Integrand size = 30, antiderivative size = 543 \[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {a+b x+c x^2}}{a d x}+\frac {b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2} d}+\frac {e \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d^2}-\frac {f \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )}}+\frac {f \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:
-(c*x^2+b*x+a)^(1/2)/a/d/x+1/2*b*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+ a)^(1/2))/a^(3/2)/d+e*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a ^(1/2)/d^2-1/2*f*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2))*arctanh(1/4*(4*a*f-b*(e- (-4*d*f+e^2)^(1/2))+2*(b*f-c*(e-(-4*d*f+e^2)^(1/2)))*x)*2^(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)/(c*x^2+b*x+a)^(1/2 ))*2^(1/2)/d^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)+1/2*f*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2 ))*arctanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e+(-4*d*f+e^2)^(1 /2)))*x)*2^(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)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/d^2/(-4*d*f+e^2)^(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)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.98 (sec) , antiderivative size = 423, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {-\frac {d \sqrt {a+x (b+c x)}}{a x}+\frac {(b d+2 a e) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{3/2}}+\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-b d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a e f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 \sqrt {c} e^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 \sqrt {c} d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+e f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-2 b \sqrt {c} d+a \sqrt {c} e+4 c d \text {$\#$1}+b e \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{d^2} \] Input:
Integrate[1/(x^2*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
Output:
(-((d*Sqrt[a + x*(b + c*x)])/(a*x)) + ((b*d + 2*a*e)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/a^(3/2) + 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*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - b*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a* e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*Sqrt[c]*e^2*Log[-(S qrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*Sqrt[c]*d*f*Log[-(Sqrt[c]*x ) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 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 = 3.04 (sec) , antiderivative size = 543, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-d f+e^2+e f x}{d^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )}-\frac {e}{d^2 x \sqrt {a+b x+c x^2}}+\frac {1}{d x^2 \sqrt {a+b x+c x^2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2} d}-\frac {f \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {e \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d^2}-\frac {\sqrt {a+b x+c x^2}}{a d x}\) |
Input:
Int[1/(x^2*Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
Output:
-(Sqrt[a + b*x + c*x^2]/(a*d*x)) + (b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[ a + b*x + c*x^2])])/(2*a^(3/2)*d) + (e*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt [a + b*x + c*x^2])])/(Sqrt[a]*d^2) - (f*(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*(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*c*d *f - b*e*f + 2*a*f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]])
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 = 2.99 (sec) , antiderivative size = 927, normalized size of antiderivative = 1.71
method | result | size |
risch | \(-\frac {\sqrt {c \,x^{2}+b x +a}}{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 f a \left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 f b \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +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 {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}-\frac {2 f \left (e +\sqrt {-4 d f +e^{2}}\right ) a \sqrt {2}\, \ln \left (\frac {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 f b \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +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 {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}}{2 a d}\) | \(927\) |
default | \(-\frac {4 f \left (-\frac {\sqrt {c \,x^{2}+b x +a}}{a x}+\frac {b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right )}-\frac {4 f^{2} \sqrt {2}\, \ln \left (\frac {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 f b \sqrt {-4 d f +e^{2}}-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {f b \sqrt {-4 d f +e^{2}}-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}+\frac {4 f^{2} \sqrt {2}\, \ln \left (\frac {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}+\frac {\left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}\, \sqrt {4 c {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2}+\frac {4 \left (-c \sqrt {-4 d f +e^{2}}+f b -c e \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 f b \sqrt {-4 d f +e^{2}}+2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-2 b e f -4 d f c +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-f b \sqrt {-4 d f +e^{2}}+\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-b e f -2 d f c +c \,e^{2}}{f^{2}}}}+\frac {16 f^{2} e \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 )^{2} \left (e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {a}}\) | \(955\) |
Input:
int(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
-(c*x^2+b*x+a)^(1/2)/a/d/x-1/2/a/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*(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)/(e+(-4*d*f+e^2)^(1/2 ))*2^(1/2)/((-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f- 2*d*f*c+c*e^2)/f^2)^(1/2)*ln(((-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)* c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2+1/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e)*( x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*((-f*b*(-4*d*f+e^2)^(1/2)+(-4* d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x+1/2*(e+ (-4*d*f+e^2)^(1/2))/f)^2+4/f*(-c*(-4*d*f+e^2)^(1/2)+f*b-c*e)*(x+1/2*(e+(-4 *d*f+e^2)^(1/2))/f)+2*(-f*b*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a* f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))-2*f *(e+(-4*d*f+e^2)^(1/2))*a/(-4*d*f+e^2)^(1/2)/(-e+(-4*d*f+e^2)^(1/2))*2^(1/ 2)/((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c *e^2)/f^2)^(1/2)*ln(((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^ 2-b*e*f-2*d*f*c+c*e^2)/f^2+(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x-1/2/f*(-e+( -4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2 )*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)*(4*c*(x-1/2/f*(-e+(-4*d*f+e^ 2)^(1/2)))^2+4*(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x-1/2/f*(-e+(-4*d*f+e^2)^ (1/2)))+2*(f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d *f*c+c*e^2)/f^2)^(1/2))/(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))))
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^{2} \sqrt {a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:
integrate(1/x**2/(c*x**2+b*x+a)**(1/2)/(f*x**2+e*x+d),x)
Output:
Integral(1/(x**2*sqrt(a + b*x + c*x**2)*(d + e*x + f*x**2)), x)
\[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int { \frac {1}{\sqrt {c x^{2} + b x + a} {\left (f x^{2} + e x + d\right )} x^{2}} \,d x } \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="maxima")
Output:
integrate(1/(sqrt(c*x^2 + b*x + a)*(f*x^2 + e*x + d)*x^2), x)
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^2\,\sqrt {c\,x^2+b\,x+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:
int(1/(x^2*(a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)),x)
Output:
int(1/(x^2*(a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)), x)
\[ \int \frac {1}{x^2 \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^{2} \sqrt {c \,x^{2}+b x +a}\, \left (f \,x^{2}+e x +d \right )}d x \] Input:
int(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)
Output:
int(1/x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)