Integrand size = 43, antiderivative size = 44 \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a b x+(-a-b) x^2+x^3}}{\sqrt {d} (a-x)}\right )}{\sqrt {d}} \]
Time = 15.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {x (-a+x) (-b+x)}}{\sqrt {d} (a-x)}\right )}{\sqrt {d}} \]
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.
| \(\displaystyle \int \frac {a b-2 a x+x^2}{\sqrt {x (x-a) (x-b)} \left (a d-x (b+d)+x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt {-x (a+b)+a b+x^2} \int \frac {x^2-2 a x+a b}{\sqrt {x} \sqrt {x^2-(a+b) x+a b} \left (x^2-(b+d) x+a d\right )}dx}{\sqrt {x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {-x (a+b)+a b+x^2} \int \frac {x^2-2 a x+a b}{\sqrt {x^2-(a+b) x+a b} \left (x^2-(b+d) x+a d\right )}d\sqrt {x}}{\sqrt {x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {-x (a+b)+a b+x^2} \int \left (\frac {a (b-d)-(2 a-b-d) x}{\sqrt {x^2-(a+b) x+a b} \left (x^2+(-b-d) x+a d\right )}+\frac {1}{\sqrt {x^2-(a+b) x+a b}}\right )d\sqrt {x}}{\sqrt {x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {-x (a+b)+a b+x^2} \int \left (\frac {a (b-d)-(2 a-b-d) x}{\sqrt {x^2-(a+b) x+a b} \left (x^2-(b+d) x+a d\right )}+\frac {1}{\sqrt {x^2-(a+b) x+a b}}\right )d\sqrt {x}}{\sqrt {x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \frac {2 \sqrt {x} \sqrt {-x (a+b)+a b+x^2} \int \left (\frac {a (b-d)-(2 a-b-d) x}{\sqrt {x^2-(a+b) x+a b} \left (x^2-(b+d) x+a d\right )}+\frac {1}{\sqrt {x^2-(a+b) x+a b}}\right )d\sqrt {x}}{\sqrt {x (a-x) (b-x)}}\) |
3.6.66.3.1 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[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 1.45 (sec) , antiderivative size = 2291, normalized size of antiderivative = 52.07
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2291\) |
| elliptic | \(\text {Expression too large to display}\) | \(2296\) |
-2*b*(-(-b+x)/b)^(1/2)*((-a+x)/(-a+b))^(1/2)*(x/b)^(1/2)/(a*b*x-a*x^2-b*x^ 2+x^3)^(1/2)*EllipticF((-(-b+x)/b)^(1/2),(b/(-a+b))^(1/2))+4/(-4*a*d+b^2+2 *b*d+d^2)^(1/2)*b*(1-x/b)^(1/2)*(-1/(-a+b)*a+1/(-a+b)*x)^(1/2)*(x/b)^(1/2) /(a*b*x-a*x^2-b*x^2+x^3)^(1/2)/(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/ 2))*EllipticPi((-(-b+x)/b)^(1/2),b/(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2) ^(1/2)),(b/(-a+b))^(1/2))*a*d+2*b*(1-x/b)^(1/2)*(-1/(-a+b)*a+1/(-a+b)*x)^( 1/2)*(x/b)^(1/2)/(a*b*x-a*x^2-b*x^2+x^3)^(1/2)/(1/2*b-1/2*d-1/2*(-4*a*d+b^ 2+2*b*d+d^2)^(1/2))*EllipticPi((-(-b+x)/b)^(1/2),b/(1/2*b-1/2*d-1/2*(-4*a* d+b^2+2*b*d+d^2)^(1/2)),(b/(-a+b))^(1/2))*a-1/(-4*a*d+b^2+2*b*d+d^2)^(1/2) *b^3*(1-x/b)^(1/2)*(-1/(-a+b)*a+1/(-a+b)*x)^(1/2)*(x/b)^(1/2)/(a*b*x-a*x^2 -b*x^2+x^3)^(1/2)/(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2))*EllipticP i((-(-b+x)/b)^(1/2),b/(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2)),(b/(- a+b))^(1/2))-2/(-4*a*d+b^2+2*b*d+d^2)^(1/2)*b^2*(1-x/b)^(1/2)*(-1/(-a+b)*a +1/(-a+b)*x)^(1/2)*(x/b)^(1/2)/(a*b*x-a*x^2-b*x^2+x^3)^(1/2)/(1/2*b-1/2*d- 1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2))*EllipticPi((-(-b+x)/b)^(1/2),b/(1/2*b-1/ 2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2)),(b/(-a+b))^(1/2))*d-b^2*(1-x/b)^(1/2 )*(-1/(-a+b)*a+1/(-a+b)*x)^(1/2)*(x/b)^(1/2)/(a*b*x-a*x^2-b*x^2+x^3)^(1/2) /(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2))*EllipticPi((-(-b+x)/b)^(1/ 2),b/(1/2*b-1/2*d-1/2*(-4*a*d+b^2+2*b*d+d^2)^(1/2)),(b/(-a+b))^(1/2))-1/(- 4*a*d+b^2+2*b*d+d^2)^(1/2)*b*(1-x/b)^(1/2)*(-1/(-a+b)*a+1/(-a+b)*x)^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (35) = 70\).
Time = 0.41 (sec) , antiderivative size = 224, normalized size of antiderivative = 5.09 \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\left [\frac {\log \left (\frac {a^{2} d^{2} - 2 \, {\left (b - 3 \, d\right )} x^{3} + x^{4} + {\left (b^{2} - 6 \, {\left (a + b\right )} d + d^{2}\right )} x^{2} + 4 \, \sqrt {a b x - {\left (a + b\right )} x^{2} + x^{3}} {\left (a d + {\left (b - d\right )} x - x^{2}\right )} \sqrt {d} + 2 \, {\left (3 \, a b d - a d^{2}\right )} x}{a^{2} d^{2} - 2 \, {\left (b + d\right )} x^{3} + x^{4} + {\left (b^{2} + 2 \, {\left (a + b\right )} d + d^{2}\right )} x^{2} - 2 \, {\left (a b d + a d^{2}\right )} x}\right )}{2 \, \sqrt {d}}, \frac {\sqrt {-d} \arctan \left (-\frac {\sqrt {a b x - {\left (a + b\right )} x^{2} + x^{3}} {\left (a d + {\left (b - d\right )} x - x^{2}\right )} \sqrt {-d}}{2 \, {\left (a b d x - {\left (a + b\right )} d x^{2} + d x^{3}\right )}}\right )}{d}\right ] \]
[1/2*log((a^2*d^2 - 2*(b - 3*d)*x^3 + x^4 + (b^2 - 6*(a + b)*d + d^2)*x^2 + 4*sqrt(a*b*x - (a + b)*x^2 + x^3)*(a*d + (b - d)*x - x^2)*sqrt(d) + 2*(3 *a*b*d - a*d^2)*x)/(a^2*d^2 - 2*(b + d)*x^3 + x^4 + (b^2 + 2*(a + b)*d + d ^2)*x^2 - 2*(a*b*d + a*d^2)*x))/sqrt(d), sqrt(-d)*arctan(-1/2*sqrt(a*b*x - (a + b)*x^2 + x^3)*(a*d + (b - d)*x - x^2)*sqrt(-d)/(a*b*d*x - (a + b)*d* x^2 + d*x^3))/d]
Timed out. \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume((d+b)^2-4*a*d>0)', see `assume?` for more
\[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\int { \frac {a b - 2 \, a x + x^{2}}{\sqrt {{\left (a - x\right )} {\left (b - x\right )} x} {\left (a d - {\left (b + d\right )} x + x^{2}\right )}} \,d x } \]
Time = 5.14 (sec) , antiderivative size = 465, normalized size of antiderivative = 10.57 \[ \int \frac {a b-2 a x+x^2}{\sqrt {x (-a+x) (-b+x)} \left (a d-(b+d) x+x^2\right )} \, dx=\frac {2\,b\,\sqrt {\frac {x}{b}}\,\sqrt {\frac {b-x}{b}}\,\sqrt {\frac {a-x}{a-b}}\,\Pi \left (\frac {b}{\frac {b}{2}-\frac {d}{2}+\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}};\mathrm {asin}\left (\sqrt {\frac {b-x}{b}}\right )\middle |-\frac {b}{a-b}\right )\,\left (\left (\frac {b}{2}+\frac {d}{2}-\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}\right )\,\left (b-2\,a+d\right )+a\,b-a\,d\right )}{\sqrt {x^3+\left (-a-b\right )\,x^2+a\,b\,x}\,\left (\frac {b}{2}-\frac {d}{2}+\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}\right )\,\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}-\frac {2\,b\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {b-x}{b}}\right )\middle |-\frac {b}{a-b}\right )\,\sqrt {\frac {x}{b}}\,\sqrt {\frac {b-x}{b}}\,\sqrt {\frac {a-x}{a-b}}}{\sqrt {x^3+\left (-a-b\right )\,x^2+a\,b\,x}}+\frac {2\,b\,\sqrt {\frac {x}{b}}\,\sqrt {\frac {b-x}{b}}\,\sqrt {\frac {a-x}{a-b}}\,\Pi \left (-\frac {b}{\frac {d}{2}-\frac {b}{2}+\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}};\mathrm {asin}\left (\sqrt {\frac {b-x}{b}}\right )\middle |-\frac {b}{a-b}\right )\,\left (\left (\frac {b}{2}+\frac {d}{2}+\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}\right )\,\left (b-2\,a+d\right )+a\,b-a\,d\right )}{\sqrt {x^3+\left (-a-b\right )\,x^2+a\,b\,x}\,\left (\frac {d}{2}-\frac {b}{2}+\frac {\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}}{2}\right )\,\sqrt {b^2+2\,b\,d+d^2-4\,a\,d}} \]
(2*b*(x/b)^(1/2)*((b - x)/b)^(1/2)*((a - x)/(a - b))^(1/2)*ellipticPi(b/(b /2 - d/2 + (2*b*d - 4*a*d + b^2 + d^2)^(1/2)/2), asin(((b - x)/b)^(1/2)), -b/(a - b))*((b/2 + d/2 - (2*b*d - 4*a*d + b^2 + d^2)^(1/2)/2)*(b - 2*a + d) + a*b - a*d))/((x^3 - x^2*(a + b) + a*b*x)^(1/2)*(b/2 - d/2 + (2*b*d - 4*a*d + b^2 + d^2)^(1/2)/2)*(2*b*d - 4*a*d + b^2 + d^2)^(1/2)) - (2*b*elli pticF(asin(((b - x)/b)^(1/2)), -b/(a - b))*(x/b)^(1/2)*((b - x)/b)^(1/2)*( (a - x)/(a - b))^(1/2))/(x^3 - x^2*(a + b) + a*b*x)^(1/2) + (2*b*(x/b)^(1/ 2)*((b - x)/b)^(1/2)*((a - x)/(a - b))^(1/2)*ellipticPi(-b/(d/2 - b/2 + (2 *b*d - 4*a*d + b^2 + d^2)^(1/2)/2), asin(((b - x)/b)^(1/2)), -b/(a - b))*( (b/2 + d/2 + (2*b*d - 4*a*d + b^2 + d^2)^(1/2)/2)*(b - 2*a + d) + a*b - a* d))/((x^3 - x^2*(a + b) + a*b*x)^(1/2)*(d/2 - b/2 + (2*b*d - 4*a*d + b^2 + d^2)^(1/2)/2)*(2*b*d - 4*a*d + b^2 + d^2)^(1/2))