\(\int \frac {x^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx\) [13]

Optimal result

Integrand size = 25, antiderivative size = 255 \[ \int \frac {x^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx=\frac {((2 c d-b e) (4 c d+b e)-2 c e (2 c d+b e) x) \sqrt {a+b x+c x^2}}{8 c^2 e^3}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)-8 c^2 d e (b d-a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} e^4}+\frac {d^2 \sqrt {c d^2-b d e+a e^2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^4} \] Output:

1/8*((-b*e+2*c*d)*(b*e+4*c*d)-2*c*e*(b*e+2*c*d)*x)*(c*x^2+b*x+a)^(1/2)/c^2 
/e^3+1/3*(c*x^2+b*x+a)^(3/2)/c/e-1/16*(16*c^3*d^3-b^3*e^3-2*b*c*e^2*(-2*a* 
e+b*d)-8*c^2*d*e*(-a*e+b*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^( 
1/2))/c^(5/2)/e^4+d^2*(a*e^2-b*d*e+c*d^2)^(1/2)*arctanh(1/2*(b*d-2*a*e+(-b 
*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^4
 

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.95 \[ \int \frac {x^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx=\frac {2 \sqrt {c} e \sqrt {a+x (b+c x)} \left (-3 b^2 e^2+2 c e (-3 b d+4 a e+b e x)+4 c^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+96 c^{5/2} d^2 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )+3 \left (16 c^3 d^3-b^3 e^3-2 b c e^2 (b d-2 a e)+8 c^2 d e (-b d+a e)\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{48 c^{5/2} e^4} \] Input:

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

Output:

(2*Sqrt[c]*e*Sqrt[a + x*(b + c*x)]*(-3*b^2*e^2 + 2*c*e*(-3*b*d + 4*a*e + b 
*e*x) + 4*c^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) + 96*c^(5/2)*d^2*Sqrt[-(c*d^2 
) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x*(b + c*x)])/Sq 
rt[-(c*d^2) + e*(b*d - a*e)]] + 3*(16*c^3*d^3 - b^3*e^3 - 2*b*c*e^2*(b*d - 
 2*a*e) + 8*c^2*d*e*(-(b*d) + a*e))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*( 
b + c*x)]])/(48*c^(5/2)*e^4)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {1267, 27, 1231, 27, 1269, 1092, 219, 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.

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

\(\Big \downarrow \) 1267

\(\displaystyle \frac {\int -\frac {3 e (b d+(2 c d+b e) x) \sqrt {c x^2+b x+a}}{2 (d+e x)}dx}{3 c e^2}+\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\int \frac {(b d+(2 c d+b e) x) \sqrt {c x^2+b x+a}}{d+e x}dx}{2 c e}\)

\(\Big \downarrow \) 1231

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {-\frac {\int -\frac {d (2 c d-b e) \left (e b^2+4 c d b-4 a c e\right )+\left (16 c^3 d^3-8 c^2 e (b d-a e) d-b^3 e^3-2 b c e^2 (b d-2 a e)\right ) x}{2 (d+e x) \sqrt {c x^2+b x+a}}dx}{4 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\int \frac {d (2 c d-b e) \left (e b^2+4 c d b-4 a c e\right )+\left (16 c^3 d^3-8 c^2 e (b d-a e) d-b^3 e^3-2 b c e^2 (b d-2 a e)\right ) x}{(d+e x) \sqrt {c x^2+b x+a}}dx}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\frac {\left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{e}-\frac {16 c^2 d^2 \left (a e^2-b d e+c d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\frac {2 \left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{e}-\frac {16 c^2 d^2 \left (a e^2-b d e+c d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right )}{\sqrt {c} e}-\frac {16 c^2 d^2 \left (a e^2-b d e+c d^2\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\frac {32 c^2 d^2 \left (a e^2-b d e+c d^2\right ) \int \frac {1}{4 \left (c d^2-b e d+a e^2\right )-\frac {(b d-2 a e+(2 c d-b e) x)^2}{c x^2+b x+a}}d\left (-\frac {b d-2 a e+(2 c d-b e) x}{\sqrt {c x^2+b x+a}}\right )}{e}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right )}{\sqrt {c} e}}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2}}{3 c e}-\frac {\frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-8 c^2 d e (b d-a e)-2 b c e^2 (b d-2 a e)-b^3 e^3+16 c^3 d^3\right )}{\sqrt {c} e}-\frac {16 c^2 d^2 \sqrt {a e^2-b d e+c d^2} \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e}}{8 c e^2}-\frac {\sqrt {a+b x+c x^2} ((2 c d-b e) (b e+4 c d)-2 c e x (b e+2 c d))}{4 c e^2}}{2 c e}\)

Input:

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

Output:

(a + b*x + c*x^2)^(3/2)/(3*c*e) - (-1/4*(((2*c*d - b*e)*(4*c*d + b*e) - 2* 
c*e*(2*c*d + b*e)*x)*Sqrt[a + b*x + c*x^2])/(c*e^2) + (((16*c^3*d^3 - b^3* 
e^3 - 2*b*c*e^2*(b*d - 2*a*e) - 8*c^2*d*e*(b*d - a*e))*ArcTanh[(b + 2*c*x) 
/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*e) - (16*c^2*d^2*Sqrt[c*d^2 
- b*d*e + a*e^2]*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b 
*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/e)/(8*c*e^2))/(2*c*e)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 1231

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1267

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b 
*x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m 
 + n + 2*p + 1))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m 
 + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - g^n*(d 
+ e*x)^(n - 2)*(b*d*e*(p + 1) + a*e^2*(m + n - 1) - c*d^2*(m + n + 2*p + 1) 
 - e*(2*c*d - b*e)*(m + n + p)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, p}, x] && IGtQ[n, 1] && IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.38

method	result	size
risch	\(\frac {\left (8 c^{2} e^{2} x^{2}+2 e^{2} x b c -12 c^{2} d e x +8 a c \,e^{2}-3 b^{2} e^{2}-6 b c d e +24 c^{2} d^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{24 c^{2} e^{3}}-\frac {\frac {\left (4 a b c \,e^{3}+8 d \,e^{2} a \,c^{2}-b^{3} e^{3}-2 d \,e^{2} b^{2} c -8 d^{2} e b \,c^{2}+16 d^{3} c^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}+\frac {16 d^{2} \left (a \,e^{2}-b d e +c \,d^{2}\right ) c^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{16 e^{3} c^{2}}\)	\(353\)
default	\(\frac {\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}}{e}+\frac {d^{2} \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (a \,e^{2}-b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{3}}-\frac {d \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{e^{2}}\)	\(495\)

Input:

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

Output:

1/24*(8*c^2*e^2*x^2+2*b*c*e^2*x-12*c^2*d*e*x+8*a*c*e^2-3*b^2*e^2-6*b*c*d*e 
+24*c^2*d^2)/c^2*(c*x^2+b*x+a)^(1/2)/e^3-1/16/e^3/c^2*((4*a*b*c*e^3+8*a*c^ 
2*d*e^2-b^3*e^3-2*b^2*c*d*e^2-8*b*c^2*d^2*e+16*c^3*d^3)/e*ln((1/2*b+c*x)/c 
^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+16*d^2*(a*e^2-b*d*e+c*d^2)*c^2/e^2/((a 
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*( 
x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e 
)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))
 

Fricas [A] (verification not implemented)

Time = 101.97 (sec) , antiderivative size = 1459, normalized size of antiderivative = 5.72 \[ \int \frac {x^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/96*(48*sqrt(c*d^2 - b*d*e + a*e^2)*c^3*d^2*log((8*a*b*d*e - 8*a^2*e^2 - 
 (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*s 
qrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b 
*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d 
*e*x + d^2)) + 3*(16*c^3*d^3 - 8*b*c^2*d^2*e - 2*(b^2*c - 4*a*c^2)*d*e^2 - 
 (b^3 - 4*a*b*c)*e^3)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^ 
2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*c^3*e^3*x^2 + 24*c^3*d^2* 
e - 6*b*c^2*d*e^2 - (3*b^2*c - 8*a*c^2)*e^3 - 2*(6*c^3*d*e^2 - b*c^2*e^3)* 
x)*sqrt(c*x^2 + b*x + a))/(c^3*e^4), 1/96*(96*sqrt(-c*d^2 + b*d*e - a*e^2) 
*c^3*d^2*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqrt(c*x^2 + b*x + a)*(b 
*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b* 
c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) + 3*(16*c^3*d^3 - 
 8*b*c^2*d^2*e - 2*(b^2*c - 4*a*c^2)*d*e^2 - (b^3 - 4*a*b*c)*e^3)*sqrt(c)* 
log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt( 
c) - 4*a*c) + 4*(8*c^3*e^3*x^2 + 24*c^3*d^2*e - 6*b*c^2*d*e^2 - (3*b^2*c - 
 8*a*c^2)*e^3 - 2*(6*c^3*d*e^2 - b*c^2*e^3)*x)*sqrt(c*x^2 + b*x + a))/(c^3 
*e^4), 1/48*(24*sqrt(c*d^2 - b*d*e + a*e^2)*c^3*d^2*log((8*a*b*d*e - 8*a^2 
*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 
 - 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c 
*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e^2*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^2*(c*x^2+b*x+a)^(1/2)/(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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F(-2)]

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

integrate(x^2*(c*x^2+b*x+a)^(1/2)/(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. \[ \int \frac {x^2 \sqrt {a+b x+c x^2}}{d+e x} \, dx=\int \frac {x^2\,\sqrt {c\,x^2+b\,x+a}}{d+e\,x} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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