\(\int (c+d x)^2 \sqrt {a+b x^2} (A+B x+C x^2) \, dx\) [19]

Optimal result

Integrand size = 29, antiderivative size = 264 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {\left (8 A b^2 c^2+a \left (a C d^2-2 b \left (c^2 C+2 B c d+A d^2\right )\right )\right ) x \sqrt {a+b x^2}}{16 b^2}+\frac {(b c (B c+2 A d)-a d (2 c C+B d)) \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac {\left (a C d^2-2 b \left (c^2 C+2 B c d+A d^2\right )\right ) x \left (a+b x^2\right )^{3/2}}{8 b^2}+\frac {C d^2 x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {d (2 c C+B d) \left (a+b x^2\right )^{5/2}}{5 b^2}+\frac {a \left (8 A b^2 c^2+a \left (a C d^2-2 b \left (c^2 C+2 B c d+A d^2\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

1/16*(8*A*b^2*c^2+a*(a*C*d^2-2*b*(A*d^2+2*B*c*d+C*c^2)))*x*(b*x^2+a)^(1/2) 
/b^2+1/3*(b*c*(2*A*d+B*c)-a*d*(B*d+2*C*c))*(b*x^2+a)^(3/2)/b^2-1/8*(a*C*d^ 
2-2*b*(A*d^2+2*B*c*d+C*c^2))*x*(b*x^2+a)^(3/2)/b^2+1/6*C*d^2*x^3*(b*x^2+a) 
^(3/2)/b+1/5*d*(B*d+2*C*c)*(b*x^2+a)^(5/2)/b^2+1/16*a*(8*A*b^2*c^2+a*(a*C* 
d^2-2*b*(A*d^2+2*B*c*d+C*c^2)))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.93 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {\sqrt {a+b x^2} \left (-a^2 d (64 c C+32 B d+15 C d x)+2 a b \left (5 A d (16 c+3 d x)+C x \left (15 c^2+16 c d x+5 d^2 x^2\right )+B \left (40 c^2+30 c d x+8 d^2 x^2\right )\right )+4 b^2 x \left (5 A \left (6 c^2+8 c d x+3 d^2 x^2\right )+x \left (2 B \left (10 c^2+15 c d x+6 d^2 x^2\right )+C x \left (15 c^2+24 c d x+10 d^2 x^2\right )\right )\right )\right )}{240 b^2}-\frac {a \left (2 A b \left (4 b c^2-a d^2\right )+a \left (a C d^2-2 b c (c C+2 B d)\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{5/2}} \] Input:

Integrate[(c + d*x)^2*Sqrt[a + b*x^2]*(A + B*x + C*x^2),x]
 

Output:

(Sqrt[a + b*x^2]*(-(a^2*d*(64*c*C + 32*B*d + 15*C*d*x)) + 2*a*b*(5*A*d*(16 
*c + 3*d*x) + C*x*(15*c^2 + 16*c*d*x + 5*d^2*x^2) + B*(40*c^2 + 30*c*d*x + 
 8*d^2*x^2)) + 4*b^2*x*(5*A*(6*c^2 + 8*c*d*x + 3*d^2*x^2) + x*(2*B*(10*c^2 
 + 15*c*d*x + 6*d^2*x^2) + C*x*(15*c^2 + 24*c*d*x + 10*d^2*x^2)))))/(240*b 
^2) - (a*(2*A*b*(4*b*c^2 - a*d^2) + a*(a*C*d^2 - 2*b*c*(c*C + 2*B*d)))*Log 
[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(5/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2185, 27, 687, 27, 676, 211, 224, 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.

\(\displaystyle \int \sqrt {a+b x^2} (c+d x)^2 \left (A+B x+C x^2\right ) \, dx\)

\(\Big \downarrow \) 2185

\(\displaystyle \frac {\int 3 d (c+d x)^2 ((2 A b-a C) d-b (c C-2 B d) x) \sqrt {b x^2+a}dx}{6 b d^2}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (c+d x)^2 ((2 A b-a C) d-b (c C-2 B d) x) \sqrt {b x^2+a}dx}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 687

\(\displaystyle \frac {\frac {\int b (c+d x) \left (d (10 A b c-3 a C c-4 a B d)+\left (5 (2 A b-a C) d^2-2 b c (c C-2 B d)\right ) x\right ) \sqrt {b x^2+a}dx}{5 b}-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int (c+d x) \left (d (10 A b c-3 a C c-4 a B d)+\left (5 (2 A b-a C) d^2-2 b c (c C-2 B d)\right ) x\right ) \sqrt {b x^2+a}dx-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 d \left (2 A b \left (4 b c^2-a d^2\right )+a \left (a C d^2-2 b c (2 B d+c C)\right )\right ) \int \sqrt {b x^2+a}dx}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (B d+2 c C)+b c \left (-10 A d^2-2 B c d+c^2 C\right )\right )}{3 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (5 d^2 (2 A b-a C)-2 b c (c C-2 B d)\right )}{4 b}\right )-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 d \left (2 A b \left (4 b c^2-a d^2\right )+a \left (a C d^2-2 b c (2 B d+c C)\right )\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (B d+2 c C)+b c \left (-10 A d^2-2 B c d+c^2 C\right )\right )}{3 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (5 d^2 (2 A b-a C)-2 b c (c C-2 B d)\right )}{4 b}\right )-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 d \left (2 A b \left (4 b c^2-a d^2\right )+a \left (a C d^2-2 b c (2 B d+c C)\right )\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (B d+2 c C)+b c \left (-10 A d^2-2 B c d+c^2 C\right )\right )}{3 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (5 d^2 (2 A b-a C)-2 b c (c C-2 B d)\right )}{4 b}\right )-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 d \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (2 A b \left (4 b c^2-a d^2\right )+a \left (a C d^2-2 b c (2 B d+c C)\right )\right )}{4 b}-\frac {2 \left (a+b x^2\right )^{3/2} \left (2 a d^2 (B d+2 c C)+b c \left (-10 A d^2-2 B c d+c^2 C\right )\right )}{3 b}+\frac {d x \left (a+b x^2\right )^{3/2} \left (5 d^2 (2 A b-a C)-2 b c (c C-2 B d)\right )}{4 b}\right )-\frac {1}{5} \left (a+b x^2\right )^{3/2} (c+d x)^2 (c C-2 B d)}{2 b d}+\frac {C \left (a+b x^2\right )^{3/2} (c+d x)^3}{6 b d}\)

Input:

Int[(c + d*x)^2*Sqrt[a + b*x^2]*(A + B*x + C*x^2),x]
 

Output:

(C*(c + d*x)^3*(a + b*x^2)^(3/2))/(6*b*d) + (-1/5*((c*C - 2*B*d)*(c + d*x) 
^2*(a + b*x^2)^(3/2)) + ((-2*(2*a*d^2*(2*c*C + B*d) + b*c*(c^2*C - 2*B*c*d 
 - 10*A*d^2))*(a + b*x^2)^(3/2))/(3*b) + (d*(5*(2*A*b - a*C)*d^2 - 2*b*c*( 
c*C - 2*B*d))*x*(a + b*x^2)^(3/2))/(4*b) + (5*d*(2*A*b*(4*b*c^2 - a*d^2) + 
 a*(a*C*d^2 - 2*b*c*(c*C + 2*B*d)))*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(S 
qrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*b))/5)/(2*b*d)
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 687

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2)) 
), x] + Simp[1/(c*(m + 2*p + 2))   Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*Simp 
[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x 
] /; FreeQ[{a, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && 
 (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && Eq 
Q[f, 0])
 

rule 2185

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x 
)^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p 
)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d 
, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && 
True) &&  !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 
 1/2, 0]))
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.02

method	result	size
default	\(A \,c^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+\frac {c \left (2 A d +B c \right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+d \left (B d +2 C c \right ) \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+C \,d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\)	\(268\)
risch	\(\frac {\left (40 C \,d^{2} b^{2} x^{5}+48 B \,b^{2} d^{2} x^{4}+96 C \,b^{2} c d \,x^{4}+60 A \,b^{2} d^{2} x^{3}+120 B \,b^{2} c d \,x^{3}+10 a C \,d^{2} b \,x^{3}+60 C \,b^{2} c^{2} x^{3}+160 A \,b^{2} c d \,x^{2}+16 B a \,d^{2} b \,x^{2}+80 B \,b^{2} c^{2} x^{2}+32 C a c d b \,x^{2}+30 A a \,d^{2} x b +120 A \,b^{2} c^{2} x +60 B a c d x b -15 C \,a^{2} d^{2} x +30 C a \,c^{2} x b +160 A a b c d -32 B \,a^{2} d^{2}+80 B a b \,c^{2}-64 C \,a^{2} c d \right ) \sqrt {b \,x^{2}+a}}{240 b^{2}}-\frac {a \left (2 A a b \,d^{2}-8 A \,b^{2} c^{2}+4 B a c d b -a^{2} C \,d^{2}+2 a b \,c^{2} C \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\)	\(284\)

Input:

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

Output:

A*c^2*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)))+ 
1/3*c*(2*A*d+B*c)/b*(b*x^2+a)^(3/2)+d*(B*d+2*C*c)*(1/5*x^2*(b*x^2+a)^(3/2) 
/b-2/15*a/b^2*(b*x^2+a)^(3/2))+(A*d^2+2*B*c*d+C*c^2)*(1/4*x*(b*x^2+a)^(3/2 
)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1 
/2))))+C*d^2*(1/6*x^3*(b*x^2+a)^(3/2)/b-1/2*a/b*(1/4*x*(b*x^2+a)^(3/2)/b-1 
/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))) 
))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.25 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\left [\frac {15 \, {\left (4 \, B a^{2} b c d + 2 \, {\left (C a^{2} b - 4 \, A a b^{2}\right )} c^{2} - {\left (C a^{3} - 2 \, A a^{2} b\right )} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, C b^{3} d^{2} x^{5} + 80 \, B a b^{2} c^{2} - 32 \, B a^{2} b d^{2} + 48 \, {\left (2 \, C b^{3} c d + B b^{3} d^{2}\right )} x^{4} + 10 \, {\left (6 \, C b^{3} c^{2} + 12 \, B b^{3} c d + {\left (C a b^{2} + 6 \, A b^{3}\right )} d^{2}\right )} x^{3} - 32 \, {\left (2 \, C a^{2} b - 5 \, A a b^{2}\right )} c d + 16 \, {\left (5 \, B b^{3} c^{2} + B a b^{2} d^{2} + 2 \, {\left (C a b^{2} + 5 \, A b^{3}\right )} c d\right )} x^{2} + 15 \, {\left (4 \, B a b^{2} c d + 2 \, {\left (C a b^{2} + 4 \, A b^{3}\right )} c^{2} - {\left (C a^{2} b - 2 \, A a b^{2}\right )} d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, \frac {15 \, {\left (4 \, B a^{2} b c d + 2 \, {\left (C a^{2} b - 4 \, A a b^{2}\right )} c^{2} - {\left (C a^{3} - 2 \, A a^{2} b\right )} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (40 \, C b^{3} d^{2} x^{5} + 80 \, B a b^{2} c^{2} - 32 \, B a^{2} b d^{2} + 48 \, {\left (2 \, C b^{3} c d + B b^{3} d^{2}\right )} x^{4} + 10 \, {\left (6 \, C b^{3} c^{2} + 12 \, B b^{3} c d + {\left (C a b^{2} + 6 \, A b^{3}\right )} d^{2}\right )} x^{3} - 32 \, {\left (2 \, C a^{2} b - 5 \, A a b^{2}\right )} c d + 16 \, {\left (5 \, B b^{3} c^{2} + B a b^{2} d^{2} + 2 \, {\left (C a b^{2} + 5 \, A b^{3}\right )} c d\right )} x^{2} + 15 \, {\left (4 \, B a b^{2} c d + 2 \, {\left (C a b^{2} + 4 \, A b^{3}\right )} c^{2} - {\left (C a^{2} b - 2 \, A a b^{2}\right )} d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \] Input:

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

Output:

[1/480*(15*(4*B*a^2*b*c*d + 2*(C*a^2*b - 4*A*a*b^2)*c^2 - (C*a^3 - 2*A*a^2 
*b)*d^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*C 
*b^3*d^2*x^5 + 80*B*a*b^2*c^2 - 32*B*a^2*b*d^2 + 48*(2*C*b^3*c*d + B*b^3*d 
^2)*x^4 + 10*(6*C*b^3*c^2 + 12*B*b^3*c*d + (C*a*b^2 + 6*A*b^3)*d^2)*x^3 - 
32*(2*C*a^2*b - 5*A*a*b^2)*c*d + 16*(5*B*b^3*c^2 + B*a*b^2*d^2 + 2*(C*a*b^ 
2 + 5*A*b^3)*c*d)*x^2 + 15*(4*B*a*b^2*c*d + 2*(C*a*b^2 + 4*A*b^3)*c^2 - (C 
*a^2*b - 2*A*a*b^2)*d^2)*x)*sqrt(b*x^2 + a))/b^3, 1/240*(15*(4*B*a^2*b*c*d 
 + 2*(C*a^2*b - 4*A*a*b^2)*c^2 - (C*a^3 - 2*A*a^2*b)*d^2)*sqrt(-b)*arctan( 
sqrt(-b)*x/sqrt(b*x^2 + a)) + (40*C*b^3*d^2*x^5 + 80*B*a*b^2*c^2 - 32*B*a^ 
2*b*d^2 + 48*(2*C*b^3*c*d + B*b^3*d^2)*x^4 + 10*(6*C*b^3*c^2 + 12*B*b^3*c* 
d + (C*a*b^2 + 6*A*b^3)*d^2)*x^3 - 32*(2*C*a^2*b - 5*A*a*b^2)*c*d + 16*(5* 
B*b^3*c^2 + B*a*b^2*d^2 + 2*(C*a*b^2 + 5*A*b^3)*c*d)*x^2 + 15*(4*B*a*b^2*c 
*d + 2*(C*a*b^2 + 4*A*b^3)*c^2 - (C*a^2*b - 2*A*a*b^2)*d^2)*x)*sqrt(b*x^2 
+ a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.83 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {C d^{2} x^{5}}{6} + \frac {x^{4} \left (B b d^{2} + 2 C b c d\right )}{5 b} + \frac {x^{3} \left (A b d^{2} + 2 B b c d + \frac {C a d^{2}}{6} + C b c^{2}\right )}{4 b} + \frac {x^{2} \cdot \left (2 A b c d + B a d^{2} + B b c^{2} + 2 C a c d - \frac {4 a \left (B b d^{2} + 2 C b c d\right )}{5 b}\right )}{3 b} + \frac {x \left (A a d^{2} + A b c^{2} + 2 B a c d + C a c^{2} - \frac {3 a \left (A b d^{2} + 2 B b c d + \frac {C a d^{2}}{6} + C b c^{2}\right )}{4 b}\right )}{2 b} + \frac {2 A a c d + B a c^{2} - \frac {2 a \left (2 A b c d + B a d^{2} + B b c^{2} + 2 C a c d - \frac {4 a \left (B b d^{2} + 2 C b c d\right )}{5 b}\right )}{3 b}}{b}\right ) + \left (A a c^{2} - \frac {a \left (A a d^{2} + A b c^{2} + 2 B a c d + C a c^{2} - \frac {3 a \left (A b d^{2} + 2 B b c d + \frac {C a d^{2}}{6} + C b c^{2}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (A c^{2} x + \frac {C d^{2} x^{5}}{5} + \frac {x^{4} \left (B d^{2} + 2 C c d\right )}{4} + \frac {x^{3} \left (A d^{2} + 2 B c d + C c^{2}\right )}{3} + \frac {x^{2} \cdot \left (2 A c d + B c^{2}\right )}{2}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((d*x+c)**2*(b*x**2+a)**(1/2)*(C*x**2+B*x+A),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(C*d**2*x**5/6 + x**4*(B*b*d**2 + 2*C*b*c*d)/( 
5*b) + x**3*(A*b*d**2 + 2*B*b*c*d + C*a*d**2/6 + C*b*c**2)/(4*b) + x**2*(2 
*A*b*c*d + B*a*d**2 + B*b*c**2 + 2*C*a*c*d - 4*a*(B*b*d**2 + 2*C*b*c*d)/(5 
*b))/(3*b) + x*(A*a*d**2 + A*b*c**2 + 2*B*a*c*d + C*a*c**2 - 3*a*(A*b*d**2 
 + 2*B*b*c*d + C*a*d**2/6 + C*b*c**2)/(4*b))/(2*b) + (2*A*a*c*d + B*a*c**2 
 - 2*a*(2*A*b*c*d + B*a*d**2 + B*b*c**2 + 2*C*a*c*d - 4*a*(B*b*d**2 + 2*C* 
b*c*d)/(5*b))/(3*b))/b) + (A*a*c**2 - a*(A*a*d**2 + A*b*c**2 + 2*B*a*c*d + 
 C*a*c**2 - 3*a*(A*b*d**2 + 2*B*b*c*d + C*a*d**2/6 + C*b*c**2)/(4*b))/(2*b 
))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), 
(x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), (sqrt(a)*(A*c**2*x + C*d**2*x** 
5/5 + x**4*(B*d**2 + 2*C*c*d)/4 + x**3*(A*d**2 + 2*B*c*d + C*c**2)/3 + x** 
2*(2*A*c*d + B*c**2)/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.16 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C d^{2} x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} A c^{2} x - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a d^{2} x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C a^{2} d^{2} x}{16 \, b^{2}} + \frac {A a c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + \frac {C a^{3} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c^{2}}{3 \, b} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A c d}{3 \, b} + \frac {{\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{2}}{5 \, b} + \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} x}{4 \, b} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} \sqrt {b x^{2} + a} a x}{8 \, b} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} a}{15 \, b^{2}} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(1/2)*(C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

1/6*(b*x^2 + a)^(3/2)*C*d^2*x^3/b + 1/2*sqrt(b*x^2 + a)*A*c^2*x - 1/8*(b*x 
^2 + a)^(3/2)*C*a*d^2*x/b^2 + 1/16*sqrt(b*x^2 + a)*C*a^2*d^2*x/b^2 + 1/2*A 
*a*c^2*arcsinh(b*x/sqrt(a*b))/sqrt(b) + 1/16*C*a^3*d^2*arcsinh(b*x/sqrt(a* 
b))/b^(5/2) + 1/3*(b*x^2 + a)^(3/2)*B*c^2/b + 2/3*(b*x^2 + a)^(3/2)*A*c*d/ 
b + 1/5*(2*C*c*d + B*d^2)*(b*x^2 + a)^(3/2)*x^2/b + 1/4*(C*c^2 + 2*B*c*d + 
 A*d^2)*(b*x^2 + a)^(3/2)*x/b - 1/8*(C*c^2 + 2*B*c*d + A*d^2)*sqrt(b*x^2 + 
 a)*a*x/b - 1/8*(C*c^2 + 2*B*c*d + A*d^2)*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/ 
2) - 2/15*(2*C*c*d + B*d^2)*(b*x^2 + a)^(3/2)*a/b^2
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.19 \[ \int (c+d x)^2 \sqrt {a+b x^2} \left (A+B x+C x^2\right ) \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, C d^{2} x + \frac {6 \, {\left (2 \, C b^{4} c d + B b^{4} d^{2}\right )}}{b^{4}}\right )} x + \frac {5 \, {\left (6 \, C b^{4} c^{2} + 12 \, B b^{4} c d + C a b^{3} d^{2} + 6 \, A b^{4} d^{2}\right )}}{b^{4}}\right )} x + \frac {8 \, {\left (5 \, B b^{4} c^{2} + 2 \, C a b^{3} c d + 10 \, A b^{4} c d + B a b^{3} d^{2}\right )}}{b^{4}}\right )} x + \frac {15 \, {\left (2 \, C a b^{3} c^{2} + 8 \, A b^{4} c^{2} + 4 \, B a b^{3} c d - C a^{2} b^{2} d^{2} + 2 \, A a b^{3} d^{2}\right )}}{b^{4}}\right )} x + \frac {16 \, {\left (5 \, B a b^{3} c^{2} - 4 \, C a^{2} b^{2} c d + 10 \, A a b^{3} c d - 2 \, B a^{2} b^{2} d^{2}\right )}}{b^{4}}\right )} + \frac {{\left (2 \, C a^{2} b c^{2} - 8 \, A a b^{2} c^{2} + 4 \, B a^{2} b c d - C a^{3} d^{2} + 2 \, A a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(1/2)*(C*x^2+B*x+A),x, algorithm="giac")
 

Output:

1/240*sqrt(b*x^2 + a)*((2*((4*(5*C*d^2*x + 6*(2*C*b^4*c*d + B*b^4*d^2)/b^4 
)*x + 5*(6*C*b^4*c^2 + 12*B*b^4*c*d + C*a*b^3*d^2 + 6*A*b^4*d^2)/b^4)*x + 
8*(5*B*b^4*c^2 + 2*C*a*b^3*c*d + 10*A*b^4*c*d + B*a*b^3*d^2)/b^4)*x + 15*( 
2*C*a*b^3*c^2 + 8*A*b^4*c^2 + 4*B*a*b^3*c*d - C*a^2*b^2*d^2 + 2*A*a*b^3*d^ 
2)/b^4)*x + 16*(5*B*a*b^3*c^2 - 4*C*a^2*b^2*c*d + 10*A*a*b^3*c*d - 2*B*a^2 
*b^2*d^2)/b^4) + 1/16*(2*C*a^2*b*c^2 - 8*A*a*b^2*c^2 + 4*B*a^2*b*c*d - C*a 
^3*d^2 + 2*A*a^2*b*d^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

int((a + b*x^2)^(1/2)*(c + d*x)^2*(A + B*x + C*x^2),x)
 

Output:

int((a + b*x^2)^(1/2)*(c + d*x)^2*(A + B*x + C*x^2), x)
 

Reduce [F]

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

int((d*x+c)^2*(b*x^2+a)^(1/2)*(C*x^2+B*x+A),x)
 

Output:

int((d*x+c)^2*(b*x^2+a)^(1/2)*(C*x^2+B*x+A),x)