\(\int \frac {A+B x^2}{(b x^2+c x^4)^3} \, dx\) [91]

Optimal result

Integrand size = 21, antiderivative size = 140 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{5 b^3 x^5}-\frac {b B-3 A c}{3 b^4 x^3}+\frac {3 c (b B-2 A c)}{b^5 x}+\frac {c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac {c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{5 b^3 x^5}-\frac {b B-3 A c}{3 b^4 x^3}+\frac {3 c (b B-2 A c)}{b^5 x}+\frac {c^2 (b B-A c) x}{4 b^4 \left (b+c x^2\right )^2}+\frac {c^2 (11 b B-15 A c) x}{8 b^5 \left (b+c x^2\right )}+\frac {7 c^{3/2} (5 b B-9 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{11/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2026, 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.

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

\(\Big \downarrow \) 2026

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2333

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c^2*(b*B - A*c)*x)/(4*b^4*(b + c*x^2)^2) + (c^2*(((11*b*B - 15*A*c)*x)/(2 
*b^5*(b + c*x^2)) + ((-8*A)/(5*b^2*c^2*x^5) - (8*(b*B - 3*A*c))/(3*b^3*c^2 
*x^3) + (24*(b*B - 2*A*c))/(b^4*c*x) + (7*(5*b*B - 9*A*c)*ArcTan[(Sqrt[c]* 
x)/Sqrt[b]])/(b^(9/2)*Sqrt[c]))/(2*b)))/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 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

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 = 0.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.85

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.04 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {210 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{8} + 350 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{6} - 48 \, A b^{4} + 112 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{4} - 16 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x^{2} - 105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{9} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{7} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{240 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}, \frac {105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{8} + 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{6} - 24 \, A b^{4} + 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{4} - 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x^{2} + 105 \, {\left ({\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{9} + 2 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{7} + {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{120 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.86 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {7 \sqrt {- \frac {c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right ) \log {\left (- \frac {7 b^{6} \sqrt {- \frac {c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right )}{- 63 A c^{3} + 35 B b c^{2}} + x \right )}}{16} + \frac {7 \sqrt {- \frac {c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right ) \log {\left (\frac {7 b^{6} \sqrt {- \frac {c^{3}}{b^{11}}} \left (- 9 A c + 5 B b\right )}{- 63 A c^{3} + 35 B b c^{2}} + x \right )}}{16} + \frac {- 24 A b^{4} + x^{8} \left (- 945 A c^{4} + 525 B b c^{3}\right ) + x^{6} \left (- 1575 A b c^{3} + 875 B b^{2} c^{2}\right ) + x^{4} \left (- 504 A b^{2} c^{2} + 280 B b^{3} c\right ) + x^{2} \cdot \left (72 A b^{3} c - 40 B b^{4}\right )}{120 b^{7} x^{5} + 240 b^{6} c x^{7} + 120 b^{5} c^{2} x^{9}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {105 \, {\left (5 \, B b c^{3} - 9 \, A c^{4}\right )} x^{8} + 175 \, {\left (5 \, B b^{2} c^{2} - 9 \, A b c^{3}\right )} x^{6} - 24 \, A b^{4} + 56 \, {\left (5 \, B b^{3} c - 9 \, A b^{2} c^{2}\right )} x^{4} - 8 \, {\left (5 \, B b^{4} - 9 \, A b^{3} c\right )} x^{2}}{120 \, {\left (b^{5} c^{2} x^{9} + 2 \, b^{6} c x^{7} + b^{7} x^{5}\right )}} + \frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} \] Input:

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

Output:

1/120*(105*(5*B*b*c^3 - 9*A*c^4)*x^8 + 175*(5*B*b^2*c^2 - 9*A*b*c^3)*x^6 - 
 24*A*b^4 + 56*(5*B*b^3*c - 9*A*b^2*c^2)*x^4 - 8*(5*B*b^4 - 9*A*b^3*c)*x^2 
)/(b^5*c^2*x^9 + 2*b^6*c*x^7 + b^7*x^5) + 7/8*(5*B*b*c^2 - 9*A*c^3)*arctan 
(c*x/sqrt(b*c))/(sqrt(b*c)*b^5)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {7 \, {\left (5 \, B b c^{2} - 9 \, A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{5}} + \frac {11 \, B b c^{3} x^{3} - 15 \, A c^{4} x^{3} + 13 \, B b^{2} c^{2} x - 17 \, A b c^{3} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{5}} + \frac {45 \, B b c x^{4} - 90 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 15 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{5} x^{5}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.87 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {A}{5\,b}-\frac {x^2\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {35\,c^2\,x^6\,\left (9\,A\,c-5\,B\,b\right )}{24\,b^4}+\frac {7\,c^3\,x^8\,\left (9\,A\,c-5\,B\,b\right )}{8\,b^5}+\frac {7\,c\,x^4\,\left (9\,A\,c-5\,B\,b\right )}{15\,b^3}}{b^2\,x^5+2\,b\,c\,x^7+c^2\,x^9}-\frac {7\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (9\,A\,c-5\,B\,b\right )}{8\,b^{11/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.97 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c^{2} x^{5}-1890 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{3} x^{7}-945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{4} x^{9}+525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4} c \,x^{5}+1050 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c^{2} x^{7}+525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{3} x^{9}-24 a \,b^{5}+72 a \,b^{4} c \,x^{2}-504 a \,b^{3} c^{2} x^{4}-1575 a \,b^{2} c^{3} x^{6}-945 a b \,c^{4} x^{8}-40 b^{6} x^{2}+280 b^{5} c \,x^{4}+875 b^{4} c^{2} x^{6}+525 b^{3} c^{3} x^{8}}{120 b^{6} x^{5} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

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

Output:

( - 945*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c**2*x**5 - 1 
890*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**3*x**7 - 945*sqrt 
(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**4*x**9 + 525*sqrt(c)*sqrt(b 
)*atan((c*x)/(sqrt(c)*sqrt(b)))*b**4*c*x**5 + 1050*sqrt(c)*sqrt(b)*atan((c 
*x)/(sqrt(c)*sqrt(b)))*b**3*c**2*x**7 + 525*sqrt(c)*sqrt(b)*atan((c*x)/(sq 
rt(c)*sqrt(b)))*b**2*c**3*x**9 - 24*a*b**5 + 72*a*b**4*c*x**2 - 504*a*b**3 
*c**2*x**4 - 1575*a*b**2*c**3*x**6 - 945*a*b*c**4*x**8 - 40*b**6*x**2 + 28 
0*b**5*c*x**4 + 875*b**4*c**2*x**6 + 525*b**3*c**3*x**8)/(120*b**6*x**5*(b 
**2 + 2*b*c*x**2 + c**2*x**4))