\(\int x^2 \sqrt {a+b x^2} (A+B x+C x^2+D x^3) \, dx\) [62]

Optimal result

Integrand size = 30, antiderivative size = 194 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a (2 A b-a C) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 A b-a C) x^3 \sqrt {a+b x^2}}{8 b}-\frac {a (b B-a D) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {C x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {(b B-2 a D) \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {D \left (a+b x^2\right )^{7/2}}{7 b^3}-\frac {a^2 (2 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

1/16*a*(2*A*b-C*a)*x*(b*x^2+a)^(1/2)/b^2+1/8*(2*A*b-C*a)*x^3*(b*x^2+a)^(1/ 
2)/b-1/3*a*(B*b-D*a)*(b*x^2+a)^(3/2)/b^3+1/6*C*x^3*(b*x^2+a)^(3/2)/b+1/5*( 
B*b-2*D*a)*(b*x^2+a)^(5/2)/b^3+1/7*D*(b*x^2+a)^(7/2)/b^3-1/16*a^2*(2*A*b-C 
*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.73 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {a+b x^2} \left (128 a^3 D+4 b^3 x^3 \left (105 A+84 B x+70 C x^2+60 D x^3\right )-a^2 b (224 B+x (105 C+64 D x))+2 a b^2 x \left (105 A+x \left (56 B+35 C x+24 D x^2\right )\right )\right )-105 a^2 \sqrt {b} (-2 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{1680 b^3} \] Input:

Integrate[x^2*Sqrt[a + b*x^2]*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

(Sqrt[a + b*x^2]*(128*a^3*D + 4*b^3*x^3*(105*A + 84*B*x + 70*C*x^2 + 60*D* 
x^3) - a^2*b*(224*B + x*(105*C + 64*D*x)) + 2*a*b^2*x*(105*A + x*(56*B + 3 
5*C*x + 24*D*x^2))) - 105*a^2*Sqrt[b]*(-2*A*b + a*C)*Log[-(Sqrt[b]*x) + Sq 
rt[a + b*x^2]])/(1680*b^3)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {2340, 2340, 27, 533, 533, 25, 27, 455, 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.

\(\displaystyle \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx\)

\(\Big \downarrow \) 2340

\(\displaystyle \frac {\int x^2 \sqrt {b x^2+a} \left (7 b C x^2+(7 b B-4 a D) x+7 A b\right )dx}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 2340

\(\displaystyle \frac {\frac {\int 3 b x^2 (7 (2 A b-a C)+2 (7 b B-4 a D) x) \sqrt {b x^2+a}dx}{6 b}+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \int x^2 (7 (2 A b-a C)+2 (7 b B-4 a D) x) \sqrt {b x^2+a}dx+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\int x (4 a (7 b B-4 a D)-35 b (2 A b-a C) x) \sqrt {b x^2+a}dx}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {-\frac {\int -a b (35 (2 A b-a C)+16 (7 b B-4 a D) x) \sqrt {b x^2+a}dx}{4 b}-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {\int a b (35 (2 A b-a C)+16 (7 b B-4 a D) x) \sqrt {b x^2+a}dx}{4 b}-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {1}{4} a \int (35 (2 A b-a C)+16 (7 b B-4 a D) x) \sqrt {b x^2+a}dx-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {1}{4} a \left (35 (2 A b-a C) \int \sqrt {b x^2+a}dx+\frac {16 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{3 b}\right )-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {1}{4} a \left (35 (2 A b-a C) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {16 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{3 b}\right )-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {1}{4} a \left (35 (2 A b-a C) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {16 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{3 b}\right )-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2 x^2 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{5 b}-\frac {\frac {1}{4} a \left (35 (2 A b-a C) \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {16 \left (a+b x^2\right )^{3/2} (7 b B-4 a D)}{3 b}\right )-\frac {35}{4} x \left (a+b x^2\right )^{3/2} (2 A b-a C)}{5 b}\right )+\frac {7}{6} C x^3 \left (a+b x^2\right )^{3/2}}{7 b}+\frac {D x^4 \left (a+b x^2\right )^{3/2}}{7 b}\)

Input:

Int[x^2*Sqrt[a + b*x^2]*(A + B*x + C*x^2 + D*x^3),x]
 

Output:

(D*x^4*(a + b*x^2)^(3/2))/(7*b) + ((7*C*x^3*(a + b*x^2)^(3/2))/6 + ((2*(7* 
b*B - 4*a*D)*x^2*(a + b*x^2)^(3/2))/(5*b) - ((-35*(2*A*b - a*C)*x*(a + b*x 
^2)^(3/2))/4 + (a*((16*(7*b*B - 4*a*D)*(a + b*x^2)^(3/2))/(3*b) + 35*(2*A* 
b - a*C)*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]) 
/(2*Sqrt[b]))))/4)/(5*b))/2)/(7*b)
 

Defintions of rubi rules used

rule 25

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

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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.23

method	result	size
default	\(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+C \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+D \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{7 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )}{7 b}\right )\)	\(238\)

Input:

int(x^2*(b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
 

Output:

A*(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))))+B*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15*a/b^2*(b*x 
^2+a)^(3/2))+C*(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) 
))))+D*(1/7*x^4*(b*x^2+a)^(3/2)/b-4/7*a/b*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15* 
a/b^2*(b*x^2+a)^(3/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.68 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\left [-\frac {105 \, {\left (C a^{3} - 2 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (240 \, D b^{3} x^{6} + 280 \, C b^{3} x^{5} + 48 \, {\left (D a b^{2} + 7 \, B b^{3}\right )} x^{4} + 128 \, D a^{3} - 224 \, B a^{2} b + 70 \, {\left (C a b^{2} + 6 \, A b^{3}\right )} x^{3} - 16 \, {\left (4 \, D a^{2} b - 7 \, B a b^{2}\right )} x^{2} - 105 \, {\left (C a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{3360 \, b^{3}}, -\frac {105 \, {\left (C a^{3} - 2 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (240 \, D b^{3} x^{6} + 280 \, C b^{3} x^{5} + 48 \, {\left (D a b^{2} + 7 \, B b^{3}\right )} x^{4} + 128 \, D a^{3} - 224 \, B a^{2} b + 70 \, {\left (C a b^{2} + 6 \, A b^{3}\right )} x^{3} - 16 \, {\left (4 \, D a^{2} b - 7 \, B a b^{2}\right )} x^{2} - 105 \, {\left (C a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{1680 \, b^{3}}\right ] \] Input:

integrate(x^2*(b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
 

Output:

[-1/3360*(105*(C*a^3 - 2*A*a^2*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a) 
*sqrt(b)*x - a) - 2*(240*D*b^3*x^6 + 280*C*b^3*x^5 + 48*(D*a*b^2 + 7*B*b^3 
)*x^4 + 128*D*a^3 - 224*B*a^2*b + 70*(C*a*b^2 + 6*A*b^3)*x^3 - 16*(4*D*a^2 
*b - 7*B*a*b^2)*x^2 - 105*(C*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^3, - 
1/1680*(105*(C*a^3 - 2*A*a^2*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a) 
) - (240*D*b^3*x^6 + 280*C*b^3*x^5 + 48*(D*a*b^2 + 7*B*b^3)*x^4 + 128*D*a^ 
3 - 224*B*a^2*b + 70*(C*a*b^2 + 6*A*b^3)*x^3 - 16*(4*D*a^2*b - 7*B*a*b^2)* 
x^2 - 105*(C*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.18 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} - \frac {a \left (A a - \frac {3 a \left (A b + \frac {C a}{6}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {C x^{5}}{6} + \frac {D x^{6}}{7} - \frac {2 a \left (B a - \frac {4 a \left (B b + \frac {D a}{7}\right )}{5 b}\right )}{3 b^{2}} + \frac {x^{4} \left (B b + \frac {D a}{7}\right )}{5 b} + \frac {x^{3} \left (A b + \frac {C a}{6}\right )}{4 b} + \frac {x^{2} \left (B a - \frac {4 a \left (B b + \frac {D a}{7}\right )}{5 b}\right )}{3 b} + \frac {x \left (A a - \frac {3 a \left (A b + \frac {C a}{6}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4} + \frac {C x^{5}}{5} + \frac {D x^{6}}{6}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(x**2*(b*x**2+a)**(1/2)*(D*x**3+C*x**2+B*x+A),x)
 

Output:

Piecewise((-a*(A*a - 3*a*(A*b + C*a/6)/(4*b))*Piecewise((log(2*sqrt(b)*sqr 
t(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/ 
(2*b) + sqrt(a + b*x**2)*(C*x**5/6 + D*x**6/7 - 2*a*(B*a - 4*a*(B*b + D*a/ 
7)/(5*b))/(3*b**2) + x**4*(B*b + D*a/7)/(5*b) + x**3*(A*b + C*a/6)/(4*b) + 
 x**2*(B*a - 4*a*(B*b + D*a/7)/(5*b))/(3*b) + x*(A*a - 3*a*(A*b + C*a/6)/( 
4*b))/(2*b)), Ne(b, 0)), (sqrt(a)*(A*x**3/3 + B*x**4/4 + C*x**5/5 + D*x**6 
/6), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.10 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{4}}{7 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x^{3}}{6 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a x^{2}}{35 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x}{4 \, b} - \frac {\sqrt {b x^{2} + a} A a x}{8 \, b} + \frac {C a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} D a^{2}}{105 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a}{15 \, b^{2}} \] Input:

integrate(x^2*(b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
 

Output:

1/7*(b*x^2 + a)^(3/2)*D*x^4/b + 1/6*(b*x^2 + a)^(3/2)*C*x^3/b - 4/35*(b*x^ 
2 + a)^(3/2)*D*a*x^2/b^2 + 1/5*(b*x^2 + a)^(3/2)*B*x^2/b - 1/8*(b*x^2 + a) 
^(3/2)*C*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*C*a^2*x/b^2 + 1/4*(b*x^2 + a)^(3/2 
)*A*x/b - 1/8*sqrt(b*x^2 + a)*A*a*x/b + 1/16*C*a^3*arcsinh(b*x/sqrt(a*b))/ 
b^(5/2) - 1/8*A*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 8/105*(b*x^2 + a)^(3/ 
2)*D*a^2/b^3 - 2/15*(b*x^2 + a)^(3/2)*B*a/b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.91 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{1680} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, D x + 7 \, C\right )} x + \frac {6 \, {\left (D a b^{4} + 7 \, B b^{5}\right )}}{b^{5}}\right )} x + \frac {35 \, {\left (C a b^{4} + 6 \, A b^{5}\right )}}{b^{5}}\right )} x - \frac {8 \, {\left (4 \, D a^{2} b^{3} - 7 \, B a b^{4}\right )}}{b^{5}}\right )} x - \frac {105 \, {\left (C a^{2} b^{3} - 2 \, A a b^{4}\right )}}{b^{5}}\right )} x + \frac {32 \, {\left (4 \, D a^{3} b^{2} - 7 \, B a^{2} b^{3}\right )}}{b^{5}}\right )} - \frac {{\left (C a^{3} - 2 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:

integrate(x^2*(b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
 

Output:

1/1680*sqrt(b*x^2 + a)*((2*((4*(5*(6*D*x + 7*C)*x + 6*(D*a*b^4 + 7*B*b^5)/ 
b^5)*x + 35*(C*a*b^4 + 6*A*b^5)/b^5)*x - 8*(4*D*a^2*b^3 - 7*B*a*b^4)/b^5)* 
x - 105*(C*a^2*b^3 - 2*A*a*b^4)/b^5)*x + 32*(4*D*a^3*b^2 - 7*B*a^2*b^3)/b^ 
5) - 1/16*(C*a^3 - 2*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/ 
2)
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\int x^2\,\sqrt {b\,x^2+a}\,\left (A+B\,x+C\,x^2+x^3\,D\right ) \,d x \] Input:

int(x^2*(a + b*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D),x)
 

Output:

int(x^2*(a + b*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.35 \[ \int x^2 \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {128 \sqrt {b \,x^{2}+a}\, a^{3} d +210 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x -224 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}-105 \sqrt {b \,x^{2}+a}\, a^{2} b c x -64 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}+420 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{3}+112 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}+70 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+48 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+336 \sqrt {b \,x^{2}+a}\, b^{4} x^{4}+280 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}+240 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{6}-210 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b +105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c}{1680 b^{3}} \] Input:

int(x^2*(b*x^2+a)^(1/2)*(D*x^3+C*x^2+B*x+A),x)
 

Output:

(128*sqrt(a + b*x**2)*a**3*d + 210*sqrt(a + b*x**2)*a**2*b**2*x - 224*sqrt 
(a + b*x**2)*a**2*b**2 - 105*sqrt(a + b*x**2)*a**2*b*c*x - 64*sqrt(a + b*x 
**2)*a**2*b*d*x**2 + 420*sqrt(a + b*x**2)*a*b**3*x**3 + 112*sqrt(a + b*x** 
2)*a*b**3*x**2 + 70*sqrt(a + b*x**2)*a*b**2*c*x**3 + 48*sqrt(a + b*x**2)*a 
*b**2*d*x**4 + 336*sqrt(a + b*x**2)*b**4*x**4 + 280*sqrt(a + b*x**2)*b**3* 
c*x**5 + 240*sqrt(a + b*x**2)*b**3*d*x**6 - 210*sqrt(b)*log((sqrt(a + b*x* 
*2) + sqrt(b)*x)/sqrt(a))*a**3*b + 105*sqrt(b)*log((sqrt(a + b*x**2) + sqr 
t(b)*x)/sqrt(a))*a**3*c)/(1680*b**3)