\(\int \frac {(a+b x^2) (c+d x^2)}{(e+f x^2)^4} \, dx\) [207]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {(b e-a f) (d e-c f) x}{6 e f^2 \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \] Output:

1/6*(-a*f+b*e)*(-c*f+d*e)*x/e/f^2/(f*x^2+e)^3-1/24*(b*e*(-c*f+7*d*e)-a*f*( 
5*c*f+d*e))*x/e^2/f^2/(f*x^2+e)^2+1/16*(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*x/e 
^3/f^2/(f*x^2+e)+1/16*(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*arctan(f^(1/2)*x/e^( 
1/2))/e^(7/2)/f^(5/2)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {(b e-a f) (d e-c f) x}{6 e f^2 \left (e+f x^2\right )^3}+\frac {(b e (-7 d e+c f)+a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \] Input:

Integrate[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]
 

Output:

((b*e - a*f)*(d*e - c*f)*x)/(6*e*f^2*(e + f*x^2)^3) + ((b*e*(-7*d*e + c*f) 
 + a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + 
a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f*( 
d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {401, 25, 298, 215, 218}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx\)

\(\Big \downarrow \) 401

\(\displaystyle -\frac {\int -\frac {3 b (d e+c f) x^2+a (d e+5 c f)}{\left (f x^2+e\right )^3}dx}{6 e f}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 b (d e+c f) x^2+a (d e+5 c f)}{\left (f x^2+e\right )^3}dx}{6 e f}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {3 (a f (5 c f+d e)+b e (c f+d e)) \int \frac {1}{\left (f x^2+e\right )^2}dx}{4 e f}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {3 (a f (5 c f+d e)+b e (c f+d e)) \left (\frac {\int \frac {1}{f x^2+e}dx}{2 e}+\frac {x}{2 e \left (e+f x^2\right )}\right )}{4 e f}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} \sqrt {f}}+\frac {x}{2 e \left (e+f x^2\right )}\right ) (a f (5 c f+d e)+b e (c f+d e))}{4 e f}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{4 e f \left (e+f x^2\right )^2}}{6 e f}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3}\)

Input:

Int[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]
 

Output:

-1/6*((d*e - c*f)*x*(a + b*x^2))/(e*f*(e + f*x^2)^3) + (-1/4*((3*b*e*(d*e 
+ c*f) - a*f*(d*e + 5*c*f))*x)/(e*f*(e + f*x^2)^2) + (3*(b*e*(d*e + c*f) + 
 a*f*(d*e + 5*c*f))*(x/(2*e*(e + f*x^2)) + ArcTan[(Sqrt[f]*x)/Sqrt[e]]/(2* 
e^(3/2)*Sqrt[f])))/(4*e*f))/(6*e*f)
 

Defintions of rubi rules used

rule 25

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

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 401

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 e^{3} f^{2} \sqrt {e f}}\)	\(163\)
risch	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) a c}{32 \sqrt {-e f}\, e^{3}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a d}{32 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b c}{32 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b d}{32 \sqrt {-e f}\, f^{2} e}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) a c}{32 \sqrt {-e f}\, e^{3}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a d}{32 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b c}{32 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b d}{32 \sqrt {-e f}\, f^{2} e}\)	\(331\)

Input:

int((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x,method=_RETURNVERBOSE)
 

Output:

(1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3*x^5+1/6*(5*a*c*f^2+a*d*e*f+b 
*c*e*f-b*d*e^2)/e^2/f*x^3+1/16*(11*a*c*f^2-a*d*e*f-b*c*e*f-b*d*e^2)/e/f^2* 
x)/(f*x^2+e)^3+1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3/f^2/(e*f)^(1/2 
)*arctan(f*x/(e*f)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 642, normalized size of antiderivative = 3.71 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\left [\frac {6 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 16 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} - 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) - 6 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{96 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}, \frac {3 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 8 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} + 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) - 3 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{48 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}\right ] \] Input:

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="fricas")
 

Output:

[1/96*(6*(b*d*e^3*f^3 + 5*a*c*e*f^5 + (b*c + a*d)*e^2*f^4)*x^5 - 16*(b*d*e 
^4*f^2 - 5*a*c*e^2*f^4 - (b*c + a*d)*e^3*f^3)*x^3 - 3*(b*d*e^5 + 5*a*c*e^3 
*f^2 + (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e^4 
*f + 3*(b*d*e^3*f^2 + 5*a*c*e*f^4 + (b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4* 
f + 5*a*c*e^2*f^3 + (b*c + a*d)*e^3*f^2)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sq 
rt(-e*f)*x - e)/(f*x^2 + e)) - 6*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c + a*d) 
*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e^5*f^5*x^4 + 3*e^6*f^4*x^2 + e^7*f^3), 1/48 
*(3*(b*d*e^3*f^3 + 5*a*c*e*f^5 + (b*c + a*d)*e^2*f^4)*x^5 - 8*(b*d*e^4*f^2 
 - 5*a*c*e^2*f^4 - (b*c + a*d)*e^3*f^3)*x^3 + 3*(b*d*e^5 + 5*a*c*e^3*f^2 + 
 (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e^4*f + 3 
*(b*d*e^3*f^2 + 5*a*c*e*f^4 + (b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4*f + 5* 
a*c*e^2*f^3 + (b*c + a*d)*e^3*f^2)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) - 
3*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c + a*d)*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e 
^5*f^5*x^4 + 3*e^6*f^4*x^2 + e^7*f^3)]
 

Sympy [A] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.81 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (- e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a c f^{4} + 3 a d e f^{3} + 3 b c e f^{3} + 3 b d e^{2} f^{2}\right ) + x^{3} \cdot \left (40 a c e f^{3} + 8 a d e^{2} f^{2} + 8 b c e^{2} f^{2} - 8 b d e^{3} f\right ) + x \left (33 a c e^{2} f^{2} - 3 a d e^{3} f - 3 b c e^{3} f - 3 b d e^{4}\right )}{48 e^{6} f^{2} + 144 e^{5} f^{3} x^{2} + 144 e^{4} f^{4} x^{4} + 48 e^{3} f^{5} x^{6}} \] Input:

integrate((b*x**2+a)*(d*x**2+c)/(f*x**2+e)**4,x)
 

Output:

-sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(-e** 
4*f**2*sqrt(-1/(e**7*f**5)) + x)/32 + sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a 
*d*e*f + b*c*e*f + b*d*e**2)*log(e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32 + 
(x**5*(15*a*c*f**4 + 3*a*d*e*f**3 + 3*b*c*e*f**3 + 3*b*d*e**2*f**2) + x**3 
*(40*a*c*e*f**3 + 8*a*d*e**2*f**2 + 8*b*c*e**2*f**2 - 8*b*d*e**3*f) + x*(3 
3*a*c*e**2*f**2 - 3*a*d*e**3*f - 3*b*c*e**3*f - 3*b*d*e**4))/(48*e**6*f**2 
 + 144*e**5*f**3*x**2 + 144*e**4*f**4*x**4 + 48*e**3*f**5*x**6)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {{\left (b d e^{2} + b c e f + a d e f + 5 \, a c f^{2}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{2}} + \frac {3 \, b d e^{2} f^{2} x^{5} + 3 \, b c e f^{3} x^{5} + 3 \, a d e f^{3} x^{5} + 15 \, a c f^{4} x^{5} - 8 \, b d e^{3} f x^{3} + 8 \, b c e^{2} f^{2} x^{3} + 8 \, a d e^{2} f^{2} x^{3} + 40 \, a c e f^{3} x^{3} - 3 \, b d e^{4} x - 3 \, b c e^{3} f x - 3 \, a d e^{3} f x + 33 \, a c e^{2} f^{2} x}{48 \, {\left (f x^{2} + e\right )}^{3} e^{3} f^{2}} \] Input:

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="giac")
 

Output:

1/16*(b*d*e^2 + b*c*e*f + a*d*e*f + 5*a*c*f^2)*arctan(f*x/sqrt(e*f))/(sqrt 
(e*f)*e^3*f^2) + 1/48*(3*b*d*e^2*f^2*x^5 + 3*b*c*e*f^3*x^5 + 3*a*d*e*f^3*x 
^5 + 15*a*c*f^4*x^5 - 8*b*d*e^3*f*x^3 + 8*b*c*e^2*f^2*x^3 + 8*a*d*e^2*f^2* 
x^3 + 40*a*c*e*f^3*x^3 - 3*b*d*e^4*x - 3*b*c*e^3*f*x - 3*a*d*e^3*f*x + 33* 
a*c*e^2*f^2*x)/((f*x^2 + e)^3*e^3*f^2)
 

Mupad [B] (verification not implemented)

Time = 1.76 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^3}-\frac {x\,\left (b\,d\,e^2-11\,a\,c\,f^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e\,f^2}+\frac {x^3\,\left (5\,a\,c\,f^2-b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{6\,e^2\,f}}{e^3+3\,e^2\,f\,x^2+3\,e\,f^2\,x^4+f^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^{7/2}\,f^{5/2}} \] Input:

int(((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x)
 

Output:

((x^5*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/(16*e^3) - (x*(b*d*e^2 - 
11*a*c*f^2 + a*d*e*f + b*c*e*f))/(16*e*f^2) + (x^3*(5*a*c*f^2 - b*d*e^2 + 
a*d*e*f + b*c*e*f))/(6*e^2*f))/(e^3 + f^3*x^6 + 3*e^2*f*x^2 + 3*e*f^2*x^4) 
 + (atan((f^(1/2)*x)/e^(1/2))*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/( 
16*e^(7/2)*f^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.57 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x)
 

Output:

(15*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*e**3*f**2 + 45*sqrt( 
f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*e**2*f**3*x**2 + 45*sqrt(f)*s 
qrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*e*f**4*x**4 + 15*sqrt(f)*sqrt(e)* 
atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*f**5*x**6 + 3*sqrt(f)*sqrt(e)*atan((f*x) 
/(sqrt(f)*sqrt(e)))*a*d*e**4*f + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqr 
t(e)))*a*d*e**3*f**2*x**2 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)) 
)*a*d*e**2*f**3*x**4 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*d 
*e*f**4*x**6 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*e**4*f 
+ 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*e**3*f**2*x**2 + 9*s 
qrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*e**2*f**3*x**4 + 3*sqrt(f 
)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*e*f**4*x**6 + 3*sqrt(f)*sqrt(e 
)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*d*e**5 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(s 
qrt(f)*sqrt(e)))*b*d*e**4*f*x**2 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*s 
qrt(e)))*b*d*e**3*f**2*x**4 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e 
)))*b*d*e**2*f**3*x**6 + 33*a*c*e**3*f**3*x + 40*a*c*e**2*f**4*x**3 + 15*a 
*c*e*f**5*x**5 - 3*a*d*e**4*f**2*x + 8*a*d*e**3*f**3*x**3 + 3*a*d*e**2*f** 
4*x**5 - 3*b*c*e**4*f**2*x + 8*b*c*e**3*f**3*x**3 + 3*b*c*e**2*f**4*x**5 - 
 3*b*d*e**5*f*x - 8*b*d*e**4*f**2*x**3 + 3*b*d*e**3*f**3*x**5)/(48*e**4*f* 
*3*(e**3 + 3*e**2*f*x**2 + 3*e*f**2*x**4 + f**3*x**6))