3.24.40 \(\int \frac {x (5-4 (1+k) x+3 k x^2)}{\sqrt [3]{(1-x) x (1-k x)} (-b+(b+b k) x-b k x^2+x^5)} \, dx\) [2340]

3.24.40.1 Optimal result

Integrand size = 56, antiderivative size = 185 \[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 x^2+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (x^2-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (x^4+\sqrt [3]{b} x^2 \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]

output

3^(1/2)*arctan(3^(1/2)*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1/3)/(2*x^2+b^(1/3)*( 
x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(2/3)+ln(x^2-b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1 
/3))/b^(2/3)-1/2*ln(x^4+b^(1/3)*x^2*(x+(-1-k)*x^2+k*x^3)^(1/3)+b^(2/3)*(x+ 
(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)
 

3.24.40.2 Mathematica [F]

\[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx \]

input

Integrate[(x*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b 
 + (b + b*k)*x - b*k*x^2 + x^5)),x]
 

output

Integrate[(x*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b 
 + (b + b*k)*x - b*k*x^2 + x^5)), x]
 

3.24.40.3 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.

\(\displaystyle \int \frac {x \left (3 k x^2-4 (k+1) x+5\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b k x^2+x (b k+b)-b+x^5\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {x^{2/3} \left (3 k x^2-4 (k+1) x+5\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{2/3} \left (3 k x^2-4 (k+1) x+5\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {x^{4/3} \left (3 k x^2-4 (k+1) x+5\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {3 k x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}+\frac {4 (-k-1) x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}+\frac {5 x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (5 \int \frac {x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}d\sqrt [3]{x}-4 (k+1) \int \frac {x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}d\sqrt [3]{x}+3 k \int \frac {x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-x^5+b k x^2-b (k+1) x+b\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\)

input

Int[(x*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b + (b 
+ b*k)*x - b*k*x^2 + x^5)),x]
 

output

$Aborted
 

3.24.40.3.1 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] 
]
 

3.24.40.4 Maple [F]

\[\int \frac {x \left (5-4 \left (1+k \right ) x +3 k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-b +\left (b k +b \right ) x -b k \,x^{2}+x^{5}\right )}d x\]

input

int(x*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(b*k+b)*x-b*k*x^2 
+x^5),x)
 

output

int(x*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(b*k+b)*x-b*k*x^2 
+x^5),x)
 

3.24.40.5 Fricas [F(-1)]

Timed out. \[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\text {Timed out} \]

input

integrate(x*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(b*k+b)*x-b 
*k*x^2+x^5),x, algorithm="fricas")
 

output

Timed out
 

3.24.40.6 Sympy [F]

\[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\int \frac {x \left (3 k x^{2} - 4 k x - 4 x + 5\right )}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (- b k x^{2} + b k x + b x - b + x^{5}\right )}\, dx \]

input

integrate(x*(5-4*(1+k)*x+3*k*x**2)/((1-x)*x*(-k*x+1))**(1/3)/(-b+(b*k+b)*x 
-b*k*x**2+x**5),x)
 

output

Integral(x*(3*k*x**2 - 4*k*x - 4*x + 5)/((x*(x - 1)*(k*x - 1))**(1/3)*(-b* 
k*x**2 + b*k*x + b*x - b + x**5)), x)
 

3.24.40.7 Maxima [F]

\[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\int { \frac {{\left (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x}{{\left (x^{5} - b k x^{2} + {\left (b k + b\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(x*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(b*k+b)*x-b 
*k*x^2+x^5),x, algorithm="maxima")
 

output

integrate((3*k*x^2 - 4*(k + 1)*x + 5)*x/((x^5 - b*k*x^2 + (b*k + b)*x - b) 
*((k*x - 1)*(x - 1)*x)^(1/3)), x)
 

3.24.40.8 Giac [F]

\[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\int { \frac {{\left (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x}{{\left (x^{5} - b k x^{2} + {\left (b k + b\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(x*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(b*k+b)*x-b 
*k*x^2+x^5),x, algorithm="giac")
 

output

integrate((3*k*x^2 - 4*(k + 1)*x + 5)*x/((x^5 - b*k*x^2 + (b*k + b)*x - b) 
*((k*x - 1)*(x - 1)*x)^(1/3)), x)
 

3.24.40.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx=\int -\frac {x\,\left (3\,k\,x^2-4\,x\,\left (k+1\right )+5\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (-x^5+b\,k\,x^2+\left (-b-b\,k\right )\,x+b\right )} \,d x \]

input

int(-(x*(3*k*x^2 - 4*x*(k + 1) + 5))/((x*(k*x - 1)*(x - 1))^(1/3)*(b - x*( 
b + b*k) - x^5 + b*k*x^2)),x)
 

output

int(-(x*(3*k*x^2 - 4*x*(k + 1) + 5))/((x*(k*x - 1)*(x - 1))^(1/3)*(b - x*( 
b + b*k) - x^5 + b*k*x^2)), x)