Integrand size = 77, antiderivative size = 116 \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\frac {4 \left (-x+x^2\right )}{\left (x-x^2-k x^2+k x^3\right )^{3/4}}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right ) \]
4*(x^2-x)/(k*x^3-k*x^2-x^2+x)^(3/4)+2*d^(1/4)*arctan(d^(1/4)*(x+(-1-k)*x^2 +k*x^3)^(3/4)/(-1+x)/x)-2*d^(1/4)*arctanh(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(3/ 4)/(-1+x)/x)
\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \]
Integrate[((-1 + x)*x*(-1 - 2*(-1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^ (3/4)*(-1 + k*x)*(-d + (1 + 3*d*k)*x - (1 + 3*d*k^2)*x^2 + d*k^3*x^3)),x]
Integrate[((-1 + x)*x*(-1 - 2*(-1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^ (3/4)*(-1 + k*x)*(-d + (1 + 3*d*k)*x - (1 + 3*d*k^2)*x^2 + d*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.
| \(\displaystyle \int \frac {(x-1) x \left (k x^2-2 (k-1) x-1\right )}{((1-x) x (1-k x))^{3/4} (k x-1) \left (d k^3 x^3-x^2 \left (3 d k^2+1\right )+x (3 d k+1)-d\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {(1-x) \sqrt [4]{x} \left (-k x^2-2 (1-k) x+1\right )}{(1-k x) \left (k x^2-(k+1) x+1\right )^{3/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}dx}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {(1-x) x \left (-k x^2-2 (1-k) x+1\right )}{(1-k x) \left (k x^2-(k+1) x+1\right )^{3/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \int \frac {\sqrt [4]{1-x} x \left (-k x^2-2 (1-k) x+1\right )}{(1-k x)^{7/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \int \left (\frac {\sqrt [4]{1-x}}{d k^2 (1-k x)^{7/4}}-\frac {\sqrt [4]{1-x} \left (\left (d (5-2 k) k^2+1\right ) x^2-(d k (k+3)+1) x+d\right )}{d k^2 (1-k x)^{7/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}\right )d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4} \left (-\frac {\int \frac {\sqrt [4]{1-x}}{(1-k x)^{7/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}d\sqrt [4]{x}}{k^2}+\frac {(d k (k+3)+1) \int \frac {\sqrt [4]{1-x} x}{(1-k x)^{7/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}d\sqrt [4]{x}}{d k^2}-\left (\frac {1}{d k^2}-2 k+5\right ) \int \frac {\sqrt [4]{1-x} x^2}{(1-k x)^{7/4} \left (-d k^3 x^3+\left (3 d k^2+1\right ) x^2-(3 d k+1) x+d\right )}d\sqrt [4]{x}+\frac {\sqrt [4]{x} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{4},\frac {7}{4},\frac {5}{4},x,k x\right )}{d k^2}\right )}{((1-x) x (1-k x))^{3/4}}\) |
Int[((-1 + x)*x*(-1 - 2*(-1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^(3/4)* (-1 + k*x)*(-d + (1 + 3*d*k)*x - (1 + 3*d*k^2)*x^2 + d*k^3*x^3)),x]
3.18.16.3.1 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]
\[\int \frac {\left (-1+x \right ) x \left (-1-2 \left (-1+k \right ) x +k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (k x -1\right ) \left (-d +\left (3 d k +1\right ) x -\left (3 d \,k^{2}+1\right ) x^{2}+d \,k^{3} x^{3}\right )}d x\]
int((-1+x)*x*(-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(k*x-1)/(-d+(3 *d*k+1)*x-(3*d*k^2+1)*x^2+d*k^3*x^3),x)
int((-1+x)*x*(-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(k*x-1)/(-d+(3 *d*k+1)*x-(3*d*k^2+1)*x^2+d*k^3*x^3),x)
Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\text {Timed out} \]
integrate((-1+x)*x*(-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(k*x-1)/ (-d+(3*d*k+1)*x-(3*d*k^2+1)*x^2+d*k^3*x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\text {Timed out} \]
integrate((-1+x)*x*(-1-2*(-1+k)*x+k*x**2)/((1-x)*x*(-k*x+1))**(3/4)/(k*x-1 )/(-d+(3*d*k+1)*x-(3*d*k**2+1)*x**2+d*k**3*x**3),x)
\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int { \frac {{\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )} x}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (k x - 1\right )}} \,d x } \]
integrate((-1+x)*x*(-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(k*x-1)/ (-d+(3*d*k+1)*x-(3*d*k^2+1)*x^2+d*k^3*x^3),x, algorithm="maxima")
integrate((k*x^2 - 2*(k - 1)*x - 1)*(x - 1)*x/((d*k^3*x^3 - (3*d*k^2 + 1)* x^2 + (3*d*k + 1)*x - d)*((k*x - 1)*(x - 1)*x)^(3/4)*(k*x - 1)), x)
\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int { \frac {{\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )} x}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (k x - 1\right )}} \,d x } \]
integrate((-1+x)*x*(-1-2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(k*x-1)/ (-d+(3*d*k+1)*x-(3*d*k^2+1)*x^2+d*k^3*x^3),x, algorithm="giac")
integrate((k*x^2 - 2*(k - 1)*x - 1)*(x - 1)*x/((d*k^3*x^3 - (3*d*k^2 + 1)* x^2 + (3*d*k + 1)*x - d)*((k*x - 1)*(x - 1)*x)^(3/4)*(k*x - 1)), x)
Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int \frac {x\,\left (x-1\right )\,\left (2\,x\,\left (k-1\right )-k\,x^2+1\right )}{\left (k\,x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (d+x^2\,\left (3\,d\,k^2+1\right )-x\,\left (3\,d\,k+1\right )-d\,k^3\,x^3\right )} \,d x \]
int((x*(x - 1)*(2*x*(k - 1) - k*x^2 + 1))/((k*x - 1)*(x*(k*x - 1)*(x - 1)) ^(3/4)*(d + x^2*(3*d*k^2 + 1) - x*(3*d*k + 1) - d*k^3*x^3)),x)