Integrand size = 30, antiderivative size = 736 \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {a+b x+c x^2}}{d x}-\frac {b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a} d}+\frac {\sqrt {a} e \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}+\frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{d}-\frac {b e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {f \left (2 c d^2-b d \left (e+\sqrt {e^2-4 d f}\right )+a \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )}}+\frac {f \left (2 c d^2-b d \left (e-\sqrt {e^2-4 d f}\right )+a \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )}} \]
-1/2*b*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/d/a^(1/2)+e*arct anh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))*a^(1/2)/d^2-1/2*b*e*arctanh (1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/d^2/c^(1/2)-1/2*(-b*e+2*c*d)*a rctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/d^2/c^(1/2)+arctanh(1/2* (2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*c^(1/2)/d-(c*x^2+b*x+a)^(1/2)/d/x-1 /2*f*arctanh(1/4*(4*a*f+2*x*(b*f-c*(e-(-4*d*f+e^2)^(1/2)))-b*(e-(-4*d*f+e^ 2)^(1/2)))*2^(1/2)/(c*x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2-(-b*f+ c*e)*(-4*d*f+e^2)^(1/2))^(1/2))*(2*c*d^2-b*d*(e+(-4*d*f+e^2)^(1/2))+a*(e^2 -2*d*f+e*(-4*d*f+e^2)^(1/2)))/d^2*2^(1/2)/(-4*d*f+e^2)^(1/2)/(f*(2*a*f-b*( e-(-4*d*f+e^2)^(1/2)))+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)+1/2*f*arc tanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*x*(b*f-c*(e+(-4*d*f+e^2)^(1/2)) ))*2^(1/2)/(c*x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4 *d*f+e^2)^(1/2))^(1/2))*(2*c*d^2-b*d*(e-(-4*d*f+e^2)^(1/2))+a*(e^2-2*d*f-e *(-4*d*f+e^2)^(1/2)))/d^2*2^(1/2)/(-4*d*f+e^2)^(1/2)/(c*(e^2-2*d*f+e*(-4*d *f+e^2)^(1/2))+f*(2*a*f-b*(e+(-4*d*f+e^2)^(1/2))))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.69 (sec) , antiderivative size = 582, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\frac {-\frac {d \sqrt {a+x (b+c x)}}{x}+\frac {(-b d+2 a e) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {-b c d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+b^2 d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a b e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+a^2 e f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+2 c^{3/2} d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 b \sqrt {c} d e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a \sqrt {c} e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a \sqrt {c} d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+b d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a e f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{d^2} \]
(-((d*Sqrt[a + x*(b + c*x)])/x) + ((-(b*d) + 2*a*e)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/Sqrt[a] + RootSum[b^2*d - a*b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 + b*e*#1^2 - 2*a*f*#1^ 2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (-(b*c*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b *x + c*x^2] - #1]) + b^2*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1 ] - a*b*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a^2*e*f*Log[- (Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*c^(3/2)*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*b*Sqrt[c]*d*e*Log[-(Sqrt[c]*x) + Sqr t[a + b*x + c*x^2] - #1]*#1 + 2*a*Sqrt[c]*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*a*Sqrt[c]*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + b*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^ 2 - a*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(2*b*Sqrt[c ]*d - a*Sqrt[c]*e - 4*c*d*#1 - b*e*#1 + 2*a*f*#1 + 3*Sqrt[c]*e*#1^2 - 2*f* #1^3) & ])/d^2
Time = 2.86 (sec) , antiderivative size = 736, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {7279, 2009}
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.
| \(\displaystyle \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\sqrt {a+b x+c x^2} \left (-d f+e^2+e f x\right )}{d^2 \left (d+e x+f x^2\right )}-\frac {e \sqrt {a+b x+c x^2}}{d^2 x}+\frac {\sqrt {a+b x+c x^2}}{d x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {f \left (a \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (\sqrt {e^2-4 d f}+e\right )+2 c d^2\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (a \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )-b d \left (e-\sqrt {e^2-4 d f}\right )+2 c d^2\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (\sqrt {e^2-4 d f}+e\right )\right )+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\sqrt {a} e \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d^2}-\frac {b e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} d^2}-\frac {b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a} d}+\frac {\sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{d}-\frac {\sqrt {a+b x+c x^2}}{d x}\) |
-(Sqrt[a + b*x + c*x^2]/(d*x)) - (b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[a]*d) + (Sqrt[a]*e*ArcTanh[(2*a + b*x)/(2*Sqrt[a ]*Sqrt[a + b*x + c*x^2])])/d^2 + (Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S qrt[a + b*x + c*x^2])])/d - (b*e*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b *x + c*x^2])])/(2*Sqrt[c]*d^2) - ((2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqr t[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*d^2) - (f*(2*c*d^2 - b*d*(e + Sqr t[e^2 - 4*d*f]) + a*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))*ArcTanh[(4*a*f - b*(e - Sqrt[e^2 - 4*d*f]) + 2*(b*f - c*(e - Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt [2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f] ]*Sqrt[a + b*x + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2* d*f - e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e - Sqrt[e^2 - 4*d*f]))]) + (f* (2*c*d^2 - b*d*(e - Sqrt[e^2 - 4*d*f]) + a*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d *f]))*ArcTanh[(4*a*f - b*(e + Sqrt[e^2 - 4*d*f]) + 2*(b*f - c*(e + Sqrt[e^ 2 - 4*d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/(Sqrt[2]*d^2*Sqrt[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e + Sqrt [e^2 - 4*d*f]))])
3.2.13.3.1 Defintions of rubi rules used
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]
Time = 0.95 (sec) , antiderivative size = 976, normalized size of antiderivative = 1.33
| method | result | size |
| risch | \(-\frac {\sqrt {c \,x^{2}+b x +a}}{d x}-\frac {\frac {4 f \left (2 a e -b d \right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {a}}+\frac {2 \left (f a \sqrt {-4 d f +e^{2}}-c d \sqrt {-4 d f +e^{2}}-a e f +2 b d f -c d e \right ) \sqrt {2}\, \ln \left (\frac {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c +\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 b f \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {\frac {-b f \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}}-\frac {2 \left (f a \sqrt {-4 d f +e^{2}}-c d \sqrt {-4 d f +e^{2}}+a e f -2 b d f +c d e \right ) \sqrt {2}\, \ln \left (\frac {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c +\frac {4 \left (c \sqrt {-4 d f +e^{2}}+b f -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 b f \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {\frac {b f \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 c d f +c \,e^{2}}{f^{2}}}}}{2 d}\) | \(976\) |
| default | \(\text {Expression too large to display}\) | \(1906\) |
-(c*x^2+b*x+a)^(1/2)/d/x-1/2/d*(4*f*(2*a*e-b*d)/(-e+(-4*d*f+e^2)^(1/2))/(e +(-4*d*f+e^2)^(1/2))/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x) +2*(f*a*(-4*d*f+e^2)^(1/2)-c*d*(-4*d*f+e^2)^(1/2)-a*e*f+2*b*d*f-c*d*e)/(-4 *d*f+e^2)^(1/2)/(e+(-4*d*f+e^2)^(1/2))*2^(1/2)/((-b*f*(-4*d*f+e^2)^(1/2)+( -4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-b*f*(- 4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2+1 /f*(-c*(-4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^ (1/2)*((-b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d *f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c+4/f*(-c*(-4*d *f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-b*f*(-4*d*f+e^ 2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2))/( x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))-2*(f*a*(-4*d*f+e^2)^(1/2)-c*d*(-4*d*f+e^2 )^(1/2)+a*e*f-2*b*d*f+c*d*e)/(-4*d*f+e^2)^(1/2)/(-e+(-4*d*f+e^2)^(1/2))*2^ (1/2)/((b*f*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d* f+c*e^2)/f^2)^(1/2)*ln(((b*f*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a *f^2-b*e*f-2*c*d*f+c*e^2)/f^2+(c*(-4*d*f+e^2)^(1/2)+b*f-c*e)/f*(x-1/2/f*(- e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((b*f*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^( 1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x-1/2/f*(-e+(-4*d*f+e ^2)^(1/2)))^2*c+4*(c*(-4*d*f+e^2)^(1/2)+b*f-c*e)/f*(x-1/2/f*(-e+(-4*d*f+e^ 2)^(1/2)))+2*(b*f*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e...
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} + e x + d\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {a+b x+c x^2}}{x^2 \left (d+e x+f x^2\right )} \, dx=\int \frac {\sqrt {c\,x^2+b\,x+a}}{x^2\,\left (f\,x^2+e\,x+d\right )} \,d x \]