\(\int \frac {x^{14} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [77]

Optimal result

Integrand size = 24, antiderivative size = 140 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 b (2 b B-A c) x}{c^5}-\frac {(3 b B-A c) x^3}{3 c^4}+\frac {B x^5}{5 c^3}-\frac {b^3 (b B-A c) x}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (17 b B-13 A c) x}{8 c^5 \left (b+c x^2\right )}-\frac {7 b^{3/2} (9 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}} \] Output:

3*b*(-A*c+2*B*b)*x/c^5-1/3*(-A*c+3*B*b)*x^3/c^4+1/5*B*x^5/c^3-1/4*b^3*(-A* 
c+B*b)*x/c^5/(c*x^2+b)^2+1/8*b^2*(-13*A*c+17*B*b)*x/c^5/(c*x^2+b)-7/8*b^(3 
/2)*(-5*A*c+9*B*b)*arctan(c^(1/2)*x/b^(1/2))/c^(11/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.95 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x \left (945 b^4 B-525 b^3 c \left (A-3 B x^2\right )+8 c^4 x^6 \left (5 A+3 B x^2\right )-8 b c^3 x^4 \left (35 A+9 B x^2\right )+7 b^2 c^2 x^2 \left (-125 A+72 B x^2\right )\right )}{120 c^5 \left (b+c x^2\right )^2}-\frac {7 b^{3/2} (9 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 c^{11/2}} \] Input:

Integrate[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {9, 360, 25, 2345, 2341, 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.

\(\displaystyle \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 360

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2345

\(\displaystyle \frac {\frac {b^2 x (17 b B-13 A c)}{2 \left (b+c x^2\right )}-\frac {\int \frac {-8 b B c^3 x^6+8 b c^2 (2 b B-A c) x^4-8 b^2 c (3 b B-2 A c) x^2+b^3 (15 b B-11 A c)}{c x^2+b}dx}{2 b}}{4 c^5}-\frac {b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {\frac {b^2 x (17 b B-13 A c)}{2 \left (b+c x^2\right )}-\frac {\int \left (-8 b B c^2 x^4+8 b c (3 b B-A c) x^2-24 b^2 (2 b B-A c)+\frac {7 \left (9 b^4 B-5 A b^3 c\right )}{c x^2+b}\right )dx}{2 b}}{4 c^5}-\frac {b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^2 x (17 b B-13 A c)}{2 \left (b+c x^2\right )}-\frac {\frac {7 b^{5/2} (9 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {c}}-24 b^2 x (2 b B-A c)+\frac {8}{3} b c x^3 (3 b B-A c)-\frac {8}{5} b B c^2 x^5}{2 b}}{4 c^5}-\frac {b^3 x (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}\)

Input:

Int[(x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*(b^3*(b*B - A*c)*x)/(c^5*(b + c*x^2)^2) + ((b^2*(17*b*B - 13*A*c)*x)/ 
(2*(b + c*x^2)) - (-24*b^2*(2*b*B - A*c)*x + (8*b*c*(3*b*B - A*c)*x^3)/3 - 
 (8*b*B*c^2*x^5)/5 + (7*b^(5/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b] 
])/Sqrt[c])/(2*b))/(4*c^5)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 25

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

rule 360

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[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[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 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -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 = 0.44 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {-\frac {1}{5} B \,c^{2} x^{5}-\frac {1}{3} A \,c^{2} x^{3}+B b c \,x^{3}+3 A b c x -6 x B \,b^{2}}{c^{5}}+\frac {b^{2} \left (\frac {\left (-\frac {13}{8} A \,c^{2}+\frac {17}{8} B b c \right ) x^{3}-\frac {b \left (11 A c -15 B b \right ) x}{8}}{\left (c \,x^{2}+b \right )^{2}}+\frac {7 \left (5 A c -9 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{5}}\)	\(119\)
risch	\(\frac {B \,x^{5}}{5 c^{3}}+\frac {A \,x^{3}}{3 c^{3}}-\frac {B b \,x^{3}}{c^{4}}-\frac {3 A b x}{c^{4}}+\frac {6 x B \,b^{2}}{c^{5}}+\frac {\left (-\frac {13}{8} A \,b^{2} c^{2}+\frac {17}{8} B \,b^{3} c \right ) x^{3}-\frac {b^{3} \left (11 A c -15 B b \right ) x}{8}}{c^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {35 \sqrt {-b c}\, b \ln \left (-\sqrt {-b c}\, x +b \right ) A}{16 c^{5}}-\frac {63 \sqrt {-b c}\, b^{2} \ln \left (-\sqrt {-b c}\, x +b \right ) B}{16 c^{6}}-\frac {35 \sqrt {-b c}\, b \ln \left (\sqrt {-b c}\, x +b \right ) A}{16 c^{5}}+\frac {63 \sqrt {-b c}\, b^{2} \ln \left (\sqrt {-b c}\, x +b \right ) B}{16 c^{6}}\)	\(200\)

Input:

int(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/c^5*(-1/5*B*c^2*x^5-1/3*A*c^2*x^3+B*b*c*x^3+3*A*b*c*x-6*x*B*b^2)+b^2/c^ 
5*(((-13/8*A*c^2+17/8*B*b*c)*x^3-1/8*b*(11*A*c-15*B*b)*x)/(c*x^2+b)^2+7/8* 
(5*A*c-9*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.97 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {48 \, B c^{4} x^{9} - 16 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 112 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 350 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 210 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{240 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}, \frac {24 \, B c^{4} x^{9} - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{7} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{5} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{3} - 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{4} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c\right )} x}{120 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}}\right ] \] Input:

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.80 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^{5}}{5 c^{3}} + x^{3} \left (\frac {A}{3 c^{3}} - \frac {B b}{c^{4}}\right ) + x \left (- \frac {3 A b}{c^{4}} + \frac {6 B b^{2}}{c^{5}}\right ) + \frac {7 \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log {\left (- \frac {7 c^{5} \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} - \frac {7 \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right ) \log {\left (\frac {7 c^{5} \sqrt {- \frac {b^{3}}{c^{11}}} \left (- 5 A c + 9 B b\right )}{- 35 A b c + 63 B b^{2}} + x \right )}}{16} + \frac {x^{3} \left (- 13 A b^{2} c^{2} + 17 B b^{3} c\right ) + x \left (- 11 A b^{3} c + 15 B b^{4}\right )}{8 b^{2} c^{5} + 16 b c^{6} x^{2} + 8 c^{7} x^{4}} \] Input:

integrate(x**14*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

B*x**5/(5*c**3) + x**3*(A/(3*c**3) - B*b/c**4) + x*(-3*A*b/c**4 + 6*B*b**2 
/c**5) + 7*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)*log(-7*c**5*sqrt(-b**3/c**11 
)*(-5*A*c + 9*B*b)/(-35*A*b*c + 63*B*b**2) + x)/16 - 7*sqrt(-b**3/c**11)*( 
-5*A*c + 9*B*b)*log(7*c**5*sqrt(-b**3/c**11)*(-5*A*c + 9*B*b)/(-35*A*b*c + 
 63*B*b**2) + x)/16 + (x**3*(-13*A*b**2*c**2 + 17*B*b**3*c) + x*(-11*A*b** 
3*c + 15*B*b**4))/(8*b**2*c**5 + 16*b*c**6*x**2 + 8*c**7*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (17 \, B b^{3} c - 13 \, A b^{2} c^{2}\right )} x^{3} + {\left (15 \, B b^{4} - 11 \, A b^{3} c\right )} x}{8 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} - \frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{5}} + \frac {3 \, B c^{2} x^{5} - 5 \, {\left (3 \, B b c - A c^{2}\right )} x^{3} + 45 \, {\left (2 \, B b^{2} - A b c\right )} x}{15 \, c^{5}} \] Input:

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

1/8*((17*B*b^3*c - 13*A*b^2*c^2)*x^3 + (15*B*b^4 - 11*A*b^3*c)*x)/(c^7*x^4 
 + 2*b*c^6*x^2 + b^2*c^5) - 7/8*(9*B*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c) 
)/(sqrt(b*c)*c^5) + 1/15*(3*B*c^2*x^5 - 5*(3*B*b*c - A*c^2)*x^3 + 45*(2*B* 
b^2 - A*b*c)*x)/c^5
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} c^{5}} + \frac {17 \, B b^{3} c x^{3} - 13 \, A b^{2} c^{2} x^{3} + 15 \, B b^{4} x - 11 \, A b^{3} c x}{8 \, {\left (c x^{2} + b\right )}^{2} c^{5}} + \frac {3 \, B c^{12} x^{5} - 15 \, B b c^{11} x^{3} + 5 \, A c^{12} x^{3} + 90 \, B b^{2} c^{10} x - 45 \, A b c^{11} x}{15 \, c^{15}} \] Input:

integrate(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

-7/8*(9*B*b^3 - 5*A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/8*(17 
*B*b^3*c*x^3 - 13*A*b^2*c^2*x^3 + 15*B*b^4*x - 11*A*b^3*c*x)/((c*x^2 + b)^ 
2*c^5) + 1/15*(3*B*c^12*x^5 - 15*B*b*c^11*x^3 + 5*A*c^12*x^3 + 90*B*b^2*c^ 
10*x - 45*A*b*c^11*x)/c^15
 

Mupad [B] (verification not implemented)

Time = 9.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.26 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x\,\left (\frac {15\,B\,b^4}{8}-\frac {11\,A\,b^3\,c}{8}\right )-x^3\,\left (\frac {13\,A\,b^2\,c^2}{8}-\frac {17\,B\,b^3\,c}{8}\right )}{b^2\,c^5+2\,b\,c^6\,x^2+c^7\,x^4}-x\,\left (\frac {3\,b\,\left (\frac {A}{c^3}-\frac {3\,B\,b}{c^4}\right )}{c}+\frac {3\,B\,b^2}{c^5}\right )+x^3\,\left (\frac {A}{3\,c^3}-\frac {B\,b}{c^4}\right )+\frac {B\,x^5}{5\,c^3}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,x\,\left (5\,A\,c-9\,B\,b\right )}{9\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-9\,B\,b\right )}{8\,c^{11/2}} \] Input:

int((x^14*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(x*((15*B*b^4)/8 - (11*A*b^3*c)/8) - x^3*((13*A*b^2*c^2)/8 - (17*B*b^3*c)/ 
8))/(b^2*c^5 + c^7*x^4 + 2*b*c^6*x^2) - x*((3*b*(A/c^3 - (3*B*b)/c^4))/c + 
 (3*B*b^2)/c^5) + x^3*(A/(3*c^3) - (B*b)/c^4) + (B*x^5)/(5*c^3) - (7*b^(3/ 
2)*atan((b^(3/2)*c^(1/2)*x*(5*A*c - 9*B*b))/(9*B*b^3 - 5*A*b^2*c))*(5*A*c 
- 9*B*b))/(8*c^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.90 \[ \int \frac {x^{14} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{3} c +1050 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c^{2} x^{2}+525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{3} x^{4}-945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{5}-1890 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4} c \,x^{2}-945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c^{2} x^{4}-525 a \,b^{3} c^{2} x -875 a \,b^{2} c^{3} x^{3}-280 a b \,c^{4} x^{5}+40 a \,c^{5} x^{7}+945 b^{5} c x +1575 b^{4} c^{2} x^{3}+504 b^{3} c^{3} x^{5}-72 b^{2} c^{4} x^{7}+24 b \,c^{5} x^{9}}{120 c^{6} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

int(x^14*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

(525*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**3*c + 1050*sqrt(c) 
*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c**2*x**2 + 525*sqrt(c)*sqrt 
(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**3*x**4 - 945*sqrt(c)*sqrt(b)*atan 
((c*x)/(sqrt(c)*sqrt(b)))*b**5 - 1890*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)* 
sqrt(b)))*b**4*c*x**2 - 945*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))* 
b**3*c**2*x**4 - 525*a*b**3*c**2*x - 875*a*b**2*c**3*x**3 - 280*a*b*c**4*x 
**5 + 40*a*c**5*x**7 + 945*b**5*c*x + 1575*b**4*c**2*x**3 + 504*b**3*c**3* 
x**5 - 72*b**2*c**4*x**7 + 24*b*c**5*x**9)/(120*c**6*(b**2 + 2*b*c*x**2 + 
c**2*x**4))