\(\int (a+b x^2)^{3/2} (A+B x+C x^2+D x^3) \, dx\) [54]

Optimal result

Integrand size = 27, antiderivative size = 163 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a (6 A b-a C) x \sqrt {a+b x^2}}{16 b}+\frac {(6 A b-a C) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(b B-a D) \left (a+b x^2\right )^{5/2}}{5 b^2}+\frac {C x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {D \left (a+b x^2\right )^{7/2}}{7 b^2}+\frac {a^2 (6 A b-a C) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \] Output:

1/16*a*(6*A*b-C*a)*x*(b*x^2+a)^(1/2)/b+1/24*(6*A*b-C*a)*x*(b*x^2+a)^(3/2)/ 
b+1/5*(B*b-D*a)*(b*x^2+a)^(5/2)/b^2+1/6*C*x*(b*x^2+a)^(5/2)/b+1/7*D*(b*x^2 
+a)^(7/2)/b^2+1/16*a^2*(6*A*b-C*a)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3 
/2)
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.87 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {a+b x^2} \left (-96 a^3 D+4 b^3 x^3 \left (105 A+84 B x+70 C x^2+60 D x^3\right )+3 a^2 b (112 B+x (35 C+16 D x))+2 a b^2 x \left (525 A+x \left (336 B+245 C x+192 D x^2\right )\right )\right )+105 a^2 \sqrt {b} (-6 A b+a C) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{1680 b^2} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-96*a^3*D + 4*b^3*x^3*(105*A + 84*B*x + 70*C*x^2 + 60*D* 
x^3) + 3*a^2*b*(112*B + x*(35*C + 16*D*x)) + 2*a*b^2*x*(525*A + x*(336*B + 
 245*C*x + 192*D*x^2))) + 105*a^2*Sqrt[b]*(-6*A*b + a*C)*Log[-(Sqrt[b]*x) 
+ Sqrt[a + b*x^2]])/(1680*b^2)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2346, 2346, 27, 455, 211, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(D*x^2*(a + b*x^2)^(5/2))/(7*b) + ((7*C*x*(a + b*x^2)^(5/2))/6 + ((6*(7*b* 
B - 2*a*D)*(a + b*x^2)^(5/2))/(5*b) + 7*(6*A*b - a*C)*((x*(a + b*x^2)^(3/2 
))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2 
]])/(2*Sqrt[b])))/4))/6)/(7*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 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.10

method	result	size
default	\(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+\frac {B \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}+C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+D \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )\)	\(180\)

Input:

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

Output:

A*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(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))))+1/5*B*(b*x^2+a)^(5/2)/b+C*(1/6*x*(b*x^2+a)^(5/2) 
/b-1/6*a/b*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(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^2*(b*x^2+a)^(5/2)/b-2/35*a/b^ 
2*(b*x^2+a)^(5/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.01 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\left [-\frac {105 \, {\left (C a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (240 \, D b^{3} x^{6} + 280 \, C b^{3} x^{5} + 48 \, {\left (8 \, D a b^{2} + 7 \, B b^{3}\right )} x^{4} - 96 \, D a^{3} + 336 \, B a^{2} b + 70 \, {\left (7 \, C a b^{2} + 6 \, A b^{3}\right )} x^{3} + 48 \, {\left (D a^{2} b + 14 \, B a b^{2}\right )} x^{2} + 105 \, {\left (C a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{3360 \, b^{2}}, \frac {105 \, {\left (C a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (240 \, D b^{3} x^{6} + 280 \, C b^{3} x^{5} + 48 \, {\left (8 \, D a b^{2} + 7 \, B b^{3}\right )} x^{4} - 96 \, D a^{3} + 336 \, B a^{2} b + 70 \, {\left (7 \, C a b^{2} + 6 \, A b^{3}\right )} x^{3} + 48 \, {\left (D a^{2} b + 14 \, B a b^{2}\right )} x^{2} + 105 \, {\left (C a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{1680 \, b^{2}}\right ] \] Input:

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

Output:

[-1/3360*(105*(C*a^3 - 6*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*(8*D*a*b^2 + 7*B*b 
^3)*x^4 - 96*D*a^3 + 336*B*a^2*b + 70*(7*C*a*b^2 + 6*A*b^3)*x^3 + 48*(D*a^ 
2*b + 14*B*a*b^2)*x^2 + 105*(C*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^2 
, 1/1680*(105*(C*a^3 - 6*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*(8*D*a*b^2 + 7*B*b^3)*x^4 - 96*D 
*a^3 + 336*B*a^2*b + 70*(7*C*a*b^2 + 6*A*b^3)*x^3 + 48*(D*a^2*b + 14*B*a*b 
^2)*x^2 + 105*(C*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^2]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (143) = 286\).

Time = 0.45 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.86 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {C b x^{5}}{6} + \frac {D b x^{6}}{7} + \frac {x^{4} \left (B b^{2} + \frac {8 D a b}{7}\right )}{5 b} + \frac {x^{3} \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b} + \frac {x^{2} \cdot \left (2 B a b + D a^{2} - \frac {4 a \left (B b^{2} + \frac {8 D a b}{7}\right )}{5 b}\right )}{3 b} + \frac {x \left (2 A a b + C a^{2} - \frac {3 a \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b}\right )}{2 b} + \frac {B a^{2} - \frac {2 a \left (2 B a b + D a^{2} - \frac {4 a \left (B b^{2} + \frac {8 D a b}{7}\right )}{5 b}\right )}{3 b}}{b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + C a^{2} - \frac {3 a \left (A b^{2} + \frac {7 C a b}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(a + b*x**2)*(C*b*x**5/6 + D*b*x**6/7 + x**4*(B*b**2 + 8*D* 
a*b/7)/(5*b) + x**3*(A*b**2 + 7*C*a*b/6)/(4*b) + x**2*(2*B*a*b + D*a**2 - 
4*a*(B*b**2 + 8*D*a*b/7)/(5*b))/(3*b) + x*(2*A*a*b + C*a**2 - 3*a*(A*b**2 
+ 7*C*a*b/6)/(4*b))/(2*b) + (B*a**2 - 2*a*(2*B*a*b + D*a**2 - 4*a*(B*b**2 
+ 8*D*a*b/7)/(5*b))/(3*b))/b) + (A*a**2 - a*(2*A*a*b + C*a**2 - 3*a*(A*b** 
2 + 7*C*a*b/6)/(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)), (a**( 
3/2)*(A*x + B*x**2/2 + C*x**3/3 + D*x**4/4), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.01 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} D x^{2}}{7 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} C a^{2} x}{16 \, b} - \frac {C a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} D a}{35 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, b} \] Input:

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

Output:

1/7*(b*x^2 + a)^(5/2)*D*x^2/b + 1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 
 + a)*A*a*x + 1/6*(b*x^2 + a)^(5/2)*C*x/b - 1/24*(b*x^2 + a)^(3/2)*C*a*x/b 
 - 1/16*sqrt(b*x^2 + a)*C*a^2*x/b - 1/16*C*a^3*arcsinh(b*x/sqrt(a*b))/b^(3 
/2) + 3/8*A*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 2/35*(b*x^2 + a)^(5/2)*D* 
a/b^2 + 1/5*(b*x^2 + a)^(5/2)*B/b
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{1680} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, D b x + 7 \, C b\right )} x + \frac {6 \, {\left (8 \, D a b^{5} + 7 \, B b^{6}\right )}}{b^{5}}\right )} x + \frac {35 \, {\left (7 \, C a b^{5} + 6 \, A b^{6}\right )}}{b^{5}}\right )} x + \frac {24 \, {\left (D a^{2} b^{4} + 14 \, B a b^{5}\right )}}{b^{5}}\right )} x + \frac {105 \, {\left (C a^{2} b^{4} + 10 \, A a b^{5}\right )}}{b^{5}}\right )} x - \frac {48 \, {\left (2 \, D a^{3} b^{3} - 7 \, B a^{2} b^{4}\right )}}{b^{5}}\right )} + \frac {{\left (C a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \] Input:

integrate((b*x^2+a)^(3/2)*(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*b*x + 7*C*b)*x + 6*(8*D*a*b^5 + 7*B 
*b^6)/b^5)*x + 35*(7*C*a*b^5 + 6*A*b^6)/b^5)*x + 24*(D*a^2*b^4 + 14*B*a*b^ 
5)/b^5)*x + 105*(C*a^2*b^4 + 10*A*a*b^5)/b^5)*x - 48*(2*D*a^3*b^3 - 7*B*a^ 
2*b^4)/b^5) + 1/16*(C*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a 
)))/b^(3/2)
 

Mupad [F(-1)]

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

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

Output:

int((a + b*x^2)^(3/2)*(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 = 1.60 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {-96 \sqrt {b \,x^{2}+a}\, a^{3} d +1050 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x +336 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}+105 \sqrt {b \,x^{2}+a}\, a^{2} b c x +48 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}+420 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{3}+672 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{2}+490 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{3}+384 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+336 \sqrt {b \,x^{2}+a}\, b^{4} x^{4}+280 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{5}+240 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{6}+630 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b -105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} c}{1680 b^{2}} \] Input:

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

Output:

( - 96*sqrt(a + b*x**2)*a**3*d + 1050*sqrt(a + b*x**2)*a**2*b**2*x + 336*s 
qrt(a + b*x**2)*a**2*b**2 + 105*sqrt(a + b*x**2)*a**2*b*c*x + 48*sqrt(a + 
b*x**2)*a**2*b*d*x**2 + 420*sqrt(a + b*x**2)*a*b**3*x**3 + 672*sqrt(a + b* 
x**2)*a*b**3*x**2 + 490*sqrt(a + b*x**2)*a*b**2*c*x**3 + 384*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 + 630*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) 
+ sqrt(b)*x)/sqrt(a))*a**3*c)/(1680*b**2)