\(\int \frac {1}{(a+b x^2)^3 (c+d x^2)} \, dx\) [40]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {b x}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (3 b c-7 a d) x}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {\sqrt {b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} (b c-a d)^3}-\frac {d^{5/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=-\frac {\frac {b (-b c+a d) x \left (-5 a b c+9 a^2 d-3 b^2 c x^2+7 a b d x^2\right )}{a^2 \left (a+b x^2\right )^2}+\frac {\sqrt {b} \left (3 b^2 c^2-10 a b c d+15 a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}}-\frac {8 d^{5/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c}}}{8 (-b c+a d)^3} \] Input:

Integrate[1/((a + b*x^2)^3*(c + d*x^2)),x]
 

Output:

-1/8*((b*(-(b*c) + a*d)*x*(-5*a*b*c + 9*a^2*d - 3*b^2*c*x^2 + 7*a*b*d*x^2) 
)/(a^2*(a + b*x^2)^2) + (Sqrt[b]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*Arc 
Tan[(Sqrt[b]*x)/Sqrt[a]])/a^(5/2) - (8*d^(5/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]] 
)/Sqrt[c])/(-(b*c) + a*d)^3
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.22, 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, 25, 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.

\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {b x (3 b c-7 a d)}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\int -\frac {3 b^2 c^2-7 a b d c+8 a^2 d^2+b d (3 b c-7 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {3 b^2 c^2-7 a b d c+8 a^2 d^2+b d (3 b c-7 a d) x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}+\frac {b x (3 b c-7 a d)}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {b \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right ) \int \frac {1}{b x^2+a}dx}{b c-a d}-\frac {8 a^2 d^3 \int \frac {1}{d x^2+c}dx}{b c-a d}}{2 a (b c-a d)}+\frac {b x (3 b c-7 a d)}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (15 a^2 d^2-10 a b c d+3 b^2 c^2\right )}{\sqrt {a} (b c-a d)}-\frac {8 a^2 d^{5/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)}}{2 a (b c-a d)}+\frac {b x (3 b c-7 a d)}{2 a \left (a+b x^2\right ) (b c-a d)}}{4 a (b c-a d)}+\frac {b x}{4 a \left (a+b x^2\right )^2 (b c-a d)}\)

Input:

Int[1/((a + b*x^2)^3*(c + d*x^2)),x]
 

Output:

(b*x)/(4*a*(b*c - a*d)*(a + b*x^2)^2) + ((b*(3*b*c - 7*a*d)*x)/(2*a*(b*c - 
 a*d)*(a + b*x^2)) + ((Sqrt[b]*(3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2)*ArcTa 
n[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) - (8*a^2*d^(5/2)*ArcTan[(Sqr 
t[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(2*a*(b*c - a*d)))/(4*a*(b*c - a* 
d))
 

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 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 = 0.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {b \left (\frac {\frac {b \left (7 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right ) x^{3}}{8 a^{2}}+\frac {\left (9 a^{2} d^{2}-14 a b c d +5 b^{2} c^{2}\right ) x}{8 a}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (15 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a d -b c \right )^{3}}+\frac {d^{3} \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{\left (a d -b c \right )^{3} \sqrt {c d}}\)	\(158\)
risch	\(\text {Expression too large to display}\)	\(2285\)

Input:

int(1/(b*x^2+a)^3/(d*x^2+c),x,method=_RETURNVERBOSE)
 

Output:

-b/(a*d-b*c)^3*((1/8*b*(7*a^2*d^2-10*a*b*c*d+3*b^2*c^2)/a^2*x^3+1/8*(9*a^2 
*d^2-14*a*b*c*d+5*b^2*c^2)/a*x)/(b*x^2+a)^2+1/8*(15*a^2*d^2-10*a*b*c*d+3*b 
^2*c^2)/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))+d^3/(a*d-b*c)^3/(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. 372 vs. \(2 (139) = 278\).

Time = 0.60 (sec) , antiderivative size = 1587, normalized size of antiderivative = 9.86 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x^2+a)^3/(d*x^2+c),x, algorithm="fricas")
 

Output:

[1/16*(2*(3*b^4*c^2 - 10*a*b^3*c*d + 7*a^2*b^2*d^2)*x^3 - (3*a^2*b^2*c^2 - 
 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4*c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x 
^4 + 2*(3*a*b^3*c^2 - 10*a^2*b^2*c*d + 15*a^3*b*d^2)*x^2)*sqrt(-b/a)*log(( 
b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 8*(a^2*b^2*d^2*x^4 + 2*a^3*b* 
d^2*x^2 + a^4*d^2)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^2 + 
c)) + 2*(5*a*b^3*c^2 - 14*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 - 3*a 
^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 
3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^4 + 2*(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3* 
a^5*b^2*c*d^2 - a^6*b*d^3)*x^2), 1/16*(2*(3*b^4*c^2 - 10*a*b^3*c*d + 7*a^2 
*b^2*d^2)*x^3 - 16*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^2)*sqrt(d/c) 
*arctan(x*sqrt(d/c)) - (3*a^2*b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4 
*c^2 - 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*a^2*b^2*c* 
d + 15*a^3*b*d^2)*x^2)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^ 
2 + a)) + 2*(5*a*b^3*c^2 - 14*a^2*b^2*c*d + 9*a^3*b*d^2)*x)/(a^4*b^3*c^3 - 
 3*a^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a^7*d^3 + (a^2*b^5*c^3 - 3*a^3*b^4*c^2* 
d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^4 + 2*(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d 
+ 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x^2), 1/8*((3*b^4*c^2 - 10*a*b^3*c*d + 7*a^ 
2*b^2*d^2)*x^3 + (3*a^2*b^2*c^2 - 10*a^3*b*c*d + 15*a^4*d^2 + (3*b^4*c^2 - 
 10*a*b^3*c*d + 15*a^2*b^2*d^2)*x^4 + 2*(3*a*b^3*c^2 - 10*a^2*b^2*c*d + 15 
*a^3*b*d^2)*x^2)*sqrt(b/a)*arctan(x*sqrt(b/a)) - 4*(a^2*b^2*d^2*x^4 + 2...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x**2+a)**3/(d*x**2+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.73 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=-\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {c d}} + \frac {{\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 15 \, a^{2} b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, b^{3} c - 7 \, a b^{2} d\right )} x^{3} + {\left (5 \, a b^{2} c - 9 \, a^{2} b d\right )} x}{8 \, {\left (a^{4} b^{2} c^{2} - 2 \, a^{5} b c d + a^{6} d^{2} + {\left (a^{2} b^{4} c^{2} - 2 \, a^{3} b^{3} c d + a^{4} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a^{3} b^{3} c^{2} - 2 \, a^{4} b^{2} c d + a^{5} b d^{2}\right )} x^{2}\right )}} \] Input:

integrate(1/(b*x^2+a)^3/(d*x^2+c),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=-\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {c d}} + \frac {{\left (3 \, b^{3} c^{2} - 10 \, a b^{2} c d + 15 \, a^{2} b d^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {a b}} + \frac {3 \, b^{3} c x^{3} - 7 \, a b^{2} d x^{3} + 5 \, a b^{2} c x - 9 \, a^{2} b d x}{8 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} {\left (b x^{2} + a\right )}^{2}} \] Input:

integrate(1/(b*x^2+a)^3/(d*x^2+c),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 6033, normalized size of antiderivative = 37.47 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*x^2)^3*(c + d*x^2)),x)
 

Output:

((x^3*(3*b^3*c - 7*a*b^2*d))/(8*a^2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (x* 
(5*b^2*c - 9*a*b*d))/(8*a*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))/(a^2 + b^2*x^4 
 + 2*a*b*x^2) + (atan(((((x*(289*a^4*b^3*d^7 + 9*b^7*c^4*d^3 - 60*a*b^6*c^ 
3*d^4 - 300*a^3*b^4*c*d^6 + 190*a^2*b^5*c^2*d^5))/(32*(a^8*d^4 + a^4*b^4*c 
^4 - 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3)) - ((-c*d^5)^(1/ 
2)*((256*a^10*b^2*d^10 - 1760*a^9*b^3*c*d^9 + 96*a^2*b^10*c^8*d^2 - 800*a^ 
3*b^9*c^7*d^3 + 3040*a^4*b^8*c^6*d^4 - 6816*a^5*b^7*c^5*d^5 + 9760*a^6*b^6 
*c^4*d^6 - 9056*a^7*b^5*c^3*d^7 + 5280*a^8*b^4*c^2*d^8)/(64*(a^10*d^6 + a^ 
4*b^6*c^6 - 6*a^5*b^5*c^5*d + 15*a^6*b^4*c^4*d^2 - 20*a^7*b^3*c^3*d^3 + 15 
*a^8*b^2*c^2*d^4 - 6*a^9*b*c*d^5)) - (x*(-c*d^5)^(1/2)*(256*a^11*b^2*d^9 - 
 1280*a^10*b^3*c*d^8 + 256*a^4*b^9*c^7*d^2 - 1280*a^5*b^8*c^6*d^3 + 2304*a 
^6*b^7*c^5*d^4 - 1280*a^7*b^6*c^4*d^5 - 1280*a^8*b^5*c^3*d^6 + 2304*a^9*b^ 
4*c^2*d^7))/(64*(b^3*c^4 - a^3*c*d^3 + 3*a^2*b*c^2*d^2 - 3*a*b^2*c^3*d)*(a 
^8*d^4 + a^4*b^4*c^4 - 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 
))))/(2*(b^3*c^4 - a^3*c*d^3 + 3*a^2*b*c^2*d^2 - 3*a*b^2*c^3*d)))*(-c*d^5) 
^(1/2)*1i)/(2*(b^3*c^4 - a^3*c*d^3 + 3*a^2*b*c^2*d^2 - 3*a*b^2*c^3*d)) + ( 
((x*(289*a^4*b^3*d^7 + 9*b^7*c^4*d^3 - 60*a*b^6*c^3*d^4 - 300*a^3*b^4*c*d^ 
6 + 190*a^2*b^5*c^2*d^5))/(32*(a^8*d^4 + a^4*b^4*c^4 - 4*a^5*b^3*c^3*d + 6 
*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3)) + ((-c*d^5)^(1/2)*((256*a^10*b^2*d^10 - 
 1760*a^9*b^3*c*d^9 + 96*a^2*b^10*c^8*d^2 - 800*a^3*b^9*c^7*d^3 + 3040*...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 571, normalized size of antiderivative = 3.55 \[ \int \frac {1}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} c \,d^{2}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,c^{2} d -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b c \,d^{2} x^{2}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{3}+20 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{2} d \,x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c \,d^{2} x^{4}-6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{3} x^{2}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{2} d \,x^{4}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} c^{3} x^{4}+8 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{5} d^{2}+16 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{4} b \,d^{2} x^{2}+8 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} b^{2} d^{2} x^{4}-9 a^{4} b c \,d^{2} x +14 a^{3} b^{2} c^{2} d x -7 a^{3} b^{2} c \,d^{2} x^{3}-5 a^{2} b^{3} c^{3} x +10 a^{2} b^{3} c^{2} d \,x^{3}-3 a \,b^{4} c^{3} x^{3}}{8 a^{3} c \left (a^{3} b^{2} d^{3} x^{4}-3 a^{2} b^{3} c \,d^{2} x^{4}+3 a \,b^{4} c^{2} d \,x^{4}-b^{5} c^{3} x^{4}+2 a^{4} b \,d^{3} x^{2}-6 a^{3} b^{2} c \,d^{2} x^{2}+6 a^{2} b^{3} c^{2} d \,x^{2}-2 a \,b^{4} c^{3} x^{2}+a^{5} d^{3}-3 a^{4} b c \,d^{2}+3 a^{3} b^{2} c^{2} d -a^{2} b^{3} c^{3}\right )} \] Input:

int(1/(b*x^2+a)^3/(d*x^2+c),x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*c*d**2 + 10*sqrt 
(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c**2*d - 30*sqrt(b)*sqrt( 
a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c*d**2*x**2 - 3*sqrt(b)*sqrt(a)*at 
an((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**3 + 20*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a**2*b**2*c**2*d*x**2 - 15*sqrt(b)*sqrt(a)*atan((b*x)/ 
(sqrt(b)*sqrt(a)))*a**2*b**2*c*d**2*x**4 - 6*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a*b**3*c**3*x**2 + 10*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b) 
*sqrt(a)))*a*b**3*c**2*d*x**4 - 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt 
(a)))*b**4*c**3*x**4 + 8*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a** 
5*d**2 + 16*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**4*b*d**2*x**2 
 + 8*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**3*b**2*d**2*x**4 - 9 
*a**4*b*c*d**2*x + 14*a**3*b**2*c**2*d*x - 7*a**3*b**2*c*d**2*x**3 - 5*a** 
2*b**3*c**3*x + 10*a**2*b**3*c**2*d*x**3 - 3*a*b**4*c**3*x**3)/(8*a**3*c*( 
a**5*d**3 - 3*a**4*b*c*d**2 + 2*a**4*b*d**3*x**2 + 3*a**3*b**2*c**2*d - 6* 
a**3*b**2*c*d**2*x**2 + a**3*b**2*d**3*x**4 - a**2*b**3*c**3 + 6*a**2*b**3 
*c**2*d*x**2 - 3*a**2*b**3*c*d**2*x**4 - 2*a*b**4*c**3*x**2 + 3*a*b**4*c** 
2*d*x**4 - b**5*c**3*x**4))