\(\int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx\) [2815]

Optimal result

Integrand size = 31, antiderivative size = 278 \[ \int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx=\frac {(-1+x)^{4/3} (1+3 x)^{2/3} \left (-\frac {21 \sqrt [3]{3} \left (-16 \sqrt [3]{1+3 x}+7 (1+3 x)^{4/3}\right )}{64 (-3+3 x)^{4/3}}+\frac {3 \sqrt [3]{3} \left (-80 \sqrt [3]{1+3 x}+23 (1+3 x)^{4/3}\right )}{64 (-3+3 x)^{4/3}}+\frac {\sqrt {3} \arctan \left (\frac {3^{5/6} \sqrt [3]{1+3 x}}{2\ 2^{2/3} \sqrt [3]{-3+3 x}+\sqrt [3]{3} \sqrt [3]{1+3 x}}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (6^{2/3} \sqrt [3]{-3+3 x}-3 \sqrt [3]{1+3 x}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (2 \sqrt [3]{6} (-3+3 x)^{2/3}+6^{2/3} \sqrt [3]{-3+3 x} \sqrt [3]{1+3 x}+3 (1+3 x)^{2/3}\right )}{8 \sqrt [3]{2}}\right )}{\left ((-1+x)^2 (1+3 x)\right )^{2/3}} \]

[Out]

(-1+x)^(4/3)*(1+3*x)^(2/3)*(-21/64*3^(1/3)*(-16*(1+3*x)^(1/3)+7*(1+3*x)^(4/3))/(-3+3*x)^(4/3)+3/64*3^(1/3)*(-8
0*(1+3*x)^(1/3)+23*(1+3*x)^(4/3))/(-3+3*x)^(4/3)+1/8*3^(1/2)*arctan(3^(5/6)*(1+3*x)^(1/3)/(2*2^(2/3)*(-3+3*x)^
(1/3)+3^(1/3)*(1+3*x)^(1/3)))*2^(2/3)-1/8*ln(6^(2/3)*(-3+3*x)^(1/3)-3*(1+3*x)^(1/3))*2^(2/3)+1/16*ln(2*6^(1/3)
*(-3+3*x)^(2/3)+6^(2/3)*(-3+3*x)^(1/3)*(1+3*x)^(1/3)+3*(1+3*x)^(2/3))*2^(2/3))/((-1+x)^2*(1+3*x))^(2/3)

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {6874, 2106, 2102, 103, 163, 62, 93, 21, 37, 49, 52} \[ \int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx=-\frac {\sqrt {3} \sqrt [3]{3 x^3-5 x^2+x+1} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{1-x}}{\sqrt {3} \sqrt [3]{3 x+1}}\right )}{4 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}}-\frac {9 \sqrt [3]{3 x^3-5 x^2+x+1} (3 x+1)}{32 (1-x)^2}+\frac {3 \sqrt [3]{3 x^3-5 x^2+x+1}}{8 (1-x)}+\frac {\sqrt [3]{3 x^3-5 x^2+x+1} \log (x-5)}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}}-\frac {3 \sqrt [3]{3 x^3-5 x^2+x+1} \log \left (-\frac {4}{3} \sqrt [3]{1-x}-\frac {2}{3} \sqrt [3]{2} \sqrt [3]{3 x+1}\right )}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{3 x+1}} \]

[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 49

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 52

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 62

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

Rule 93

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 103

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 163

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 2102

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 2106

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \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} \text {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} \text {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} \text {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} \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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} \arctan \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 (2^{2/3} \sqrt [3]{1-x}+\sqrt [3]{1+3 x}\right )}{8 \sqrt [3]{2} (1-x)^{2/3} \sqrt [3]{1+3 x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.70 \[ \int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx=\frac {1}{32} \sqrt [3]{(-1+x)^2 (1+3 x)} \left (\frac {3-39 x}{(-1+x)^2}+\frac {4\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt [3]{-1+x}+\sqrt [3]{2+6 x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )}{(-1+x)^{2/3} \sqrt [3]{1+3 x}}-\frac {4\ 2^{2/3} \log \left (-2+\frac {\sqrt [3]{2+6 x}}{\sqrt [3]{-1+x}}\right )}{(-1+x)^{2/3} \sqrt [3]{1+3 x}}+\frac {2\ 2^{2/3} \log \left (4+\frac {2 \sqrt [3]{2+6 x}}{\sqrt [3]{-1+x}}+\frac {(2+6 x)^{2/3}}{(-1+x)^{2/3}}\right )}{(-1+x)^{2/3} \sqrt [3]{1+3 x}}\right ) \]

[In]

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

[Out]

(((-1 + x)^2*(1 + 3*x))^(1/3)*((3 - 39*x)/(-1 + x)^2 + (4*2^(2/3)*Sqrt[3]*ArcTan[((-1 + x)^(1/3) + (2 + 6*x)^(
1/3))/(Sqrt[3]*(-1 + x)^(1/3))])/((-1 + x)^(2/3)*(1 + 3*x)^(1/3)) - (4*2^(2/3)*Log[-2 + (2 + 6*x)^(1/3)/(-1 +
x)^(1/3)])/((-1 + x)^(2/3)*(1 + 3*x)^(1/3)) + (2*2^(2/3)*Log[4 + (2*(2 + 6*x)^(1/3))/(-1 + x)^(1/3) + (2 + 6*x
)^(2/3)/(-1 + x)^(2/3)])/((-1 + x)^(2/3)*(1 + 3*x)^(1/3))))/32

Maple [C] (warning: unable to verify)

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

Time = 1.64 (sec) , antiderivative size = 1546, normalized size of antiderivative = 5.56

method	result	size
trager	\(\text {Expression too large to display}\)	\(1546\)

[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)+1/8*RootOf(_Z^3+4)*ln((-192*RootOf(RootOf(_Z^3+4)^2+48*_Z*Root
Of(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2+23040*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*Roo
tOf(_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-23040*RootOf(
RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x+576*(3*x^3-5*x^2+x+1)^(2/3)*RootOf(RootO
f(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^2-9*RootOf(_Z^3+4)^2*x^2+1080*RootOf(RootOf(_Z^3+4)
^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x^2-48*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)*x+2016*(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+22*RootOf(_Z^3+4)^2*x-2640*RootOf(R
ootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x-90*(3*x^3-5*x^2+x+1)^(2/3)+48*(3*x^3-5*x^2+x+
1)^(1/3)*RootOf(_Z^3+4)-2016*(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)-1
3*RootOf(_Z^3+4)^2+1560*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4))/(-5+x)/(-1+x))
-1/8*ln(-(960*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2-18432*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+18432*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3
+4)^3*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-
125*RootOf(_Z^3+4)^2*x^2+2400*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x^2-96*(3
*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)*x-8640*(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+190*RootOf(_Z^3+4)^2*x-3648*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^
3+4)*x+84*(3*x^3-5*x^2+x+1)^(2/3)+96*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)+8640*(3*x^3-5*x^2+x+1)^(1/3)*RootO
f(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)-65*RootOf(_Z^3+4)^2+1248*RootOf(RootOf(_Z^3+4)^2+48*_Z*Root
Of(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4))/(-5+x)/(-1+x))*RootOf(_Z^3+4)-6*ln(-(960*RootOf(RootOf(_Z^3+4)^2+48*_Z*R
ootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^4*x^2-18432*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+18432*Root
Of(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)^2*RootOf(_Z^3+4)^3*x+1152*(3*x^3-5*x^2+x+1)^(2/3)*RootOf(R
ootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)^2-125*RootOf(_Z^3+4)^2*x^2+2400*RootOf(RootOf(_
Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x^2-96*(3*x^3-5*x^2+x+1)^(1/3)*RootOf(_Z^3+4)*x-8640*(
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+190*RootOf(_Z^3+4)^2*x-3648*R
ootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4)*x+84*(3*x^3-5*x^2+x+1)^(2/3)+96*(3*x^3-5
*x^2+x+1)^(1/3)*RootOf(_Z^3+4)+8640*(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)-65*RootOf(_Z^3+4)^2+1248*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)*RootOf(_Z^3+4))/(-5+x)/
(-1+x))*RootOf(RootOf(_Z^3+4)^2+48*_Z*RootOf(_Z^3+4)+2304*_Z^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.85 \[ \int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx=-\frac {4 \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} {\left (x - 1\right )} + 2 \cdot 2^{\frac {1}{6}} \left (-1\right )^{\frac {2}{3}} {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (x - 1\right )}}\right ) + 2 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 2 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x + 1\right )} - {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {2}{3}}}{x^{2} - 2 \, x + 1}\right ) - 4 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x - 1\right )} + {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}}}{x - 1}\right ) + 3 \, {\left (3 \, x^{3} - 5 \, x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (13 \, x - 1\right )}}{32 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(-7+x) \sqrt [3]{1+x-5 x^2+3 x^3}}{(-5+x) (-1+x)^3} \, dx=\int \frac {\left (x-7\right )\,{\left (3\,x^3-5\,x^2+x+1\right )}^{1/3}}{{\left (x-1\right )}^3\,\left (x-5\right )} \,d x \]

[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)