\(\int \frac {A+B x+C x^2+D x^3}{(a+b x^2)^{9/2}} \, dx\) [60]

Optimal result

Integrand size = 27, antiderivative size = 145 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {a (b B-a D)-b (A b-a C) x}{7 a b^2 \left (a+b x^2\right )^{7/2}}-\frac {7 a^2 D-b (6 A b+a C) x}{35 a^2 b^2 \left (a+b x^2\right )^{5/2}}+\frac {4 (6 A b+a C) x}{105 a^3 b \left (a+b x^2\right )^{3/2}}+\frac {8 (6 A b+a C) x}{105 a^4 b \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-6 a^5 D+48 A b^5 x^7+35 a^3 b^2 x \left (3 A+C x^2\right )+8 a b^4 x^5 \left (21 A+C x^2\right )+14 a^2 b^3 x^3 \left (15 A+2 C x^2\right )-3 a^4 b \left (5 B+7 D x^2\right )}{105 a^4 b^2 \left (a+b x^2\right )^{7/2}} \] Input:

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

Output:

(-6*a^5*D + 48*A*b^5*x^7 + 35*a^3*b^2*x*(3*A + C*x^2) + 8*a*b^4*x^5*(21*A 
+ C*x^2) + 14*a^2*b^3*x^3*(15*A + 2*C*x^2) - 3*a^4*b*(5*B + 7*D*x^2))/(105 
*a^4*b^2*(a + b*x^2)^(7/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2345, 25, 27, 454, 209, 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.

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

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 454

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

-1/7*(a*(B - (a*D)/b) - (A*b - a*C)*x)/(a*b*(a + b*x^2)^(7/2)) + (-1/5*(7* 
a^2*D - b*(6*A*b + a*C)*x)/(a*b*(a + b*x^2)^(5/2)) + (4*(6*A*b + a*C)*(x/( 
3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/(5*a))/(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 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[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[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {48 A \,b^{5} x^{7}+8 C a \,b^{4} x^{7}+168 A a \,b^{4} x^{5}+28 C \,a^{2} b^{3} x^{5}+210 a^{2} A \,b^{3} x^{3}+35 C \,a^{3} b^{2} x^{3}-21 D x^{2} a^{4} b +105 a^{3} b^{2} A x -15 a^{4} b B -6 D a^{5}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b^{2}}\)	\(117\)
trager	\(\frac {48 A \,b^{5} x^{7}+8 C a \,b^{4} x^{7}+168 A a \,b^{4} x^{5}+28 C \,a^{2} b^{3} x^{5}+210 a^{2} A \,b^{3} x^{3}+35 C \,a^{3} b^{2} x^{3}-21 D x^{2} a^{4} b +105 a^{3} b^{2} A x -15 a^{4} b B -6 D a^{5}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b^{2}}\)	\(117\)
orering	\(\frac {48 A \,b^{5} x^{7}+8 C a \,b^{4} x^{7}+168 A a \,b^{4} x^{5}+28 C \,a^{2} b^{3} x^{5}+210 a^{2} A \,b^{3} x^{3}+35 C \,a^{3} b^{2} x^{3}-21 D x^{2} a^{4} b +105 a^{3} b^{2} A x -15 a^{4} b B -6 D a^{5}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4} b^{2}}\)	\(117\)
default	\(A \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )-\frac {B}{7 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+C \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )+D \left (-\frac {x^{2}}{5 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {2 a}{35 b^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\right )\)	\(224\)

Input:

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

Output:

1/105*(48*A*b^5*x^7+8*C*a*b^4*x^7+168*A*a*b^4*x^5+28*C*a^2*b^3*x^5+210*A*a 
^2*b^3*x^3+35*C*a^3*b^2*x^3-21*D*a^4*b*x^2+105*A*a^3*b^2*x-15*B*a^4*b-6*D* 
a^5)/(b*x^2+a)^(7/2)/a^4/b^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {{\left (21 \, D a^{4} b x^{2} - 8 \, {\left (C a b^{4} + 6 \, A b^{5}\right )} x^{7} - 105 \, A a^{3} b^{2} x + 6 \, D a^{5} + 15 \, B a^{4} b - 28 \, {\left (C a^{2} b^{3} + 6 \, A a b^{4}\right )} x^{5} - 35 \, {\left (C a^{3} b^{2} + 6 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{4} b^{6} x^{8} + 4 \, a^{5} b^{5} x^{6} + 6 \, a^{6} b^{4} x^{4} + 4 \, a^{7} b^{3} x^{2} + a^{8} b^{2}\right )}} \] Input:

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

Output:

-1/105*(21*D*a^4*b*x^2 - 8*(C*a*b^4 + 6*A*b^5)*x^7 - 105*A*a^3*b^2*x + 6*D 
*a^5 + 15*B*a^4*b - 28*(C*a^2*b^3 + 6*A*a*b^4)*x^5 - 35*(C*a^3*b^2 + 6*A*a 
^2*b^3)*x^3)*sqrt(b*x^2 + a)/(a^4*b^6*x^8 + 4*a^5*b^5*x^6 + 6*a^6*b^4*x^4 
+ 4*a^7*b^3*x^2 + a^8*b^2)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1266 vs. \(2 (134) = 268\).

Time = 28.82 (sec) , antiderivative size = 2064, normalized size of antiderivative = 14.23 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

A*(35*a**14*x/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt 
(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2 
)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a 
) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12* 
sqrt(1 + b*x**2/a)) + 175*a**13*b*x**3/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 
210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + 
 b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b* 
*4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 
 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 371*a**12*b**2*x**5/(35*a** 
(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525* 
a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + 
b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b** 
5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 
 429*a**11*b**3*x**7/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x* 
*2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a 
**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b 
*x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6 
*x**12*sqrt(1 + b*x**2/a)) + 286*a**10*b**4*x**9/(35*a**(37/2)*sqrt(1 + b* 
x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x** 
4*sqrt(1 + b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {D x^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {16 \, A x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {C x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, C x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, C x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {2 \, D a}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {B}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left ({\left (4 \, x^{2} {\left (\frac {2 \, {\left (C a b^{5} + 6 \, A b^{6}\right )} x^{2}}{a^{4} b^{3}} + \frac {7 \, {\left (C a^{2} b^{4} + 6 \, A a b^{5}\right )}}{a^{4} b^{3}}\right )} + \frac {35 \, {\left (C a^{3} b^{3} + 6 \, A a^{2} b^{4}\right )}}{a^{4} b^{3}}\right )} x - \frac {21 \, D}{b}\right )} x + \frac {105 \, A}{a}\right )} x - \frac {3 \, {\left (2 \, D a^{5} b + 5 \, B a^{4} b^{2}\right )}}{a^{4} b^{3}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {16\,A\,x}{35\,a^4\,\sqrt {b\,x^2+a}}-\frac {\left (7\,b\,x^2+2\,a\right )\,D}{35\,b^2\,{\left (b\,x^2+a\right )}^{7/2}}-\frac {B}{7\,b\,{\left (b\,x^2+a\right )}^{7/2}}+\frac {8\,A\,x}{35\,a^3\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {6\,A\,x}{35\,a^2\,{\left (b\,x^2+a\right )}^{5/2}}+\frac {A\,x}{7\,a\,{\left (b\,x^2+a\right )}^{7/2}}-\frac {C\,x}{7\,b\,{\left (b\,x^2+a\right )}^{7/2}}+\frac {8\,C\,x}{105\,a^3\,b\,\sqrt {b\,x^2+a}}+\frac {4\,C\,x}{105\,a^2\,b\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {C\,x}{35\,a\,b\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.31 \[ \int \frac {A+B x+C x^2+D x^3}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-6 \sqrt {b \,x^{2}+a}\, a^{4} d +105 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x -15 \sqrt {b \,x^{2}+a}\, a^{3} b^{2}-21 \sqrt {b \,x^{2}+a}\, a^{3} b d \,x^{2}+210 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{3}+35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c \,x^{3}+168 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{5}+28 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{5}+48 \sqrt {b \,x^{2}+a}\, b^{5} x^{7}+8 \sqrt {b \,x^{2}+a}\, b^{4} c \,x^{7}-48 \sqrt {b}\, a^{4} b -8 \sqrt {b}\, a^{4} c -192 \sqrt {b}\, a^{3} b^{2} x^{2}-32 \sqrt {b}\, a^{3} b c \,x^{2}-288 \sqrt {b}\, a^{2} b^{3} x^{4}-48 \sqrt {b}\, a^{2} b^{2} c \,x^{4}-192 \sqrt {b}\, a \,b^{4} x^{6}-32 \sqrt {b}\, a \,b^{3} c \,x^{6}-48 \sqrt {b}\, b^{5} x^{8}-8 \sqrt {b}\, b^{4} c \,x^{8}}{105 a^{3} b^{2} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

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

Output:

( - 6*sqrt(a + b*x**2)*a**4*d + 105*sqrt(a + b*x**2)*a**3*b**2*x - 15*sqrt 
(a + b*x**2)*a**3*b**2 - 21*sqrt(a + b*x**2)*a**3*b*d*x**2 + 210*sqrt(a + 
b*x**2)*a**2*b**3*x**3 + 35*sqrt(a + b*x**2)*a**2*b**2*c*x**3 + 168*sqrt(a 
 + b*x**2)*a*b**4*x**5 + 28*sqrt(a + b*x**2)*a*b**3*c*x**5 + 48*sqrt(a + b 
*x**2)*b**5*x**7 + 8*sqrt(a + b*x**2)*b**4*c*x**7 - 48*sqrt(b)*a**4*b - 8* 
sqrt(b)*a**4*c - 192*sqrt(b)*a**3*b**2*x**2 - 32*sqrt(b)*a**3*b*c*x**2 - 2 
88*sqrt(b)*a**2*b**3*x**4 - 48*sqrt(b)*a**2*b**2*c*x**4 - 192*sqrt(b)*a*b* 
*4*x**6 - 32*sqrt(b)*a*b**3*c*x**6 - 48*sqrt(b)*b**5*x**8 - 8*sqrt(b)*b**4 
*c*x**8)/(105*a**3*b**2*(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))