\(\int \frac {(a+b x^2)^2}{(c+d x^2) (e+f x^2)^3} \, dx\) [238]

Optimal result

Integrand size = 28, antiderivative size = 274 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=-\frac {(b e-a f)^2 x}{4 e f (d e-c f) \left (e+f x^2\right )^2}+\frac {(b e-a f) (b e (d e-5 c f)+a f (7 d e-3 c f)) x}{8 e^2 f (d e-c f)^2 \left (e+f x^2\right )}+\frac {\sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (d e-c f)^3}+\frac {\left (b^2 e^2 \left (d^2 e^2-6 c d e f-3 c^2 f^2\right )+2 a b e f \left (3 d^2 e^2+6 c d e f-c^2 f^2\right )-a^2 f^2 \left (15 d^2 e^2-10 c d e f+3 c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{8 e^{5/2} f^{3/2} (d e-c f)^3} \] Output:

-1/4*(-a*f+b*e)^2*x/e/f/(-c*f+d*e)/(f*x^2+e)^2+1/8*(-a*f+b*e)*(b*e*(-5*c*f 
+d*e)+a*f*(-3*c*f+7*d*e))*x/e^2/f/(-c*f+d*e)^2/(f*x^2+e)+d^(1/2)*(-a*d+b*c 
)^2*arctan(d^(1/2)*x/c^(1/2))/c^(1/2)/(-c*f+d*e)^3+1/8*(b^2*e^2*(-3*c^2*f^ 
2-6*c*d*e*f+d^2*e^2)+2*a*b*e*f*(-c^2*f^2+6*c*d*e*f+3*d^2*e^2)-a^2*f^2*(3*c 
^2*f^2-10*c*d*e*f+15*d^2*e^2))*arctan(f^(1/2)*x/e^(1/2))/e^(5/2)/f^(3/2)/( 
-c*f+d*e)^3
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\frac {1}{8} \left (-\frac {2 (b e-a f)^2 x}{e f (d e-c f) \left (e+f x^2\right )^2}+\frac {(b e-a f) (b e (d e-5 c f)+a f (7 d e-3 c f)) x}{e^2 f (d e-c f)^2 \left (e+f x^2\right )}-\frac {8 \sqrt {d} (b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (-d e+c f)^3}+\frac {\left (b^2 e^2 \left (d^2 e^2-6 c d e f-3 c^2 f^2\right )+a^2 f^2 \left (-15 d^2 e^2+10 c d e f-3 c^2 f^2\right )+2 a b e f \left (3 d^2 e^2+6 c d e f-c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{e^{5/2} f^{3/2} (d e-c f)^3}\right ) \] Input:

Integrate[(a + b*x^2)^2/((c + d*x^2)*(e + f*x^2)^3),x]
 

Output:

((-2*(b*e - a*f)^2*x)/(e*f*(d*e - c*f)*(e + f*x^2)^2) + ((b*e - a*f)*(b*e* 
(d*e - 5*c*f) + a*f*(7*d*e - 3*c*f))*x)/(e^2*f*(d*e - c*f)^2*(e + f*x^2)) 
- (8*Sqrt[d]*(b*c - a*d)^2*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(-(d*e) + 
 c*f)^3) + ((b^2*e^2*(d^2*e^2 - 6*c*d*e*f - 3*c^2*f^2) + a^2*f^2*(-15*d^2* 
e^2 + 10*c*d*e*f - 3*c^2*f^2) + 2*a*b*e*f*(3*d^2*e^2 + 6*c*d*e*f - c^2*f^2 
))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(e^(5/2)*f^(3/2)*(d*e - c*f)^3))/8
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {419, 25, 397, 218, 401, 27, 298, 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.

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

\(\Big \downarrow \) 419

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 401

\(\displaystyle \frac {\frac {x \left (a+b x^2\right ) (b e-a f) (d e-c f)}{4 e \left (e+f x^2\right )^2}-\frac {\int \frac {f \left (b (a f (5 d e-c f)-b e (d e+3 c f)) x^2+a (a f (7 d e-3 c f)-b e (3 d e+c f))\right )}{\left (f x^2+e\right )^2}dx}{4 e f}}{(d e-c f)^2}-\frac {d (b c-a d) \left (\frac {(b e-a f) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} \sqrt {f} (d e-c f)}-\frac {(b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d} (d e-c f)}\right )}{(d e-c f)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {x \left (a+b x^2\right ) (b e-a f) (d e-c f)}{4 e \left (e+f x^2\right )^2}-\frac {\int \frac {b (a f (5 d e-c f)-b e (d e+3 c f)) x^2+a (a f (7 d e-3 c f)-b e (3 d e+c f))}{\left (f x^2+e\right )^2}dx}{4 e}}{(d e-c f)^2}-\frac {d (b c-a d) \left (\frac {(b e-a f) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} \sqrt {f} (d e-c f)}-\frac {(b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d} (d e-c f)}\right )}{(d e-c f)^2}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {x \left (a+b x^2\right ) (b e-a f) (d e-c f)}{4 e \left (e+f x^2\right )^2}-\frac {\frac {\left (a^2 f^2 (7 d e-3 c f)+2 a b e f (d e-c f)-b^2 e^2 (3 c f+d e)\right ) \int \frac {1}{f x^2+e}dx}{2 e f}-\frac {x (b e-a f) (a f (7 d e-3 c f)-b e (3 c f+d e))}{2 e f \left (e+f x^2\right )}}{4 e}}{(d e-c f)^2}-\frac {d (b c-a d) \left (\frac {(b e-a f) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} \sqrt {f} (d e-c f)}-\frac {(b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d} (d e-c f)}\right )}{(d e-c f)^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {x \left (a+b x^2\right ) (b e-a f) (d e-c f)}{4 e \left (e+f x^2\right )^2}-\frac {\frac {\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a^2 f^2 (7 d e-3 c f)+2 a b e f (d e-c f)-b^2 e^2 (3 c f+d e)\right )}{2 e^{3/2} f^{3/2}}-\frac {x (b e-a f) (a f (7 d e-3 c f)-b e (3 c f+d e))}{2 e f \left (e+f x^2\right )}}{4 e}}{(d e-c f)^2}-\frac {d (b c-a d) \left (\frac {(b e-a f) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} \sqrt {f} (d e-c f)}-\frac {(b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d} (d e-c f)}\right )}{(d e-c f)^2}\)

Input:

Int[(a + b*x^2)^2/((c + d*x^2)*(e + f*x^2)^3),x]
 

Output:

-((d*(b*c - a*d)*(-(((b*c - a*d)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*Sqr 
t[d]*(d*e - c*f))) + ((b*e - a*f)*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(Sqrt[e]*Sq 
rt[f]*(d*e - c*f))))/(d*e - c*f)^2) + (((b*e - a*f)*(d*e - c*f)*x*(a + b*x 
^2))/(4*e*(e + f*x^2)^2) - (-1/2*((b*e - a*f)*(a*f*(7*d*e - 3*c*f) - b*e*( 
d*e + 3*c*f))*x)/(e*f*(e + f*x^2)) + ((a^2*f^2*(7*d*e - 3*c*f) + 2*a*b*e*f 
*(d*e - c*f) - b^2*e^2*(d*e + 3*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(2*e^(3 
/2)*f^(3/2)))/(4*e))/(d*e - c*f)^2
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

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]
 

rule 419

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*( 
x_)^2), x_Symbol] :> Simp[b*((b*e - a*f)/(b*c - a*d)^2)   Int[(c + d*x^2)^( 
q + 2)*((e + f*x^2)^(r - 1)/(a + b*x^2)), x], x] - Simp[1/(b*c - a*d)^2   I 
nt[(c + d*x^2)^q*(e + f*x^2)^(r - 1)*(2*b*c*d*e - a*d^2*e - b*c^2*f + d^2*( 
b*e - a*f)*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[q, -1] && Gt 
Q[r, 1]
 

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {\frac {\frac {\left (3 a^{2} c^{2} f^{4}-10 a^{2} c d e \,f^{3}+7 a^{2} d^{2} e^{2} f^{2}+2 a b \,c^{2} e \,f^{3}+4 a b c d \,e^{2} f^{2}-6 a b \,d^{2} e^{3} f -5 b^{2} c^{2} e^{2} f^{2}+6 b^{2} c d \,e^{3} f -b^{2} d^{2} e^{4}\right ) x^{3}}{8 e^{2}}+\frac {\left (5 a^{2} c^{2} f^{4}-14 a^{2} c d e \,f^{3}+9 a^{2} d^{2} e^{2} f^{2}-2 a b \,c^{2} e \,f^{3}+12 a b c d \,e^{2} f^{2}-10 a b \,d^{2} e^{3} f -3 b^{2} c^{2} e^{2} f^{2}+2 b^{2} c d \,e^{3} f +b^{2} d^{2} e^{4}\right ) x}{8 e f}}{\left (f \,x^{2}+e \right )^{2}}+\frac {\left (3 a^{2} c^{2} f^{4}-10 a^{2} c d e \,f^{3}+15 a^{2} d^{2} e^{2} f^{2}+2 a b \,c^{2} e \,f^{3}-12 a b c d \,e^{2} f^{2}-6 a b \,d^{2} e^{3} f +3 b^{2} c^{2} e^{2} f^{2}+6 b^{2} c d \,e^{3} f -b^{2} d^{2} e^{4}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 e^{2} f \sqrt {e f}}}{\left (c f -d e \right )^{3}}-\frac {d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{\left (c f -d e \right )^{3} \sqrt {c d}}\)	\(432\)
risch	\(\text {Expression too large to display}\)	\(8600\)

Input:

int((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^3,x,method=_RETURNVERBOSE)
 

Output:

1/(c*f-d*e)^3*((1/8*(3*a^2*c^2*f^4-10*a^2*c*d*e*f^3+7*a^2*d^2*e^2*f^2+2*a* 
b*c^2*e*f^3+4*a*b*c*d*e^2*f^2-6*a*b*d^2*e^3*f-5*b^2*c^2*e^2*f^2+6*b^2*c*d* 
e^3*f-b^2*d^2*e^4)/e^2*x^3+1/8*(5*a^2*c^2*f^4-14*a^2*c*d*e*f^3+9*a^2*d^2*e 
^2*f^2-2*a*b*c^2*e*f^3+12*a*b*c*d*e^2*f^2-10*a*b*d^2*e^3*f-3*b^2*c^2*e^2*f 
^2+2*b^2*c*d*e^3*f+b^2*d^2*e^4)/e/f*x)/(f*x^2+e)^2+1/8*(3*a^2*c^2*f^4-10*a 
^2*c*d*e*f^3+15*a^2*d^2*e^2*f^2+2*a*b*c^2*e*f^3-12*a*b*c*d*e^2*f^2-6*a*b*d 
^2*e^3*f+3*b^2*c^2*e^2*f^2+6*b^2*c*d*e^3*f-b^2*d^2*e^4)/e^2/f/(e*f)^(1/2)* 
arctan(f*x/(e*f)^(1/2)))-d*(a^2*d^2-2*a*b*c*d+b^2*c^2)/(c*f-d*e)^3/(c*d)^( 
1/2)*arctan(x*d/(c*d)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 788 vs. \(2 (252) = 504\).

Time = 18.60 (sec) , antiderivative size = 3241, normalized size of antiderivative = 11.83 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**2/(d*x**2+c)/(f*x**2+e)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^3,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 = 434, normalized size of antiderivative = 1.58 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\frac {{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \sqrt {c d}} + \frac {{\left (b^{2} d^{2} e^{4} - 6 \, b^{2} c d e^{3} f + 6 \, a b d^{2} e^{3} f - 3 \, b^{2} c^{2} e^{2} f^{2} + 12 \, a b c d e^{2} f^{2} - 15 \, a^{2} d^{2} e^{2} f^{2} - 2 \, a b c^{2} e f^{3} + 10 \, a^{2} c d e f^{3} - 3 \, a^{2} c^{2} f^{4}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \, {\left (d^{3} e^{5} f - 3 \, c d^{2} e^{4} f^{2} + 3 \, c^{2} d e^{3} f^{3} - c^{3} e^{2} f^{4}\right )} \sqrt {e f}} + \frac {b^{2} d e^{3} f x^{3} - 5 \, b^{2} c e^{2} f^{2} x^{3} + 6 \, a b d e^{2} f^{2} x^{3} + 2 \, a b c e f^{3} x^{3} - 7 \, a^{2} d e f^{3} x^{3} + 3 \, a^{2} c f^{4} x^{3} - b^{2} d e^{4} x - 3 \, b^{2} c e^{3} f x + 10 \, a b d e^{3} f x - 2 \, a b c e^{2} f^{2} x - 9 \, a^{2} d e^{2} f^{2} x + 5 \, a^{2} c e f^{3} x}{8 \, {\left (d^{2} e^{4} f - 2 \, c d e^{3} f^{2} + c^{2} e^{2} f^{3}\right )} {\left (f x^{2} + e\right )}^{2}} \] Input:

integrate((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^3,x, algorithm="giac")
 

Output:

(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(d*x/sqrt(c*d))/((d^3*e^3 - 3*c* 
d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*sqrt(c*d)) + 1/8*(b^2*d^2*e^4 - 6*b^2 
*c*d*e^3*f + 6*a*b*d^2*e^3*f - 3*b^2*c^2*e^2*f^2 + 12*a*b*c*d*e^2*f^2 - 15 
*a^2*d^2*e^2*f^2 - 2*a*b*c^2*e*f^3 + 10*a^2*c*d*e*f^3 - 3*a^2*c^2*f^4)*arc 
tan(f*x/sqrt(e*f))/((d^3*e^5*f - 3*c*d^2*e^4*f^2 + 3*c^2*d*e^3*f^3 - c^3*e 
^2*f^4)*sqrt(e*f)) + 1/8*(b^2*d*e^3*f*x^3 - 5*b^2*c*e^2*f^2*x^3 + 6*a*b*d* 
e^2*f^2*x^3 + 2*a*b*c*e*f^3*x^3 - 7*a^2*d*e*f^3*x^3 + 3*a^2*c*f^4*x^3 - b^ 
2*d*e^4*x - 3*b^2*c*e^3*f*x + 10*a*b*d*e^3*f*x - 2*a*b*c*e^2*f^2*x - 9*a^2 
*d*e^2*f^2*x + 5*a^2*c*e*f^3*x)/((d^2*e^4*f - 2*c*d*e^3*f^2 + c^2*e^2*f^3) 
*(f*x^2 + e)^2)
 

Mupad [B] (verification not implemented)

Time = 5.79 (sec) , antiderivative size = 13717, normalized size of antiderivative = 50.06 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Too large to display} \] Input:

int((a + b*x^2)^2/((c + d*x^2)*(e + f*x^2)^3),x)
                                                                                    
                                                                                    
 

Output:

((x^3*(3*a^2*c*f^3 + b^2*d*e^3 - 7*a^2*d*e*f^2 - 5*b^2*c*e^2*f + 2*a*b*c*e 
*f^2 + 6*a*b*d*e^2*f))/(8*e^2*(c^2*f^2 + d^2*e^2 - 2*c*d*e*f)) - (x*(b^2*d 
*e^3 - 5*a^2*c*f^3 + 9*a^2*d*e*f^2 + 3*b^2*c*e^2*f + 2*a*b*c*e*f^2 - 10*a* 
b*d*e^2*f))/(8*e*(c^2*f^3 + d^2*e^2*f - 2*c*d*e*f^2)))/(e^2 + f^2*x^4 + 2* 
e*f*x^2) + (atan(((((x*(b^4*d^7*e^8 + 9*a^4*c^4*d^3*f^8 + 289*a^4*d^7*e^4* 
f^4 + 6*a^2*b^2*d^7*e^6*f^2 + 190*a^4*c^2*d^5*e^2*f^6 + 30*b^4*c^2*d^5*e^6 
*f^2 + 36*b^4*c^3*d^4*e^5*f^3 + 73*b^4*c^4*d^3*e^4*f^4 + 12*a*b^3*d^7*e^7* 
f - 12*b^4*c*d^6*e^7*f - 180*a^3*b*d^7*e^5*f^3 - 300*a^4*c*d^6*e^3*f^5 - 6 
0*a^4*c^3*d^4*e*f^7 - 48*a*b^3*c*d^6*e^6*f^2 - 496*a^3*b*c*d^6*e^4*f^4 + 1 
2*a^3*b*c^4*d^3*e*f^7 - 184*a*b^3*c^2*d^5*e^5*f^3 - 304*a*b^3*c^3*d^4*e^4* 
f^4 + 12*a*b^3*c^4*d^3*e^3*f^5 + 344*a^2*b^2*c*d^6*e^5*f^3 + 264*a^3*b*c^2 
*d^5*e^3*f^5 - 112*a^3*b*c^3*d^4*e^2*f^6 + 468*a^2*b^2*c^2*d^5*e^4*f^4 - 7 
2*a^2*b^2*c^3*d^4*e^3*f^5 + 22*a^2*b^2*c^4*d^3*e^2*f^6))/(32*(d^4*e^8*f + 
c^4*e^4*f^5 - 4*c*d^3*e^7*f^2 - 4*c^3*d*e^5*f^4 + 6*c^2*d^2*e^6*f^3)) - (( 
(256*a^2*d^10*e^10*f^3 + 5280*a^2*c^2*d^8*e^8*f^5 - 9056*a^2*c^3*d^7*e^7*f 
^6 + 9760*a^2*c^4*d^6*e^6*f^7 - 6816*a^2*c^5*d^5*e^5*f^8 + 3040*a^2*c^6*d^ 
4*e^4*f^9 - 800*a^2*c^7*d^3*e^3*f^10 + 96*a^2*c^8*d^2*e^2*f^11 - 96*b^2*c^ 
2*d^8*e^10*f^3 - 96*b^2*c^3*d^7*e^9*f^4 + 800*b^2*c^4*d^6*e^8*f^5 - 1440*b 
^2*c^5*d^5*e^7*f^6 + 1248*b^2*c^6*d^4*e^6*f^7 - 544*b^2*c^7*d^3*e^5*f^8 + 
96*b^2*c^8*d^2*e^4*f^9 - 1760*a^2*c*d^9*e^9*f^4 + 32*b^2*c*d^9*e^11*f^2...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1587, normalized size of antiderivative = 5.79 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^3,x)
 

Output:

( - 8*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*d**2*e**5*f**2 - 
16*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*d**2*e**4*f**3*x**2 
- 8*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*d**2*e**3*f**4*x**4 
 + 16*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c*d*e**5*f**2 + 32 
*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c*d*e**4*f**3*x**2 + 16 
*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b*c*d*e**3*f**4*x**4 - 8* 
sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**2*e**5*f**2 - 16*sqr 
t(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**2*e**4*f**3*x**2 - 8*sq 
rt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**2*c**2*e**3*f**4*x**4 + 3*s 
qrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**3*e**2*f**4 + 6*sqrt( 
f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**3*e*f**5*x**2 + 3*sqrt(f) 
*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**3*f**6*x**4 - 10*sqrt(f)*sq 
rt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**2*d*e**3*f**3 - 20*sqrt(f)*sqr 
t(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**2*d*e**2*f**4*x**2 - 10*sqrt(f) 
*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c**2*d*e*f**5*x**4 + 15*sqrt(f 
)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c*d**2*e**4*f**2 + 30*sqrt(f) 
*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c*d**2*e**3*f**3*x**2 + 15*sqr 
t(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*c*d**2*e**2*f**4*x**4 + 2* 
sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b*c**3*e**3*f**3 + 4*sqrt( 
f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b*c**3*e**2*f**4*x**2 + 2*sq...