3.26.61 \(\int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} (b-(1+2 b) x+(b+k) x^2)} \, dx\) [2561]

3.26.61.1 Optimal result

Integrand size = 52, antiderivative size = 217 \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x-2 k x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{b^{2/3}}+\frac {\log \left (-x+k x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{b^{2/3}}-\frac {\log \left (x^2-2 k x^3+k^2 x^4+\left (\sqrt [3]{b} x-\sqrt [3]{b} k x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 b^{2/3}} \]

output

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

3.26.61.2 Mathematica [F]

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

input

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

output

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

3.26.61.3 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\frac {\left (2 b (1-k)+(1-2 k) \sqrt {-4 b k+4 b+1}+2 k^2-2 k+1\right ) \int \frac {1}{\left (-2 b+2 (b+k) x-\sqrt {-4 k b+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b+k}-\frac {\left (2 b (1-k)-(1-2 k) \sqrt {-4 b k+4 b+1}+2 k^2-2 k+1\right ) \int \frac {1}{\left (-2 b+2 (b+k) x+\sqrt {-4 k b+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b+k}-\frac {(1-2 k) (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{(b+k) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\)

input

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

output

$Aborted
 

3.26.61.3.1 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]
 

3.26.61.4 Maple [F]

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

input

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

output

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

3.26.61.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.26.61.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.26.61.7 Maxima [F]

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

input

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

output

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

3.26.61.8 Giac [F]

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

input

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

output

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

3.26.61.9 Mupad [F(-1)]

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

input

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

output

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