\(\int \frac {1}{x^4 (a+b x^2)^2 (c+d x^2)} \, dx\) [691]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {5 b c-2 a d}{6 a^2 c (b c-a d) x^3}+\frac {5 b^2 c^2-2 a b c d-2 a^2 d^2}{2 a^3 c^2 (b c-a d) x}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right )}+\frac {b^{5/2} (5 b c-7 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} (b c-a d)^2}+\frac {d^{7/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)^2} \] Output:

-1/6*(-2*a*d+5*b*c)/a^2/c/(-a*d+b*c)/x^3+1/2*(-2*a^2*d^2-2*a*b*c*d+5*b^2*c 
^2)/a^3/c^2/(-a*d+b*c)/x+1/2*b/a/(-a*d+b*c)/x^3/(b*x^2+a)+1/2*b^(5/2)*(-7* 
a*d+5*b*c)*arctan(b^(1/2)*x/a^(1/2))/a^(7/2)/(-a*d+b*c)^2+d^(7/2)*arctan(d 
^(1/2)*x/c^(1/2))/c^(5/2)/(-a*d+b*c)^2
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {1}{3 a^2 c x^3}+\frac {2 b c+a d}{a^3 c^2 x}-\frac {b^3 x}{2 a^3 (-b c+a d) \left (a+b x^2\right )}-\frac {b^{5/2} (-5 b c+7 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{7/2} (-b c+a d)^2}+\frac {d^{7/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)^2} \] Input:

Integrate[1/(x^4*(a + b*x^2)^2*(c + d*x^2)),x]
 

Output:

-1/3*1/(a^2*c*x^3) + (2*b*c + a*d)/(a^3*c^2*x) - (b^3*x)/(2*a^3*(-(b*c) + 
a*d)*(a + b*x^2)) - (b^(5/2)*(-5*b*c + 7*a*d)*ArcTan[(Sqrt[b]*x)/Sqrt[a]]) 
/(2*a^(7/2)*(-(b*c) + a*d)^2) + (d^(7/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^( 
5/2)*(b*c - a*d)^2)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {374, 25, 445, 27, 445, 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.

\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 445

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

\(\Big \downarrow \) 397

\(\displaystyle \frac {-\frac {-\frac {\frac {2 a^3 d^4 \int \frac {1}{d x^2+c}dx}{b c-a d}+\frac {b^3 c^2 (5 b c-7 a d) \int \frac {1}{b x^2+a}dx}{b c-a d}}{a c}-\frac {\frac {5 b^2 c}{a}-\frac {2 a d^2}{c}-2 b d}{x}}{a c}-\frac {5 b c-2 a d}{3 a c x^3}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) (b c-a d)}\)

\(\Big \downarrow \) 218

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

Input:

Int[1/(x^4*(a + b*x^2)^2*(c + d*x^2)),x]
 

Output:

b/(2*a*(b*c - a*d)*x^3*(a + b*x^2)) + (-1/3*(5*b*c - 2*a*d)/(a*c*x^3) - (- 
(((5*b^2*c)/a - 2*b*d - (2*a*d^2)/c)/x) - ((b^(5/2)*c^2*(5*b*c - 7*a*d)*Ar 
cTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*(b*c - a*d)) + (2*a^3*d^(7/2)*ArcTan[( 
Sqrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)))/(a*c))/(a*c))/(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 374

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
 + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - 
a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, 
 c, d, e, m, 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 445

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {b^{3} \left (\frac {\left (\frac {a d}{2}-\frac {b c}{2}\right ) x}{b \,x^{2}+a}+\frac {\left (7 a d -5 b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3} \left (a d -b c \right )^{2}}-\frac {1}{3 a^{2} c \,x^{3}}-\frac {-a d -2 b c}{x \,a^{3} c^{2}}+\frac {d^{4} \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{c^{2} \left (a d -b c \right )^{2} \sqrt {c d}}\)	\(128\)
risch	\(\text {Expression too large to display}\)	\(1263\)

Input:

int(1/x^4/(b*x^2+a)^2/(d*x^2+c),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 1281, normalized size of antiderivative = 6.78 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")
 

Output:

[-1/12*(4*a^2*b^2*c^3 - 8*a^3*b*c^2*d + 4*a^4*c*d^2 - 6*(5*b^4*c^3 - 7*a*b 
^3*c^2*d + 2*a^3*b*d^3)*x^4 - 4*(5*a*b^3*c^3 - 7*a^2*b^2*c^2*d - a^3*b*c*d 
^2 + 3*a^4*d^3)*x^2 + 3*((5*b^4*c^3 - 7*a*b^3*c^2*d)*x^5 + (5*a*b^3*c^3 - 
7*a^2*b^2*c^2*d)*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 
 + a)) - 6*(a^3*b*d^3*x^5 + a^4*d^3*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqr 
t(-d/c) - c)/(d*x^2 + c)))/((a^3*b^3*c^4 - 2*a^4*b^2*c^3*d + a^5*b*c^2*d^2 
)*x^5 + (a^4*b^2*c^4 - 2*a^5*b*c^3*d + a^6*c^2*d^2)*x^3), -1/12*(4*a^2*b^2 
*c^3 - 8*a^3*b*c^2*d + 4*a^4*c*d^2 - 6*(5*b^4*c^3 - 7*a*b^3*c^2*d + 2*a^3* 
b*d^3)*x^4 - 4*(5*a*b^3*c^3 - 7*a^2*b^2*c^2*d - a^3*b*c*d^2 + 3*a^4*d^3)*x 
^2 - 12*(a^3*b*d^3*x^5 + a^4*d^3*x^3)*sqrt(d/c)*arctan(x*sqrt(d/c)) + 3*(( 
5*b^4*c^3 - 7*a*b^3*c^2*d)*x^5 + (5*a*b^3*c^3 - 7*a^2*b^2*c^2*d)*x^3)*sqrt 
(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/((a^3*b^3*c^4 - 2* 
a^4*b^2*c^3*d + a^5*b*c^2*d^2)*x^5 + (a^4*b^2*c^4 - 2*a^5*b*c^3*d + a^6*c^ 
2*d^2)*x^3), -1/6*(2*a^2*b^2*c^3 - 4*a^3*b*c^2*d + 2*a^4*c*d^2 - 3*(5*b^4* 
c^3 - 7*a*b^3*c^2*d + 2*a^3*b*d^3)*x^4 - 2*(5*a*b^3*c^3 - 7*a^2*b^2*c^2*d 
- a^3*b*c*d^2 + 3*a^4*d^3)*x^2 - 3*((5*b^4*c^3 - 7*a*b^3*c^2*d)*x^5 + (5*a 
*b^3*c^3 - 7*a^2*b^2*c^2*d)*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)) - 3*(a^3*b* 
d^3*x^5 + a^4*d^3*x^3)*sqrt(-d/c)*log((d*x^2 + 2*c*x*sqrt(-d/c) - c)/(d*x^ 
2 + c)))/((a^3*b^3*c^4 - 2*a^4*b^2*c^3*d + a^5*b*c^2*d^2)*x^5 + (a^4*b^2*c 
^4 - 2*a^5*b*c^3*d + a^6*c^2*d^2)*x^3), -1/6*(2*a^2*b^2*c^3 - 4*a^3*b*c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x**4/(b*x**2+a)**2/(d*x**2+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt {c d}} + \frac {{\left (5 \, b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \sqrt {a b}} - \frac {2 \, a^{2} b c^{2} - 2 \, a^{3} c d - 3 \, {\left (5 \, b^{3} c^{2} - 2 \, a b^{2} c d - 2 \, a^{2} b d^{2}\right )} x^{4} - 2 \, {\left (5 \, a b^{2} c^{2} - 2 \, a^{2} b c d - 3 \, a^{3} d^{2}\right )} x^{2}}{6 \, {\left ({\left (a^{3} b^{2} c^{3} - a^{4} b c^{2} d\right )} x^{5} + {\left (a^{4} b c^{3} - a^{5} c^{2} d\right )} x^{3}\right )}} \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c),x, algorithm="maxima")
 

Output:

d^4*arctan(d*x/sqrt(c*d))/((b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*sqrt(c*d) 
) + 1/2*(5*b^4*c - 7*a*b^3*d)*arctan(b*x/sqrt(a*b))/((a^3*b^2*c^2 - 2*a^4* 
b*c*d + a^5*d^2)*sqrt(a*b)) - 1/6*(2*a^2*b*c^2 - 2*a^3*c*d - 3*(5*b^3*c^2 
- 2*a*b^2*c*d - 2*a^2*b*d^2)*x^4 - 2*(5*a*b^2*c^2 - 2*a^2*b*c*d - 3*a^3*d^ 
2)*x^2)/((a^3*b^2*c^3 - a^4*b*c^2*d)*x^5 + (a^4*b*c^3 - a^5*c^2*d)*x^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} \sqrt {c d}} + \frac {b^{3} x}{2 \, {\left (a^{3} b c - a^{4} d\right )} {\left (b x^{2} + a\right )}} + \frac {{\left (5 \, b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \sqrt {a b}} + \frac {6 \, b c x^{2} + 3 \, a d x^{2} - a c}{3 \, a^{3} c^{2} x^{3}} \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")
 

Output:

d^4*arctan(d*x/sqrt(c*d))/((b^2*c^4 - 2*a*b*c^3*d + a^2*c^2*d^2)*sqrt(c*d) 
) + 1/2*b^3*x/((a^3*b*c - a^4*d)*(b*x^2 + a)) + 1/2*(5*b^4*c - 7*a*b^3*d)* 
arctan(b*x/sqrt(a*b))/((a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*sqrt(a*b)) + 
1/3*(6*b*c*x^2 + 3*a*d*x^2 - a*c)/(a^3*c^2*x^3)
 

Mupad [B] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 4654, normalized size of antiderivative = 24.62 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

int(1/(x^4*(a + b*x^2)^2*(c + d*x^2)),x)
 

Output:

((x^2*(3*a*d + 5*b*c))/(3*a^2*c^2) - 1/(3*a*c) + (x^4*(2*a^2*b*d^2 - 5*b^3 
*c^2 + 2*a*b^2*c*d))/(2*a^3*c^2*(a*d - b*c)))/(a*x^3 + b*x^5) - (atan((((( 
x*(400*a^9*b^12*c^15*d^3 - 2320*a^10*b^11*c^14*d^4 + 5344*a^11*b^10*c^13*d 
^5 - 6112*a^12*b^9*c^12*d^6 + 3472*a^13*b^8*c^11*d^7 - 784*a^14*b^7*c^10*d 
^8 + 64*a^15*b^6*c^9*d^9 - 192*a^16*b^5*c^8*d^10 + 192*a^17*b^4*c^7*d^11 - 
 64*a^18*b^3*c^6*d^12))/2 + ((-c^5*d^7)^(1/2)*((x*(-c^5*d^7)^(1/2)*(256*a^ 
15*b^10*c^18*d^2 - 1536*a^16*b^9*c^17*d^3 + 3584*a^17*b^8*c^16*d^4 - 3584* 
a^18*b^7*c^15*d^5 + 3584*a^20*b^5*c^13*d^7 - 3584*a^21*b^4*c^12*d^8 + 1536 
*a^22*b^3*c^11*d^9 - 256*a^23*b^2*c^10*d^10))/(4*(b^2*c^7 + a^2*c^5*d^2 - 
2*a*b*c^6*d)) - 160*a^12*b^11*c^17*d^2 + 1024*a^13*b^10*c^16*d^3 - 2720*a^ 
14*b^9*c^15*d^4 + 3840*a^15*b^8*c^14*d^5 - 3104*a^16*b^7*c^13*d^6 + 1600*a 
^17*b^6*c^12*d^7 - 864*a^18*b^5*c^11*d^8 + 640*a^19*b^4*c^10*d^9 - 320*a^2 
0*b^3*c^9*d^10 + 64*a^21*b^2*c^8*d^11))/(2*(b^2*c^7 + a^2*c^5*d^2 - 2*a*b* 
c^6*d)))*(-c^5*d^7)^(1/2)*1i)/(b^2*c^7 + a^2*c^5*d^2 - 2*a*b*c^6*d) + (((x 
*(400*a^9*b^12*c^15*d^3 - 2320*a^10*b^11*c^14*d^4 + 5344*a^11*b^10*c^13*d^ 
5 - 6112*a^12*b^9*c^12*d^6 + 3472*a^13*b^8*c^11*d^7 - 784*a^14*b^7*c^10*d^ 
8 + 64*a^15*b^6*c^9*d^9 - 192*a^16*b^5*c^8*d^10 + 192*a^17*b^4*c^7*d^11 - 
64*a^18*b^3*c^6*d^12))/2 + ((-c^5*d^7)^(1/2)*((x*(-c^5*d^7)^(1/2)*(256*a^1 
5*b^10*c^18*d^2 - 1536*a^16*b^9*c^17*d^3 + 3584*a^17*b^8*c^16*d^4 - 3584*a 
^18*b^7*c^15*d^5 + 3584*a^20*b^5*c^13*d^7 - 3584*a^21*b^4*c^12*d^8 + 15...
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.95 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {-21 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{3} d \,x^{3}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{4} x^{3}-21 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{3} d \,x^{5}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} c^{4} x^{5}+6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{5} d^{3} x^{3}+6 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{4} b \,d^{3} x^{5}-2 a^{5} c^{2} d^{2}+6 a^{5} c \,d^{3} x^{2}+4 a^{4} b \,c^{3} d -2 a^{4} b \,c^{2} d^{2} x^{2}+6 a^{4} b c \,d^{3} x^{4}-2 a^{3} b^{2} c^{4}-14 a^{3} b^{2} c^{3} d \,x^{2}+10 a^{2} b^{3} c^{4} x^{2}-21 a^{2} b^{3} c^{3} d \,x^{4}+15 a \,b^{4} c^{4} x^{4}}{6 a^{4} c^{3} x^{3} \left (a^{2} b \,d^{2} x^{2}-2 a \,b^{2} c d \,x^{2}+b^{3} c^{2} x^{2}+a^{3} d^{2}-2 a^{2} b c d +a \,b^{2} c^{2}\right )} \] Input:

int(1/x^4/(b*x^2+a)^2/(d*x^2+c),x)
 

Output:

( - 21*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**3*d*x**3 
 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c**4*x**3 - 21* 
sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c**3*d*x**5 + 15*sqrt 
(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4*c**4*x**5 + 6*sqrt(d)*sqrt( 
c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**5*d**3*x**3 + 6*sqrt(d)*sqrt(c)*atan(( 
d*x)/(sqrt(d)*sqrt(c)))*a**4*b*d**3*x**5 - 2*a**5*c**2*d**2 + 6*a**5*c*d** 
3*x**2 + 4*a**4*b*c**3*d - 2*a**4*b*c**2*d**2*x**2 + 6*a**4*b*c*d**3*x**4 
- 2*a**3*b**2*c**4 - 14*a**3*b**2*c**3*d*x**2 + 10*a**2*b**3*c**4*x**2 - 2 
1*a**2*b**3*c**3*d*x**4 + 15*a*b**4*c**4*x**4)/(6*a**4*c**3*x**3*(a**3*d** 
2 - 2*a**2*b*c*d + a**2*b*d**2*x**2 + a*b**2*c**2 - 2*a*b**2*c*d*x**2 + b* 
*3*c**2*x**2))