [A]  time = 0.46, size = 37, normalized size = 1.12 \begin {gather*} \frac {4}{4389} \, {\left (693 \, x^{4} + 154 \, x^{3} - 1029 \, x^{2} - 549 \, x + 731\right )} {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/4389*(693*x^4 + 154*x^3 - 1029*x^2 - 549*x + 731)*(x^3 - 3*x^2 + 3*x - 1)^(1/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 30, normalized size = 0.91

method	result	size
risch	\(\frac {4 \left (\left (-1+x \right )^{3}\right )^{\frac {1}{4}} \left (693 x^{4}+154 x^{3}-1029 x^{2}-549 x +731\right )}{4389}\)	\(30\)
gosper	\(\frac {4 \left (-1+x \right ) \left (693 x^{3}+847 x^{2}-182 x -731\right ) \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{4}}}{4389}\)	\(36\)
trager	\(\left (\frac {12}{19} x^{4}+\frac {8}{57} x^{3}-\frac {196}{209} x^{2}-\frac {732}{1463} x +\frac {2924}{4389}\right ) \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{4}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/4389*((-1+x)^3)^(1/4)*(693*x^4+154*x^3-1029*x^2-549*x+731)

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maxima [B]  time = 0.34, size = 74, normalized size = 2.24 \begin {gather*} \frac {12}{7315} \, {\left (385 \, x^{4} - 77 \, x^{3} - 84 \, x^{2} - 96 \, x - 128\right )} {\left (x - 1\right )}^{\frac {3}{4}} + \frac {4}{1155} \, {\left (77 \, x^{3} - 21 \, x^{2} - 24 \, x - 32\right )} {\left (x - 1\right )}^{\frac {3}{4}} - \frac {8}{77} \, {\left (7 \, x^{2} - 3 \, x - 4\right )} {\left (x - 1\right )}^{\frac {3}{4}} - \frac {4}{7} \, {\left (x - 1\right )}^{\frac {7}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12/7315*(385*x^4 - 77*x^3 - 84*x^2 - 96*x - 128)*(x - 1)^(3/4) + 4/1155*(77*x^3 - 21*x^2 - 24*x - 32)*(x - 1)^
(3/4) - 8/77*(7*x^2 - 3*x - 4)*(x - 1)^(3/4) - 4/7*(x - 1)^(7/4)

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mupad [B]  time = 0.20, size = 35, normalized size = 1.06 \begin {gather*} -\frac {4\,\left (x-1\right )\,{\left (x^3-3\,x^2+3\,x-1\right )}^{1/4}\,\left (-693\,x^3-847\,x^2+182\,x+731\right )}{4389} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(x - 1)*(3*x - 3*x^2 + x^3 - 1)^(1/4)*(182*x - 847*x^2 - 693*x^3 + 731))/4389

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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