\(\int x \sqrt {a+b x^2} (A+B x+C x^2+D x^3) \, dx\) [63]

Optimal result

Integrand size = 28, antiderivative size = 167 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a (2 b B-a D) x \sqrt {a+b x^2}}{16 b^2}+\frac {(2 b B-a D) x^3 \sqrt {a+b x^2}}{8 b}+\frac {(A b-a C) \left (a+b x^2\right )^{3/2}}{3 b^2}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {C \left (a+b x^2\right )^{5/2}}{5 b^2}-\frac {a^2 (2 b B-a D) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.77 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (-a^2 (32 C+15 D x)+4 b^2 x^2 (20 A+x (15 B+2 x (6 C+5 D x)))+2 a b (40 A+x (15 B+x (8 C+5 D x)))\right )-15 a^2 (-2 b B+a D) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{240 b^{5/2}} \] Input:

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

Output:

(Sqrt[b]*Sqrt[a + b*x^2]*(-(a^2*(32*C + 15*D*x)) + 4*b^2*x^2*(20*A + x*(15 
*B + 2*x*(6*C + 5*D*x))) + 2*a*b*(40*A + x*(15*B + x*(8*C + 5*D*x)))) - 15 
*a^2*(-2*b*B + a*D)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(240*b^(5/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {2340, 27, 2340, 27, 533, 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.

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 x \left (a+b x^2\right )^{3/2} (2 b B-a D)}{4 b}-\frac {\int (5 a (2 b B-a D)-8 b (5 A b-2 a C) x) \sqrt {b x^2+a}dx}{4 b}\right )+\frac {2}{5} C x^2 \left (a+b x^2\right )^{3/2}}{2 b}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 x \left (a+b x^2\right )^{3/2} (2 b B-a D)}{4 b}-\frac {5 a (2 b B-a D) \int \sqrt {b x^2+a}dx-\frac {8}{3} \left (a+b x^2\right )^{3/2} (5 A b-2 a C)}{4 b}\right )+\frac {2}{5} C x^2 \left (a+b x^2\right )^{3/2}}{2 b}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 x \left (a+b x^2\right )^{3/2} (2 b B-a D)}{4 b}-\frac {5 a (2 b B-a D) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {8}{3} \left (a+b x^2\right )^{3/2} (5 A b-2 a C)}{4 b}\right )+\frac {2}{5} C x^2 \left (a+b x^2\right )^{3/2}}{2 b}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 x \left (a+b x^2\right )^{3/2} (2 b B-a D)}{4 b}-\frac {5 a (2 b B-a D) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {8}{3} \left (a+b x^2\right )^{3/2} (5 A b-2 a C)}{4 b}\right )+\frac {2}{5} C x^2 \left (a+b x^2\right )^{3/2}}{2 b}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{5} \left (\frac {5 x \left (a+b x^2\right )^{3/2} (2 b B-a D)}{4 b}-\frac {5 a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) (2 b B-a D)-\frac {8}{3} \left (a+b x^2\right )^{3/2} (5 A b-2 a C)}{4 b}\right )+\frac {2}{5} C x^2 \left (a+b x^2\right )^{3/2}}{2 b}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}\)

Input:

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

Output:

(D*x^3*(a + b*x^2)^(3/2))/(6*b) + ((2*C*x^2*(a + b*x^2)^(3/2))/5 + ((5*(2* 
b*B - a*D)*x*(a + b*x^2)^(3/2))/(4*b) - ((-8*(5*A*b - 2*a*C)*(a + b*x^2)^( 
3/2))/3 + 5*a*(2*b*B - a*D)*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x 
)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*b))/5)/(2*b)
 

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 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 = 0.56 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {A \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 b}+C \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )+D \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(194\)

Input:

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

Output:

1/3*A*(b*x^2+a)^(3/2)/b+C*(1/5*x^2*(b*x^2+a)^(3/2)/b-2/15*a/b^2*(b*x^2+a)^ 
(3/2))+D*(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)))))+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 = 0.09 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.72 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\left [-\frac {15 \, {\left (D a^{3} - 2 \, B a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (40 \, D b^{3} x^{5} + 48 \, C b^{3} x^{4} - 32 \, C a^{2} b + 80 \, A a b^{2} + 10 \, {\left (D a b^{2} + 6 \, B b^{3}\right )} x^{3} + 16 \, {\left (C a b^{2} + 5 \, A b^{3}\right )} x^{2} - 15 \, {\left (D a^{2} b - 2 \, B a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, -\frac {15 \, {\left (D a^{3} - 2 \, B a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, D b^{3} x^{5} + 48 \, C b^{3} x^{4} - 32 \, C a^{2} b + 80 \, A a b^{2} + 10 \, {\left (D a b^{2} + 6 \, B b^{3}\right )} x^{3} + 16 \, {\left (C a b^{2} + 5 \, A b^{3}\right )} x^{2} - 15 \, {\left (D a^{2} b - 2 \, B a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.17 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} - \frac {a \left (B a - \frac {3 a \left (B b + \frac {D a}{6}\right )}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {C x^{4}}{5} + \frac {D x^{5}}{6} + \frac {x^{3} \left (B b + \frac {D a}{6}\right )}{4 b} + \frac {x^{2} \left (A b + \frac {C a}{5}\right )}{3 b} + \frac {x \left (B a - \frac {3 a \left (B b + \frac {D a}{6}\right )}{4 b}\right )}{2 b} + \frac {A a - \frac {2 a \left (A b + \frac {C a}{5}\right )}{3 b}}{b}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3} + \frac {C x^{4}}{4} + \frac {D x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.04 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{3}}{6 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x^{2}}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} D a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x}{4 \, b} - \frac {\sqrt {b x^{2} + a} B a x}{8 \, b} + \frac {D a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} C a}{15 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.89 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, D x + 6 \, C\right )} x + \frac {5 \, {\left (D a b^{3} + 6 \, B b^{4}\right )}}{b^{4}}\right )} x + \frac {8 \, {\left (C a b^{3} + 5 \, A b^{4}\right )}}{b^{4}}\right )} x - \frac {15 \, {\left (D a^{2} b^{2} - 2 \, B a b^{3}\right )}}{b^{4}}\right )} x - \frac {16 \, {\left (2 \, C a^{2} b^{2} - 5 \, A a b^{3}\right )}}{b^{4}}\right )} - \frac {{\left (D a^{3} - 2 \, B a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.36 \[ \int x \sqrt {a+b x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {80 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}-32 \sqrt {b \,x^{2}+a}\, a^{2} b c -15 \sqrt {b \,x^{2}+a}\, a^{2} b d x +80 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}+30 \sqrt {b \,x^{2}+a}\, a \,b^{3} x +16 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}+10 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+60 \sqrt {b \,x^{2}+a}\, b^{4} x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{4}+40 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{5}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d -30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2}}{240 b^{3}} \] Input:

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

Output:

(80*sqrt(a + b*x**2)*a**2*b**2 - 32*sqrt(a + b*x**2)*a**2*b*c - 15*sqrt(a 
+ b*x**2)*a**2*b*d*x + 80*sqrt(a + b*x**2)*a*b**3*x**2 + 30*sqrt(a + b*x** 
2)*a*b**3*x + 16*sqrt(a + b*x**2)*a*b**2*c*x**2 + 10*sqrt(a + b*x**2)*a*b* 
*2*d*x**3 + 60*sqrt(a + b*x**2)*b**4*x**3 + 48*sqrt(a + b*x**2)*b**3*c*x** 
4 + 40*sqrt(a + b*x**2)*b**3*d*x**5 + 15*sqrt(b)*log((sqrt(a + b*x**2) + s 
qrt(b)*x)/sqrt(a))*a**3*d - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/ 
sqrt(a))*a**2*b**2)/(240*b**3)