3.24.90 \(\int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(2+b) x+(1+b k) x^2)} \, dx\) [2390]

3.24.90.1 Optimal result

Integrand size = 46, antiderivative size = 192 \[ \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2-2 x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-1+x+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (1-2 x+x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]

output

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

3.24.90.2 Mathematica [A] (verified)

Time = 15.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.79 \[ \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}{2-2 x+\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (-1+x+\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (1-2 x+x^2-\sqrt [3]{b} (-1+x) \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]

input

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

output

(2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*((-1 + x)*x*(-1 + k*x))^(1/3))/(2 - 2*x 
 + b^(1/3)*((-1 + x)*x*(-1 + k*x))^(1/3))] + 2*Log[-1 + x + b^(1/3)*((-1 + 
 x)*x*(-1 + k*x))^(1/3)] - Log[1 - 2*x + x^2 - b^(1/3)*(-1 + x)*((-1 + x)* 
x*(-1 + k*x))^(1/3) + b^(2/3)*((-1 + x)*x*(-1 + k*x))^(2/3)])/(2*b^(2/3))
 

3.24.90.3 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.24.90.3.1 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 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

3.24.90.4 Maple [F]

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

input

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

output

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

3.24.90.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.24.90.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.24.90.7 Maxima [F]

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

input

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

output

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

3.24.90.8 Giac [F]

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

input

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

output

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

3.24.90.9 Mupad [F(-1)]

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

input

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

output

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