\(\int \frac {1}{(a+b x^2)^2 (c+d x^2)^2} \, dx\) [699]

Optimal result

Integrand size = 19, antiderivative size = 167 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {d (b c+a d) x}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {b^{3/2} (b c-5 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)^3}+\frac {d^{3/2} (5 b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{2 c^{3/2} (b c-a d)^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {1}{2} \left (\frac {b^{3/2} (-b c+5 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2} (-b c+a d)^3}+\frac {(b c-a d) x \left (\frac {b^2}{a^2+a b x^2}+\frac {d^2}{c^2+c d x^2}\right )+\frac {d^{3/2} (5 b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{3/2}}}{(b c-a d)^3}\right ) \] Input:

Integrate[1/((a + b*x^2)^2*(c + d*x^2)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {316, 25, 402, 27, 397, 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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 218

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

Input:

Int[1/((a + b*x^2)^2*(c + d*x^2)^2),x]
 

Output:

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

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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

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 402

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 + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {b^{2} \left (\frac {\left (a d -b c \right ) x}{2 a \left (b \,x^{2}+a \right )}+\frac {\left (5 a d -b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a d -b c \right )^{3}}+\frac {d^{2} \left (\frac {\left (a d -b c \right ) x}{2 c \left (x^{2} d +c \right )}+\frac {\left (a d -5 b c \right ) \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{2 c \sqrt {c d}}\right )}{\left (a d -b c \right )^{3}}\)	\(133\)
risch	\(\text {Expression too large to display}\)	\(2124\)

Input:

int(1/(b*x^2+a)^2/(d*x^2+c)^2,x,method=_RETURNVERBOSE)
 

Output:

b^2/(a*d-b*c)^3*(1/2*(a*d-b*c)/a*x/(b*x^2+a)+1/2*(5*a*d-b*c)/a/(a*b)^(1/2) 
*arctan(b*x/(a*b)^(1/2)))+d^2/(a*d-b*c)^3*(1/2*(a*d-b*c)/c*x/(d*x^2+c)+1/2 
*(a*d-5*b*c)/c/(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. 395 vs. \(2 (143) = 286\).

Time = 0.56 (sec) , antiderivative size = 1681, normalized size of antiderivative = 10.07 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**2+a)**2/(d*x**2+c)**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (143) = 286\).

Time = 0.16 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.76 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {{\left (b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (5 \, b c d^{2} - a d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt {c d}} + \frac {{\left (b^{2} c d + a b d^{2}\right )} x^{3} + {\left (b^{2} c^{2} + a^{2} d^{2}\right )} x}{2 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} + {\left (a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} + a^{3} b c d^{3}\right )} x^{4} + {\left (a b^{3} c^{4} - a^{2} b^{2} c^{3} d - a^{3} b c^{2} d^{2} + a^{4} c d^{3}\right )} x^{2}\right )}} \] Input:

integrate(1/(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*(b^3*c - 5*a*b^2*d)*arctan(b*x/sqrt(a*b))/((a*b^3*c^3 - 3*a^2*b^2*c^2* 
d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(a*b)) + 1/2*(5*b*c*d^2 - a*d^3)*arctan(d 
*x/sqrt(c*d))/((b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*sqr 
t(c*d)) + 1/2*((b^2*c*d + a*b*d^2)*x^3 + (b^2*c^2 + a^2*d^2)*x)/(a^2*b^2*c 
^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + a^3* 
b*c*d^3)*x^4 + (a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + a^4*c*d^3)*x^2 
)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {{\left (b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} \sqrt {a b}} + \frac {{\left (5 \, b c d^{2} - a d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{2 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt {c d}} + \frac {b^{2} c d x^{3} + a b d^{2} x^{3} + b^{2} c^{2} x + a^{2} d^{2} x}{2 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} {\left (b d x^{4} + b c x^{2} + a d x^{2} + a c\right )}} \] Input:

integrate(1/(b*x^2+a)^2/(d*x^2+c)^2,x, algorithm="giac")
 

Output:

1/2*(b^3*c - 5*a*b^2*d)*arctan(b*x/sqrt(a*b))/((a*b^3*c^3 - 3*a^2*b^2*c^2* 
d + 3*a^3*b*c*d^2 - a^4*d^3)*sqrt(a*b)) + 1/2*(5*b*c*d^2 - a*d^3)*arctan(d 
*x/sqrt(c*d))/((b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*sqr 
t(c*d)) + 1/2*(b^2*c*d*x^3 + a*b*d^2*x^3 + b^2*c^2*x + a^2*d^2*x)/((a*b^2* 
c^3 - 2*a^2*b*c^2*d + a^3*c*d^2)*(b*d*x^4 + b*c*x^2 + a*d*x^2 + a*c))
 

Mupad [B] (verification not implemented)

Time = 2.62 (sec) , antiderivative size = 6183, normalized size of antiderivative = 37.02 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*x^2)^2*(c + d*x^2)^2),x)
 

Output:

((x*(a^2*d^2 + b^2*c^2))/(2*a*c*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (b*d*x^ 
3*(a*d + b*c))/(2*a*c*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(a*c + x^2*(a*d + 
b*c) + b*d*x^4) + (atan(((((x*(a^4*b^3*d^7 + b^7*c^4*d^3 - 10*a*b^6*c^3*d^ 
4 - 10*a^3*b^4*c*d^6 + 50*a^2*b^5*c^2*d^5))/(2*(a^2*b^4*c^6 + a^6*c^2*d^4 
- 4*a^3*b^3*c^5*d - 4*a^5*b*c^3*d^3 + 6*a^4*b^2*c^4*d^2)) - (((2*a*b^10*c^ 
9*d^2 + 2*a^9*b^2*c*d^10 - 20*a^2*b^9*c^8*d^3 + 80*a^3*b^8*c^7*d^4 - 172*a 
^4*b^7*c^6*d^5 + 220*a^5*b^6*c^5*d^6 - 172*a^6*b^5*c^4*d^7 + 80*a^7*b^4*c^ 
3*d^8 - 20*a^8*b^3*c^2*d^9)/(a^2*b^6*c^8 + a^8*c^2*d^6 - 6*a^3*b^5*c^7*d - 
 6*a^7*b*c^3*d^5 + 15*a^4*b^4*c^6*d^2 - 20*a^5*b^3*c^5*d^3 + 15*a^6*b^2*c^ 
4*d^4) - (x*(5*a*d - b*c)*(-a^3*b^3)^(1/2)*(16*a^2*b^9*c^9*d^2 - 80*a^3*b^ 
8*c^8*d^3 + 144*a^4*b^7*c^7*d^4 - 80*a^5*b^6*c^6*d^5 - 80*a^6*b^5*c^5*d^6 
+ 144*a^7*b^4*c^4*d^7 - 80*a^8*b^3*c^3*d^8 + 16*a^9*b^2*c^2*d^9))/(8*(a^6* 
d^3 - a^3*b^3*c^3 + 3*a^4*b^2*c^2*d - 3*a^5*b*c*d^2)*(a^2*b^4*c^6 + a^6*c^ 
2*d^4 - 4*a^3*b^3*c^5*d - 4*a^5*b*c^3*d^3 + 6*a^4*b^2*c^4*d^2)))*(5*a*d - 
b*c)*(-a^3*b^3)^(1/2))/(4*(a^6*d^3 - a^3*b^3*c^3 + 3*a^4*b^2*c^2*d - 3*a^5 
*b*c*d^2)))*(5*a*d - b*c)*(-a^3*b^3)^(1/2)*1i)/(4*(a^6*d^3 - a^3*b^3*c^3 + 
 3*a^4*b^2*c^2*d - 3*a^5*b*c*d^2)) + (((x*(a^4*b^3*d^7 + b^7*c^4*d^3 - 10* 
a*b^6*c^3*d^4 - 10*a^3*b^4*c*d^6 + 50*a^2*b^5*c^2*d^5))/(2*(a^2*b^4*c^6 + 
a^6*c^2*d^4 - 4*a^3*b^3*c^5*d - 4*a^5*b*c^3*d^3 + 6*a^4*b^2*c^4*d^2)) + (( 
(2*a*b^10*c^9*d^2 + 2*a^9*b^2*c*d^10 - 20*a^2*b^9*c^8*d^3 + 80*a^3*b^8*...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 613, normalized size of antiderivative = 3.67 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {5 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,c^{3} d +5 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,c^{2} d^{2} x^{2}-\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{4}+4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{3} d \,x^{2}+5 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c^{2} d^{2} x^{4}-\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} c^{4} x^{2}-\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} c^{3} d \,x^{4}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{4} c \,d^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{4} d^{3} x^{2}-5 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} b \,c^{2} d -4 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} b c \,d^{2} x^{2}+\sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} b \,d^{3} x^{4}-5 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b^{2} c^{2} d \,x^{2}-5 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b^{2} c \,d^{2} x^{4}+a^{4} c \,d^{3} x -a^{3} b \,c^{2} d^{2} x +a^{3} b c \,d^{3} x^{3}+a^{2} b^{2} c^{3} d x -a \,b^{3} c^{4} x -a \,b^{3} c^{3} d \,x^{3}}{2 a^{2} c^{2} \left (a^{3} b \,d^{4} x^{4}-3 a^{2} b^{2} c \,d^{3} x^{4}+3 a \,b^{3} c^{2} d^{2} x^{4}-b^{4} c^{3} d \,x^{4}+a^{4} d^{4} x^{2}-2 a^{3} b c \,d^{3} x^{2}+2 a \,b^{3} c^{3} d \,x^{2}-b^{4} c^{4} x^{2}+a^{4} c \,d^{3}-3 a^{3} b \,c^{2} d^{2}+3 a^{2} b^{2} c^{3} d -a \,b^{3} c^{4}\right )} \] Input:

int(1/(b*x^2+a)^2/(d*x^2+c)^2,x)
 

Output:

(5*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*c**3*d + 5*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*c**2*d**2*x**2 - sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**4 + 4*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a*b**2*c**3*d*x**2 + 5*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a*b**2*c**2*d**2*x**4 - sqrt(b)*sqrt(a)*atan((b*x)/(sqrt 
(b)*sqrt(a)))*b**3*c**4*x**2 - sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a) 
))*b**3*c**3*d*x**4 + sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**4*c 
*d**2 + sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**4*d**3*x**2 - 5*s 
qrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**3*b*c**2*d - 4*sqrt(d)*sqr 
t(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**3*b*c*d**2*x**2 + sqrt(d)*sqrt(c)*at 
an((d*x)/(sqrt(d)*sqrt(c)))*a**3*b*d**3*x**4 - 5*sqrt(d)*sqrt(c)*atan((d*x 
)/(sqrt(d)*sqrt(c)))*a**2*b**2*c**2*d*x**2 - 5*sqrt(d)*sqrt(c)*atan((d*x)/ 
(sqrt(d)*sqrt(c)))*a**2*b**2*c*d**2*x**4 + a**4*c*d**3*x - a**3*b*c**2*d** 
2*x + a**3*b*c*d**3*x**3 + a**2*b**2*c**3*d*x - a*b**3*c**4*x - a*b**3*c** 
3*d*x**3)/(2*a**2*c**2*(a**4*c*d**3 + a**4*d**4*x**2 - 3*a**3*b*c**2*d**2 
- 2*a**3*b*c*d**3*x**2 + a**3*b*d**4*x**4 + 3*a**2*b**2*c**3*d - 3*a**2*b* 
*2*c*d**3*x**4 - a*b**3*c**4 + 2*a*b**3*c**3*d*x**2 + 3*a*b**3*c**2*d**2*x 
**4 - b**4*c**4*x**2 - b**4*c**3*d*x**4))