Integrand size = 80, antiderivative size = 107 \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{d} x}{\sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{3/4}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{3/4}} \]
-2*arctan(d^(1/4)*x/(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)*x^3+x^4)^(1/4))/d^( 3/4)+2*arctanh(d^(1/4)*x/(-a*b^2*x+(2*a*b+b^2)*x^2+(-a-2*b)*x^3+x^4)^(1/4) )/d^(3/4)
\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx \]
Integrate[(x^2*(3*a*b^2 - 2*b*(2*a + b)*x + (a + 2*b)*x^2))/((x*(-a + x)*( -b + x)^2)^(3/4)*(a*b^2 - b*(2*a + b)*x + (a + 2*b)*x^2 + (-1 + d)*x^3)),x ]
Integrate[(x^2*(3*a*b^2 - 2*b*(2*a + b)*x + (a + 2*b)*x^2))/((x*(-a + x)*( -b + x)^2)^(3/4)*(a*b^2 - b*(2*a + b)*x + (a + 2*b)*x^2 + (-1 + 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.
\(\displaystyle \int \frac {x^2 \left (3 a b^2+x^2 (a+2 b)-2 b x (2 a+b)\right )}{\left (x (x-a) (x-b)^2\right )^{3/4} \left (a b^2+x^2 (a+2 b)-b x (2 a+b)+(d-1) x^3\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{3/4} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \frac {x^{5/4} \left (3 a b^2-2 (2 a+b) x b+(a+2 b) x^2\right )}{\left (x^3-(a+2 b) x^2+b (2 a+b) x-a b^2\right )^{3/4} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}dx}{\left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {4 x^{3/4} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \frac {x^2 \left (3 a b^2-2 (2 a+b) x b+(a+2 b) x^2\right )}{\left (x^3-(a+2 b) x^2+b (2 a+b) x-a b^2\right )^{3/4} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}d\sqrt [4]{x}}{\left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {4 x^{3/4} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \frac {x^2 \left (3 a b^2-2 (2 a+b) x b+(a+2 b) x^2\right )}{\left (-\left ((a-x) (x-b)^2\right )\right )^{3/4} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}d\sqrt [4]{x}}{\left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 2058 |
\(\displaystyle \frac {4 x^{3/4} (a-x)^{3/4} (x-b)^{3/2} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \frac {x^2 \left (3 a b^2-2 (2 a+b) x b+(a+2 b) x^2\right )}{(a-x)^{3/4} (x-b)^{3/2} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}d\sqrt [4]{x}}{\left (-\left ((a-x) (b-x)^2\right )\right )^{3/4} \left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 1387 |
\(\displaystyle \frac {4 x^{3/4} (a-x)^{3/4} (x-b)^{3/2} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \frac {x^2 ((a+2 b) x-3 a b)}{(a-x)^{3/4} \sqrt {x-b} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}d\sqrt [4]{x}}{\left (-\left ((a-x) (b-x)^2\right )\right )^{3/4} \left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4 x^{3/4} (a-x)^{3/4} (x-b)^{3/2} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \int \left (-\frac {a+2 b}{(1-d) (a-x)^{3/4} \sqrt {x-b}}-\frac {a (a+2 b) b^2-(2 a+b) (a+2 b) x b+\left (a^2+(3 d b+b) a+4 b^2\right ) x^2}{(d-1) (a-x)^{3/4} \sqrt {x-b} \left (-\left ((1-d) x^3\right )+(a+2 b) x^2-b (2 a+b) x+a b^2\right )}\right )d\sqrt [4]{x}}{\left (-\left ((a-x) (b-x)^2\right )\right )^{3/4} \left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 x^{3/4} (a-x)^{3/4} (x-b)^{3/2} \left (-a b^2-x^2 (a+2 b)+b x (2 a+b)+x^3\right )^{3/4} \left (\frac {\left (a^2+a (3 b d+b)+4 b^2\right ) \int \frac {x^2}{(a-x)^{3/4} \sqrt {x-b} \left (-\left ((1-d) x^3\right )+a \left (\frac {2 b}{a}+1\right ) x^2-2 a b \left (\frac {b}{2 a}+1\right ) x+a b^2\right )}d\sqrt [4]{x}}{1-d}+\frac {a b^2 (a+2 b) \int \frac {1}{(a-x)^{3/4} \sqrt {x-b} \left (-\left ((1-d) x^3\right )+a \left (\frac {2 b}{a}+1\right ) x^2-2 a b \left (\frac {b}{2 a}+1\right ) x+a b^2\right )}d\sqrt [4]{x}}{1-d}-\frac {b (2 a+b) (a+2 b) \int \frac {x}{(a-x)^{3/4} \sqrt {x-b} \left (-\left ((1-d) x^3\right )+a \left (\frac {2 b}{a}+1\right ) x^2-2 a b \left (\frac {b}{2 a}+1\right ) x+a b^2\right )}d\sqrt [4]{x}}{1-d}+\frac {\sqrt [4]{x} (a+2 b) \sqrt {x-b} \left (\frac {b (a-x)}{a (b-x)}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {(a-b) x}{a (b-x)}\right )}{b (1-d) (a-x)^{3/4}}\right )}{\left (-\left ((a-x) (b-x)^2\right )\right )^{3/4} \left (-\left (x (a-x) (b-x)^2\right )\right )^{3/4}}\) |
Int[(x^2*(3*a*b^2 - 2*b*(2*a + b)*x + (a + 2*b)*x^2))/((x*(-a + x)*(-b + x )^2)^(3/4)*(a*b^2 - b*(2*a + b)*x + (a + 2*b)*x^2 + (-1 + d)*x^3)),x]
3.16.51.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)* (x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^ p, x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
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]
Time = 0.92 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {d^{\frac {1}{4}} x +\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}{-d^{\frac {1}{4}} x +\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}{x \,d^{\frac {1}{4}}}\right )}{d^{\frac {3}{4}}}\) | \(84\) |
int(x^2*(3*a*b^2-2*b*(2*a+b)*x+(a+2*b)*x^2)/(x*(-a+x)*(-b+x)^2)^(3/4)/(a*b ^2-b*(2*a+b)*x+(a+2*b)*x^2+(-1+d)*x^3),x,method=_RETURNVERBOSE)
1/d^(3/4)*(ln((d^(1/4)*x+(-x*(a-x)*(b-x)^2)^(1/4))/(-d^(1/4)*x+(-x*(a-x)*( b-x)^2)^(1/4)))+2*arctan((-x*(a-x)*(b-x)^2)^(1/4)/x/d^(1/4)))
Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
integrate(x^2*(3*a*b^2-2*b*(2*a+b)*x+(a+2*b)*x^2)/(x*(-a+x)*(-b+x)^2)^(3/4 )/(a*b^2-b*(2*a+b)*x+(a+2*b)*x^2+(-1+d)*x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
integrate(x**2*(3*a*b**2-2*b*(2*a+b)*x+(a+2*b)*x**2)/(x*(-a+x)*(-b+x)**2)* *(3/4)/(a*b**2-b*(2*a+b)*x+(a+2*b)*x**2+(-1+d)*x**3),x)
\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {{\left (3 \, a b^{2} - 2 \, {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {3}{4}} {\left ({\left (d - 1\right )} x^{3} + a b^{2} - {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )}} \,d x } \]
integrate(x^2*(3*a*b^2-2*b*(2*a+b)*x+(a+2*b)*x^2)/(x*(-a+x)*(-b+x)^2)^(3/4 )/(a*b^2-b*(2*a+b)*x+(a+2*b)*x^2+(-1+d)*x^3),x, algorithm="maxima")
integrate((3*a*b^2 - 2*(2*a + b)*b*x + (a + 2*b)*x^2)*x^2/((-(a - x)*(b - x)^2*x)^(3/4)*((d - 1)*x^3 + a*b^2 - (2*a + b)*b*x + (a + 2*b)*x^2)), x)
\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {{\left (3 \, a b^{2} - 2 \, {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {3}{4}} {\left ({\left (d - 1\right )} x^{3} + a b^{2} - {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )}} \,d x } \]
integrate(x^2*(3*a*b^2-2*b*(2*a+b)*x+(a+2*b)*x^2)/(x*(-a+x)*(-b+x)^2)^(3/4 )/(a*b^2-b*(2*a+b)*x+(a+2*b)*x^2+(-1+d)*x^3),x, algorithm="giac")
integrate((3*a*b^2 - 2*(2*a + b)*b*x + (a + 2*b)*x^2)*x^2/((-(a - x)*(b - x)^2*x)^(3/4)*((d - 1)*x^3 + a*b^2 - (2*a + b)*b*x + (a + 2*b)*x^2)), x)
Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2\,\left (3\,a\,b^2+x^2\,\left (a+2\,b\right )-2\,b\,x\,\left (2\,a+b\right )\right )}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{3/4}\,\left (a\,b^2+x^2\,\left (a+2\,b\right )+x^3\,\left (d-1\right )-b\,x\,\left (2\,a+b\right )\right )} \,d x \]
int((x^2*(3*a*b^2 + x^2*(a + 2*b) - 2*b*x*(2*a + b)))/((-x*(a - x)*(b - x) ^2)^(3/4)*(a*b^2 + x^2*(a + 2*b) + x^3*(d - 1) - b*x*(2*a + b))),x)