\(\int \frac {A+B x^2+C x^4+D x^6}{x^6 (a+b x^2)^3} \, dx\) [186]

Optimal result

Integrand size = 30, antiderivative size = 200 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {6 A b^2-a (3 b B-a C)}{a^5 x}-\frac {\left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 A b^3-a \left (11 b^2 B-7 a b C+3 a^2 D\right )\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac {\left (63 A b^3-a \left (35 b^2 B-15 a b C+3 a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2} \sqrt {b}} \] Output:

-1/5*A/a^3/x^5+1/3*(3*A*b-B*a)/a^4/x^3-(6*A*b^2-a*(3*B*b-C*a))/a^5/x-1/4*( 
A*b^3-a*(B*b^2-C*a*b+D*a^2))*x/a^4/(b*x^2+a)^2-1/8*(15*A*b^3-a*(11*B*b^2-7 
*C*a*b+3*D*a^2))*x/a^5/(b*x^2+a)-1/8*(63*A*b^3-a*(35*B*b^2-15*C*a*b+3*D*a^ 
2))*arctan(b^(1/2)*x/a^(1/2))/a^(11/2)/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}+\frac {-6 A b^2+3 a b B-a^2 C}{a^5 x}+\frac {\left (-A b^3+a b^2 B-a^2 b C+a^3 D\right ) x}{4 a^4 \left (a+b x^2\right )^2}+\frac {\left (-15 A b^3+11 a b^2 B-7 a^2 b C+3 a^3 D\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac {\left (-63 A b^3+35 a b^2 B-15 a^2 b C+3 a^3 D\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2} \sqrt {b}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(x^6*(a + b*x^2)^3),x]
 

Output:

-1/5*A/(a^3*x^5) + (3*A*b - a*B)/(3*a^4*x^3) + (-6*A*b^2 + 3*a*b*B - a^2*C 
)/(a^5*x) + ((-(A*b^3) + a*b^2*B - a^2*b*C + a^3*D)*x)/(4*a^4*(a + b*x^2)^ 
2) + ((-15*A*b^3 + 11*a*b^2*B - 7*a^2*b*C + 3*a^3*D)*x)/(8*a^5*(a + b*x^2) 
) + ((-63*A*b^3 + 35*a*b^2*B - 15*a^2*b*C + 3*a^3*D)*ArcTan[(Sqrt[b]*x)/Sq 
rt[a]])/(8*a^(11/2)*Sqrt[b])
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2336, 25, 2336, 25, 2333, 2009}

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.

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

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {-\frac {\left (15 A b^3-a \left (3 D a^2-7 b C a+11 b^2 B\right )\right ) x^6}{a^3}+\frac {8 \left (3 A b^2-a (2 b B-a C)\right ) x^4}{a^2}-8 \left (\frac {2 A b}{a}-B\right ) x^2+8 A}{x^6 \left (b x^2+a\right )}dx}{2 a}-\frac {x \left (15 A b^3-a \left (3 a^2 D-7 a b C+11 b^2 B\right )\right )}{2 a^4 \left (a+b x^2\right )}}{4 a}-\frac {x \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-\frac {\left (15 A b^3-a \left (3 D a^2-7 b C a+11 b^2 B\right )\right ) x^6}{a^3}+\frac {8 \left (3 A b^2-a (2 b B-a C)\right ) x^4}{a^2}-8 \left (\frac {2 A b}{a}-B\right ) x^2+8 A}{x^6 \left (b x^2+a\right )}dx}{2 a}-\frac {x \left (15 A b^3-a \left (3 a^2 D-7 a b C+11 b^2 B\right )\right )}{2 a^4 \left (a+b x^2\right )}}{4 a}-\frac {x \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {\frac {\int \left (\frac {8 A}{a x^6}+\frac {a \left (3 D a^2-15 b C a+35 b^2 B\right )-63 A b^3}{a^3 \left (b x^2+a\right )}+\frac {8 \left (6 A b^2-a (3 b B-a C)\right )}{a^3 x^2}+\frac {8 (a B-3 A b)}{a^2 x^4}\right )dx}{2 a}-\frac {x \left (15 A b^3-a \left (3 a^2 D-7 a b C+11 b^2 B\right )\right )}{2 a^4 \left (a+b x^2\right )}}{4 a}-\frac {x \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {-\frac {8 \left (6 A b^2-a (3 b B-a C)\right )}{a^3 x}+\frac {8 (3 A b-a B)}{3 a^2 x^3}-\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (63 A b^3-a \left (3 a^2 D-15 a b C+35 b^2 B\right )\right )}{a^{7/2} \sqrt {b}}-\frac {8 A}{5 a x^5}}{2 a}-\frac {x \left (15 A b^3-a \left (3 a^2 D-7 a b C+11 b^2 B\right )\right )}{2 a^4 \left (a+b x^2\right )}}{4 a}-\frac {x \left (A b^3-a \left (a^2 D-a b C+b^2 B\right )\right )}{4 a^4 \left (a+b x^2\right )^2}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(x^6*(a + b*x^2)^3),x]
 

Output:

-1/4*((A*b^3 - a*(b^2*B - a*b*C + a^2*D))*x)/(a^4*(a + b*x^2)^2) + (-1/2*( 
(15*A*b^3 - a*(11*b^2*B - 7*a*b*C + 3*a^2*D))*x)/(a^4*(a + b*x^2)) + ((-8* 
A)/(5*a*x^5) + (8*(3*A*b - a*B))/(3*a^2*x^3) - (8*(6*A*b^2 - a*(3*b*B - a* 
C)))/(a^3*x) - ((63*A*b^3 - a*(35*b^2*B - 15*a*b*C + 3*a^2*D))*ArcTan[(Sqr 
t[b]*x)/Sqrt[a]])/(a^(7/2)*Sqrt[b]))/(2*a))/(4*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2333

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] 
&& PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2336

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {\frac {\left (\frac {15}{8} A \,b^{4}-\frac {11}{8} B a \,b^{3}+\frac {7}{8} C \,a^{2} b^{2}-\frac {3}{8} a^{3} D b \right ) x^{3}+\frac {a \left (17 b^{3} A -13 a \,b^{2} B +9 a^{2} b C -5 a^{3} D\right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (63 b^{3} A -35 a \,b^{2} B +15 a^{2} b C -3 a^{3} D\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{5}}-\frac {A}{5 a^{3} x^{5}}-\frac {-3 A b +B a}{3 a^{4} x^{3}}-\frac {6 b^{2} A -3 a b B +a^{2} C}{a^{5} x}\)	\(177\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/x^6/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/a^5*(((15/8*A*b^4-11/8*B*a*b^3+7/8*C*a^2*b^2-3/8*a^3*D*b)*x^3+1/8*a*(17 
*A*b^3-13*B*a*b^2+9*C*a^2*b-5*D*a^3)*x)/(b*x^2+a)^2+1/8*(63*A*b^3-35*B*a*b 
^2+15*C*a^2*b-3*D*a^3)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))-1/5*A/a^3/x^5- 
1/3*(-3*A*b+B*a)/a^4/x^3-(6*A*b^2-3*B*a*b+C*a^2)/a^5/x
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 628, normalized size of antiderivative = 3.14 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=\left [\frac {30 \, {\left (3 \, D a^{4} b^{2} - 15 \, C a^{3} b^{3} + 35 \, B a^{2} b^{4} - 63 \, A a b^{5}\right )} x^{8} - 48 \, A a^{5} b + 50 \, {\left (3 \, D a^{5} b - 15 \, C a^{4} b^{2} + 35 \, B a^{3} b^{3} - 63 \, A a^{2} b^{4}\right )} x^{6} - 16 \, {\left (15 \, C a^{5} b - 35 \, B a^{4} b^{2} + 63 \, A a^{3} b^{3}\right )} x^{4} - 16 \, {\left (5 \, B a^{5} b - 9 \, A a^{4} b^{2}\right )} x^{2} + 15 \, {\left ({\left (3 \, D a^{3} b^{2} - 15 \, C a^{2} b^{3} + 35 \, B a b^{4} - 63 \, A b^{5}\right )} x^{9} + 2 \, {\left (3 \, D a^{4} b - 15 \, C a^{3} b^{2} + 35 \, B a^{2} b^{3} - 63 \, A a b^{4}\right )} x^{7} + {\left (3 \, D a^{5} - 15 \, C a^{4} b + 35 \, B a^{3} b^{2} - 63 \, A a^{2} b^{3}\right )} x^{5}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{240 \, {\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\right )}}, \frac {15 \, {\left (3 \, D a^{4} b^{2} - 15 \, C a^{3} b^{3} + 35 \, B a^{2} b^{4} - 63 \, A a b^{5}\right )} x^{8} - 24 \, A a^{5} b + 25 \, {\left (3 \, D a^{5} b - 15 \, C a^{4} b^{2} + 35 \, B a^{3} b^{3} - 63 \, A a^{2} b^{4}\right )} x^{6} - 8 \, {\left (15 \, C a^{5} b - 35 \, B a^{4} b^{2} + 63 \, A a^{3} b^{3}\right )} x^{4} - 8 \, {\left (5 \, B a^{5} b - 9 \, A a^{4} b^{2}\right )} x^{2} + 15 \, {\left ({\left (3 \, D a^{3} b^{2} - 15 \, C a^{2} b^{3} + 35 \, B a b^{4} - 63 \, A b^{5}\right )} x^{9} + 2 \, {\left (3 \, D a^{4} b - 15 \, C a^{3} b^{2} + 35 \, B a^{2} b^{3} - 63 \, A a b^{4}\right )} x^{7} + {\left (3 \, D a^{5} - 15 \, C a^{4} b + 35 \, B a^{3} b^{2} - 63 \, A a^{2} b^{3}\right )} x^{5}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{120 \, {\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\right )}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/x^6/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

[1/240*(30*(3*D*a^4*b^2 - 15*C*a^3*b^3 + 35*B*a^2*b^4 - 63*A*a*b^5)*x^8 - 
48*A*a^5*b + 50*(3*D*a^5*b - 15*C*a^4*b^2 + 35*B*a^3*b^3 - 63*A*a^2*b^4)*x 
^6 - 16*(15*C*a^5*b - 35*B*a^4*b^2 + 63*A*a^3*b^3)*x^4 - 16*(5*B*a^5*b - 9 
*A*a^4*b^2)*x^2 + 15*((3*D*a^3*b^2 - 15*C*a^2*b^3 + 35*B*a*b^4 - 63*A*b^5) 
*x^9 + 2*(3*D*a^4*b - 15*C*a^3*b^2 + 35*B*a^2*b^3 - 63*A*a*b^4)*x^7 + (3*D 
*a^5 - 15*C*a^4*b + 35*B*a^3*b^2 - 63*A*a^2*b^3)*x^5)*sqrt(-a*b)*log((b*x^ 
2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^6*b^3*x^9 + 2*a^7*b^2*x^7 + a^8*b 
*x^5), 1/120*(15*(3*D*a^4*b^2 - 15*C*a^3*b^3 + 35*B*a^2*b^4 - 63*A*a*b^5)* 
x^8 - 24*A*a^5*b + 25*(3*D*a^5*b - 15*C*a^4*b^2 + 35*B*a^3*b^3 - 63*A*a^2* 
b^4)*x^6 - 8*(15*C*a^5*b - 35*B*a^4*b^2 + 63*A*a^3*b^3)*x^4 - 8*(5*B*a^5*b 
 - 9*A*a^4*b^2)*x^2 + 15*((3*D*a^3*b^2 - 15*C*a^2*b^3 + 35*B*a*b^4 - 63*A* 
b^5)*x^9 + 2*(3*D*a^4*b - 15*C*a^3*b^2 + 35*B*a^2*b^3 - 63*A*a*b^4)*x^7 + 
(3*D*a^5 - 15*C*a^4*b + 35*B*a^3*b^2 - 63*A*a^2*b^3)*x^5)*sqrt(a*b)*arctan 
(sqrt(a*b)*x/a))/(a^6*b^3*x^9 + 2*a^7*b^2*x^7 + a^8*b*x^5)]
 

Sympy [A] (verification not implemented)

Time = 52.95 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.42 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{11} b}} \left (- 63 A b^{3} + 35 B a b^{2} - 15 C a^{2} b + 3 D a^{3}\right ) \log {\left (- a^{6} \sqrt {- \frac {1}{a^{11} b}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{11} b}} \left (- 63 A b^{3} + 35 B a b^{2} - 15 C a^{2} b + 3 D a^{3}\right ) \log {\left (a^{6} \sqrt {- \frac {1}{a^{11} b}} + x \right )}}{16} + \frac {- 24 A a^{4} + x^{8} \left (- 945 A b^{4} + 525 B a b^{3} - 225 C a^{2} b^{2} + 45 D a^{3} b\right ) + x^{6} \left (- 1575 A a b^{3} + 875 B a^{2} b^{2} - 375 C a^{3} b + 75 D a^{4}\right ) + x^{4} \left (- 504 A a^{2} b^{2} + 280 B a^{3} b - 120 C a^{4}\right ) + x^{2} \cdot \left (72 A a^{3} b - 40 B a^{4}\right )}{120 a^{7} x^{5} + 240 a^{6} b x^{7} + 120 a^{5} b^{2} x^{9}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/x**6/(b*x**2+a)**3,x)
 

Output:

-sqrt(-1/(a**11*b))*(-63*A*b**3 + 35*B*a*b**2 - 15*C*a**2*b + 3*D*a**3)*lo 
g(-a**6*sqrt(-1/(a**11*b)) + x)/16 + sqrt(-1/(a**11*b))*(-63*A*b**3 + 35*B 
*a*b**2 - 15*C*a**2*b + 3*D*a**3)*log(a**6*sqrt(-1/(a**11*b)) + x)/16 + (- 
24*A*a**4 + x**8*(-945*A*b**4 + 525*B*a*b**3 - 225*C*a**2*b**2 + 45*D*a**3 
*b) + x**6*(-1575*A*a*b**3 + 875*B*a**2*b**2 - 375*C*a**3*b + 75*D*a**4) + 
 x**4*(-504*A*a**2*b**2 + 280*B*a**3*b - 120*C*a**4) + x**2*(72*A*a**3*b - 
 40*B*a**4))/(120*a**7*x**5 + 240*a**6*b*x**7 + 120*a**5*b**2*x**9)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {15 \, {\left (3 \, D a^{3} b - 15 \, C a^{2} b^{2} + 35 \, B a b^{3} - 63 \, A b^{4}\right )} x^{8} + 25 \, {\left (3 \, D a^{4} - 15 \, C a^{3} b + 35 \, B a^{2} b^{2} - 63 \, A a b^{3}\right )} x^{6} - 24 \, A a^{4} - 8 \, {\left (15 \, C a^{4} - 35 \, B a^{3} b + 63 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2}}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}} + \frac {{\left (3 \, D a^{3} - 15 \, C a^{2} b + 35 \, B a b^{2} - 63 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/x^6/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

1/120*(15*(3*D*a^3*b - 15*C*a^2*b^2 + 35*B*a*b^3 - 63*A*b^4)*x^8 + 25*(3*D 
*a^4 - 15*C*a^3*b + 35*B*a^2*b^2 - 63*A*a*b^3)*x^6 - 24*A*a^4 - 8*(15*C*a^ 
4 - 35*B*a^3*b + 63*A*a^2*b^2)*x^4 - 8*(5*B*a^4 - 9*A*a^3*b)*x^2)/(a^5*b^2 
*x^9 + 2*a^6*b*x^7 + a^7*x^5) + 1/8*(3*D*a^3 - 15*C*a^2*b + 35*B*a*b^2 - 6 
3*A*b^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, D a^{3} - 15 \, C a^{2} b + 35 \, B a b^{2} - 63 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} + \frac {3 \, D a^{3} b x^{3} - 7 \, C a^{2} b^{2} x^{3} + 11 \, B a b^{3} x^{3} - 15 \, A b^{4} x^{3} + 5 \, D a^{4} x - 9 \, C a^{3} b x + 13 \, B a^{2} b^{2} x - 17 \, A a b^{3} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{5}} - \frac {15 \, C a^{2} x^{4} - 45 \, B a b x^{4} + 90 \, A b^{2} x^{4} + 5 \, B a^{2} x^{2} - 15 \, A a b x^{2} + 3 \, A a^{2}}{15 \, a^{5} x^{5}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/x^6/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

1/8*(3*D*a^3 - 15*C*a^2*b + 35*B*a*b^2 - 63*A*b^3)*arctan(b*x/sqrt(a*b))/( 
sqrt(a*b)*a^5) + 1/8*(3*D*a^3*b*x^3 - 7*C*a^2*b^2*x^3 + 11*B*a*b^3*x^3 - 1 
5*A*b^4*x^3 + 5*D*a^4*x - 9*C*a^3*b*x + 13*B*a^2*b^2*x - 17*A*a*b^3*x)/((b 
*x^2 + a)^2*a^5) - 1/15*(15*C*a^2*x^4 - 45*B*a*b*x^4 + 90*A*b^2*x^4 + 5*B* 
a^2*x^2 - 15*A*a*b*x^2 + 3*A*a^2)/(a^5*x^5)
 

Mupad [B] (verification not implemented)

Time = 1.58 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {\frac {7\,B\,b\,x^2}{3\,a^2}-\frac {B}{3\,a}+\frac {175\,B\,b^2\,x^4}{24\,a^3}+\frac {35\,B\,b^3\,x^6}{8\,a^4}}{a^2\,x^3+2\,a\,b\,x^5+b^2\,x^7}-\frac {\frac {A}{5\,a}-\frac {3\,A\,b\,x^2}{5\,a^2}+\frac {21\,A\,b^2\,x^4}{5\,a^3}+\frac {105\,A\,b^3\,x^6}{8\,a^4}+\frac {63\,A\,b^4\,x^8}{8\,a^5}}{a^2\,x^5+2\,a\,b\,x^7+b^2\,x^9}-\frac {\frac {C}{a}+\frac {25\,C\,b\,x^2}{8\,a^2}+\frac {15\,C\,b^2\,x^4}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}+\frac {x\,D\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},3;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{a^3}-\frac {63\,A\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{11/2}}+\frac {35\,B\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{9/2}}-\frac {15\,C\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}} \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(x^6*(a + b*x^2)^3),x)
 

Output:

((7*B*b*x^2)/(3*a^2) - B/(3*a) + (175*B*b^2*x^4)/(24*a^3) + (35*B*b^3*x^6) 
/(8*a^4))/(a^2*x^3 + b^2*x^7 + 2*a*b*x^5) - (A/(5*a) - (3*A*b*x^2)/(5*a^2) 
 + (21*A*b^2*x^4)/(5*a^3) + (105*A*b^3*x^6)/(8*a^4) + (63*A*b^4*x^8)/(8*a^ 
5))/(a^2*x^5 + b^2*x^9 + 2*a*b*x^7) - (C/a + (25*C*b*x^2)/(8*a^2) + (15*C* 
b^2*x^4)/(8*a^3))/(a^2*x + b^2*x^5 + 2*a*b*x^3) + (x*D*hypergeom([1/2, 3], 
 3/2, -(b*x^2)/a))/a^3 - (63*A*b^(5/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(11 
/2)) + (35*B*b^(3/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(9/2)) - (15*C*b^(1/2 
)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.86 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} d \,x^{5}-225 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b c \,x^{5}+90 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b d \,x^{7}-420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3} x^{5}-450 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c \,x^{7}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} d \,x^{9}-840 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{4} x^{7}-225 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c \,x^{9}-420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{5} x^{9}-24 a^{5} b +32 a^{4} b^{2} x^{2}-120 a^{4} b c \,x^{4}+75 a^{4} b d \,x^{6}-224 a^{3} b^{3} x^{4}-375 a^{3} b^{2} c \,x^{6}+45 a^{3} b^{2} d \,x^{8}-700 a^{2} b^{4} x^{6}-225 a^{2} b^{3} c \,x^{8}-420 a \,b^{5} x^{8}}{120 a^{5} b \,x^{5} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/x^6/(b*x^2+a)^3,x)
 

Output:

(45*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*d*x**5 - 225*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c*x**5 + 90*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*d*x**7 - 420*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**2*b**3*x**5 - 450*sqrt(b)*sqrt(a)*atan((b*x)/(sq 
rt(b)*sqrt(a)))*a**2*b**2*c*x**7 + 45*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)* 
sqrt(a)))*a**2*b**2*d*x**9 - 840*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt( 
a)))*a*b**4*x**7 - 225*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b** 
3*c*x**9 - 420*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**5*x**9 - 2 
4*a**5*b + 32*a**4*b**2*x**2 - 120*a**4*b*c*x**4 + 75*a**4*b*d*x**6 - 224* 
a**3*b**3*x**4 - 375*a**3*b**2*c*x**6 + 45*a**3*b**2*d*x**8 - 700*a**2*b** 
4*x**6 - 225*a**2*b**3*c*x**8 - 420*a*b**5*x**8)/(120*a**5*b*x**5*(a**2 + 
2*a*b*x**2 + b**2*x**4))