Integrand size = 68, antiderivative size = 113 \[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\frac {4 \left (x-x^2-k x^2+k x^3\right )^{3/4}}{x (-1+k x)}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right ) \]
4*(k*x^3-k*x^2-x^2+x)^(3/4)/x/(k*x-1)+2*d^(1/4)*arctan(d^(1/4)*(x+(-1-k)*x ^2+k*x^3)^(1/4)/(-1+x))-2*d^(1/4)*arctanh(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(1/ 4)/(-1+x))
\[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \]
Integrate[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x) )^(1/4)*(-1 + k*x)*(-1 + (3 + d)*x - (3 + d*k)*x^2 + x^3)),x]
Integrate[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x) )^(1/4)*(-1 + k*x)*(-1 + (3 + d)*x - (3 + d*k)*x^2 + 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.
| \(\displaystyle \int \frac {(x-1)^3 \left (k x^2+2 (k-1) x-1\right )}{x \sqrt [4]{(1-x) x (1-k x)} (k x-1) \left (-x^2 (d k+3)+(d+3) x+x^3-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {(1-x)^3 \left (-k x^2+2 (1-k) x+1\right )}{x^{5/4} (1-k x) \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+(d k+3) x^2-(d+3) x+1\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {(1-x)^3 \left (-k x^2+2 (1-k) x+1\right )}{\sqrt {x} (1-k x) \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+(d k+3) x^2-(d+3) x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \frac {(1-x)^{11/4} \left (-k x^2+2 (1-k) x+1\right )}{\sqrt {x} (1-k x)^{5/4} \left (-x^3+(d k+3) x^2-(d+3) x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \left (\frac {\sqrt {x} \left (x^2-(d k+k+3) x+d-2 k+5\right ) (1-x)^{11/4}}{(1-k x)^{5/4} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}+\frac {(1-x)^{11/4}}{\sqrt {x} (1-k x)^{5/4}}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left ((d-2 k+5) \int \frac {(1-x)^{11/4} \sqrt {x}}{(1-k x)^{5/4} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}d\sqrt [4]{x}-(d k+k+3) \int \frac {(1-x)^{11/4} x^{3/2}}{(1-k x)^{5/4} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}d\sqrt [4]{x}+\int \frac {(1-x)^{11/4} x^{5/2}}{(1-k x)^{5/4} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}d\sqrt [4]{x}-\frac {\operatorname {AppellF1}\left (-\frac {1}{4},-\frac {11}{4},\frac {5}{4},\frac {3}{4},x,k x\right )}{\sqrt [4]{x}}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\) |
Int[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(1/4 )*(-1 + k*x)*(-1 + (3 + d)*x - (3 + d*k)*x^2 + x^3)),x]
3.17.83.3.1 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]
\[\int \frac {\left (-1+x \right )^{3} \left (-1+2 \left (-1+k \right ) x +k \,x^{2}\right )}{x \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (k x -1\right ) \left (-1+\left (3+d \right ) x -\left (d k +3\right ) x^{2}+x^{3}\right )}d x\]
int((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+ (3+d)*x-(d*k+3)*x^2+x^3),x)
int((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+ (3+d)*x-(d*k+3)*x^2+x^3),x)
Timed out. \[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]
integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1 )/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]
integrate((-1+x)**3*(-1+2*(-1+k)*x+k*x**2)/x/((1-x)*x*(-k*x+1))**(1/4)/(k* x-1)/(-1+(3+d)*x-(d*k+3)*x**2+x**3),x)
\[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )}^{3}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )} {\left (k x - 1\right )} x} \,d x } \]
integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1 )/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, algorithm="maxima")
-integrate((k*x^2 + 2*(k - 1)*x - 1)*(x - 1)^3/(((k*x - 1)*(x - 1)*x)^(1/4 )*((d*k + 3)*x^2 - x^3 - (d + 3)*x + 1)*(k*x - 1)*x), x)
\[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )}^{3}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )} {\left (k x - 1\right )} x} \,d x } \]
integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1 )/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, algorithm="giac")
integrate(-(k*x^2 + 2*(k - 1)*x - 1)*(x - 1)^3/(((k*x - 1)*(x - 1)*x)^(1/4 )*((d*k + 3)*x^2 - x^3 - (d + 3)*x + 1)*(k*x - 1)*x), x)
Timed out. \[ \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int \frac {{\left (x-1\right )}^3\,\left (2\,x\,\left (k-1\right )+k\,x^2-1\right )}{x\,\left (k\,x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (x^3+\left (-d\,k-3\right )\,x^2+\left (d+3\right )\,x-1\right )} \,d x \]
int(((x - 1)^3*(2*x*(k - 1) + k*x^2 - 1))/(x*(k*x - 1)*(x*(k*x - 1)*(x - 1 ))^(1/4)*(x*(d + 3) - x^2*(d*k + 3) + x^3 - 1)),x)