\(\int \frac {(a+b x^2)^3}{(c+d x^2) (e+f x^2)^2} \, dx\) [248]

Optimal result

Integrand size = 28, antiderivative size = 169 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx=\frac {b^3 x}{d f^2}+\frac {(b e-a f)^3 x}{2 e f^2 (d e-c f) \left (e+f x^2\right )}-\frac {(b c-a d)^3 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} d^{3/2} (d e-c f)^2}-\frac {(b e-a f)^2 (b e (3 d e-5 c f)+a f (3 d e-c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{5/2} (d e-c f)^2} \] Output:

b^3*x/d/f^2+1/2*(-a*f+b*e)^3*x/e/f^2/(-c*f+d*e)/(f*x^2+e)-(-a*d+b*c)^3*arc 
tan(d^(1/2)*x/c^(1/2))/c^(1/2)/d^(3/2)/(-c*f+d*e)^2-1/2*(-a*f+b*e)^2*(b*e* 
(-5*c*f+3*d*e)+a*f*(-c*f+3*d*e))*arctan(f^(1/2)*x/e^(1/2))/e^(3/2)/f^(5/2) 
/(-c*f+d*e)^2
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx=\frac {b^3 x}{d f^2}+\frac {(b e-a f)^3 x}{2 e f^2 (d e-c f) \left (e+f x^2\right )}-\frac {(b c-a d)^3 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} d^{3/2} (d e-c f)^2}-\frac {(b e-a f)^2 (b e (3 d e-5 c f)+a f (3 d e-c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{5/2} (d e-c f)^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.85, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {419, 25, 300, 401, 27, 403, 25, 299, 218, 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.

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

\(\Big \downarrow \) 419

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 401

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {x \left (a+b x^2\right )^2 (b e-a f) (d e-c f)}{2 e \left (e+f x^2\right )}-\frac {\frac {\int -\frac {a \left (b^2 (3 d e-5 c f) e^2+2 a b f (d e+3 c f) e-3 a^2 f^2 (3 d e-c f)\right )-b \left (-3 b^2 (3 d e-5 c f) e^2+2 a b f (3 d e-11 c f) e+a^2 f^2 (7 d e+3 c f)\right ) x^2}{f x^2+e}dx}{3 f}+\frac {b x \left (a+b x^2\right ) (b e (3 d e-5 c f)-a f (d e-3 c f))}{3 f}}{2 e}}{(d e-c f)^2}-\frac {d (b c-a d) \int \left (\frac {b^2 x^2}{d}-\frac {b (b c-2 a d)}{d^2}+\frac {b^2 c^2-2 a b d c+a^2 d^2}{d^2 \left (d x^2+c\right )}\right )dx}{(d e-c f)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {x \left (a+b x^2\right )^2 (b e-a f) (d e-c f)}{2 e \left (e+f x^2\right )}-\frac {\frac {b x \left (a+b x^2\right ) (b e (3 d e-5 c f)-a f (d e-3 c f))}{3 f}-\frac {\int \frac {a \left (b^2 (3 d e-5 c f) e^2+2 a b f (d e+3 c f) e-3 a^2 f^2 (3 d e-c f)\right )-b \left (-3 b^2 (3 d e-5 c f) e^2+2 a b f (3 d e-11 c f) e+a^2 f^2 (7 d e+3 c f)\right ) x^2}{f x^2+e}dx}{3 f}}{2 e}}{(d e-c f)^2}-\frac {d (b c-a d) \int \left (\frac {b^2 x^2}{d}-\frac {b (b c-2 a d)}{d^2}+\frac {b^2 c^2-2 a b d c+a^2 d^2}{d^2 \left (d x^2+c\right )}\right )dx}{(d e-c f)^2}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {x \left (a+b x^2\right )^2 (b e-a f) (d e-c f)}{2 e \left (e+f x^2\right )}-\frac {\frac {b x \left (a+b x^2\right ) (b e (3 d e-5 c f)-a f (d e-3 c f))}{3 f}-\frac {-\frac {3 (b e-a f)^2 (a f (3 d e-c f)+b e (3 d e-5 c f)) \int \frac {1}{f x^2+e}dx}{f}-\frac {b x \left (a^2 f^2 (3 c f+7 d e)+2 a b e f (3 d e-11 c f)-3 b^2 e^2 (3 d e-5 c f)\right )}{f}}{3 f}}{2 e}}{(d e-c f)^2}-\frac {d (b c-a d) \int \left (\frac {b^2 x^2}{d}-\frac {b (b c-2 a d)}{d^2}+\frac {b^2 c^2-2 a b d c+a^2 d^2}{d^2 \left (d x^2+c\right )}\right )dx}{(d e-c f)^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {x \left (a+b x^2\right )^2 (b e-a f) (d e-c f)}{2 e \left (e+f x^2\right )}-\frac {\frac {b x \left (a+b x^2\right ) (b e (3 d e-5 c f)-a f (d e-3 c f))}{3 f}-\frac {-\frac {b x \left (a^2 f^2 (3 c f+7 d e)+2 a b e f (3 d e-11 c f)-3 b^2 e^2 (3 d e-5 c f)\right )}{f}-\frac {3 (b e-a f)^2 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (3 d e-c f)+b e (3 d e-5 c f))}{\sqrt {e} f^{3/2}}}{3 f}}{2 e}}{(d e-c f)^2}-\frac {d (b c-a d) \int \left (\frac {b^2 x^2}{d}-\frac {b (b c-2 a d)}{d^2}+\frac {b^2 c^2-2 a b d c+a^2 d^2}{d^2 \left (d x^2+c\right )}\right )dx}{(d e-c f)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {x \left (a+b x^2\right )^2 (b e-a f) (d e-c f)}{2 e \left (e+f x^2\right )}-\frac {\frac {b x \left (a+b x^2\right ) (b e (3 d e-5 c f)-a f (d e-3 c f))}{3 f}-\frac {-\frac {b x \left (a^2 f^2 (3 c f+7 d e)+2 a b e f (3 d e-11 c f)-3 b^2 e^2 (3 d e-5 c f)\right )}{f}-\frac {3 (b e-a f)^2 \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (3 d e-c f)+b e (3 d e-5 c f))}{\sqrt {e} f^{3/2}}}{3 f}}{2 e}}{(d e-c f)^2}-\frac {d (b c-a d) \left (\frac {(b c-a d)^2 \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} d^{5/2}}-\frac {b x (b c-2 a d)}{d^2}+\frac {b^2 x^3}{3 d}\right )}{(d e-c f)^2}\)

Input:

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

Output:

-((d*(b*c - a*d)*(-((b*(b*c - 2*a*d)*x)/d^2) + (b^2*x^3)/(3*d) + ((b*c - a 
*d)^2*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*d^(5/2))))/(d*e - c*f)^2) + (( 
(b*e - a*f)*(d*e - c*f)*x*(a + b*x^2)^2)/(2*e*(e + f*x^2)) - ((b*(b*e*(3*d 
*e - 5*c*f) - a*f*(d*e - 3*c*f))*x*(a + b*x^2))/(3*f) - (-((b*(2*a*b*e*f*( 
3*d*e - 11*c*f) - 3*b^2*e^2*(3*d*e - 5*c*f) + a^2*f^2*(7*d*e + 3*c*f))*x)/ 
f) - (3*(b*e - a*f)^2*(b*e*(3*d*e - 5*c*f) + a*f*(3*d*e - c*f))*ArcTan[(Sq 
rt[f]*x)/Sqrt[e]])/(Sqrt[e]*f^(3/2)))/(3*f))/(2*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 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 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

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 403

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.76

method	result	size
default	\(\frac {b^{3} x}{d \,f^{2}}+\frac {\frac {\left (a^{3} c \,f^{4}-a^{3} d e \,f^{3}-3 a^{2} b c e \,f^{3}+3 a^{2} b d \,e^{2} f^{2}+3 a \,b^{2} c \,e^{2} f^{2}-3 a \,b^{2} d \,e^{3} f -b^{3} c \,e^{3} f +e^{4} b^{3} d \right ) x}{2 e \left (f \,x^{2}+e \right )}+\frac {\left (a^{3} c \,f^{4}-3 a^{3} d e \,f^{3}+3 a^{2} b c e \,f^{3}+3 a^{2} b d \,e^{2} f^{2}-9 a \,b^{2} c \,e^{2} f^{2}+3 a \,b^{2} d \,e^{3} f +5 b^{3} c \,e^{3} f -3 e^{4} b^{3} d \right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 e \sqrt {e f}}}{f^{2} \left (c f -d e \right )^{2}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{d \left (c f -d e \right )^{2} \sqrt {c d}}\)	\(297\)
risch	\(\text {Expression too large to display}\)	\(1073\)

Input:

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

Output:

b^3*x/d/f^2+1/f^2/(c*f-d*e)^2*(1/2*(a^3*c*f^4-a^3*d*e*f^3-3*a^2*b*c*e*f^3+ 
3*a^2*b*d*e^2*f^2+3*a*b^2*c*e^2*f^2-3*a*b^2*d*e^3*f-b^3*c*e^3*f+b^3*d*e^4) 
/e*x/(f*x^2+e)+1/2*(a^3*c*f^4-3*a^3*d*e*f^3+3*a^2*b*c*e*f^3+3*a^2*b*d*e^2* 
f^2-9*a*b^2*c*e^2*f^2+3*a*b^2*d*e^3*f+5*b^3*c*e^3*f-3*b^3*d*e^4)/e/(e*f)^( 
1/2)*arctan(f*x/(e*f)^(1/2)))+1/d*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3 
*c^3)/(c*f-d*e)^2/(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. 628 vs. \(2 (149) = 298\).

Time = 64.52 (sec) , antiderivative size = 2588, normalized size of antiderivative = 15.31 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((b*x^2+a)^3/(d*x^2+c)/(f*x^2+e)^2,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 = 293, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx=\frac {b^{3} x}{d f^{2}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} \sqrt {c d}} - \frac {{\left (3 \, b^{3} d e^{4} - 5 \, b^{3} c e^{3} f - 3 \, a b^{2} d e^{3} f + 9 \, a b^{2} c e^{2} f^{2} - 3 \, a^{2} b d e^{2} f^{2} - 3 \, a^{2} b c e f^{3} + 3 \, a^{3} d e f^{3} - a^{3} c f^{4}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{2 \, {\left (d^{2} e^{3} f^{2} - 2 \, c d e^{2} f^{3} + c^{2} e f^{4}\right )} \sqrt {e f}} + \frac {b^{3} e^{3} x - 3 \, a b^{2} e^{2} f x + 3 \, a^{2} b e f^{2} x - a^{3} f^{3} x}{2 \, {\left (d e^{2} f^{2} - c e f^{3}\right )} {\left (f x^{2} + e\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.74 (sec) , antiderivative size = 11519, normalized size of antiderivative = 68.16 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(b^3*x)/(d*f^2) + (atan(((((x*(9*a^2*b^4*d^6*e^6*f^2 - a^6*c^2*d^4*f^8 - 1 
3*a^6*d^6*e^2*f^6 - 4*b^6*c^6*e^2*f^6 - 9*b^6*d^6*e^8 - 36*a^3*b^3*d^6*e^5 
*f^3 + 9*a^4*b^2*d^6*e^4*f^4 - 25*b^6*c^2*d^4*e^6*f^2 + 18*a*b^5*d^6*e^7*f 
 + 6*a^6*c*d^5*e*f^7 + 30*b^6*c*d^5*e^7*f + 18*a^5*b*d^6*e^3*f^5 - 84*a*b^ 
5*c*d^5*e^6*f^2 + 24*a*b^5*c^5*d*e^2*f^6 + 36*a^5*b*c*d^5*e^2*f^6 - 6*a^5* 
b*c^2*d^4*e*f^7 + 90*a*b^5*c^2*d^4*e^5*f^3 + 42*a^2*b^4*c*d^5*e^5*f^3 + 72 
*a^3*b^3*c*d^5*e^4*f^4 - 78*a^4*b^2*c*d^5*e^3*f^5 - 111*a^2*b^4*c^2*d^4*e^ 
4*f^4 - 60*a^2*b^4*c^4*d^2*e^2*f^6 + 44*a^3*b^3*c^2*d^4*e^3*f^5 + 80*a^3*b 
^3*c^3*d^3*e^2*f^6 - 51*a^4*b^2*c^2*d^4*e^2*f^6))/(2*(d^3*e^4*f^3 - 2*c*d^ 
2*e^3*f^4 + c^2*d*e^2*f^5)) + ((-c*d^3)^(1/2)*(a*d - b*c)^3*((16*a^3*d^8*e 
^6*f^5 + 128*a^3*c^2*d^6*e^4*f^7 - 112*a^3*c^3*d^5*e^3*f^8 + 48*a^3*c^4*d^ 
4*e^2*f^9 + 112*b^3*c^2*d^6*e^7*f^4 - 208*b^3*c^3*d^5*e^6*f^5 + 192*b^3*c^ 
4*d^4*e^5*f^6 - 88*b^3*c^5*d^3*e^4*f^7 + 16*b^3*c^6*d^2*e^3*f^8 - 72*a^3*c 
*d^7*e^5*f^6 - 8*a^3*c^5*d^3*e*f^10 - 24*b^3*c*d^7*e^8*f^3 + 24*a*b^2*c*d^ 
7*e^7*f^4 - 24*a^2*b*c*d^7*e^6*f^5 - 96*a*b^2*c^2*d^6*e^6*f^5 + 144*a*b^2* 
c^3*d^5*e^5*f^6 - 96*a*b^2*c^4*d^4*e^4*f^7 + 24*a*b^2*c^5*d^3*e^3*f^8 + 96 
*a^2*b*c^2*d^6*e^5*f^6 - 144*a^2*b*c^3*d^5*e^4*f^7 + 96*a^2*b*c^4*d^4*e^3* 
f^8 - 24*a^2*b*c^5*d^3*e^2*f^9)/(4*(d^4*e^5*f^3 - 3*c*d^3*e^4*f^4 - c^3*d* 
e^2*f^6 + 3*c^2*d^2*e^3*f^5)) - (x*(-c*d^3)^(1/2)*(a*d - b*c)^3*(16*d^8*e^ 
7*f^5 - 48*c*d^7*e^6*f^6 + 32*c^2*d^6*e^5*f^7 + 32*c^3*d^5*e^4*f^8 - 48...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1064, normalized size of antiderivative = 6.30 \[ \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right ) \left (e+f x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**3*d**3*e**3*f**3 + 2*s 
qrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**3*d**3*e**2*f**4*x**2 - 6* 
sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*b*c*d**2*e**3*f**3 - 6* 
sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*b*c*d**2*e**2*f**4*x**2 
 + 6*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b**2*c**2*d*e**3*f**3 
 + 6*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a*b**2*c**2*d*e**2*f**4 
*x**2 - 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**3*c**3*e**3*f** 
3 - 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*b**3*c**3*e**2*f**4*x* 
*2 + sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**3*c**2*d**2*e*f**4 + 
 sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**3*c**2*d**2*f**5*x**2 - 
3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**3*c*d**3*e**2*f**3 - 3* 
sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**3*c*d**3*e*f**4*x**2 + 3* 
sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*b*c**2*d**2*e**2*f**3 + 
 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*b*c**2*d**2*e*f**4*x 
**2 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*b*c*d**3*e**3*f 
**2 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a**2*b*c*d**3*e**2*f 
**3*x**2 - 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b**2*c**2*d** 
2*e**3*f**2 - 9*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b**2*c**2* 
d**2*e**2*f**3*x**2 + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b* 
*2*c*d**3*e**4*f + 3*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*b*...