\(\int \frac {(a+b x^2) (c+d x^2)^2}{(e+f x^2)^4} \, dx\) [213]

Optimal result

Integrand size = 26, antiderivative size = 244 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx=-\frac {(b e-a f) (d e-c f)^2 x}{6 e f^3 \left (e+f x^2\right )^3}+\frac {(d e-c f) (b e (13 d e-c f)-a f (7 d e+5 c f)) x}{24 e^2 f^3 \left (e+f x^2\right )^2}-\frac {\left (b e \left (11 d^2 e^2-2 c d e f-c^2 f^2\right )-a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) x}{16 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{7/2}} \] Output:

-1/6*(-a*f+b*e)*(-c*f+d*e)^2*x/e/f^3/(f*x^2+e)^3+1/24*(-c*f+d*e)*(b*e*(-c* 
f+13*d*e)-a*f*(5*c*f+7*d*e))*x/e^2/f^3/(f*x^2+e)^2-1/16*(b*e*(-c^2*f^2-2*c 
*d*e*f+11*d^2*e^2)-a*f*(5*c^2*f^2+2*c*d*e*f+d^2*e^2))*x/e^3/f^3/(f*x^2+e)+ 
1/16*(b*e*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+a*f*(5*c^2*f^2+2*c*d*e*f+d^2*e^2)) 
*arctan(f^(1/2)*x/e^(1/2))/e^(7/2)/f^(7/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx=-\frac {(b e-a f) (d e-c f)^2 x}{6 e f^3 \left (e+f x^2\right )^3}+\frac {(d e-c f) (b e (13 d e-c f)-a f (7 d e+5 c f)) x}{24 e^2 f^3 \left (e+f x^2\right )^2}+\frac {\left (b e \left (-11 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) x}{16 e^3 f^3 \left (e+f x^2\right )}+\frac {\left (b e \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+a f \left (d^2 e^2+2 c d e f+5 c^2 f^2\right )\right ) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{7/2}} \] Input:

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

Output:

-1/6*((b*e - a*f)*(d*e - c*f)^2*x)/(e*f^3*(e + f*x^2)^3) + ((d*e - c*f)*(b 
*e*(13*d*e - c*f) - a*f*(7*d*e + 5*c*f))*x)/(24*e^2*f^3*(e + f*x^2)^2) + ( 
(b*e*(-11*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^ 
2*f^2))*x)/(16*e^3*f^3*(e + f*x^2)) + ((b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f 
^2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/ 
(16*e^(7/2)*f^(7/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {401, 25, 401, 25, 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.

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

\(\Big \downarrow \) 401

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 401

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

-1/6*((b*e - a*f)*x*(c + d*x^2)^2)/(e*f*(e + f*x^2)^3) + (-1/4*((d*e*(5*b* 
e + a*f) - c*f*(b*e + 5*a*f))*x*(c + d*x^2))/(e*f*(e + f*x^2)^2) + (-1/2*( 
(a*f*(3*d^2*e^2 + 4*c*d*e*f - 15*c^2*f^2) + b*e*(15*d^2*e^2 - 4*c*d*e*f - 
3*c^2*f^2))*x)/(e*f*(e + f*x^2)) + (3*(b*e*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^ 
2) + a*f*(d^2*e^2 + 2*c*d*e*f + 5*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/( 
2*e^(3/2)*f^(3/2)))/(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 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 = 0.54 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\frac {\left (5 a \,c^{2} f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -11 e^{3} b \,d^{2}\right ) x^{5}}{16 e^{3} f}+\frac {\left (5 a \,c^{2} f^{3}+2 a c d e \,f^{2}-a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 e^{3} b \,d^{2}\right ) x^{3}}{6 e^{2} f^{2}}+\frac {\left (11 a \,c^{2} f^{3}-2 a c d e \,f^{2}-a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 e^{3} b \,d^{2}\right ) x}{16 f^{3} e}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 a \,c^{2} f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f +5 e^{3} b \,d^{2}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 e^{3} f^{3} \sqrt {e f}}\)	\(289\)
risch	\(\frac {\frac {\left (5 a \,c^{2} f^{3}+2 a c d e \,f^{2}+a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}+2 b c d \,e^{2} f -11 e^{3} b \,d^{2}\right ) x^{5}}{16 e^{3} f}+\frac {\left (5 a \,c^{2} f^{3}+2 a c d e \,f^{2}-a \,d^{2} e^{2} f +b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 e^{3} b \,d^{2}\right ) x^{3}}{6 e^{2} f^{2}}+\frac {\left (11 a \,c^{2} f^{3}-2 a c d e \,f^{2}-a \,d^{2} e^{2} f -b \,c^{2} e \,f^{2}-2 b c d \,e^{2} f -5 e^{3} b \,d^{2}\right ) x}{16 f^{3} e}}{\left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) a \,c^{2}}{32 \sqrt {-e f}\, e^{3}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a c d}{16 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a \,d^{2}}{32 \sqrt {-e f}\, f^{2} e}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b \,c^{2}}{32 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b c d}{16 \sqrt {-e f}\, f^{2} e}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) b \,d^{2}}{32 \sqrt {-e f}\, f^{3}}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) a \,c^{2}}{32 \sqrt {-e f}\, e^{3}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a c d}{16 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a \,d^{2}}{32 \sqrt {-e f}\, f^{2} e}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b \,c^{2}}{32 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b c d}{16 \sqrt {-e f}\, f^{2} e}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) b \,d^{2}}{32 \sqrt {-e f}\, f^{3}}\)	\(550\)

Input:

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

Output:

(1/16*(5*a*c^2*f^3+2*a*c*d*e*f^2+a*d^2*e^2*f+b*c^2*e*f^2+2*b*c*d*e^2*f-11* 
b*d^2*e^3)/e^3/f*x^5+1/6*(5*a*c^2*f^3+2*a*c*d*e*f^2-a*d^2*e^2*f+b*c^2*e*f^ 
2-2*b*c*d*e^2*f-5*b*d^2*e^3)/e^2/f^2*x^3+1/16*(11*a*c^2*f^3-2*a*c*d*e*f^2- 
a*d^2*e^2*f-b*c^2*e*f^2-2*b*c*d*e^2*f-5*b*d^2*e^3)/f^3/e*x)/(f*x^2+e)^3+1/ 
16*(5*a*c^2*f^3+2*a*c*d*e*f^2+a*d^2*e^2*f+b*c^2*e*f^2+2*b*c*d*e^2*f+5*b*d^ 
2*e^3)/e^3/f^3/(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. 502 vs. \(2 (228) = 456\).

Time = 0.10 (sec) , antiderivative size = 1024, normalized size of antiderivative = 4.20 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (240) = 480\).

Time = 18.07 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{e^{7} f^{7}}} \cdot \left (5 a c^{2} f^{3} + 2 a c d e f^{2} + a d^{2} e^{2} f + b c^{2} e f^{2} + 2 b c d e^{2} f + 5 b d^{2} e^{3}\right ) \log {\left (- e^{4} f^{3} \sqrt {- \frac {1}{e^{7} f^{7}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{e^{7} f^{7}}} \cdot \left (5 a c^{2} f^{3} + 2 a c d e f^{2} + a d^{2} e^{2} f + b c^{2} e f^{2} + 2 b c d e^{2} f + 5 b d^{2} e^{3}\right ) \log {\left (e^{4} f^{3} \sqrt {- \frac {1}{e^{7} f^{7}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a c^{2} f^{5} + 6 a c d e f^{4} + 3 a d^{2} e^{2} f^{3} + 3 b c^{2} e f^{4} + 6 b c d e^{2} f^{3} - 33 b d^{2} e^{3} f^{2}\right ) + x^{3} \cdot \left (40 a c^{2} e f^{4} + 16 a c d e^{2} f^{3} - 8 a d^{2} e^{3} f^{2} + 8 b c^{2} e^{2} f^{3} - 16 b c d e^{3} f^{2} - 40 b d^{2} e^{4} f\right ) + x \left (33 a c^{2} e^{2} f^{3} - 6 a c d e^{3} f^{2} - 3 a d^{2} e^{4} f - 3 b c^{2} e^{3} f^{2} - 6 b c d e^{4} f - 15 b d^{2} e^{5}\right )}{48 e^{6} f^{3} + 144 e^{5} f^{4} x^{2} + 144 e^{4} f^{5} x^{4} + 48 e^{3} f^{6} x^{6}} \] Input:

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

Output:

-sqrt(-1/(e**7*f**7))*(5*a*c**2*f**3 + 2*a*c*d*e*f**2 + a*d**2*e**2*f + b* 
c**2*e*f**2 + 2*b*c*d*e**2*f + 5*b*d**2*e**3)*log(-e**4*f**3*sqrt(-1/(e**7 
*f**7)) + x)/32 + sqrt(-1/(e**7*f**7))*(5*a*c**2*f**3 + 2*a*c*d*e*f**2 + a 
*d**2*e**2*f + b*c**2*e*f**2 + 2*b*c*d*e**2*f + 5*b*d**2*e**3)*log(e**4*f* 
*3*sqrt(-1/(e**7*f**7)) + x)/32 + (x**5*(15*a*c**2*f**5 + 6*a*c*d*e*f**4 + 
 3*a*d**2*e**2*f**3 + 3*b*c**2*e*f**4 + 6*b*c*d*e**2*f**3 - 33*b*d**2*e**3 
*f**2) + x**3*(40*a*c**2*e*f**4 + 16*a*c*d*e**2*f**3 - 8*a*d**2*e**3*f**2 
+ 8*b*c**2*e**2*f**3 - 16*b*c*d*e**3*f**2 - 40*b*d**2*e**4*f) + x*(33*a*c* 
*2*e**2*f**3 - 6*a*c*d*e**3*f**2 - 3*a*d**2*e**4*f - 3*b*c**2*e**3*f**2 - 
6*b*c*d*e**4*f - 15*b*d**2*e**5))/(48*e**6*f**3 + 144*e**5*f**4*x**2 + 144 
*e**4*f**5*x**4 + 48*e**3*f**6*x**6)
 

Maxima [F(-2)]

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

integrate((b*x^2+a)*(d*x^2+c)^2/(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 = 330, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx=\frac {{\left (5 \, b d^{2} e^{3} + 2 \, b c d e^{2} f + a d^{2} e^{2} f + b c^{2} e f^{2} + 2 \, a c d e f^{2} + 5 \, a c^{2} f^{3}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{3}} - \frac {33 \, b d^{2} e^{3} f^{2} x^{5} - 6 \, b c d e^{2} f^{3} x^{5} - 3 \, a d^{2} e^{2} f^{3} x^{5} - 3 \, b c^{2} e f^{4} x^{5} - 6 \, a c d e f^{4} x^{5} - 15 \, a c^{2} f^{5} x^{5} + 40 \, b d^{2} e^{4} f x^{3} + 16 \, b c d e^{3} f^{2} x^{3} + 8 \, a d^{2} e^{3} f^{2} x^{3} - 8 \, b c^{2} e^{2} f^{3} x^{3} - 16 \, a c d e^{2} f^{3} x^{3} - 40 \, a c^{2} e f^{4} x^{3} + 15 \, b d^{2} e^{5} x + 6 \, b c d e^{4} f x + 3 \, a d^{2} e^{4} f x + 3 \, b c^{2} e^{3} f^{2} x + 6 \, a c d e^{3} f^{2} x - 33 \, a c^{2} e^{2} f^{3} x}{48 \, {\left (f x^{2} + e\right )}^{3} e^{3} f^{3}} \] Input:

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

Output:

1/16*(5*b*d^2*e^3 + 2*b*c*d*e^2*f + a*d^2*e^2*f + b*c^2*e*f^2 + 2*a*c*d*e* 
f^2 + 5*a*c^2*f^3)*arctan(f*x/sqrt(e*f))/(sqrt(e*f)*e^3*f^3) - 1/48*(33*b* 
d^2*e^3*f^2*x^5 - 6*b*c*d*e^2*f^3*x^5 - 3*a*d^2*e^2*f^3*x^5 - 3*b*c^2*e*f^ 
4*x^5 - 6*a*c*d*e*f^4*x^5 - 15*a*c^2*f^5*x^5 + 40*b*d^2*e^4*f*x^3 + 16*b*c 
*d*e^3*f^2*x^3 + 8*a*d^2*e^3*f^2*x^3 - 8*b*c^2*e^2*f^3*x^3 - 16*a*c*d*e^2* 
f^3*x^3 - 40*a*c^2*e*f^4*x^3 + 15*b*d^2*e^5*x + 6*b*c*d*e^4*f*x + 3*a*d^2* 
e^4*f*x + 3*b*c^2*e^3*f^2*x + 6*a*c*d*e^3*f^2*x - 33*a*c^2*e^2*f^3*x)/((f* 
x^2 + e)^3*e^3*f^3)
 

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3-2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-5\,b\,d^2\,e^3-a\,d^2\,e^2\,f\right )}{6\,e^2\,f^2}-\frac {x\,\left (b\,c^2\,e\,f^2-11\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2+5\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e\,f^3}+\frac {x^5\,\left (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-11\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e^3\,f}}{e^3+3\,e^2\,f\,x^2+3\,e\,f^2\,x^4+f^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (b\,c^2\,e\,f^2+5\,a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2+5\,b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}{16\,e^{7/2}\,f^{7/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 988, normalized size of antiderivative = 4.05 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^2}{\left (e+f x^2\right )^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(15*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*e**3*f**3 + 45*sq 
rt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*e**2*f**4*x**2 + 45*sqr 
t(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*e*f**5*x**4 + 15*sqrt(f) 
*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c**2*f**6*x**6 + 6*sqrt(f)*sqrt(e 
)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*c*d*e**4*f**2 + 18*sqrt(f)*sqrt(e)*atan( 
(f*x)/(sqrt(f)*sqrt(e)))*a*c*d*e**3*f**3*x**2 + 18*sqrt(f)*sqrt(e)*atan((f 
*x)/(sqrt(f)*sqrt(e)))*a*c*d*e**2*f**4*x**4 + 6*sqrt(f)*sqrt(e)*atan((f*x) 
/(sqrt(f)*sqrt(e)))*a*c*d*e*f**5*x**6 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt 
(f)*sqrt(e)))*a*d**2*e**5*f + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e 
)))*a*d**2*e**4*f**2*x**2 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)) 
)*a*d**2*e**3*f**3*x**4 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))* 
a*d**2*e**2*f**4*x**6 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b* 
c**2*e**4*f**2 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c**2*e* 
*3*f**3*x**2 + 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c**2*e**2 
*f**4*x**4 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c**2*e*f**5 
*x**6 + 6*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**5*f + 18* 
sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**4*f**2*x**2 + 18*sq 
rt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**3*f**3*x**4 + 6*sqrt( 
f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*c*d*e**2*f**4*x**6 + 15*sqrt(f) 
*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*d**2*e**6 + 45*sqrt(f)*sqrt(e)...