Integrand size = 47, antiderivative size = 153 \[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{b-x}\right )}{d^{3/4}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{b-x}\right )}{d^{3/4}} \]
-2*arctan(d^(1/4)*(a*b^3*x+(-3*a*b^2-b^3)*x^2+(3*a*b+3*b^2)*x^3+(-a-3*b)*x ^4+x^5)^(1/4)/(b-x))/d^(3/4)+2*arctanh(d^(1/4)*(a*b^3*x+(-3*a*b^2-b^3)*x^2 +(3*a*b+3*b^2)*x^3+(-a-3*b)*x^4+x^5)^(1/4)/(b-x))/d^(3/4)
Time = 15.36 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.47 \[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^3}}{-b+x}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^3}}{b-x}\right )\right )}{d^{3/4}} \]
(2*(ArcTan[(d^(1/4)*(x*(-a + x)*(-b + x)^3)^(1/4))/(-b + x)] + ArcTanh[(d^ (1/4)*(x*(-a + x)*(-b + x)^3)^(1/4))/(b - 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.
| \(\displaystyle \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (x-a) (x-b)^3} \left (-x (a d+1)+b+d x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \int \frac {x^2-2 b x+a b}{\sqrt [4]{x} \left (d x^2-(a d+1) x+b\right ) \sqrt [4]{x^4-(a+3 b) x^3+3 b (a+b) x^2-b^2 (3 a+b) x+a b^3}}dx}{\sqrt [4]{x (a-x) (b-x)^3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \int \frac {\sqrt {x} \left (x^2-2 b x+a b\right )}{\left (d x^2-(a d+1) x+b\right ) \sqrt [4]{x^4-(a+3 b) x^3+3 b (a+b) x^2-b^2 (3 a+b) x+a b^3}}d\sqrt [4]{x}}{\sqrt [4]{x (a-x) (b-x)^3}}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \int \frac {\sqrt {x} \left (x^2-2 b x+a b\right )}{\sqrt [4]{(a-x) (b-x)^3} \left (d x^2-(a d+1) x+b\right )}d\sqrt [4]{x}}{\sqrt [4]{x (a-x) (b-x)^3}}\) |
| \(\Big \downarrow \) 2058 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \int \frac {\sqrt {x} \left (x^2-2 b x+a b\right )}{\sqrt [4]{a-x} (b-x)^{3/4} \left (d x^2-(a d+1) x+b\right )}d\sqrt [4]{x}}{\sqrt [4]{(a-x) (b-x)^3} \sqrt [4]{x (a-x) (b-x)^3}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \int \left (\frac {\sqrt {x} ((a d-2 b d+1) x-b (1-a d))}{d \sqrt [4]{a-x} (b-x)^{3/4} \left (d x^2+(-a d-1) x+b\right )}+\frac {\sqrt {x}}{d \sqrt [4]{a-x} (b-x)^{3/4}}\right )d\sqrt [4]{x}}{\sqrt [4]{(a-x) (b-x)^3} \sqrt [4]{x (a-x) (b-x)^3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{a b^3-b^2 x (3 a+b)-x^3 (a+3 b)+3 b x^2 (a+b)+x^4} \left (\frac {(a d-2 b d+1) \left (-\sqrt {a^2 d^2+2 a d-4 b d+1}+a d+1\right ) \int \frac {\sqrt {x}}{\sqrt [4]{a-x} (b-x)^{3/4} \left (a d-2 x d-\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [4]{x}}{d \sqrt {a^2 d^2+2 a d-4 b d+1}}-\frac {(a d-2 b d+1) \left (\sqrt {a^2 d^2+2 a d-4 b d+1}+a d+1\right ) \int \frac {\sqrt {x}}{\sqrt [4]{a-x} (b-x)^{3/4} \left (a d-2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [4]{x}}{d \sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {2 b (1-a d) \int \frac {\sqrt {x}}{\sqrt [4]{a-x} (b-x)^{3/4} \left (a d-2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}+1\right )}d\sqrt [4]{x}}{\sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {2 b (1-a d) \int \frac {\sqrt {x}}{\sqrt [4]{a-x} (b-x)^{3/4} \left (-a d+2 x d+\sqrt {a^2 d^2+2 a d-4 b d+1}-1\right )}d\sqrt [4]{x}}{\sqrt {a^2 d^2+2 a d-4 b d+1}}+\frac {x^{3/4} \sqrt [4]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{4},\frac {3}{4},\frac {7}{4},\frac {x}{a},\frac {x}{b}\right )}{3 d \sqrt [4]{a-x} (b-x)^{3/4}}\right )}{\sqrt [4]{(a-x) (b-x)^3} \sqrt [4]{x (a-x) (b-x)^3}}\) |
3.22.4.3.1 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[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^ (r_.))^(p_), x_Symbol] :> Simp[Simp[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))] Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)^(p* r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, 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]
\[\int \frac {a b -2 b x +x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{3}\right )^{\frac {1}{4}} \left (b -\left (a d +1\right ) x +d \,x^{2}\right )}d x\]
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
integrate((a*b-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(1/4)/(b-(a*d+1)*x+d*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
integrate((a*b-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(1/4)/(b-(a*d+1)*x+d*x^2),x, algorithm="maxima")
\[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (b-(1+a d) x+d x^2\right )} \, dx=\int \frac {x^2-2\,b\,x+a\,b}{{\left (x\,\left (a-x\right )\,{\left (b-x\right )}^3\right )}^{1/4}\,\left (d\,x^2+\left (-a\,d-1\right )\,x+b\right )} \,d x \]