Integrand size = 53, antiderivative size = 75 \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )}{d^{3/4}} \]
2*arctan(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(1/4)/x)/d^(3/4)-2*arctanh(d^(1/4)*( x+(-1-k)*x^2+k*x^3)^(1/4)/x)/d^(3/4)
Time = 11.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{x}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{x}\right )\right )}{d^{3/4}} \]
Integrate[(3 - 2*(1 + k)*x + k*x^2)/(((1 - x)*x*(1 - k*x))^(1/4)*(-d + d*( 1 + k)*x - d*k*x^2 + x^3)),x]
(2*(ArcTan[(d^(1/4)*((-1 + x)*x*(-1 + k*x))^(1/4))/x] - ArcTanh[(d^(1/4)*( (-1 + x)*x*(-1 + k*x))^(1/4))/x]))/d^(3/4)
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.
| \(\displaystyle \int \frac {k x^2-2 (k+1) x+3}{\sqrt [4]{(1-x) x (1-k x)} \left (-d k x^2+d (k+1) x-d+x^3\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int -\frac {k x^2-2 (k+1) x+3}{\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {k x^2-2 (k+1) x+3}{\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {\sqrt {x} \left (k x^2-2 (k+1) x+3\right )}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \left (\frac {k x^{5/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}+\frac {2 (-k-1) x^{3/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}+\frac {3 \sqrt {x}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \left (3 \int \frac {\sqrt {x}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}-2 (k+1) \int \frac {x^{3/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}+k \int \frac {x^{5/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\) |
Int[(3 - 2*(1 + k)*x + k*x^2)/(((1 - x)*x*(1 - k*x))^(1/4)*(-d + d*(1 + k) *x - d*k*x^2 + x^3)),x]
3.10.84.3.1 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]
\[\int \frac {3-2 \left (1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-d +d \left (1+k \right ) x -d k \,x^{2}+x^{3}\right )}d x\]
Timed out. \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\text {Timed out} \]
integrate((3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-d+d*(1+k)*x-d*k*x ^2+x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\int { -\frac {k x^{2} - 2 \, {\left (k + 1\right )} x + 3}{{\left (d k x^{2} - d {\left (k + 1\right )} x - x^{3} + d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}}} \,d x } \]
integrate((3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-d+d*(1+k)*x-d*k*x ^2+x^3),x, algorithm="maxima")
-integrate((k*x^2 - 2*(k + 1)*x + 3)/((d*k*x^2 - d*(k + 1)*x - x^3 + d)*(( k*x - 1)*(x - 1)*x)^(1/4)), x)
Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (61) = 122\).
Time = 0.36 (sec) , antiderivative size = 288, normalized size of antiderivative = 3.84 \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=-\frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{3}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} - 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{3}} + \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{3}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{3}} \]
integrate((3-2*(1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-d+d*(1+k)*x-d*k*x ^2+x^3),x, algorithm="giac")
-sqrt(2)*(-d^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-1/d)^(1/4) + 2*(k/x - k/x^2 - 1/x^2 + 1/x^3)^(1/4))/(-1/d)^(1/4))/d^3 - sqrt(2)*(-d^3)^(3/4)*arc tan(-1/2*sqrt(2)*(sqrt(2)*(-1/d)^(1/4) - 2*(k/x - k/x^2 - 1/x^2 + 1/x^3)^( 1/4))/(-1/d)^(1/4))/d^3 + 1/2*sqrt(2)*(-d^3)^(3/4)*log(sqrt(2)*(k/x - k/x^ 2 - 1/x^2 + 1/x^3)^(1/4)*(-1/d)^(1/4) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3) + sqrt(-1/d))/d^3 - 1/2*sqrt(2)*(-d^3)^(3/4)*log(-sqrt(2)*(k/x - k/x^2 - 1 /x^2 + 1/x^3)^(1/4)*(-1/d)^(1/4) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3) + sqr t(-1/d))/d^3
Timed out. \[ \int \frac {3-2 (1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=-\int \frac {k\,x^2-2\,x\,\left (k+1\right )+3}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (-x^3+d\,k\,x^2-d\,\left (k+1\right )\,x+d\right )} \,d x \]
int(-(k*x^2 - 2*x*(k + 1) + 3)/((x*(k*x - 1)*(x - 1))^(1/4)*(d - x^3 - d*x *(k + 1) + d*k*x^2)),x)