\(\int \frac {A+B x^2}{x^6 (a+b x^2)^3} \, dx\) [112]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 142 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {A}{5 a^3 x^5}+\frac {3 A b-a B}{3 a^4 x^3}-\frac {3 b (2 A b-a B)}{a^5 x}-\frac {b^2 (A b-a B) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {b^2 (15 A b-11 a B) x}{8 a^5 \left (a+b x^2\right )}-\frac {7 b^{3/2} (9 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}} \] Output:

-1/5*A/a^3/x^5+1/3*(3*A*b-B*a)/a^4/x^3-3*b*(2*A*b-B*a)/a^5/x-1/4*b^2*(A*b- 
B*a)*x/a^4/(b*x^2+a)^2-1/8*b^2*(15*A*b-11*B*a)*x/a^5/(b*x^2+a)-7/8*b^(3/2) 
*(9*A*b-5*B*a)*arctan(b^(1/2)*x/a^(1/2))/a^(11/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {-945 A b^4 x^8+525 a b^3 x^6 \left (-3 A+B x^2\right )-8 a^4 \left (3 A+5 B x^2\right )+8 a^3 b x^2 \left (9 A+35 B x^2\right )+7 a^2 b^2 x^4 \left (-72 A+125 B x^2\right )}{120 a^5 x^5 \left (a+b x^2\right )^2}+\frac {7 b^{3/2} (-9 A b+5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2}} \] Input:

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

Output:

(-945*A*b^4*x^8 + 525*a*b^3*x^6*(-3*A + B*x^2) - 8*a^4*(3*A + 5*B*x^2) + 8 
*a^3*b*x^2*(9*A + 35*B*x^2) + 7*a^2*b^2*x^4*(-72*A + 125*B*x^2))/(120*a^5* 
x^5*(a + b*x^2)^2) + (7*b^(3/2)*(-9*A*b + 5*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a 
]])/(8*a^(11/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {361, 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.

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

\(\displaystyle \frac {1}{4} b^2 \left (-\frac {\int -\frac {-\frac {(15 A b-11 a B) x^6}{a^4}+\frac {8 (3 A b-2 a B) x^4}{a^3 b}-\frac {8 (2 A b-a B) x^2}{a^2 b^2}+\frac {8 A}{a b^2}}{x^6 \left (b x^2+a\right )}dx}{2 a}-\frac {x (15 A b-11 a B)}{2 a^5 \left (a+b x^2\right )}\right )-\frac {b^2 x (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} b^2 \left (\frac {\int \frac {-\frac {(15 A b-11 a B) x^6}{a^4}+\frac {8 (3 A b-2 a B) x^4}{a^3 b}-\frac {8 (2 A b-a B) x^2}{a^2 b^2}+\frac {8 A}{a b^2}}{x^6 \left (b x^2+a\right )}dx}{2 a}-\frac {x (15 A b-11 a B)}{2 a^5 \left (a+b x^2\right )}\right )-\frac {b^2 x (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2333

\(\displaystyle \frac {1}{4} b^2 \left (\frac {\int \left (\frac {8 A}{a^2 b^2 x^6}+\frac {7 (5 a B-9 A b)}{a^4 \left (b x^2+a\right )}-\frac {24 (a B-2 A b)}{a^4 b x^2}+\frac {8 (a B-3 A b)}{a^3 b^2 x^4}\right )dx}{2 a}-\frac {x (15 A b-11 a B)}{2 a^5 \left (a+b x^2\right )}\right )-\frac {b^2 x (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} b^2 \left (\frac {-\frac {7 (9 A b-5 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{9/2} \sqrt {b}}-\frac {24 (2 A b-a B)}{a^4 b x}+\frac {8 (3 A b-a B)}{3 a^3 b^2 x^3}-\frac {8 A}{5 a^2 b^2 x^5}}{2 a}-\frac {x (15 A b-11 a B)}{2 a^5 \left (a+b x^2\right )}\right )-\frac {b^2 x (A b-a B)}{4 a^4 \left (a+b x^2\right )^2}\)

Input:

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

Output:

-1/4*(b^2*(A*b - a*B)*x)/(a^4*(a + b*x^2)^2) + (b^2*(-1/2*((15*A*b - 11*a* 
B)*x)/(a^5*(a + b*x^2)) + ((-8*A)/(5*a^2*b^2*x^5) + (8*(3*A*b - a*B))/(3*a 
^3*b^2*x^3) - (24*(2*A*b - a*B))/(a^4*b*x) - (7*(9*A*b - 5*a*B)*ArcTan[(Sq 
rt[b]*x)/Sqrt[a]])/(a^(9/2)*Sqrt[b]))/(2*a)))/4
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 361
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

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 = 0.42 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84

method result size
default \(-\frac {b^{2} \left (\frac {\left (\frac {15}{8} b^{2} A -\frac {11}{8} a b B \right ) x^{3}+\frac {a \left (17 A b -13 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (9 A b -5 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}-\frac {A}{5 a^{3} x^{5}}-\frac {-3 A b +B a}{3 a^{4} x^{3}}-\frac {3 b \left (2 A b -B a \right )}{a^{5} x}\) \(119\)
risch \(\frac {-\frac {7 b^{3} \left (9 A b -5 B a \right ) x^{8}}{8 a^{5}}-\frac {35 b^{2} \left (9 A b -5 B a \right ) x^{6}}{24 a^{4}}-\frac {7 b \left (9 A b -5 B a \right ) x^{4}}{15 a^{3}}+\frac {\left (9 A b -5 B a \right ) x^{2}}{15 a^{2}}-\frac {A}{5 a}}{x^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {63 \sqrt {-a b}\, b^{2} \ln \left (-b x +\sqrt {-a b}\right ) A}{16 a^{6}}-\frac {35 \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right ) B}{16 a^{5}}-\frac {63 \sqrt {-a b}\, b^{2} \ln \left (-b x -\sqrt {-a b}\right ) A}{16 a^{6}}+\frac {35 \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right ) B}{16 a^{5}}\) \(205\)

Input:

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

Output:

-1/a^5*b^2*(((15/8*b^2*A-11/8*a*b*B)*x^3+1/8*a*(17*A*b-13*B*a)*x)/(b*x^2+a 
)^2+7/8*(9*A*b-5*B*a)/(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^3-3*b*(2*A*b-B*a)/a^5/x
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.00 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\left [\frac {210 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 350 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} + 112 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 16 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2} - 105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{9} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{7} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{5}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{240 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}}, \frac {105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 24 \, A a^{4} + 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2} + 105 \, {\left ({\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{9} + 2 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{7} + {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{5}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}}\right ] \] Input:

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

Output:

[1/240*(210*(5*B*a*b^3 - 9*A*b^4)*x^8 + 350*(5*B*a^2*b^2 - 9*A*a*b^3)*x^6 
- 48*A*a^4 + 112*(5*B*a^3*b - 9*A*a^2*b^2)*x^4 - 16*(5*B*a^4 - 9*A*a^3*b)* 
x^2 - 105*((5*B*a*b^3 - 9*A*b^4)*x^9 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^7 + ( 
5*B*a^3*b - 9*A*a^2*b^2)*x^5)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a 
)/(b*x^2 + a)))/(a^5*b^2*x^9 + 2*a^6*b*x^7 + a^7*x^5), 1/120*(105*(5*B*a*b 
^3 - 9*A*b^4)*x^8 + 175*(5*B*a^2*b^2 - 9*A*a*b^3)*x^6 - 24*A*a^4 + 56*(5*B 
*a^3*b - 9*A*a^2*b^2)*x^4 - 8*(5*B*a^4 - 9*A*a^3*b)*x^2 + 105*((5*B*a*b^3 
- 9*A*b^4)*x^9 + 2*(5*B*a^2*b^2 - 9*A*a*b^3)*x^7 + (5*B*a^3*b - 9*A*a^2*b^ 
2)*x^5)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^5*b^2*x^9 + 2*a^6*b*x^7 + a^7*x^ 
5)]
 

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=- \frac {7 \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right ) \log {\left (- \frac {7 a^{6} \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right )}{- 63 A b^{3} + 35 B a b^{2}} + x \right )}}{16} + \frac {7 \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right ) \log {\left (\frac {7 a^{6} \sqrt {- \frac {b^{3}}{a^{11}}} \left (- 9 A b + 5 B a\right )}{- 63 A b^{3} + 35 B a b^{2}} + x \right )}}{16} + \frac {- 24 A a^{4} + x^{8} \left (- 945 A b^{4} + 525 B a b^{3}\right ) + x^{6} \left (- 1575 A a b^{3} + 875 B a^{2} b^{2}\right ) + x^{4} \left (- 504 A a^{2} b^{2} + 280 B a^{3} b\right ) + x^{2} \cdot \left (72 A a^{3} b - 40 B a^{4}\right )}{120 a^{7} x^{5} + 240 a^{6} b x^{7} + 120 a^{5} b^{2} x^{9}} \] Input:

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

Output:

-7*sqrt(-b**3/a**11)*(-9*A*b + 5*B*a)*log(-7*a**6*sqrt(-b**3/a**11)*(-9*A* 
b + 5*B*a)/(-63*A*b**3 + 35*B*a*b**2) + x)/16 + 7*sqrt(-b**3/a**11)*(-9*A* 
b + 5*B*a)*log(7*a**6*sqrt(-b**3/a**11)*(-9*A*b + 5*B*a)/(-63*A*b**3 + 35* 
B*a*b**2) + x)/16 + (-24*A*a**4 + x**8*(-945*A*b**4 + 525*B*a*b**3) + x**6 
*(-1575*A*a*b**3 + 875*B*a**2*b**2) + x**4*(-504*A*a**2*b**2 + 280*B*a**3* 
b) + x**2*(72*A*a**3*b - 40*B*a**4))/(120*a**7*x**5 + 240*a**6*b*x**7 + 12 
0*a**5*b**2*x**9)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {105 \, {\left (5 \, B a b^{3} - 9 \, A b^{4}\right )} x^{8} + 175 \, {\left (5 \, B a^{2} b^{2} - 9 \, A a b^{3}\right )} x^{6} - 24 \, A a^{4} + 56 \, {\left (5 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (5 \, B a^{4} - 9 \, A a^{3} b\right )} x^{2}}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}} + \frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \] Input:

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

Output:

1/120*(105*(5*B*a*b^3 - 9*A*b^4)*x^8 + 175*(5*B*a^2*b^2 - 9*A*a*b^3)*x^6 - 
 24*A*a^4 + 56*(5*B*a^3*b - 9*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) + 7/8*(5*B*a*b^2 - 9*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 = 135, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {7 \, {\left (5 \, B a b^{2} - 9 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} + \frac {11 \, B a b^{3} x^{3} - 15 \, A b^{4} x^{3} + 13 \, B a^{2} b^{2} x - 17 \, A a b^{3} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{5}} + \frac {45 \, B a b x^{4} - 90 \, A b^{2} x^{4} - 5 \, B a^{2} x^{2} + 15 \, A a b x^{2} - 3 \, A a^{2}}{15 \, a^{5} x^{5}} \] Input:

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

Output:

7/8*(5*B*a*b^2 - 9*A*b^3)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/8*(11* 
B*a*b^3*x^3 - 15*A*b^4*x^3 + 13*B*a^2*b^2*x - 17*A*a*b^3*x)/((b*x^2 + a)^2 
*a^5) + 1/15*(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 = 0.47 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=-\frac {\frac {A}{5\,a}-\frac {x^2\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^2}+\frac {35\,b^2\,x^6\,\left (9\,A\,b-5\,B\,a\right )}{24\,a^4}+\frac {7\,b^3\,x^8\,\left (9\,A\,b-5\,B\,a\right )}{8\,a^5}+\frac {7\,b\,x^4\,\left (9\,A\,b-5\,B\,a\right )}{15\,a^3}}{a^2\,x^5+2\,a\,b\,x^7+b^2\,x^9}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (9\,A\,b-5\,B\,a\right )}{8\,a^{11/2}} \] Input:

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

Output:

- (A/(5*a) - (x^2*(9*A*b - 5*B*a))/(15*a^2) + (35*b^2*x^6*(9*A*b - 5*B*a)) 
/(24*a^4) + (7*b^3*x^8*(9*A*b - 5*B*a))/(8*a^5) + (7*b*x^4*(9*A*b - 5*B*a) 
)/(15*a^3))/(a^2*x^5 + b^2*x^9 + 2*a*b*x^7) - (7*b^(3/2)*atan((b^(1/2)*x)/ 
a^(1/2))*(9*A*b - 5*B*a))/(8*a^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^3} \, dx=\frac {-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{5}-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{7}-6 a^{4}+14 a^{3} b \,x^{2}-70 a^{2} b^{2} x^{4}-105 a \,b^{3} x^{6}}{30 a^{5} x^{5} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

( - 105*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*x**5 - 105*sq 
rt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*x**7 - 6*a**4 + 14*a**3*b 
*x**2 - 70*a**2*b**2*x**4 - 105*a*b**3*x**6)/(30*a**5*x**5*(a + b*x**2))