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

Optimal. Leaf size=33 \[ \frac {4 \left ((x-1)^3\right )^{3/4} \left (135 x^3+245 x^2+158 x+47\right )}{585 (x-1)^2} \]

________________________________________________________________________________________

Rubi [B]  time = 0.19, antiderivative size = 70, normalized size of antiderivative = 2.12, 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 (1-x)^4}{13 \sqrt [4]{(x-1)^3}}+\frac {36 (1-x)^2}{5 \sqrt [4]{(x-1)^3}}-\frac {4 (1-x)}{\sqrt [4]{(x-1)^3}}+\frac {40}{9} \left ((x-1)^3\right )^{3/4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - x))/((-1 + x)^3)^(1/4) + (36*(1 - x)^2)/(5*((-1 + x)^3)^(1/4)) + (12*(1 - x)^4)/(13*((-1 + x)^3)^(1/4
)) + (40*((-1 + x)^3)^(3/4))/9

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 \frac {-1-2 x+x^2+3 x^3}{\sqrt [4]{-1+3 x-3 x^2+x^3}} \, dx &=\int \left (-\frac {1}{\sqrt [4]{-1+3 x-3 x^2+x^3}}-\frac {2 x}{\sqrt [4]{-1+3 x-3 x^2+x^3}}+\frac {x^2}{\sqrt [4]{-1+3 x-3 x^2+x^3}}+\frac {3 x^3}{\sqrt [4]{-1+3 x-3 x^2+x^3}}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\sqrt [4]{-1+3 x-3 x^2+x^3}} \, dx\right )+3 \int \frac {x^3}{\sqrt [4]{-1+3 x-3 x^2+x^3}} \, dx-\int \frac {1}{\sqrt [4]{-1+3 x-3 x^2+x^3}} \, dx+\int \frac {x^2}{\sqrt [4]{-1+3 x-3 x^2+x^3}} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1+x}{\sqrt [4]{x^3}} \, dx,x,-1+x\right )\right )+3 \operatorname {Subst}\left (\int \frac {(1+x)^3}{\sqrt [4]{x^3}} \, dx,x,-1+x\right )-\operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{x^3}} \, dx,x,-1+x\right )+\operatorname {Subst}\left (\int \frac {(1+x)^2}{\sqrt [4]{x^3}} \, dx,x,-1+x\right )\\ &=-\frac {(-1+x)^{3/4} \operatorname {Subst}\left (\int \frac {1}{x^{3/4}} \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}+\frac {(-1+x)^{3/4} \operatorname {Subst}\left (\int \frac {(1+x)^2}{x^{3/4}} \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}-\frac {\left (2 (-1+x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1+x}{x^{3/4}} \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}+\frac {\left (3 (-1+x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {(1+x)^3}{x^{3/4}} \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}\\ &=\frac {4 (1-x)}{\sqrt [4]{(-1+x)^3}}+\frac {(-1+x)^{3/4} \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/4}}+2 \sqrt [4]{x}+x^{5/4}\right ) \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}-\frac {\left (2 (-1+x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/4}}+\sqrt [4]{x}\right ) \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}+\frac {\left (3 (-1+x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/4}}+3 \sqrt [4]{x}+3 x^{5/4}+x^{9/4}\right ) \, dx,x,-1+x\right )}{\sqrt [4]{(-1+x)^3}}\\ &=-\frac {4 (1-x)}{\sqrt [4]{(-1+x)^3}}+\frac {36 (1-x)^2}{5 \sqrt [4]{(-1+x)^3}}+\frac {12 (1-x)^4}{13 \sqrt [4]{(-1+x)^3}}+\frac {40}{9} \left ((-1+x)^3\right )^{3/4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 31, normalized size = 0.94 \begin {gather*} \frac {4 (x-1) \left (135 x^3+245 x^2+158 x+47\right )}{585 \sqrt [4]{(x-1)^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1 + x)*(47 + 158*x + 245*x^2 + 135*x^3))/(585*((-1 + x)^3)^(1/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 4.92, size = 57, normalized size = 1.73 \begin {gather*} \frac {4 \left (585 \sqrt [4]{-1+x}+1053 (-1+x)^{5/4}+650 (-1+x)^{9/4}+135 (-1+x)^{13/4}\right ) \left ((-1+x)^3\right )^{3/4}}{585 (-1+x)^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(585*(-1 + x)^(1/4) + 1053*(-1 + x)^(5/4) + 650*(-1 + x)^(9/4) + 135*(-1 + x)^(13/4))*((-1 + x)^3)^(3/4))/(
585*(-1 + x)^(9/4))

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 42, normalized size = 1.27 \begin {gather*} \frac {4 \, {\left (135 \, x^{3} + 245 \, x^{2} + 158 \, x + 47\right )} {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}^{\frac {3}{4}}}{585 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/585*(135*x^3 + 245*x^2 + 158*x + 47)*(x^3 - 3*x^2 + 3*x - 1)^(3/4)/(x^2 - 2*x + 1)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 28, normalized size = 0.85

method	result	size
risch	\(\frac {4 \left (-1+x \right ) \left (135 x^{3}+245 x^{2}+158 x +47\right )}{585 \left (\left (-1+x \right )^{3}\right )^{\frac {1}{4}}}\)	\(28\)
gosper	\(\frac {4 \left (-1+x \right ) \left (135 x^{3}+245 x^{2}+158 x +47\right )}{585 \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {1}{4}}}\)	\(36\)
trager	\(\frac {4 \left (135 x^{3}+245 x^{2}+158 x +47\right ) \left (x^{3}-3 x^{2}+3 x -1\right )^{\frac {3}{4}}}{585 \left (-1+x \right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/585*(-1+x)*(135*x^3+245*x^2+158*x+47)/((-1+x)^3)^(1/4)

________________________________________________________________________________________

maxima [B]  time = 0.34, size = 72, normalized size = 2.18 \begin {gather*} \frac {4 \, {\left (15 \, x^{4} + 5 \, x^{3} + 12 \, x^{2} + 96 \, x - 128\right )}}{65 \, {\left (x - 1\right )}^{\frac {3}{4}}} + \frac {4 \, {\left (5 \, x^{3} + 3 \, x^{2} + 24 \, x - 32\right )}}{45 \, {\left (x - 1\right )}^{\frac {3}{4}}} - \frac {8 \, {\left (x^{2} + 3 \, x - 4\right )}}{5 \, {\left (x - 1\right )}^{\frac {3}{4}}} - 4 \, {\left (x - 1\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/65*(15*x^4 + 5*x^3 + 12*x^2 + 96*x - 128)/(x - 1)^(3/4) + 4/45*(5*x^3 + 3*x^2 + 24*x - 32)/(x - 1)^(3/4) - 8
/5*(x^2 + 3*x - 4)/(x - 1)^(3/4) - 4*(x - 1)^(1/4)

________________________________________________________________________________________

mupad [B]  time = 0.26, size = 41, normalized size = 1.24 \begin {gather*} \frac {{\left (x^3-3\,x^2+3\,x-1\right )}^{3/4}\,\left (\frac {12\,x^3}{13}+\frac {196\,x^2}{117}+\frac {632\,x}{585}+\frac {188}{585}\right )}{x^2-2\,x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*x - 3*x^2 + x^3 - 1)^(3/4)*((632*x)/585 + (196*x^2)/117 + (12*x^3)/13 + 188/585))/(x^2 - 2*x + 1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________