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

Optimal result

Integrand size = 37, antiderivative size = 706 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=-\frac {b \left (A \left (2 a b^2 c e f^2 (7 d e-5 c f)+a^3 d f^3 (7 d e-3 c f)-8 b^3 e^2 (d e-c f)^2-a^2 b f^2 \left (14 d^2 e^2-3 c d e f-3 c^2 f^2\right )\right )-a B e \left (a^2 d f^2 (3 d e+c f)-a b f \left (10 d^2 e^2-3 c d e f+c^2 f^2\right )-2 b^2 e \left (4 d^2 e^2-13 c d e f+7 c^2 f^2\right )\right )\right ) x}{8 a (b c-a d) e^2 (b e-a f)^3 (d e-c f)^2 \sqrt {a+b x^2}}-\frac {f (B e-A f) x}{4 e (b e-a f) (d e-c f) \sqrt {a+b x^2} \left (e+f x^2\right )^2}+\frac {f (4 b e (A f (3 d e-2 c f)-B e (2 d e-c f))-a f (A f (7 d e-3 c f)-B e (3 d e+c f))) x}{8 e^2 (b e-a f)^2 (d e-c f)^2 \sqrt {a+b x^2} \left (e+f x^2\right )}+\frac {d^3 (B c-A d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{3/2} (d e-c f)^3}+\frac {f \left (4 a b e f \left (B e \left (3 d^2 e^2+3 c d e f-2 c^2 f^2\right )-A f \left (12 d^2 e^2-11 c d e f+3 c^2 f^2\right )\right )-a^2 f^2 \left (B e \left (3 d^2 e^2+6 c d e f-c^2 f^2\right )-A f \left (15 d^2 e^2-10 c d e f+3 c^2 f^2\right )\right )-8 b^2 e^2 \left (B e \left (3 d^2 e^2-3 c d e f+c^2 f^2\right )-A f \left (6 d^2 e^2-8 c d e f+3 c^2 f^2\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {b e-a f} x}{\sqrt {e} \sqrt {a+b x^2}}\right )}{8 e^{5/2} (b e-a f)^{7/2} (d e-c f)^3} \] Output:

-1/8*b*(A*(2*a*b^2*c*e*f^2*(-5*c*f+7*d*e)+a^3*d*f^3*(-3*c*f+7*d*e)-8*b^3*e 
^2*(-c*f+d*e)^2-a^2*b*f^2*(-3*c^2*f^2-3*c*d*e*f+14*d^2*e^2))-a*B*e*(a^2*d* 
f^2*(c*f+3*d*e)-a*b*f*(c^2*f^2-3*c*d*e*f+10*d^2*e^2)-2*b^2*e*(7*c^2*f^2-13 
*c*d*e*f+4*d^2*e^2)))*x/a/(-a*d+b*c)/e^2/(-a*f+b*e)^3/(-c*f+d*e)^2/(b*x^2+ 
a)^(1/2)-1/4*f*(-A*f+B*e)*x/e/(-a*f+b*e)/(-c*f+d*e)/(b*x^2+a)^(1/2)/(f*x^2 
+e)^2+1/8*f*(4*b*e*(A*f*(-2*c*f+3*d*e)-B*e*(-c*f+2*d*e))-a*f*(A*f*(-3*c*f+ 
7*d*e)-B*e*(c*f+3*d*e)))*x/e^2/(-a*f+b*e)^2/(-c*f+d*e)^2/(b*x^2+a)^(1/2)/( 
f*x^2+e)+d^3*(-A*d+B*c)*arctanh((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a)^(1/2) 
)/c^(1/2)/(-a*d+b*c)^(3/2)/(-c*f+d*e)^3+1/8*f*(4*a*b*e*f*(B*e*(-2*c^2*f^2+ 
3*c*d*e*f+3*d^2*e^2)-A*f*(3*c^2*f^2-11*c*d*e*f+12*d^2*e^2))-a^2*f^2*(B*e*( 
-c^2*f^2+6*c*d*e*f+3*d^2*e^2)-A*f*(3*c^2*f^2-10*c*d*e*f+15*d^2*e^2))-8*b^2 
*e^2*(B*e*(c^2*f^2-3*c*d*e*f+3*d^2*e^2)-A*f*(3*c^2*f^2-8*c*d*e*f+6*d^2*e^2 
)))*arctanh((-a*f+b*e)^(1/2)*x/e^(1/2)/(b*x^2+a)^(1/2))/e^(5/2)/(-a*f+b*e) 
^(7/2)/(-c*f+d*e)^3
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 20.69 (sec) , antiderivative size = 2913, normalized size of antiderivative = 4.13 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Result too large to show} \] Input:

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

Output:

(d^2*(-(B*c) + A*d)*x*(-15*c*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] - 10* 
d*x^2*Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 15*c*ArcTanh[Sqrt[((b*c - 
a*d)*x^2)/(c*(a + b*x^2))]] + 10*d*x^2*ArcTanh[Sqrt[((b*c - a*d)*x^2)/(c*( 
a + b*x^2))]] + 2*c*(((b*c - a*d)*x^2)/(c*(a + b*x^2)))^(5/2)*Hypergeometr 
ic2F1[2, 5/2, 7/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + 2*d*x^2*(((b*c - a 
*d)*x^2)/(c*(a + b*x^2)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, ((b*c - a*d 
)*x^2)/(c*(a + b*x^2))]))/(5*a*c^2*(d*e - c*f)^3*(((b*c - a*d)*x^2)/(c*(a 
+ b*x^2)))^(3/2)*Sqrt[a + b*x^2]*(1 + (b*x^2)/a)) + (d*(-(B*c) + A*d)*f*x* 
(-15*e*Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2))] - 10*f*x^2*Sqrt[((b*e - a*f 
)*x^2)/(e*(a + b*x^2))] + 15*e*ArcTanh[Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^ 
2))]] + 10*f*x^2*ArcTanh[Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2))]] + 2*e*(( 
(b*e - a*f)*x^2)/(e*(a + b*x^2)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, ((b 
*e - a*f)*x^2)/(e*(a + b*x^2))] + 2*f*x^2*(((b*e - a*f)*x^2)/(e*(a + b*x^2 
)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, ((b*e - a*f)*x^2)/(e*(a + b*x^2)) 
]))/(5*a*e^2*(-(d*e) + c*f)^3*(((b*e - a*f)*x^2)/(e*(a + b*x^2)))^(3/2)*Sq 
rt[a + b*x^2]*(1 + (b*x^2)/a)) + ((B*c - A*d)*f*x*(-2625*Sqrt[((b*e - a*f) 
*x^2)/(e*(a + b*x^2))] - (5250*f*x^2*Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2) 
)])/e - (2310*f^2*x^4*Sqrt[((b*e - a*f)*x^2)/(e*(a + b*x^2))])/e^2 + 70*(( 
(b*e - a*f)*x^2)/(e*(a + b*x^2)))^(3/2) + (560*f*x^2*(((b*e - a*f)*x^2)/(e 
*(a + b*x^2)))^(3/2))/e + (280*f^2*x^4*(((b*e - a*f)*x^2)/(e*(a + b*x^2...
 

Rubi [A] (verified)

Time = 1.81 (sec) , antiderivative size = 736, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {7276, 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.

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 f \left (a^2 f^2-4 a b e f+8 b^2 e^2\right ) (B e-A f) \text {arctanh}\left (\frac {x \sqrt {b e-a f}}{\sqrt {e} \sqrt {a+b x^2}}\right )}{8 e^{5/2} (b e-a f)^{7/2} (d e-c f)}+\frac {d^3 (B c-A d) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{3/2} (d e-c f)^3}-\frac {f^2 (4 b e-a f) (B c-A d) \text {arctanh}\left (\frac {x \sqrt {b e-a f}}{\sqrt {e} \sqrt {a+b x^2}}\right )}{2 e^{3/2} (b e-a f)^{5/2} (d e-c f)^2}-\frac {d f^2 (B c-A d) \text {arctanh}\left (\frac {x \sqrt {b e-a f}}{\sqrt {e} \sqrt {a+b x^2}}\right )}{\sqrt {e} (b e-a f)^{3/2} (d e-c f)^3}-\frac {b d^2 x (B c-A d)}{a \sqrt {a+b x^2} (b c-a d) (d e-c f)^3}+\frac {f x \sqrt {a+b x^2} (4 b e-a f) (3 a f+2 b e) (B e-A f)}{8 a e^2 \left (e+f x^2\right ) (b e-a f)^3 (d e-c f)}-\frac {f^2 x (B c-A d)}{2 e \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f) (d e-c f)^2}+\frac {b d f x (B c-A d)}{a \sqrt {a+b x^2} (b e-a f) (d e-c f)^3}+\frac {b f x (a f+2 b e) (B c-A d)}{2 a e \sqrt {a+b x^2} (b e-a f)^2 (d e-c f)^2}+\frac {b x (a f+4 b e) (B e-A f)}{4 a e \sqrt {a+b x^2} \left (e+f x^2\right ) (b e-a f)^2 (d e-c f)}-\frac {f x (B e-A f)}{4 e \sqrt {a+b x^2} \left (e+f x^2\right )^2 (b e-a f) (d e-c f)}\)

Input:

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

Output:

-((b*d^2*(B*c - A*d)*x)/(a*(b*c - a*d)*(d*e - c*f)^3*Sqrt[a + b*x^2])) + ( 
b*d*(B*c - A*d)*f*x)/(a*(b*e - a*f)*(d*e - c*f)^3*Sqrt[a + b*x^2]) + (b*(B 
*c - A*d)*f*(2*b*e + a*f)*x)/(2*a*e*(b*e - a*f)^2*(d*e - c*f)^2*Sqrt[a + b 
*x^2]) - (f*(B*e - A*f)*x)/(4*e*(b*e - a*f)*(d*e - c*f)*Sqrt[a + b*x^2]*(e 
 + f*x^2)^2) - ((B*c - A*d)*f^2*x)/(2*e*(b*e - a*f)*(d*e - c*f)^2*Sqrt[a + 
 b*x^2]*(e + f*x^2)) + (b*(4*b*e + a*f)*(B*e - A*f)*x)/(4*a*e*(b*e - a*f)^ 
2*(d*e - c*f)*Sqrt[a + b*x^2]*(e + f*x^2)) + (f*(4*b*e - a*f)*(2*b*e + 3*a 
*f)*(B*e - A*f)*x*Sqrt[a + b*x^2])/(8*a*e^2*(b*e - a*f)^3*(d*e - c*f)*(e + 
 f*x^2)) + (d^3*(B*c - A*d)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + 
b*x^2])])/(Sqrt[c]*(b*c - a*d)^(3/2)*(d*e - c*f)^3) - (d*(B*c - A*d)*f^2*A 
rcTanh[(Sqrt[b*e - a*f]*x)/(Sqrt[e]*Sqrt[a + b*x^2])])/(Sqrt[e]*(b*e - a*f 
)^(3/2)*(d*e - c*f)^3) - ((B*c - A*d)*f^2*(4*b*e - a*f)*ArcTanh[(Sqrt[b*e 
- a*f]*x)/(Sqrt[e]*Sqrt[a + b*x^2])])/(2*e^(3/2)*(b*e - a*f)^(5/2)*(d*e - 
c*f)^2) - (3*f*(B*e - A*f)*(8*b^2*e^2 - 4*a*b*e*f + a^2*f^2)*ArcTanh[(Sqrt 
[b*e - a*f]*x)/(Sqrt[e]*Sqrt[a + b*x^2])])/(8*e^(5/2)*(b*e - a*f)^(7/2)*(d 
*e - c*f))
 

Defintions of rubi rules used

rule 2009

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

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [A] (verified)

Time = 3.35 (sec) , antiderivative size = 953, normalized size of antiderivative = 1.35

method	result	size
pseudoelliptic	\(\text {Expression too large to display}\)	\(953\)
default	\(\text {Expression too large to display}\)	\(5470\)

Input:

int((B*x^2+A)/(b*x^2+a)^(3/2)/(d*x^2+c)/(f*x^2+e)^3,x,method=_RETURNVERBOS 
E)
 

Output:

5/8/(b*x^2+a)^(1/2)/((a*d-b*c)*c)^(1/2)/((a*f-b*e)*e)^(1/2)*(-3/5*(f^2*(f^ 
2*(A*f+1/3*B*e)*c^2-10/3*(A*f+3/5*B*e)*f*e*d*c+5*A*d^2*e^2*f-B*d^2*e^3)*a^ 
2-4*b*(f^2*(A*f+2/3*B*e)*c^2+d*(-11/3*e*A*f^2-B*e^2*f)*c+4*A*d^2*e^2*f-B*d 
^2*e^3)*f*e*a+8*b^2*(f^2*(A*f-1/3*B*e)*c^2-8/3*f*e*d*(A*f-3/8*B*e)*c+2*A*d 
^2*e^2*f-B*d^2*e^3)*e^2)*a*((a*d-b*c)*c)^(1/2)*(a*d-b*c)*(b*x^2+a)^(1/2)*f 
*(f*x^2+e)^2*arctan(e*(b*x^2+a)^(1/2)/x/((a*f-b*e)*e)^(1/2))+((a*f-b*e)*e) 
^(1/2)*(8/5*(b*x^2+a)^(1/2)*a*d^3*e^2*(f*x^2+e)^2*(a*f-b*e)^3*(A*d-B*c)*ar 
ctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2))+x*((f*(-1/5*B*e^2+f*(1/5*x^2 
*B+A)*e+3/5*f^2*x^2*A)*c-9/5*(-5/9*B*e^2+f*(-1/3*x^2*B+A)*e+7/9*f^2*x^2*A) 
*e*d)*f^3*d*a^4-(f^2*(-1/5*B*e^2+f*(1/5*x^2*B+A)*e+3/5*f^2*x^2*A)*c^2+3/5* 
(-B*e^3+f*(-2/3*x^2*B+A)*e^2-2/3*x^2*(1/2*x^2*B+A)*f^2*e-A*f^3*x^4)*f*d*c- 
16/5*(-3/4*B*e^3+f*(-5/16*x^2*B+A)*e^2+5/16*x^2*(3/5*x^2*B+A)*f^2*e-7/16*A 
*f^3*x^4)*e*d^2)*b*f^2*a^3+12/5*b^2*((-2/3*B*e^3+f*(-5/12*x^2*B+A)*e^2+5/1 
2*x^2*f^2*(-1/5*x^2*B+A)*e-1/4*A*f^3*x^4)*f*c^2-4/3*e*d*(-3/4*B*e^3+f*(-13 
/16*x^2*B+A)*e^2+17/16*x^2*f^2*(-3/17*x^2*B+A)*e+3/16*A*f^3*x^4)*c+4/3*x^2 
*e^2*d^2*(-3/4*B*e^2+f*(-5/8*x^2*B+A)*e+7/8*f^2*x^2*A))*f^2*a^2+12/5*b^3*( 
(-2/3*B*e^3-2*B*e^2*f*x^2+f^2*x^2*(-7/6*x^2*B+A)*e+5/6*A*f^3*x^4)*f^2*c^2- 
4/3*f*e*d*(-B*e^3-11/4*B*e^2*f*x^2+f^2*x^2*(-13/8*x^2*B+A)*e+7/8*A*f^3*x^4 
)*c-2/3*B*d^2*e^3*(f*x^2+e)^2)*e*a+8/5*A*b^4*e^2*(f*x^2+e)^2*(c*f-d*e)^2)* 
((a*d-b*c)*c)^(1/2)*(c*f-d*e)))/(c*f-d*e)^3/(a*f-b*e)^3/(a*d-b*c)/e^2/(...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\int { \frac {B x^{2} + A}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} {\left (d x^{2} + c\right )} {\left (f x^{2} + e\right )}^{3}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2528 vs. \(2 (671) = 1342\).

Time = 8.73 (sec) , antiderivative size = 2528, normalized size of antiderivative = 3.58 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-(B*a*b^3 - A*b^4)*x/((a*b^4*c*e^3 - a^2*b^3*d*e^3 - 3*a^2*b^3*c*e^2*f + 3 
*a^3*b^2*d*e^2*f + 3*a^3*b^2*c*e*f^2 - 3*a^4*b*d*e*f^2 - a^4*b*c*f^3 + a^5 
*d*f^3)*sqrt(b*x^2 + a)) - (B*sqrt(b)*c*d^3 - A*sqrt(b)*d^4)*arctan(1/2*(( 
sqrt(b)*x - sqrt(b*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/ 
((b*c*d^3*e^3 - a*d^4*e^3 - 3*b*c^2*d^2*e^2*f + 3*a*c*d^3*e^2*f + 3*b*c^3* 
d*e*f^2 - 3*a*c^2*d^2*e*f^2 - b*c^4*f^3 + a*c^3*d*f^3)*sqrt(-b^2*c^2 + a*b 
*c*d)) + 1/8*(24*B*b^(5/2)*d^2*e^5*f - 24*B*b^(5/2)*c*d*e^4*f^2 - 12*B*a*b 
^(3/2)*d^2*e^4*f^2 - 48*A*b^(5/2)*d^2*e^4*f^2 + 8*B*b^(5/2)*c^2*e^3*f^3 - 
12*B*a*b^(3/2)*c*d*e^3*f^3 + 64*A*b^(5/2)*c*d*e^3*f^3 + 3*B*a^2*sqrt(b)*d^ 
2*e^3*f^3 + 48*A*a*b^(3/2)*d^2*e^3*f^3 + 8*B*a*b^(3/2)*c^2*e^2*f^4 - 24*A* 
b^(5/2)*c^2*e^2*f^4 + 6*B*a^2*sqrt(b)*c*d*e^2*f^4 - 44*A*a*b^(3/2)*c*d*e^2 
*f^4 - 15*A*a^2*sqrt(b)*d^2*e^2*f^4 - B*a^2*sqrt(b)*c^2*e*f^5 + 12*A*a*b^( 
3/2)*c^2*e*f^5 + 10*A*a^2*sqrt(b)*c*d*e*f^5 - 3*A*a^2*sqrt(b)*c^2*f^6)*arc 
tan(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*f + 2*b*e - a*f)/sqrt(-b^2*e^2 + 
a*b*e*f))/((b^3*d^3*e^8 - 3*b^3*c*d^2*e^7*f - 3*a*b^2*d^3*e^7*f + 3*b^3*c^ 
2*d*e^6*f^2 + 9*a*b^2*c*d^2*e^6*f^2 + 3*a^2*b*d^3*e^6*f^2 - b^3*c^3*e^5*f^ 
3 - 9*a*b^2*c^2*d*e^5*f^3 - 9*a^2*b*c*d^2*e^5*f^3 - a^3*d^3*e^5*f^3 + 3*a* 
b^2*c^3*e^4*f^4 + 9*a^2*b*c^2*d*e^4*f^4 + 3*a^3*c*d^2*e^4*f^4 - 3*a^2*b*c^ 
3*e^3*f^5 - 3*a^3*c^2*d*e^3*f^5 + a^3*c^3*e^2*f^6)*sqrt(-b^2*e^2 + a*b*e*f 
)) + 1/4*(16*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*b^(5/2)*d*e^4*f^2 - 8*(s...
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 12786, normalized size of antiderivative = 18.11 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (e+f x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(16*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x** 
2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*d**3*e**5*f**4 + 32*sqrt(c 
)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt( 
d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*d**3*e**4*f**5*x**2 + 16*sqrt(c)*sqr 
t(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sq 
rt(b)*x)/(sqrt(c)*sqrt(b)))*a**4*d**3*e**3*f**6*x**4 - 80*sqrt(c)*sqrt(a*d 
 - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b) 
*x)/(sqrt(c)*sqrt(b)))*a**3*b*d**3*e**6*f**3 - 160*sqrt(c)*sqrt(a*d - b*c) 
*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sq 
rt(c)*sqrt(b)))*a**3*b*d**3*e**5*f**4*x**2 - 80*sqrt(c)*sqrt(a*d - b*c)*at 
an((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt( 
c)*sqrt(b)))*a**3*b*d**3*e**4*f**5*x**4 + 144*sqrt(c)*sqrt(a*d - b*c)*atan 
((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c) 
*sqrt(b)))*a**2*b**2*d**3*e**7*f**2 + 288*sqrt(c)*sqrt(a*d - b*c)*atan((sq 
rt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqr 
t(b)))*a**2*b**2*d**3*e**6*f**3*x**2 + 144*sqrt(c)*sqrt(a*d - b*c)*atan((s 
qrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sq 
rt(b)))*a**2*b**2*d**3*e**5*f**4*x**4 - 112*sqrt(c)*sqrt(a*d - b*c)*atan(( 
sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*s 
qrt(b)))*a*b**3*d**3*e**8*f - 224*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*...