\(\int \frac {(a+b x^2) (c+d x^2)^2}{(e+f x^2)^3} \, dx\) [212]

Optimal result

Integrand size = 26, antiderivative size = 183 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx=\frac {b d^2 x}{f^3}-\frac {(b e-a f) (d e-c f)^2 x}{4 e f^3 \left (e+f x^2\right )^2}+\frac {(d e-c f) (b e (9 d e-c f)-a f (5 d e+3 c f)) x}{8 e^2 f^3 \left (e+f x^2\right )}-\frac {\left (b e \left (15 d^2 e^2-6 c d e f-c^2 f^2\right )-a f \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{8 e^{5/2} f^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx=\frac {b d^2 x}{f^3}-\frac {(b e-a f) (d e-c f)^2 x}{4 e f^3 \left (e+f x^2\right )^2}+\frac {(d e-c f) (b e (9 d e-c f)-a f (5 d e+3 c f)) x}{8 e^2 f^3 \left (e+f x^2\right )}-\frac {\left (b e \left (15 d^2 e^2-6 c d e f-c^2 f^2\right )-a f \left (3 d^2 e^2+2 c d e f+3 c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{8 e^{5/2} f^{7/2}} \] Input:

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

Output:

(b*d^2*x)/f^3 - ((b*e - a*f)*(d*e - c*f)^2*x)/(4*e*f^3*(e + f*x^2)^2) + (( 
d*e - c*f)*(b*e*(9*d*e - c*f) - a*f*(5*d*e + 3*c*f))*x)/(8*e^2*f^3*(e + f* 
x^2)) - ((b*e*(15*d^2*e^2 - 6*c*d*e*f - c^2*f^2) - a*f*(3*d^2*e^2 + 2*c*d* 
e*f + 3*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(8*e^(5/2)*f^(7/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {401, 25, 401, 299, 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.

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

\(\Big \downarrow \) 401

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 401

\(\displaystyle \frac {-\frac {\int \frac {c (a f (d e-3 c f)-b e (5 d e+c f))-d (b e (15 d e-c f)-3 a f (d e+c f)) x^2}{f x^2+e}dx}{2 e f}-\frac {x \left (c+d x^2\right ) (b e (5 d e-c f)-a f (3 c f+d e))}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{4 e f \left (e+f x^2\right )^2}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {-\frac {\frac {\left (b e \left (-c^2 f^2-6 c d e f+15 d^2 e^2\right )-a f \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right ) \int \frac {1}{f x^2+e}dx}{f}-\frac {d x (b e (15 d e-c f)-3 a f (c f+d e))}{f}}{2 e f}-\frac {x \left (c+d x^2\right ) (b e (5 d e-c f)-a f (3 c f+d e))}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{4 e f \left (e+f x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {-\frac {\frac {\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^2 f^2-6 c d e f+15 d^2 e^2\right )-a f \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{\sqrt {e} f^{3/2}}-\frac {d x (b e (15 d e-c f)-3 a f (c f+d e))}{f}}{2 e f}-\frac {x \left (c+d x^2\right ) (b e (5 d e-c f)-a f (3 c f+d e))}{2 e f \left (e+f x^2\right )}}{4 e f}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{4 e f \left (e+f x^2\right )^2}\)

Input:

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

Output:

-1/4*((b*e - a*f)*x*(c + d*x^2)^2)/(e*f*(e + f*x^2)^2) + (-1/2*((b*e*(5*d* 
e - c*f) - a*f*(d*e + 3*c*f))*x*(c + d*x^2))/(e*f*(e + f*x^2)) - (-((d*(b* 
e*(15*d*e - c*f) - 3*a*f*(d*e + c*f))*x)/f) + ((b*e*(15*d^2*e^2 - 6*c*d*e* 
f - c^2*f^2) - a*f*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^2))*ArcTan[(Sqrt[f]*x) 
/Sqrt[e]])/(Sqrt[e]*f^(3/2)))/(2*e*f))/(4*e*f)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 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 = 0.55 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {b \,d^{2} x}{f^{3}}+\frac {\frac {\frac {f \left (3 a \,c^{2} f^{3}+2 a c d e \,f^{2}-5 a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-10 b c d \,e^{2} f +9 e^{3} b \,d^{2}\right ) x^{3}}{8 e^{2}}+\frac {\left (5 a \,c^{2} f^{3}-2 a c d e \,f^{2}-3 a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-6 b c d \,e^{2} f +7 e^{3} b \,d^{2}\right ) x}{8 e}}{\left (f \,x^{2}+e \right )^{2}}+\frac {\left (3 a \,c^{2} f^{3}+2 a c d e \,f^{2}+3 a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+6 b c d \,e^{2} f -15 e^{3} b \,d^{2}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 e^{2} \sqrt {e f}}}{f^{3}}\)	\(230\)
risch	\(\frac {b \,d^{2} x}{f^{3}}+\frac {\frac {f \left (3 a \,c^{2} f^{3}+2 a c d e \,f^{2}-5 a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-10 b c d \,e^{2} f +9 e^{3} b \,d^{2}\right ) x^{3}}{8 e^{2}}+\frac {\left (5 a \,c^{2} f^{3}-2 a c d e \,f^{2}-3 a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-6 b c d \,e^{2} f +7 e^{3} b \,d^{2}\right ) x}{8 e}}{f^{3} \left (f \,x^{2}+e \right )^{2}}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) a \,c^{2}}{16 \sqrt {-e f}\, e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a c d}{8 f \sqrt {-e f}\, e}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) a \,d^{2}}{16 f^{2} \sqrt {-e f}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b \,c^{2}}{16 f \sqrt {-e f}\, e}-\frac {3 \ln \left (f x +\sqrt {-e f}\right ) b c d}{8 f^{2} \sqrt {-e f}}+\frac {15 e \ln \left (f x +\sqrt {-e f}\right ) b \,d^{2}}{16 f^{3} \sqrt {-e f}}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) a \,c^{2}}{16 \sqrt {-e f}\, e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a c d}{8 f \sqrt {-e f}\, e}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) a \,d^{2}}{16 f^{2} \sqrt {-e f}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b \,c^{2}}{16 f \sqrt {-e f}\, e}+\frac {3 \ln \left (-f x +\sqrt {-e f}\right ) b c d}{8 f^{2} \sqrt {-e f}}-\frac {15 e \ln \left (-f x +\sqrt {-e f}\right ) b \,d^{2}}{16 f^{3} \sqrt {-e f}}\)	\(481\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (169) = 338\).

Time = 0.09 (sec) , antiderivative size = 777, normalized size of antiderivative = 4.25 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(16*b*d^2*e^3*f^3*x^5 + 2*(25*b*d^2*e^4*f^2 + 3*a*c^2*e*f^5 - 5*(2*b 
*c*d + a*d^2)*e^3*f^3 + (b*c^2 + 2*a*c*d)*e^2*f^4)*x^3 + (15*b*d^2*e^5 - 3 
*a*c^2*e^2*f^3 - 3*(2*b*c*d + a*d^2)*e^4*f - (b*c^2 + 2*a*c*d)*e^3*f^2 + ( 
15*b*d^2*e^3*f^2 - 3*a*c^2*f^5 - 3*(2*b*c*d + a*d^2)*e^2*f^3 - (b*c^2 + 2* 
a*c*d)*e*f^4)*x^4 + 2*(15*b*d^2*e^4*f - 3*a*c^2*e*f^4 - 3*(2*b*c*d + a*d^2 
)*e^3*f^2 - (b*c^2 + 2*a*c*d)*e^2*f^3)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt 
(-e*f)*x - e)/(f*x^2 + e)) + 2*(15*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 - 3*(2*b* 
c*d + a*d^2)*e^4*f^2 - (b*c^2 + 2*a*c*d)*e^3*f^3)*x)/(e^3*f^6*x^4 + 2*e^4* 
f^5*x^2 + e^5*f^4), 1/8*(8*b*d^2*e^3*f^3*x^5 + (25*b*d^2*e^4*f^2 + 3*a*c^2 
*e*f^5 - 5*(2*b*c*d + a*d^2)*e^3*f^3 + (b*c^2 + 2*a*c*d)*e^2*f^4)*x^3 - (1 
5*b*d^2*e^5 - 3*a*c^2*e^2*f^3 - 3*(2*b*c*d + a*d^2)*e^4*f - (b*c^2 + 2*a*c 
*d)*e^3*f^2 + (15*b*d^2*e^3*f^2 - 3*a*c^2*f^5 - 3*(2*b*c*d + a*d^2)*e^2*f^ 
3 - (b*c^2 + 2*a*c*d)*e*f^4)*x^4 + 2*(15*b*d^2*e^4*f - 3*a*c^2*e*f^4 - 3*( 
2*b*c*d + a*d^2)*e^3*f^2 - (b*c^2 + 2*a*c*d)*e^2*f^3)*x^2)*sqrt(e*f)*arcta 
n(sqrt(e*f)*x/e) + (15*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 - 3*(2*b*c*d + a*d^2) 
*e^4*f^2 - (b*c^2 + 2*a*c*d)*e^3*f^3)*x)/(e^3*f^6*x^4 + 2*e^4*f^5*x^2 + e^ 
5*f^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (177) = 354\).

Time = 4.59 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.19 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx=\frac {b d^{2} x}{f^{3}} - \frac {\sqrt {- \frac {1}{e^{5} f^{7}}} \cdot \left (3 a c^{2} f^{3} + 2 a c d e f^{2} + 3 a d^{2} e^{2} f + b c^{2} e f^{2} + 6 b c d e^{2} f - 15 b d^{2} e^{3}\right ) \log {\left (- e^{3} f^{3} \sqrt {- \frac {1}{e^{5} f^{7}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{e^{5} f^{7}}} \cdot \left (3 a c^{2} f^{3} + 2 a c d e f^{2} + 3 a d^{2} e^{2} f + b c^{2} e f^{2} + 6 b c d e^{2} f - 15 b d^{2} e^{3}\right ) \log {\left (e^{3} f^{3} \sqrt {- \frac {1}{e^{5} f^{7}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a c^{2} f^{4} + 2 a c d e f^{3} - 5 a d^{2} e^{2} f^{2} + b c^{2} e f^{3} - 10 b c d e^{2} f^{2} + 9 b d^{2} e^{3} f\right ) + x \left (5 a c^{2} e f^{3} - 2 a c d e^{2} f^{2} - 3 a d^{2} e^{3} f - b c^{2} e^{2} f^{2} - 6 b c d e^{3} f + 7 b d^{2} e^{4}\right )}{8 e^{4} f^{3} + 16 e^{3} f^{4} x^{2} + 8 e^{2} f^{5} x^{4}} \] Input:

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

Output:

b*d**2*x/f**3 - sqrt(-1/(e**5*f**7))*(3*a*c**2*f**3 + 2*a*c*d*e*f**2 + 3*a 
*d**2*e**2*f + b*c**2*e*f**2 + 6*b*c*d*e**2*f - 15*b*d**2*e**3)*log(-e**3* 
f**3*sqrt(-1/(e**5*f**7)) + x)/16 + sqrt(-1/(e**5*f**7))*(3*a*c**2*f**3 + 
2*a*c*d*e*f**2 + 3*a*d**2*e**2*f + b*c**2*e*f**2 + 6*b*c*d*e**2*f - 15*b*d 
**2*e**3)*log(e**3*f**3*sqrt(-1/(e**5*f**7)) + x)/16 + (x**3*(3*a*c**2*f** 
4 + 2*a*c*d*e*f**3 - 5*a*d**2*e**2*f**2 + b*c**2*e*f**3 - 10*b*c*d*e**2*f* 
*2 + 9*b*d**2*e**3*f) + x*(5*a*c**2*e*f**3 - 2*a*c*d*e**2*f**2 - 3*a*d**2* 
e**3*f - b*c**2*e**2*f**2 - 6*b*c*d*e**3*f + 7*b*d**2*e**4))/(8*e**4*f**3 
+ 16*e**3*f**4*x**2 + 8*e**2*f**5*x**4)
 

Maxima [F(-2)]

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

integrate((b*x^2+a)*(d*x^2+c)^2/(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 = 0.12 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx=\frac {b d^{2} x}{f^{3}} - \frac {{\left (15 \, b d^{2} e^{3} - 6 \, b c d e^{2} f - 3 \, a d^{2} e^{2} f - b c^{2} e f^{2} - 2 \, a c d e f^{2} - 3 \, a c^{2} f^{3}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \, \sqrt {e f} e^{2} f^{3}} + \frac {9 \, b d^{2} e^{3} f x^{3} - 10 \, b c d e^{2} f^{2} x^{3} - 5 \, a d^{2} e^{2} f^{2} x^{3} + b c^{2} e f^{3} x^{3} + 2 \, a c d e f^{3} x^{3} + 3 \, a c^{2} f^{4} x^{3} + 7 \, b d^{2} e^{4} x - 6 \, b c d e^{3} f x - 3 \, a d^{2} e^{3} f x - b c^{2} e^{2} f^{2} x - 2 \, a c d e^{2} f^{2} x + 5 \, a c^{2} e f^{3} x}{8 \, {\left (f x^{2} + e\right )}^{2} e^{2} f^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (b\,c^2\,e\,f^2+3\,a\,c^2\,f^3+6\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-15\,b\,d^2\,e^3+3\,a\,d^2\,e^2\,f\right )}{8\,e^{5/2}\,f^{7/2}}-\frac {\frac {x\,\left (b\,c^2\,e\,f^2-5\,a\,c^2\,f^3+6\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-7\,b\,d^2\,e^3+3\,a\,d^2\,e^2\,f\right )}{8\,e}-\frac {x^3\,\left (b\,c^2\,e\,f^3+3\,a\,c^2\,f^4-10\,b\,c\,d\,e^2\,f^2+2\,a\,c\,d\,e\,f^3+9\,b\,d^2\,e^3\,f-5\,a\,d^2\,e^2\,f^2\right )}{8\,e^2}}{e^2\,f^3+2\,e\,f^4\,x^2+f^5\,x^4}+\frac {b\,d^2\,x}{f^3} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 717, normalized size of antiderivative = 3.92 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*e**2*f**3 + 6*sqrt 
(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*e*f**4*x**2 + 3*sqrt(f)*s 
qrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*f**5*x**4 + 2*sqrt(f)*sqrt(e)* 
atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*d*e**3*f**2 + 4*sqrt(f)*sqrt(e)*atan((f* 
x)/(sqrt(f)*sqrt(e)))*a*c*d*e**2*f**3*x**2 + 2*sqrt(f)*sqrt(e)*atan((f*x)/ 
(sqrt(f)*sqrt(e)))*a*c*d*e*f**4*x**4 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt( 
f)*sqrt(e)))*a*d**2*e**4*f + 6*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e) 
))*a*d**2*e**3*f**2*x**2 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e))) 
*a*d**2*e**2*f**3*x**4 + sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c 
**2*e**3*f**2 + 2*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c**2*e** 
2*f**3*x**2 + sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c**2*e*f**4* 
x**4 + 6*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**4*f + 12*s 
qrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**3*f**2*x**2 + 6*sqrt 
(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**2*f**3*x**4 - 15*sqrt(f 
)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*d**2*e**5 - 30*sqrt(f)*sqrt(e)*a 
tan((f*x)/(sqrt(f)*sqrt(e)))*b*d**2*e**4*f*x**2 - 15*sqrt(f)*sqrt(e)*atan( 
(f*x)/(sqrt(f)*sqrt(e)))*b*d**2*e**3*f**2*x**4 + 5*a*c**2*e**2*f**4*x + 3* 
a*c**2*e*f**5*x**3 - 2*a*c*d*e**3*f**3*x + 2*a*c*d*e**2*f**4*x**3 - 3*a*d* 
*2*e**4*f**2*x - 5*a*d**2*e**3*f**3*x**3 - b*c**2*e**3*f**3*x + b*c**2*e** 
2*f**4*x**3 - 6*b*c*d*e**4*f**2*x - 10*b*c*d*e**3*f**3*x**3 + 15*b*d**2...