Integrand size = 90, antiderivative size = 115 \[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\frac {4 \left (x-x^2-k x^2+k x^3\right )^{3/4}}{(-1+x) x}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right ) \]
4*(k*x^3-k*x^2-x^2+x)^(3/4)/(-1+x)/x+2*d^(1/4)*arctan(d^(1/4)*(x+(-1-k)*x^ 2+k*x^3)^(1/4)/(k*x-1))-2*d^(1/4)*arctanh(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(1/ 4)/(k*x-1))
\[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx \]
Integrate[((-1 - 2*(-1 + k)*x + k*x^2)*(-1 + 3*k*x - 3*k^2*x^2 + k^3*x^3)) /((-1 + x)*x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + (d + 3*k)*x - (d + 3*k^2)*x ^2 + k^3*x^3)),x]
Integrate[((-1 - 2*(-1 + k)*x + k*x^2)*(-1 + 3*k*x - 3*k^2*x^2 + k^3*x^3)) /((-1 + x)*x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + (d + 3*k)*x - (d + 3*k^2)*x ^2 + k^3*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.
| \(\displaystyle \int \frac {\left (k x^2-2 (k-1) x-1\right ) \left (k^3 x^3-3 k^2 x^2+3 k x-1\right )}{(x-1) x \sqrt [4]{(1-x) x (1-k x)} \left (-x^2 \left (d+3 k^2\right )+x (d+3 k)+k^3 x^3-1\right )} \, dx\) |
| \(\Big \downarrow \) 2006 |
| \(\displaystyle \int \frac {(k x-1)^3 \left (k x^2-2 (k-1) x-1\right )}{(x-1) x \sqrt [4]{(1-x) x (1-k x)} \left (-x^2 \left (d+3 k^2\right )+x (d+3 k)+k^3 x^3-1\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {(1-k x)^3 \left (-k x^2-2 (1-k) x+1\right )}{(1-x) x^{5/4} \sqrt [4]{k x^2-(k+1) x+1} \left (-k^3 x^3+\left (3 k^2+d\right ) x^2-(d+3 k) x+1\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {(1-k x)^3 \left (-k x^2-2 (1-k) x+1\right )}{(1-x) \sqrt {x} \sqrt [4]{k x^2-(k+1) x+1} \left (-k^3 x^3+\left (3 k^2+d\right ) x^2-(d+3 k) x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \frac {(1-k x)^{11/4} \left (-k x^2-2 (1-k) x+1\right )}{(1-x)^{5/4} \sqrt {x} \left (-k^3 x^3+\left (3 k^2+d\right ) x^2-(d+3 k) x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \left (\frac {(1-k x)^{11/4}}{(1-x)^{5/4} \sqrt {x}}+\frac {\sqrt {x} \left (x^2 k^3+5 k+d-\left (3 k^2+k+d\right ) x-2\right ) (1-k x)^{11/4}}{(1-x)^{5/4} \left (-k^3 x^3+d \left (\frac {3 k^2}{d}+1\right ) x^2-d \left (\frac {3 k}{d}+1\right ) x+1\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (-(-d-5 k+2) \int \frac {\sqrt {x} (1-k x)^{11/4}}{(1-x)^{5/4} \left (-k^3 x^3+d \left (\frac {3 k^2}{d}+1\right ) x^2-d \left (\frac {3 k}{d}+1\right ) x+1\right )}d\sqrt [4]{x}+k^3 \int \frac {x^{5/2} (1-k x)^{11/4}}{(1-x)^{5/4} \left (-k^3 x^3+d \left (\frac {3 k^2}{d}+1\right ) x^2-d \left (\frac {3 k}{d}+1\right ) x+1\right )}d\sqrt [4]{x}-\left (d+3 k^2+k\right ) \int \frac {x^{3/2} (1-k x)^{11/4}}{(1-x)^{5/4} \left (-k^3 x^3+d \left (\frac {3 k^2}{d}+1\right ) x^2-d \left (\frac {3 k}{d}+1\right ) x+1\right )}d\sqrt [4]{x}-\frac {\operatorname {AppellF1}\left (-\frac {1}{4},\frac {5}{4},-\frac {11}{4},\frac {3}{4},x,k x\right )}{\sqrt [4]{x}}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\) |
Int[((-1 - 2*(-1 + k)*x + k*x^2)*(-1 + 3*k*x - 3*k^2*x^2 + k^3*x^3))/((-1 + x)*x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + (d + 3*k)*x - (d + 3*k^2)*x^2 + k ^3*x^3)),x]
3.18.9.3.1 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[(u_.)*(Px_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^Expon[Px , x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; PolyQ[Px, x] && GtQ[Expon[P x, x], 1] && NeQ[Coeff[Px, x, 0], 0] && !MatchQ[Px, (a_.)*(v_)^Expon[Px, x ] /; FreeQ[a, x] && LinearQ[v, x]]
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]
\[\int \frac {\left (-1-2 \left (-1+k \right ) x +k \,x^{2}\right ) \left (k^{3} x^{3}-3 k^{2} x^{2}+3 k x -1\right )}{\left (-1+x \right ) x \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-1+\left (d +3 k \right ) x -\left (3 k^{2}+d \right ) x^{2}+k^{3} x^{3}\right )}d x\]
int((-1-2*(-1+k)*x+k*x^2)*(k^3*x^3-3*k^2*x^2+3*k*x-1)/(-1+x)/x/((1-x)*x*(- k*x+1))^(1/4)/(-1+(d+3*k)*x-(3*k^2+d)*x^2+k^3*x^3),x)
int((-1-2*(-1+k)*x+k*x^2)*(k^3*x^3-3*k^2*x^2+3*k*x-1)/(-1+x)/x/((1-x)*x*(- k*x+1))^(1/4)/(-1+(d+3*k)*x-(3*k^2+d)*x^2+k^3*x^3),x)
Timed out. \[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\text {Timed out} \]
integrate((-1-2*(-1+k)*x+k*x^2)*(k^3*x^3-3*k^2*x^2+3*k*x-1)/(-1+x)/x/((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="fri cas")
Timed out. \[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\text {Timed out} \]
integrate((-1-2*(-1+k)*x+k*x**2)*(k**3*x**3-3*k**2*x**2+3*k*x-1)/(-1+x)/x/ ((1-x)*x*(-k*x+1))**(1/4)/(-1+(d+3*k)*x-(3*k**2+d)*x**2+k**3*x**3),x)
\[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\int { \frac {{\left (k^{3} x^{3} - 3 \, k^{2} x^{2} + 3 \, k x - 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (k^{3} x^{3} - {\left (3 \, k^{2} + d\right )} x^{2} + {\left (d + 3 \, k\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]
integrate((-1-2*(-1+k)*x+k*x^2)*(k^3*x^3-3*k^2*x^2+3*k*x-1)/(-1+x)/x/((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="max ima")
integrate((k^3*x^3 - 3*k^2*x^2 + 3*k*x - 1)*(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 - 1)*x), x)
\[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\int { \frac {{\left (k^{3} x^{3} - 3 \, k^{2} x^{2} + 3 \, k x - 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (k^{3} x^{3} - {\left (3 \, k^{2} + d\right )} x^{2} + {\left (d + 3 \, k\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left (x - 1\right )} x} \,d x } \]
integrate((-1-2*(-1+k)*x+k*x^2)*(k^3*x^3-3*k^2*x^2+3*k*x-1)/(-1+x)/x/((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="gia c")
integrate((k^3*x^3 - 3*k^2*x^2 + 3*k*x - 1)*(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 - 1)*x), x)
Timed out. \[ \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx=\int \frac {\left (2\,x\,\left (k-1\right )-k\,x^2+1\right )\,\left (-k^3\,x^3+3\,k^2\,x^2-3\,k\,x+1\right )}{x\,\left (x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (k^3\,x^3-x^2\,\left (3\,k^2+d\right )+x\,\left (d+3\,k\right )-1\right )} \,d x \]
int(((2*x*(k - 1) - k*x^2 + 1)*(3*k^2*x^2 - k^3*x^3 - 3*k*x + 1))/(x*(x - 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)