3.12.12 \(\int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} (-1+(d+3 k) x-(d+3 k^2) x^2+k^3 x^3)} \, dx\) [1112]

3.12.12.1 Optimal result

Integrand size = 61, antiderivative size = 83 \[ \int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )}{d^{3/4}} \]

output

2*arctan(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(1/4)/(k*x-1))/d^(3/4)-2*arctanh(d^( 
1/4)*(x+(-1-k)*x^2+k*x^3)^(1/4)/(k*x-1))/d^(3/4)
 

3.12.12.2 Mathematica [A] (verified)

Time = 11.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72 \[ \int \frac {-1-2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} (-1+x) x}{((-1+x) x (-1+k x))^{3/4}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} (-1+x) x}{((-1+x) x (-1+k x))^{3/4}}\right )\right )}{d^{3/4}} \]

input

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

output

(2*(ArcTan[(d^(1/4)*(-1 + x)*x)/((-1 + x)*x*(-1 + k*x))^(3/4)] - ArcTanh[( 
d^(1/4)*(-1 + x)*x)/((-1 + x)*x*(-1 + k*x))^(3/4)]))/d^(3/4)
 

3.12.12.3 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.12.12.3.1 Defintions of rubi rules used

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] 
]
 

3.12.12.4 Maple [F]

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

input

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

output

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

3.12.12.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.12.12.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.12.12.7 Maxima [F]

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

input

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

output

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

3.12.12.8 Giac [F]

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

input

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

output

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

3.12.12.9 Mupad [F(-1)]

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

input

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

output

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