Integrand size = 70, antiderivative size = 116 \[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\frac {4 \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}+2 \sqrt [4]{d} \arctan \left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right ) \]
4*(x+(-1-k)*x^2+k*x^3)^(1/4)/(-1+x)+2*d^(1/4)*arctan((-d^(1/4)+d^(1/4)*x)/ (x+(-1-k)*x^2+k*x^3)^(1/4))-2*d^(1/4)*arctanh((-d^(1/4)+d^(1/4)*x)/(x+(-1- k)*x^2+k*x^3)^(1/4))
\[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx \]
Integrate[(x*(-1 + k*x)*(-1 + 2*(-1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x* (1 - k*x))^(3/4)*(-d + (1 + 3*d)*x - (3*d + k)*x^2 + d*x^3)),x]
Integrate[(x*(-1 + k*x)*(-1 + 2*(-1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x* (1 - k*x))^(3/4)*(-d + (1 + 3*d)*x - (3*d + k)*x^2 + d*x^3)), x]
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.
| \(\displaystyle \int \frac {x (k x-1) \left (k x^2+2 (k-1) x-1\right )}{(x-1) ((1-x) x (1-k x))^{3/4} \left (-x^2 (3 d+k)+d x^3+(3 d+1) x-d\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {\sqrt [4]{x} (1-k x) \left (-k x^2+2 (1-k) x+1\right )}{(1-x) \left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+(3 d+k) x^2-(3 d+1) x+d\right )}dx}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {x (1-k x) \left (-k x^2+2 (1-k) x+1\right )}{(1-x) \left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+(3 d+k) x^2-(3 d+1) x+d\right )}d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \int \frac {x \sqrt [4]{1-k x} \left (-k x^2+2 (1-k) x+1\right )}{(1-x)^{7/4} \left (-d x^3+(3 d+k) x^2-(3 d+1) x+d\right )}d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \int \left (\frac {k \sqrt [4]{1-k x}}{d (1-x)^{7/4}}-\frac {\sqrt [4]{1-k x} \left (-\left (\left (d (2-5 k)-k^2\right ) x^2\right )-(3 k d+d+k) x+d k\right )}{d (1-x)^{7/4} \left (-d x^3+(3 d+k) x^2-(3 d+1) x+d\right )}\right )d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \left (\frac {\left (d (2-5 k)-k^2\right ) \int \frac {x^2 \sqrt [4]{1-k x}}{(1-x)^{7/4} \left (-d x^3+3 d \left (\frac {k}{3 d}+1\right ) x^2-(3 d+1) x+d\right )}d\sqrt [4]{x}}{d}-k \int \frac {\sqrt [4]{1-k x}}{(1-x)^{7/4} \left (-d x^3+3 d \left (\frac {k}{3 d}+1\right ) x^2-(3 d+1) x+d\right )}d\sqrt [4]{x}+\frac {(3 d k+d+k) \int \frac {x \sqrt [4]{1-k x}}{(1-x)^{7/4} \left (-d x^3+3 d \left (\frac {k}{3 d}+1\right ) x^2-(3 d+1) x+d\right )}d\sqrt [4]{x}}{d}+\frac {k \sqrt [4]{x} \operatorname {AppellF1}\left (\frac {1}{4},\frac {7}{4},-\frac {1}{4},\frac {5}{4},x,k x\right )}{d}\right )}{((1-x) x (1-k x))^{3/4}}\) |
Int[(x*(-1 + k*x)*(-1 + 2*(-1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k *x))^(3/4)*(-d + (1 + 3*d)*x - (3*d + k)*x^2 + d*x^3)),x]
3.18.15.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
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]
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]
\[\int \frac {x \left (k x -1\right ) \left (-1+2 \left (-1+k \right ) x +k \,x^{2}\right )}{\left (-1+x \right ) \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-d +\left (1+3 d \right ) x -\left (3 d +k \right ) x^{2}+d \,x^{3}\right )}d x\]
int(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/(-d+(1 +3*d)*x-(3*d+k)*x^2+d*x^3),x)
int(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/(-d+(1 +3*d)*x-(3*d+k)*x^2+d*x^3),x)
Timed out. \[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\text {Timed out} \]
integrate(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/ (-d+(1+3*d)*x-(3*d+k)*x^2+d*x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\text {Timed out} \]
integrate(x*(k*x-1)*(-1+2*(-1+k)*x+k*x**2)/(-1+x)/((1-x)*x*(-k*x+1))**(3/4 )/(-d+(1+3*d)*x-(3*d+k)*x**2+d*x**3),x)
\[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\int { \frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left (d x^{3} - {\left (3 \, d + k\right )} x^{2} + {\left (3 \, d + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (x - 1\right )}} \,d x } \]
integrate(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/ (-d+(1+3*d)*x-(3*d+k)*x^2+d*x^3),x, algorithm="maxima")
integrate((k*x^2 + 2*(k - 1)*x - 1)*(k*x - 1)*x/((d*x^3 - (3*d + k)*x^2 + (3*d + 1)*x - d)*((k*x - 1)*(x - 1)*x)^(3/4)*(x - 1)), x)
\[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\int { \frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left (d x^{3} - {\left (3 \, d + k\right )} x^{2} + {\left (3 \, d + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (x - 1\right )}} \,d x } \]
integrate(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/ (-d+(1+3*d)*x-(3*d+k)*x^2+d*x^3),x, algorithm="giac")
integrate((k*x^2 + 2*(k - 1)*x - 1)*(k*x - 1)*x/((d*x^3 - (3*d + k)*x^2 + (3*d + 1)*x - d)*((k*x - 1)*(x - 1)*x)^(3/4)*(x - 1)), x)
Timed out. \[ \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=-\int \frac {x\,\left (k\,x-1\right )\,\left (2\,x\,\left (k-1\right )+k\,x^2-1\right )}{\left (x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (-d\,x^3+\left (3\,d+k\right )\,x^2+\left (-3\,d-1\right )\,x+d\right )} \,d x \]
int(-(x*(k*x - 1)*(2*x*(k - 1) + k*x^2 - 1))/((x - 1)*(x*(k*x - 1)*(x - 1) )^(3/4)*(d - d*x^3 + x^2*(3*d + k) - x*(3*d + 1))),x)