3.29.15 \(\int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx\)

Optimal. Leaf size=278 \[ \frac {(x-1)^{4/3} (3 x+1)^{2/3} \left (-\frac {21 \sqrt [3]{3} \left (7 (3 x+1)^{4/3}-16 \sqrt [3]{3 x+1}\right )}{64 (3 x-3)^{4/3}}+\frac {3 \sqrt [3]{3} \left (23 (3 x+1)^{4/3}-80 \sqrt [3]{3 x+1}\right )}{64 (3 x-3)^{4/3}}-\frac {\log \left (6^{2/3} \sqrt [3]{3 x-3}-3 \sqrt [3]{3 x+1}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (2 \sqrt [3]{6} (3 x-3)^{2/3}+6^{2/3} \sqrt [3]{3 x+1} \sqrt [3]{3 x-3}+3 (3 x+1)^{2/3}\right )}{8 \sqrt [3]{2}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {3^{5/6} \sqrt [3]{3 x+1}}{2\ 2^{2/3} \sqrt [3]{3 x-3}+\sqrt [3]{3} \sqrt [3]{3 x+1}}\right )}{4 \sqrt [3]{2}}\right )}{\left ((x-1)^2 (3 x+1)\right )^{2/3}} \]

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Rubi [A]  time = 1.45, antiderivative size = 273, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {6742, 2081, 2077, 101, 157, 60, 91, 21, 37, 47, 50} \begin {gather*} -\frac {9 \sqrt [3]{3 x^3-5 x^2+x+1} (3 x+1)}{32 (1-x)^2}+\frac {3 \sqrt [3]{3 x^3-5 x^2+x+1}}{8 (1-x)}+\frac {\sqrt [3]{3 x^3-5 x^2+x+1} \log (x-5)}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}}-\frac {3 \sqrt [3]{3 x^3-5 x^2+x+1} \log \left (-\frac {4}{3} \sqrt [3]{1-x}-\frac {2}{3} \sqrt [3]{2} \sqrt [3]{3 x+1}\right )}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}}-\frac {\sqrt {3} \sqrt [3]{3 x^3-5 x^2+x+1} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{3 x+1}}\right )}{4 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-7 + x)*(1 + x - 5*x^2 + 3*x^3)^(1/3))/((-5 + x)*(-1 + x)^3),x]

[Out]

(3*(1 + x - 5*x^2 + 3*x^3)^(1/3))/(8*(1 - x)) - (9*(1 + 3*x)*(1 + x - 5*x^2 + 3*x^3)^(1/3))/(32*(1 - x)^2) - (
Sqrt[3]*(1 + x - 5*x^2 + 3*x^3)^(1/3)*ArcTan[1/Sqrt[3] - (2*2^(2/3)*(1 - x)^(1/3))/(Sqrt[3]*(1 + 3*x)^(1/3))])
/(4*2^(1/3)*(1 - x)^(2/3)*(1 + 3*x)^(1/3)) + ((1 + x - 5*x^2 + 3*x^3)^(1/3)*Log[-5 + x])/(8*2^(1/3)*(1 - x)^(2
/3)*(1 + 3*x)^(1/3)) - (3*(1 + x - 5*x^2 + 3*x^3)^(1/3)*Log[(-4*(1 - x)^(1/3))/3 - (2*2^(1/3)*(1 + 3*x)^(1/3))
/3])/(8*2^(1/3)*(1 - x)^(2/3)*(1 + 3*x)^(1/3))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

Rule 91

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 2077

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/
((3*a - b*x)^p*(3*a + 2*b*x)^(2*p)), Int[(e + f*x)^m*(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b,
d, e, f, m, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0] &&  !IntegerQ[p]

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 {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx &=\int \left (-\frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{32 (-5+x)}+\frac {3 \sqrt [3]{1+x-5 x^2+3 x^3}}{2 (-1+x)^3}+\frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{8 (-1+x)^2}+\frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{32 (-1+x)}\right ) \, dx\\ &=-\left (\frac {1}{32} \int \frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{-5+x} \, dx\right )+\frac {1}{32} \int \frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{-1+x} \, dx+\frac {1}{8} \int \frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{(-1+x)^2} \, dx+\frac {3}{2} \int \frac {\sqrt [3]{1+x-5 x^2+3 x^3}}{(-1+x)^3} \, dx\\ &=-\left (\frac {1}{32} \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{243}-\frac {16 x}{9}+3 x^3}}{-\frac {40}{9}+x} \, dx,x,-\frac {5}{9}+x\right )\right )+\frac {1}{32} \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{243}-\frac {16 x}{9}+3 x^3}}{-\frac {4}{9}+x} \, dx,x,-\frac {5}{9}+x\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{243}-\frac {16 x}{9}+3 x^3}}{\left (-\frac {4}{9}+x\right )^2} \, dx,x,-\frac {5}{9}+x\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{243}-\frac {16 x}{9}+3 x^3}}{\left (-\frac {4}{9}+x\right )^3} \, dx,x,-\frac {5}{9}+x\right )\\ &=-\frac {\left (9 \sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {128}{81}-\frac {32 x}{9}\right )^{2/3} \sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{-\frac {40}{9}+x} \, dx,x,-\frac {5}{9}+x\right )}{512\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (9 \sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {128}{81}-\frac {32 x}{9}\right )^{2/3} \sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{-\frac {4}{9}+x} \, dx,x,-\frac {5}{9}+x\right )}{512\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (9 \sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {128}{81}-\frac {32 x}{9}\right )^{2/3} \sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{\left (-\frac {4}{9}+x\right )^2} \, dx,x,-\frac {5}{9}+x\right )}{128\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (27 \sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {128}{81}-\frac {32 x}{9}\right )^{2/3} \sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{\left (-\frac {4}{9}+x\right )^3} \, dx,x,-\frac {5}{9}+x\right )}{32\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}\\ &=-\frac {1}{32} \sqrt [3]{1+x-5 x^2+3 x^3}+\frac {\left (4 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{\left (\frac {128}{81}-\frac {32 x}{9}\right )^{4/3}} \, dx,x,-\frac {5}{9}+x\right )}{3\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}-\frac {\left (512 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{\left (\frac {128}{81}-\frac {32 x}{9}\right )^{7/3}} \, dx,x,-\frac {5}{9}+x\right )}{9\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (9 \sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\frac {65536}{6561}+\frac {20480 x}{729}}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}} \left (-\frac {40}{9}+x\right ) \left (\frac {128}{81}+\frac {16 x}{9}\right )^{2/3}} \, dx,x,-\frac {5}{9}+x\right )}{512\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}-\frac {\left (\sqrt [3]{3} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{\frac {128}{81}+\frac {16 x}{9}}}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}}} \, dx,x,-\frac {5}{9}+x\right )}{16\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}\\ &=\frac {3 \sqrt [3]{1+x-5 x^2+3 x^3}}{8 (1-x)}-\frac {9 (1+3 x) \sqrt [3]{1+x-5 x^2+3 x^3}}{32 (1-x)^2}-\frac {\left (2 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}} \left (\frac {128}{81}+\frac {16 x}{9}\right )^{2/3}} \, dx,x,-\frac {5}{9}+x\right )}{27\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}-\frac {\left (2 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}} \left (\frac {128}{81}+\frac {16 x}{9}\right )^{2/3}} \, dx,x,-\frac {5}{9}+x\right )}{3\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (20 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}} \left (\frac {128}{81}+\frac {16 x}{9}\right )^{2/3}} \, dx,x,-\frac {5}{9}+x\right )}{27\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\left (32 \sqrt [3]{2} \sqrt [3]{1+x-5 x^2+3 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {128}{81}-\frac {32 x}{9}} \left (-\frac {40}{9}+x\right ) \left (\frac {128}{81}+\frac {16 x}{9}\right )^{2/3}} \, dx,x,-\frac {5}{9}+x\right )}{9\ 3^{2/3} (1-x)^{2/3} \sqrt [3]{1+3 x}}\\ &=\frac {3 \sqrt [3]{1+x-5 x^2+3 x^3}}{8 (1-x)}-\frac {9 (1+3 x) \sqrt [3]{1+x-5 x^2+3 x^3}}{32 (1-x)^2}-\frac {\sqrt {3} \sqrt [3]{1+x-5 x^2+3 x^3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{1+3 x}}\right )}{4 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{1+3 x}}+\frac {\sqrt [3]{1+x-5 x^2+3 x^3} \log (5-x)}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{1+3 x}}-\frac {3 \sqrt [3]{1+x-5 x^2+3 x^3} \log \left (6 \sqrt [3]{1-x}+3 \sqrt [3]{2} \sqrt [3]{1+3 x}\right )}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{1+3 x}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 56, normalized size = 0.20 \begin {gather*} \frac {3 \left (8 (x-1)^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {4 (x-1)}{3 x+1}\right )-39 x^2-10 x+1\right )}{32 \left ((x-1)^2 (3 x+1)\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-7 + x)*(1 + x - 5*x^2 + 3*x^3)^(1/3))/((-5 + x)*(-1 + x)^3),x]

[Out]

(3*(1 - 10*x - 39*x^2 + 8*(-1 + x)^2*Hypergeometric2F1[2/3, 1, 5/3, (4*(-1 + x))/(1 + 3*x)]))/(32*((-1 + x)^2*
(1 + 3*x))^(2/3))

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IntegrateAlgebraic [A]  time = 16.25, size = 218, normalized size = 0.78 \begin {gather*} \frac {\sqrt [3]{-1+x} (1+3 x)^{2/3} \sqrt [3]{(-1+x)^2 (1+3 x)} \left (-\frac {3 \sqrt [3]{1+3 x} \left (4+\frac {3 (1+3 x)}{-1+x}\right )}{32 \sqrt [3]{-1+x}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {\sqrt [3]{2} \sqrt [3]{1+3 x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (-2+\frac {\sqrt [3]{2} \sqrt [3]{1+3 x}}{\sqrt [3]{-1+x}}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (4+\frac {2 \sqrt [3]{2} \sqrt [3]{1+3 x}}{\sqrt [3]{-1+x}}+\frac {2^{2/3} (1+3 x)^{2/3}}{(-1+x)^{2/3}}\right )}{8 \sqrt [3]{2}}\right )}{-1-2 x+3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-7 + x)*(1 + x - 5*x^2 + 3*x^3)^(1/3))/((-5 + x)*(-1 + x)^3),x]

[Out]

((-1 + x)^(1/3)*(1 + 3*x)^(2/3)*((-1 + x)^2*(1 + 3*x))^(1/3)*((-3*(1 + 3*x)^(1/3)*(4 + (3*(1 + 3*x))/(-1 + x))
)/(32*(-1 + x)^(1/3)) + (Sqrt[3]*ArcTan[1/Sqrt[3] + (2^(1/3)*(1 + 3*x)^(1/3))/(Sqrt[3]*(-1 + x)^(1/3))])/(4*2^
(1/3)) - Log[-2 + (2^(1/3)*(1 + 3*x)^(1/3))/(-1 + x)^(1/3)]/(4*2^(1/3)) + Log[4 + (2*2^(1/3)*(1 + 3*x)^(1/3))/
(-1 + x)^(1/3) + (2^(2/3)*(1 + 3*x)^(2/3))/(-1 + x)^(2/3)]/(8*2^(1/3))))/(-1 - 2*x + 3*x^2)

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fricas [A]  time = 0.51, size = 236, normalized size = 0.85 \begin {gather*} -\frac {4 \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} {\left (x - 1\right )} + 2 \cdot 2^{\frac {1}{6}} \left (-1\right )^{\frac {2}{3}} {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (x - 1\right )}}\right ) + 2 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 2 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x + 1\right )} - {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {2}{3}}}{x^{2} - 2 \, x + 1}\right ) - 4 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x - 1\right )} + {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}}}{x - 1}\right ) + 3 \, {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (13 \, x - 1\right )}}{32 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7+x)*(3*x^3-5*x^2+x+1)^(1/3)/(-5+x)/(-1+x)^3,x, algorithm="fricas")

[Out]

-1/32*(4*sqrt(3)*2^(2/3)*(-1)^(1/3)*(x^2 - 2*x + 1)*arctan(1/6*sqrt(3)*2^(1/6)*(2^(5/6)*(x - 1) + 2*2^(1/6)*(-
1)^(2/3)*(3*x^3 - 5*x^2 + x + 1)^(1/3))/(x - 1)) + 2*2^(2/3)*(-1)^(1/3)*(x^2 - 2*x + 1)*log(-(2^(2/3)*(-1)^(1/
3)*(3*x^3 - 5*x^2 + x + 1)^(1/3)*(x - 1) - 2*2^(1/3)*(-1)^(2/3)*(x^2 - 2*x + 1) - (3*x^3 - 5*x^2 + x + 1)^(2/3
))/(x^2 - 2*x + 1)) - 4*2^(2/3)*(-1)^(1/3)*(x^2 - 2*x + 1)*log((2^(2/3)*(-1)^(1/3)*(x - 1) + (3*x^3 - 5*x^2 +
x + 1)^(1/3))/(x - 1)) + 3*(3*x^3 - 5*x^2 + x + 1)^(1/3)*(13*x - 1))/(x^2 - 2*x + 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x - 7\right )}}{{\left (x - 1\right )}^{3} {\left (x - 5\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7+x)*(3*x^3-5*x^2+x+1)^(1/3)/(-5+x)/(-1+x)^3,x, algorithm="giac")

[Out]

integrate((3*x^3 - 5*x^2 + x + 1)^(1/3)*(x - 7)/((x - 1)^3*(x - 5)), x)

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maple [C]  time = 2.99, size = 1563, normalized size = 5.62

method	result	size
trager	\(\text {Expression too large to display}\)	\(1563\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-7+x)*(3*x^3-5*x^2+x+1)^(1/3)/(-5+x)/(-1+x)^3,x,method=_RETURNVERBOSE)

[Out]

-3/32*(13*x-1)/(-1+x)^2*(3*x^3-5*x^2+x+1)^(1/3)+6*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*ln(-
(32256*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x^2+192*RootOf(RootOf(_Z^3+4
)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2-32256*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+23
04*_Z^2)^2*RootOf(_Z^3+4)^3*x-192*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x+5
76*(3*x^3-5*x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^2-4200*RootO
f(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x^2-25*RootOf(_Z^3+4)^2*x^2+2304*(3*x^3-5*x^
2+x+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*x-42*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+
4)*x+6384*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x+38*RootOf(_Z^3+4)^2*x+42*(3
*x^3-5*x^2+x+1)^(2/3)-2304*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)+42*
(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)-2184*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^
3+4)-13*RootOf(_Z^3+4)^2)/(-5+x)/(-1+x))-1/8*ln((-64512*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2
)^2*RootOf(_Z^3+4)^3*x^2-960*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2+6451
2*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x+960*RootOf(RootOf(_Z^3+4)^2+48*
_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x+1152*(3*x^3-5*x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*Roo
tOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^2-3024*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z
^3+4)*x^2-45*RootOf(_Z^3+4)^2*x^2+4608*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+23
04*_Z^2)*x+180*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)*x+7392*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304
*_Z^2)*RootOf(_Z^3+4)*x+110*RootOf(_Z^3+4)^2*x-180*(3*x^3-5*x^2+x+1)^(2/3)-4608*(3*x^3-5*x^2+x+1)^(1/3)*RootOf
(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)-180*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)-4368*RootOf(RootO
f(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)-65*RootOf(_Z^3+4)^2)/(-5+x)/(-1+x))*RootOf(_Z^3+4)-
6*ln((-64512*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x^2-960*RootOf(RootOf(
_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2+64512*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3
+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x+960*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)
^4*x+1152*(3*x^3-5*x^2+x+1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^2-302
4*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x^2-45*RootOf(_Z^3+4)^2*x^2+4608*(3*x
^3-5*x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*x+180*(3*x^3-5*x^2+x+1)^(1/3)*Root
Of(_Z^3+4)*x+7392*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x+110*RootOf(_Z^3+4)^
2*x-180*(3*x^3-5*x^2+x+1)^(2/3)-4608*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304
*_Z^2)-180*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)-4368*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)
*RootOf(_Z^3+4)-65*RootOf(_Z^3+4)^2)/(-5+x)/(-1+x))*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x - 7\right )}}{{\left (x - 1\right )}^{3} {\left (x - 5\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7+x)*(3*x^3-5*x^2+x+1)^(1/3)/(-5+x)/(-1+x)^3,x, algorithm="maxima")

[Out]

integrate((3*x^3 - 5*x^2 + x + 1)^(1/3)*(x - 7)/((x - 1)^3*(x - 5)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (x-7\right )\,{\left (3\,x^3-5\,x^2+x+1\right )}^{1/3}}{{\left (x-1\right )}^3\,\left (x-5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 7)*(x - 5*x^2 + 3*x^3 + 1)^(1/3))/((x - 1)^3*(x - 5)),x)

[Out]

int(((x - 7)*(x - 5*x^2 + 3*x^3 + 1)^(1/3))/((x - 1)^3*(x - 5)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{\left (x - 1\right )^{2} \left (3 x + 1\right )} \left (x - 7\right )}{\left (x - 5\right ) \left (x - 1\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-7+x)*(3*x**3-5*x**2+x+1)**(1/3)/(-5+x)/(-1+x)**3,x)

[Out]

Integral(((x - 1)**2*(3*x + 1))**(1/3)*(x - 7)/((x - 5)*(x - 1)**3), x)

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