\(\int \frac {x^2 (A+B x+C x^2+D x^3)}{(a+b x^2)^3} \, dx\) [55]

Optimal result

Integrand size = 28, antiderivative size = 134 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {a (b B-a D)-b (A b-a C) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {4 a (b B-2 a D)-b (A b-5 a C) x}{8 a b^3 \left (a+b x^2\right )}+\frac {(A b+3 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2}}+\frac {D \log \left (a+b x^2\right )}{2 b^3} \] Output:

1/4*(a*(B*b-D*a)-b*(A*b-C*a)*x)/b^3/(b*x^2+a)^2-1/8*(4*a*(B*b-2*D*a)-b*(A* 
b-5*C*a)*x)/a/b^3/(b*x^2+a)+1/8*(A*b+3*C*a)*arctan(b^(1/2)*x/a^(1/2))/a^(3 
/2)/b^(5/2)+1/2*D*ln(b*x^2+a)/b^3
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {-2 a^2 D-2 A b^2 x+2 a b (B+C x)}{\left (a+b x^2\right )^2}+\frac {8 a^2 D+A b^2 x-a b (4 B+5 C x)}{a \left (a+b x^2\right )}+\frac {\sqrt {b} (A b+3 a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}+4 D \log \left (a+b x^2\right )}{8 b^3} \] Input:

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

Output:

((-2*a^2*D - 2*A*b^2*x + 2*a*b*(B + C*x))/(a + b*x^2)^2 + (8*a^2*D + A*b^2 
*x - a*b*(4*B + 5*C*x))/(a*(a + b*x^2)) + (Sqrt[b]*(A*b + 3*a*C)*ArcTan[(S 
qrt[b]*x)/Sqrt[a]])/a^(3/2) + 4*D*Log[a + b*x^2])/(8*b^3)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2335, 25, 2335, 25, 27, 452, 218, 240}

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.

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

\(\Big \downarrow \) 2335

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2335

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 452

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 240

\(\displaystyle \frac {\frac {\frac {(3 a C+A b) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {4 a D \log \left (a+b x^2\right )}{b}}{2 b}-\frac {x (-2 x (b B-3 a D)+3 a C+A b)}{2 b \left (a+b x^2\right )}}{4 a b}-\frac {x^2 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}\)

Input:

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

Output:

-1/4*(x^2*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(a*b*(a + b*x^2)^2) + (-1/2*( 
x*(A*b + 3*a*C - 2*(b*B - 3*a*D)*x))/(b*(a + b*x^2)) + (((A*b + 3*a*C)*Arc 
Tan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]) + (4*a*D*Log[a + b*x^2])/b)/(2 
*b))/(4*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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 240

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

rule 452

Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c   Int[1/ 
(a + b*x^2), x], x] + Simp[d   Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, 
 d}, x] && NeQ[b*c^2 + a*d^2, 0]
 

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 = 0.46 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\frac {\left (A b -5 C a \right ) x^{3}}{8 a b}-\frac {\left (B b -2 D a \right ) x^{2}}{2 b^{2}}-\frac {\left (A b +3 C a \right ) x}{8 b^{2}}-\frac {a \left (B b -3 D a \right )}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\frac {4 D a \ln \left (b \,x^{2}+a \right )}{b}+\frac {\left (A b +3 C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 a \,b^{2}}\)	\(123\)

Input:

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

Output:

(1/8*(A*b-5*C*a)/a/b*x^3-1/2*(B*b-2*D*a)/b^2*x^2-1/8*(A*b+3*C*a)/b^2*x-1/4 
*a*(B*b-3*D*a)/b^3)/(b*x^2+a)^2+1/8/a/b^2*(4*D*a/b*ln(b*x^2+a)+(A*b+3*C*a) 
/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.34 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {12 \, D a^{4} - 4 \, B a^{3} b - 2 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} + 8 \, {\left (2 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2} - {\left ({\left (3 \, C a b^{2} + A b^{3}\right )} x^{4} + 3 \, C a^{3} + A a^{2} b + 2 \, {\left (3 \, C a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (3 \, C a^{3} b + A a^{2} b^{2}\right )} x + 8 \, {\left (D a^{2} b^{2} x^{4} + 2 \, D a^{3} b x^{2} + D a^{4}\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, \frac {6 \, D a^{4} - 2 \, B a^{3} b - {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} + 4 \, {\left (2 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2} + {\left ({\left (3 \, C a b^{2} + A b^{3}\right )} x^{4} + 3 \, C a^{3} + A a^{2} b + 2 \, {\left (3 \, C a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (3 \, C a^{3} b + A a^{2} b^{2}\right )} x + 4 \, {\left (D a^{2} b^{2} x^{4} + 2 \, D a^{3} b x^{2} + D a^{4}\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \] Input:

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

Output:

[1/16*(12*D*a^4 - 4*B*a^3*b - 2*(5*C*a^2*b^2 - A*a*b^3)*x^3 + 8*(2*D*a^3*b 
 - B*a^2*b^2)*x^2 - ((3*C*a*b^2 + A*b^3)*x^4 + 3*C*a^3 + A*a^2*b + 2*(3*C* 
a^2*b + A*a*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + 
 a)) - 2*(3*C*a^3*b + A*a^2*b^2)*x + 8*(D*a^2*b^2*x^4 + 2*D*a^3*b*x^2 + D* 
a^4)*log(b*x^2 + a))/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3), 1/8*(6*D*a^4 
 - 2*B*a^3*b - (5*C*a^2*b^2 - A*a*b^3)*x^3 + 4*(2*D*a^3*b - B*a^2*b^2)*x^2 
 + ((3*C*a*b^2 + A*b^3)*x^4 + 3*C*a^3 + A*a^2*b + 2*(3*C*a^2*b + A*a*b^2)* 
x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (3*C*a^3*b + A*a^2*b^2)*x + 4*(D*a^ 
2*b^2*x^4 + 2*D*a^3*b*x^2 + D*a^4)*log(b*x^2 + a))/(a^2*b^5*x^4 + 2*a^3*b^ 
4*x^2 + a^4*b^3)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (121) = 242\).

Time = 10.08 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.27 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\left (\frac {D}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right ) \log {\left (x + \frac {- 8 D a^{2} + 16 a^{2} b^{3} \left (\frac {D}{2 b^{3}} - \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right )}{A b^{2} + 3 C a b} \right )} + \left (\frac {D}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right ) \log {\left (x + \frac {- 8 D a^{2} + 16 a^{2} b^{3} \left (\frac {D}{2 b^{3}} + \frac {\sqrt {- a^{3} b^{7}} \left (A b + 3 C a\right )}{16 a^{3} b^{6}}\right )}{A b^{2} + 3 C a b} \right )} + \frac {- 2 B a^{2} b + 6 D a^{3} + x^{3} \left (A b^{3} - 5 C a b^{2}\right ) + x^{2} \left (- 4 B a b^{2} + 8 D a^{2} b\right ) + x \left (- A a b^{2} - 3 C a^{2} b\right )}{8 a^{3} b^{3} + 16 a^{2} b^{4} x^{2} + 8 a b^{5} x^{4}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {6 \, D a^{3} - 2 \, B a^{2} b - {\left (5 \, C a b^{2} - A b^{3}\right )} x^{3} + 4 \, {\left (2 \, D a^{2} b - B a b^{2}\right )} x^{2} - {\left (3 \, C a^{2} b + A a b^{2}\right )} x}{8 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}} + \frac {D \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (3 \, C a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {D \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {{\left (3 \, C a + A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{2}} - \frac {{\left (5 \, C a b - A b^{2}\right )} x^{3} - 4 \, {\left (2 \, D a^{2} - B a b\right )} x^{2} + {\left (3 \, C a^{2} + A a b\right )} x - \frac {2 \, {\left (3 \, D a^{3} - B a^{2} b\right )}}{b}}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{2}} \] Input:

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

Output:

1/2*D*log(b*x^2 + a)/b^3 + 1/8*(3*C*a + A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a 
*b)*a*b^2) - 1/8*((5*C*a*b - A*b^2)*x^3 - 4*(2*D*a^2 - B*a*b)*x^2 + (3*C*a 
^2 + A*a*b)*x - 2*(3*D*a^3 - B*a^2*b)/b)/((b*x^2 + a)^2*a*b^2)
 

Mupad [B] (verification not implemented)

Time = 2.44 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.46 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {A\,x^3}{8\,a}-\frac {A\,x}{8\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {B\,x^2}{2\,b}+\frac {B\,a}{4\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {5\,C\,x^3}{8\,b}+\frac {3\,C\,a\,x}{8\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {D\,\left (\ln \left (b\,x^2+a\right )+\frac {2\,a}{b\,x^2+a}-\frac {a^2}{2\,{\left (b\,x^2+a\right )}^2}\right )}{2\,b^3}+\frac {A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,b^{3/2}}+\frac {3\,C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,b^{5/2}} \] Input:

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

Output:

((A*x^3)/(8*a) - (A*x)/(8*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) - ((B*x^2)/(2*b) 
 + (B*a)/(4*b^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - ((5*C*x^3)/(8*b) + (3*C*a* 
x)/(8*b^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) + (D*(log(a + b*x^2) + (2*a)/(a + 
b*x^2) - a^2/(2*(a + b*x^2)^2)))/(2*b^3) + (A*atan((b^(1/2)*x)/a^(1/2)))/( 
8*a^(3/2)*b^(3/2)) + (3*C*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(1/2)*b^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.07 \[ \int \frac {x^2 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} c +2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b c \,x^{2}+\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{4}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c \,x^{4}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} d +8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b d \,x^{2}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{4}+2 a^{3} d -a^{2} b^{2} x -3 a^{2} b c x +a \,b^{3} x^{3}-5 a \,b^{2} c \,x^{3}-4 a \,b^{2} d \,x^{4}+2 b^{4} x^{4}}{8 a \,b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b + 3*sqrt(b)*sqrt(a)* 
atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*c + 2*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt( 
b)*sqrt(a)))*a*b**2*x**2 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a))) 
*a*b*c*x**2 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*x**4 + 3* 
sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**2*c*x**4 + 4*log(a + b*x* 
*2)*a**3*d + 8*log(a + b*x**2)*a**2*b*d*x**2 + 4*log(a + b*x**2)*a*b**2*d* 
x**4 + 2*a**3*d - a**2*b**2*x - 3*a**2*b*c*x + a*b**3*x**3 - 5*a*b**2*c*x* 
*3 - 4*a*b**2*d*x**4 + 2*b**4*x**4)/(8*a*b**3*(a**2 + 2*a*b*x**2 + b**2*x* 
*4))