3.25.95 \(\int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} (b-(1+2 b) x+(b+k) x^2)} \, dx\) [2495]

3.25.95.1 Optimal result

Integrand size = 46, antiderivative size = 207 \[ \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3}-2 b^{2/3} x+b^{2/3} x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]

output

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

3.25.95.2 Mathematica [A] (verified)

Time = 15.49 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.70 \[ \int \frac {-1+(-1+2 k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{(-1+x) x (-1+k x)}}{-2 \sqrt [3]{b} (-1+x)+\sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (\sqrt [3]{b} (-1+x)+\sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (b^{2/3} (-1+x)^2-\sqrt [3]{b} (-1+x) \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{b}} \]

input

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

output

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

3.25.95.3 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.25.95.3.1 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]
 

3.25.95.4 Maple [F]

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

input

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

output

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

3.25.95.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.25.95.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.25.95.7 Maxima [F]

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

input

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

output

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

3.25.95.8 Giac [F]

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

input

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

output

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

3.25.95.9 Mupad [F(-1)]

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

input

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

output

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