Integrand size = 54, antiderivative size = 71 \[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}} \]
2*arctan(d^(1/4)*x/(x+(-1-k)*x^2+k*x^3)^(1/4))/d^(3/4)-2*arctanh(d^(1/4)*x /(x+(-1-k)*x^2+k*x^3)^(1/4))/d^(3/4)
\[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx \]
Integrate[(x^2*(3 - 2*(1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - k*x^2 + d*x^3)),x]
Integrate[(x^2*(3 - 2*(1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - 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.
| \(\displaystyle \int \frac {x^2 \left (k x^2-2 (k+1) x+3\right )}{((1-x) x (1-k x))^{3/4} \left (d x^3-k x^2+(k+1) x-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int -\frac {x^{5/4} \left (k x^2-2 (k+1) x+3\right )}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}dx}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {x^{5/4} \left (k x^2-2 (k+1) x+3\right )}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}dx}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {4 x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \frac {x^2 \left (k x^2-2 (k+1) x+3\right )}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {4 x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \int \left (-\frac {k^2-2 d (k+1)}{d^2 \left (k x^2-(k+1) x+1\right )^{3/4}}-\frac {k x}{d \left (k x^2-(k+1) x+1\right )^{3/4}}-\frac {-k^2-\left (k^3-3 d (k+1) k+3 d^2\right ) x^2+2 d (k+1)-\left (-k^3+2 d k^2-k^2+5 d k+2 d\right ) x}{d^2 \left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}\right )d\sqrt [4]{x}}{((1-x) x (1-k x))^{3/4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 x^{3/4} \left (k x^2-(k+1) x+1\right )^{3/4} \left (\frac {\left (3 d^2-3 d (k+1) k+k^3\right ) \int \frac {x^2}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}d\sqrt [4]{x}}{d^2}+\frac {\left (k^2-2 d (k+1)\right ) \int \frac {1}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}d\sqrt [4]{x}}{d^2}-\frac {\left (k^2 (k+1)-d \left (2 k^2+5 k+2\right )\right ) \int \frac {x}{\left (k x^2-(k+1) x+1\right )^{3/4} \left (-d x^3+k x^2-(k+1) x+1\right )}d\sqrt [4]{x}}{d^2}-\frac {(1-x)^{3/4} \sqrt [4]{x} \left (k^2-2 d (k+1)\right ) (1-k x)^{3/4} \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},\frac {3}{4},\frac {5}{4},\frac {2 k x}{k+\sqrt {k^2-2 k+1}+1},\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}\right )}{d^2 \left (k x^2-(k+1) x+1\right )^{3/4}}-\frac {k (1-x)^{3/4} x^{5/4} (1-k x)^{3/4} \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},\frac {3}{4},\frac {9}{4},\frac {2 k x}{k+\sqrt {k^2-2 k+1}+1},\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}\right )}{5 d \left (k x^2-(k+1) x+1\right )^{3/4}}\right )}{((1-x) x (1-k x))^{3/4}}\) |
Int[(x^2*(3 - 2*(1 + k)*x + k*x^2))/(((1 - x)*x*(1 - k*x))^(3/4)*(-1 + (1 + k)*x - k*x^2 + d*x^3)),x]
3.10.34.3.1 Defintions of rubi rules used
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]
\[\int \frac {x^{2} \left (3-2 \left (1+k \right ) x +k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-1+\left (1+k \right ) x -k \,x^{2}+d \,x^{3}\right )}d x\]
Timed out. \[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\text {Timed out} \]
integrate(x^2*(3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x ^2+d*x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\int { \frac {{\left (k x^{2} - 2 \, {\left (k + 1\right )} x + 3\right )} x^{2}}{{\left (d x^{3} - k x^{2} + {\left (k + 1\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}}} \,d x } \]
integrate(x^2*(3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x ^2+d*x^3),x, algorithm="maxima")
integrate((k*x^2 - 2*(k + 1)*x + 3)*x^2/((d*x^3 - k*x^2 + (k + 1)*x - 1)*( (k*x - 1)*(x - 1)*x)^(3/4)), x)
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (57) = 114\).
Time = 0.51 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.72 \[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=-\frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-d\right )^{\frac {1}{4}} + 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-d\right )^{\frac {1}{4}}}\right )}{d} - \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-d\right )^{\frac {1}{4}} - 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-d\right )^{\frac {1}{4}}}\right )}{d} - \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-d\right )^{\frac {1}{4}} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-d} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}}\right )}{2 \, d} + \frac {\sqrt {2} \left (-d\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-d\right )^{\frac {1}{4}} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-d} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}}\right )}{2 \, d} \]
integrate(x^2*(3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(3/4)/(-1+(1+k)*x-k*x ^2+d*x^3),x, algorithm="giac")
-sqrt(2)*(-d)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-d)^(1/4) + 2*(k/x - k/x^ 2 - 1/x^2 + 1/x^3)^(1/4))/(-d)^(1/4))/d - sqrt(2)*(-d)^(1/4)*arctan(-1/2*s qrt(2)*(sqrt(2)*(-d)^(1/4) - 2*(k/x - k/x^2 - 1/x^2 + 1/x^3)^(1/4))/(-d)^( 1/4))/d - 1/2*sqrt(2)*(-d)^(1/4)*log(sqrt(2)*(-d)^(1/4)*(k/x - k/x^2 - 1/x ^2 + 1/x^3)^(1/4) + sqrt(-d) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3))/d + 1/2* sqrt(2)*(-d)^(1/4)*log(-sqrt(2)*(-d)^(1/4)*(k/x - k/x^2 - 1/x^2 + 1/x^3)^( 1/4) + sqrt(-d) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3))/d
Timed out. \[ \int \frac {x^2 \left (3-2 (1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-1+(1+k) x-k x^2+d x^3\right )} \, dx=\int \frac {x^2\,\left (k\,x^2-2\,x\,\left (k+1\right )+3\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (d\,x^3-k\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]
int((x^2*(k*x^2 - 2*x*(k + 1) + 3))/((x*(k*x - 1)*(x - 1))^(3/4)*(d*x^3 + x*(k + 1) - k*x^2 - 1)),x)