\(\int \frac {x^3 (A+B x+C x^2)}{(a+b x^2)^{9/2}} \, dx\) [5]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {a (A b-a C+b B x)}{7 b^3 \left (a+b x^2\right )^{7/2}}-\frac {7 (A b-2 a C)+8 b B x}{35 b^3 \left (a+b x^2\right )^{5/2}}-\frac {35 a C-3 b B x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {2 B x}{35 a^2 b^2 \sqrt {a+b x^2}} \] Output:

1/7*a*(B*b*x+A*b-C*a)/b^3/(b*x^2+a)^(7/2)-1/35*(8*B*b*x+7*A*b-14*C*a)/b^3/ 
(b*x^2+a)^(5/2)-1/105*(-3*B*b*x+35*C*a)/a/b^3/(b*x^2+a)^(3/2)+2/35*B*x/a^2 
/b^2/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-8 a^4 C+21 a b^3 B x^5+6 b^4 B x^7-7 a^2 b^2 x^2 \left (3 A+5 C x^2\right )-2 a^3 b \left (3 A+14 C x^2\right )}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \] Input:

Integrate[(x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]
 

Output:

(-8*a^4*C + 21*a*b^3*B*x^5 + 6*b^4*B*x^7 - 7*a^2*b^2*x^2*(3*A + 5*C*x^2) - 
 2*a^3*b*(3*A + 14*C*x^2))/(105*a^2*b^3*(a + b*x^2)^(7/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2335, 25, 530, 25, 27, 454, 208}

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.

\(\displaystyle \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx\)

\(\Big \downarrow \) 2335

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 530

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 454

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

\(\Big \downarrow \) 208

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

Input:

Int[(x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x]
 

Output:

-1/7*(x^3*(a*B - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(7/2)) + ((a*(3*A*b + 4* 
a*C - 3*b*B*x))/(5*b^2*(a + b*x^2)^(5/2)) + (-1/3*(5*(3*A*b + 4*a*C) - 3*b 
*B*x)/(b*(a + b*x^2)^(3/2)) + (2*B*x)/(a*Sqrt[a + b*x^2]))/(5*b))/(7*a*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 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 454

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*d 
 - b*c*x)/(2*a*b*(p + 1)))*(a + b*x^2)^(p + 1), x] + Simp[c*((2*p + 3)/(2*a 
*(p + 1)))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && L 
tQ[p, -1] && NeQ[p, -3/2]
 

rule 530

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co 
eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po 
lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x 
)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(a 
 + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; 
FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 
1] && IntegerQ[2*p]
 

rule 2335

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq 
, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] 
+ Simp[c/(2*a*b*(p + 1))   Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu 
m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, 
 b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71

method	result	size
gosper	\(-\frac {-6 B \,x^{7} b^{4}-21 B \,x^{5} a \,b^{3}+35 C \,a^{2} b^{2} x^{4}+21 A \,a^{2} b^{2} x^{2}+28 C \,a^{3} b \,x^{2}+6 A \,a^{3} b +8 C \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}}\)	\(85\)
trager	\(-\frac {-6 B \,x^{7} b^{4}-21 B \,x^{5} a \,b^{3}+35 C \,a^{2} b^{2} x^{4}+21 A \,a^{2} b^{2} x^{2}+28 C \,a^{3} b \,x^{2}+6 A \,a^{3} b +8 C \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}}\)	\(85\)
orering	\(-\frac {-6 B \,x^{7} b^{4}-21 B \,x^{5} a \,b^{3}+35 C \,a^{2} b^{2} x^{4}+21 A \,a^{2} b^{2} x^{2}+28 C \,a^{3} b \,x^{2}+6 A \,a^{3} b +8 C \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}}\)	\(85\)
default	\(A \left (-\frac {x^{2}}{5 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {2 a}{35 b^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\right )+B \left (-\frac {x^{3}}{4 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {3 a \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )}{4 b}\right )+C \left (-\frac {x^{4}}{3 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {4 a \left (-\frac {x^{2}}{5 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {2 a}{35 b^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\right )}{3 b}\right )\)	\(217\)

Input:

int(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/105*(-6*B*b^4*x^7-21*B*a*b^3*x^5+35*C*a^2*b^2*x^4+21*A*a^2*b^2*x^2+28*C 
*a^3*b*x^2+6*A*a^3*b+8*C*a^4)/(b*x^2+a)^(7/2)/a^2/b^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.09 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (6 \, B b^{4} x^{7} + 21 \, B a b^{3} x^{5} - 35 \, C a^{2} b^{2} x^{4} - 8 \, C a^{4} - 6 \, A a^{3} b - 7 \, {\left (4 \, C a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{2} b^{7} x^{8} + 4 \, a^{3} b^{6} x^{6} + 6 \, a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{2} + a^{6} b^{3}\right )}} \] Input:

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")
 

Output:

1/105*(6*B*b^4*x^7 + 21*B*a*b^3*x^5 - 35*C*a^2*b^2*x^4 - 8*C*a^4 - 6*A*a^3 
*b - 7*(4*C*a^3*b + 3*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/(a^2*b^7*x^8 + 4*a^3 
*b^6*x^6 + 6*a^4*b^5*x^4 + 4*a^5*b^4*x^2 + a^6*b^3)
 

Sympy [A] (verification not implemented)

Time = 23.24 (sec) , antiderivative size = 660, normalized size of antiderivative = 5.50 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=A \left (\begin {cases} - \frac {2 a}{35 a^{3} b^{2} \sqrt {a + b x^{2}} + 105 a^{2} b^{3} x^{2} \sqrt {a + b x^{2}} + 105 a b^{4} x^{4} \sqrt {a + b x^{2}} + 35 b^{5} x^{6} \sqrt {a + b x^{2}}} - \frac {7 b x^{2}}{35 a^{3} b^{2} \sqrt {a + b x^{2}} + 105 a^{2} b^{3} x^{2} \sqrt {a + b x^{2}} + 105 a b^{4} x^{4} \sqrt {a + b x^{2}} + 35 b^{5} x^{6} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {9}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {7 a x^{5}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{7}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + C \left (\begin {cases} - \frac {8 a^{2}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} - \frac {28 a b x^{2}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} - \frac {35 b^{2} x^{4}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {9}{2}}} & \text {otherwise} \end {cases}\right ) \] Input:

integrate(x**3*(C*x**2+B*x+A)/(b*x**2+a)**(9/2),x)
 

Output:

A*Piecewise((-2*a/(35*a**3*b**2*sqrt(a + b*x**2) + 105*a**2*b**3*x**2*sqrt 
(a + b*x**2) + 105*a*b**4*x**4*sqrt(a + b*x**2) + 35*b**5*x**6*sqrt(a + b* 
x**2)) - 7*b*x**2/(35*a**3*b**2*sqrt(a + b*x**2) + 105*a**2*b**3*x**2*sqrt 
(a + b*x**2) + 105*a*b**4*x**4*sqrt(a + b*x**2) + 35*b**5*x**6*sqrt(a + b* 
x**2)), Ne(b, 0)), (x**4/(4*a**(9/2)), True)) + B*(7*a*x**5/(35*a**(11/2)* 
sqrt(1 + b*x**2/a) + 105*a**(9/2)*b*x**2*sqrt(1 + b*x**2/a) + 105*a**(7/2) 
*b**2*x**4*sqrt(1 + b*x**2/a) + 35*a**(5/2)*b**3*x**6*sqrt(1 + b*x**2/a)) 
+ 2*b*x**7/(35*a**(11/2)*sqrt(1 + b*x**2/a) + 105*a**(9/2)*b*x**2*sqrt(1 + 
 b*x**2/a) + 105*a**(7/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 35*a**(5/2)*b**3* 
x**6*sqrt(1 + b*x**2/a))) + C*Piecewise((-8*a**2/(105*a**3*b**3*sqrt(a + b 
*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2) + 315*a*b**5*x**4*sqrt(a + b* 
x**2) + 105*b**6*x**6*sqrt(a + b*x**2)) - 28*a*b*x**2/(105*a**3*b**3*sqrt( 
a + b*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2) + 315*a*b**5*x**4*sqrt(a 
 + b*x**2) + 105*b**6*x**6*sqrt(a + b*x**2)) - 35*b**2*x**4/(105*a**3*b**3 
*sqrt(a + b*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2) + 315*a*b**5*x**4* 
sqrt(a + b*x**2) + 105*b**6*x**6*sqrt(a + b*x**2)), Ne(b, 0)), (x**6/(6*a* 
*(9/2)), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.49 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {C x^{4}}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {B x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {4 \, C a x^{2}}{15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {A x^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {3 \, B x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, B x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, B a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {8 \, C a^{2}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} - \frac {2 \, A a}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} \] Input:

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/3*C*x^4/((b*x^2 + a)^(7/2)*b) - 1/4*B*x^3/((b*x^2 + a)^(7/2)*b) - 4/15* 
C*a*x^2/((b*x^2 + a)^(7/2)*b^2) - 1/5*A*x^2/((b*x^2 + a)^(7/2)*b) + 3/140* 
B*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*B*x/(sqrt(b*x^2 + a)*a^2*b^2) + 1/35*B* 
x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*B*a*x/((b*x^2 + a)^(7/2)*b^2) - 8/105*C 
*a^2/((b*x^2 + a)^(7/2)*b^3) - 2/35*A*a/((b*x^2 + a)^(7/2)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (3 \, {\left (\frac {2 \, B b x^{2}}{a^{2}} + \frac {7 \, B}{a}\right )} x - \frac {35 \, C}{b}\right )} x^{2} - \frac {7 \, {\left (4 \, C a^{4} b + 3 \, A a^{3} b^{2}\right )}}{a^{3} b^{3}}\right )} x^{2} - \frac {2 \, {\left (4 \, C a^{5} + 3 \, A a^{4} b\right )}}{a^{3} b^{3}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:

integrate(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x, algorithm="giac")
 

Output:

1/105*(((3*(2*B*b*x^2/a^2 + 7*B/a)*x - 35*C/b)*x^2 - 7*(4*C*a^4*b + 3*A*a^ 
3*b^2)/(a^3*b^3))*x^2 - 2*(4*C*a^5 + 3*A*a^4*b)/(a^3*b^3))/(b*x^2 + a)^(7/ 
2)
 

Mupad [B] (verification not implemented)

Time = 2.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {\frac {a\,\left (\frac {A}{7\,b}-\frac {C\,a}{7\,b^2}\right )}{b}+\frac {B\,a\,x}{7\,b^2}}{{\left (b\,x^2+a\right )}^{7/2}}-\frac {\frac {C}{3\,b^3}-\frac {B\,x}{35\,a\,b^2}}{{\left (b\,x^2+a\right )}^{3/2}}+\frac {\frac {a\,\left (\frac {C}{5\,b^2}-\frac {7\,A\,b-7\,C\,a}{35\,a\,b^2}\right )}{b}-\frac {8\,B\,x}{35\,b^2}}{{\left (b\,x^2+a\right )}^{5/2}}+\frac {2\,B\,x}{35\,a^2\,b^2\,\sqrt {b\,x^2+a}} \] Input:

int((x^3*(A + B*x + C*x^2))/(a + b*x^2)^(9/2),x)
 

Output:

((a*(A/(7*b) - (C*a)/(7*b^2)))/b + (B*a*x)/(7*b^2))/(a + b*x^2)^(7/2) - (C 
/(3*b^3) - (B*x)/(35*a*b^2))/(a + b*x^2)^(3/2) + ((a*(C/(5*b^2) - (7*A*b - 
 7*C*a)/(35*a*b^2)))/b - (8*B*x)/(35*b^2))/(a + b*x^2)^(5/2) + (2*B*x)/(35 
*a^2*b^2*(a + b*x^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.87 \[ \int \frac {x^3 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-6 \sqrt {b \,x^{2}+a}\, a^{4} b -8 \sqrt {b \,x^{2}+a}\, a^{4} c -21 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{2}-28 \sqrt {b \,x^{2}+a}\, a^{3} b c \,x^{2}-35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,x^{4}+21 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{5}+6 \sqrt {b \,x^{2}+a}\, b^{5} x^{7}-6 \sqrt {b}\, a^{4} b -24 \sqrt {b}\, a^{3} b^{2} x^{2}-36 \sqrt {b}\, a^{2} b^{3} x^{4}-24 \sqrt {b}\, a \,b^{4} x^{6}-6 \sqrt {b}\, b^{5} x^{8}}{105 a^{2} b^{3} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

int(x^3*(C*x^2+B*x+A)/(b*x^2+a)^(9/2),x)
 

Output:

( - 6*sqrt(a + b*x**2)*a**4*b - 8*sqrt(a + b*x**2)*a**4*c - 21*sqrt(a + b* 
x**2)*a**3*b**2*x**2 - 28*sqrt(a + b*x**2)*a**3*b*c*x**2 - 35*sqrt(a + b*x 
**2)*a**2*b**2*c*x**4 + 21*sqrt(a + b*x**2)*a*b**4*x**5 + 6*sqrt(a + b*x** 
2)*b**5*x**7 - 6*sqrt(b)*a**4*b - 24*sqrt(b)*a**3*b**2*x**2 - 36*sqrt(b)*a 
**2*b**3*x**4 - 24*sqrt(b)*a*b**4*x**6 - 6*sqrt(b)*b**5*x**8)/(105*a**2*b* 
*3*(a**4 + 4*a**3*b*x**2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))