\(\int \frac {x^2}{\sqrt {a+b x+c x^2} (d+e x+f x^2)} \, dx\) [142]

Optimal result

Integrand size = 30, antiderivative size = 463 \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f}-\frac {\left (e^2-2 d f-e \sqrt {e^2-4 d f}\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}-\frac {\left (2 d f-e \left (e+\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:

arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(1/2)/f-1/2*(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)/f/(-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)-1/ 
2*(2*d*f-e*(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)/f/(- 
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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.64 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.69 \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {-\frac {\log \left (f \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{\sqrt {c}}+\text {RootSum}\left [b^2 d-a b e+a^2 f-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {b d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+e \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{2 b \sqrt {c} d-a \sqrt {c} e-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+3 \sqrt {c} e \text {$\#$1}^2-2 f \text {$\#$1}^3}\&\right ]}{f} \] Input:

Integrate[x^2/(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
 

Output:

(-(Log[f*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]/Sqrt[c]) + 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*d*Log[-(Sqrt[c] 
*x) + Sqrt[a + b*x + c*x^2] - #1] - a*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + 
c*x^2] - #1] - 2*Sqrt[c]*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]* 
#1 + e*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) & ])/f
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2143, 25, 1092, 219, 1365, 1154, 219}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx\)

\(\Big \downarrow \) 2143

\(\displaystyle \frac {\int -\frac {d+e x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {\int \frac {1}{\sqrt {c x^2+b x+a}}dx}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {1}{\sqrt {c x^2+b x+a}}dx}{f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {2 \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f}-\frac {\int \frac {d+e x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}\)

\(\Big \downarrow \) 1365

\(\displaystyle \frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f}-\frac {-\frac {\left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}-\frac {\left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}}{f}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f}-\frac {\frac {2 \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}+\frac {2 \left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} f}-\frac {\frac {\left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\left (2 d f-e \left (\sqrt {e^2-4 d f}+e\right )\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}}{f}\)

Input:

Int[x^2/(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2)),x]
 

Output:

ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]/(Sqrt[c]*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]*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]]) + ((2*d*f - e*(e + Sqrt[e^2 - 4*d*f]))*ArcTan 
h[(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]*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]]))/f
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1365

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + ( 
e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Sim 
p[(2*c*g - h*(b - q))/q   Int[1/((b - q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x] 
, x] - Simp[(2*c*g - h*(b + q))/q   Int[1/((b + q + 2*c*x)*Sqrt[d + e*x + f 
*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0 
] && NeQ[e^2 - 4*d*f, 0] && PosQ[b^2 - 4*a*c]
 

rule 2143

Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_ 
.)*(x_)^2]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C 
 = Coeff[Px, x, 2]}, Simp[C/c   Int[1/Sqrt[d + e*x + f*x^2], x], x] + Simp[ 
1/c   Int[(A*c - a*C + (B*c - b*C)*x)/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x 
^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && PolyQ[Px, x, 2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(841\) vs. \(2(407)=814\).

Time = 2.76 (sec) , antiderivative size = 842, normalized size of antiderivative = 1.82

method	result	size
default	\(\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{f \sqrt {c}}+\frac {\left (e \sqrt {-4 d f +e^{2}}+2 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 )}{2 f^{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 {\left (e^{2}-2 d f +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 )}{2 f^{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}}}}\)	\(842\)

Input:

int(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

Output:

1/f*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+1/2/f^2*(e*(-4*d*f 
+e^2)^(1/2)+2*d*f-e^2)/(-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))))+1/2/f^2*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2))/( 
-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))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^{2}}{\sqrt {a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \] Input:

integrate(x**2/(c*x**2+b*x+a)**(1/2)/(f*x**2+e*x+d),x)
 

Output:

Integral(x**2/(sqrt(a + b*x + c*x**2)*(d + e*x + f*x**2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for 
 more deta
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^2}{\sqrt {c\,x^2+b\,x+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:

int(x^2/((a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)),x)
 

Output:

int(x^2/((a + b*x + c*x^2)^(1/2)*(d + e*x + f*x^2)), x)
 

Reduce [F]

\[ \int \frac {x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {x^{2}}{\sqrt {c \,x^{2}+b x +a}\, \left (f \,x^{2}+e x +d \right )}d x \] Input:

int(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)
 

Output:

int(x^2/(c*x^2+b*x+a)^(1/2)/(f*x^2+e*x+d),x)