3.24.18 \(\int \frac {x^2 (8-7 (1+k) x+6 k x^2)}{\sqrt [3]{(1-x) x (1-k x)} (-1+(1+k) x-k x^2+b x^8)} \, dx\) [2318]

3.24.18.1 Optimal result

Integrand size = 55, antiderivative size = 179 \[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b} x^3+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x^3+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3} x^6+\sqrt [3]{b} x^3 \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]

output

3^(1/2)*arctan(3^(1/2)*(x+(-1-k)*x^2+k*x^3)^(1/3)/(2*b^(1/3)*x^3+(x+(-1-k) 
*x^2+k*x^3)^(1/3)))/b^(1/3)+ln(-b^(1/3)*x^3+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^ 
(1/3)-1/2*ln(b^(2/3)*x^6+b^(1/3)*x^3*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)* 
x^2+k*x^3)^(2/3))/b^(1/3)
 

3.24.18.2 Mathematica [F]

\[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx \]

input

Integrate[(x^2*(8 - 7*(1 + k)*x + 6*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*( 
-1 + (1 + k)*x - k*x^2 + b*x^8)),x]
 

output

Integrate[(x^2*(8 - 7*(1 + k)*x + 6*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*( 
-1 + (1 + k)*x - k*x^2 + b*x^8)), x]
 

3.24.18.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 \left (6 k x^2-7 (k+1) x+8\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b x^8-k x^2+(k+1) x-1\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {x^{5/3} \left (6 k x^2-7 (k+1) x+8\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{5/3} \left (6 k x^2-7 (k+1) x+8\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{7/3} \left (6 k x^2-7 (k+1) x+8\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {6 k x^{13/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}+\frac {7 (-k-1) x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}+\frac {8 x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (8 \int \frac {x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}d\sqrt [3]{x}-7 (k+1) \int \frac {x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}d\sqrt [3]{x}+6 k \int \frac {x^{13/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^8+k x^2-(k+1) x+1\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\)

input

Int[(x^2*(8 - 7*(1 + k)*x + 6*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + ( 
1 + k)*x - k*x^2 + b*x^8)),x]
 

output

$Aborted
 

3.24.18.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7293

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

3.24.18.4 Maple [F]

\[\int \frac {x^{2} \left (8-7 \left (1+k \right ) x +6 k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+\left (1+k \right ) x -k \,x^{2}+b \,x^{8}\right )}d x\]

input

int(x^2*(8-7*(1+k)*x+6*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(1+k)*x-k*x^2+b 
*x^8),x)
 

output

int(x^2*(8-7*(1+k)*x+6*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(1+k)*x-k*x^2+b 
*x^8),x)
 

3.24.18.5 Fricas [F(-1)]

Timed out. \[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\text {Timed out} \]

input

integrate(x^2*(8-7*(1+k)*x+6*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(1+k)*x-k 
*x^2+b*x^8),x, algorithm="fricas")
 

output

Timed out
 

3.24.18.6 Sympy [F]

\[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\int \frac {x^{2} \cdot \left (6 k x^{2} - 7 k x - 7 x + 8\right )}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (b x^{8} - k x^{2} + k x + x - 1\right )}\, dx \]

input

integrate(x**2*(8-7*(1+k)*x+6*k*x**2)/((1-x)*x*(-k*x+1))**(1/3)/(-1+(1+k)* 
x-k*x**2+b*x**8),x)
 

output

Integral(x**2*(6*k*x**2 - 7*k*x - 7*x + 8)/((x*(x - 1)*(k*x - 1))**(1/3)*( 
b*x**8 - k*x**2 + k*x + x - 1)), x)
 

3.24.18.7 Maxima [F]

\[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\int { \frac {{\left (6 \, k x^{2} - 7 \, {\left (k + 1\right )} x + 8\right )} x^{2}}{{\left (b x^{8} - k x^{2} + {\left (k + 1\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(x^2*(8-7*(1+k)*x+6*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(1+k)*x-k 
*x^2+b*x^8),x, algorithm="maxima")
 

output

integrate((6*k*x^2 - 7*(k + 1)*x + 8)*x^2/((b*x^8 - k*x^2 + (k + 1)*x - 1) 
*((k*x - 1)*(x - 1)*x)^(1/3)), x)
 

3.24.18.8 Giac [F]

\[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\int { \frac {{\left (6 \, k x^{2} - 7 \, {\left (k + 1\right )} x + 8\right )} x^{2}}{{\left (b x^{8} - k x^{2} + {\left (k + 1\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(x^2*(8-7*(1+k)*x+6*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(1+k)*x-k 
*x^2+b*x^8),x, algorithm="giac")
 

output

integrate((6*k*x^2 - 7*(k + 1)*x + 8)*x^2/((b*x^8 - k*x^2 + (k + 1)*x - 1) 
*((k*x - 1)*(x - 1)*x)^(1/3)), x)
 

3.24.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (8-7 (1+k) x+6 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^8\right )} \, dx=\int \frac {x^2\,\left (6\,k\,x^2-7\,x\,\left (k+1\right )+8\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (b\,x^8-k\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]

input

int((x^2*(6*k*x^2 - 7*x*(k + 1) + 8))/((x*(k*x - 1)*(x - 1))^(1/3)*(b*x^8 
+ x*(k + 1) - k*x^2 - 1)),x)
 

output

int((x^2*(6*k*x^2 - 7*x*(k + 1) + 8))/((x*(k*x - 1)*(x - 1))^(1/3)*(b*x^8 
+ x*(k + 1) - k*x^2 - 1)), x)