\(\int \frac {1}{(a+b x^2)^2 (c+d x^2)} \, dx\) [32]

Optimal result

Integrand size = 19, antiderivative size = 108 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b x}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {\sqrt {b} (b c-3 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)^2}+\frac {d^{3/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=-\frac {b x}{2 a (-b c+a d) \left (a+b x^2\right )}-\frac {\sqrt {b} (-b c+3 a d) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} (-b c+a d)^2}+\frac {d^{3/2} \arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{\sqrt {c} (b c-a d)^2} \] Input:

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

Output:

-1/2*(b*x)/(a*(-(b*c) + a*d)*(a + b*x^2)) - (Sqrt[b]*(-(b*c) + 3*a*d)*ArcT 
an[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(3/2)*(-(b*c) + a*d)^2) + (d^(3/2)*ArcTan[(S 
qrt[d]*x)/Sqrt[c]])/(Sqrt[c]*(b*c - a*d)^2)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {316, 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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {b \left (\frac {\left (a d -b c \right ) x}{2 a \left (b \,x^{2}+a \right )}+\frac {\left (3 a d -b c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a d -b c \right )^{2}}+\frac {d^{2} \arctan \left (\frac {x d}{\sqrt {c d}}\right )}{\left (a d -b c \right )^{2} \sqrt {c d}}\)	\(95\)
risch	\(\text {Expression too large to display}\)	\(1035\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 699, normalized size of antiderivative = 6.47 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\left [-\frac {{\left (a b c - 3 \, a^{2} d + {\left (b^{2} c - 3 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 2 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} + 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) - 2 \, {\left (b^{2} c - a b d\right )} x}{4 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}}, \frac {4 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) - {\left (a b c - 3 \, a^{2} d + {\left (b^{2} c - 3 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 2 \, {\left (b^{2} c - a b d\right )} x}{4 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}}, \frac {{\left (a b c - 3 \, a^{2} d + {\left (b^{2} c - 3 \, a b d\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + {\left (a b d x^{2} + a^{2} d\right )} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} + 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) + {\left (b^{2} c - a b d\right )} x}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}}, \frac {{\left (a b c - 3 \, a^{2} d + {\left (b^{2} c - 3 \, a b d\right )} x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 2 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) + {\left (b^{2} c - a b d\right )} x}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{2} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c d}} + \frac {b x}{2 \, {\left (a^{2} b c - a^{3} d + {\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )}} + \frac {{\left (b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {d^{2} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c d}} + \frac {{\left (b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {a b}} + \frac {b x}{2 \, {\left (a b c - a^{2} d\right )} {\left (b x^{2} + a\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 3649, normalized size of antiderivative = 33.79 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-a^3*b)^(1/2)*(3*a*d - b*c)*((x*(13*a^2*b^3*d^5 + b^5*c^2*d^3 - 6 
*a*b^4*c*d^4))/(2*(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d)) - (((4*a^6*b^2*d^ 
7 - 2*a*b^7*c^5*d^2 - 18*a^5*b^3*c*d^6 + 12*a^2*b^6*c^4*d^3 - 28*a^3*b^5*c 
^3*d^4 + 32*a^4*b^4*c^2*d^5)/(a^5*d^3 - a^2*b^3*c^3 + 3*a^3*b^2*c^2*d - 3* 
a^4*b*c*d^2) - (x*(-a^3*b)^(1/2)*(3*a*d - b*c)*(16*a^7*b^2*d^7 - 48*a^6*b^ 
3*c*d^6 + 16*a^2*b^7*c^5*d^2 - 48*a^3*b^6*c^4*d^3 + 32*a^4*b^5*c^3*d^4 + 3 
2*a^5*b^4*c^2*d^5))/(8*(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d)*(a^5*d^2 + a^ 
3*b^2*c^2 - 2*a^4*b*c*d)))*(-a^3*b)^(1/2)*(3*a*d - b*c))/(4*(a^5*d^2 + a^3 
*b^2*c^2 - 2*a^4*b*c*d)))*1i)/(4*(a^5*d^2 + a^3*b^2*c^2 - 2*a^4*b*c*d)) + 
((-a^3*b)^(1/2)*(3*a*d - b*c)*((x*(13*a^2*b^3*d^5 + b^5*c^2*d^3 - 6*a*b^4* 
c*d^4))/(2*(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d)) + (((4*a^6*b^2*d^7 - 2*a 
*b^7*c^5*d^2 - 18*a^5*b^3*c*d^6 + 12*a^2*b^6*c^4*d^3 - 28*a^3*b^5*c^3*d^4 
+ 32*a^4*b^4*c^2*d^5)/(a^5*d^3 - a^2*b^3*c^3 + 3*a^3*b^2*c^2*d - 3*a^4*b*c 
*d^2) + (x*(-a^3*b)^(1/2)*(3*a*d - b*c)*(16*a^7*b^2*d^7 - 48*a^6*b^3*c*d^6 
 + 16*a^2*b^7*c^5*d^2 - 48*a^3*b^6*c^4*d^3 + 32*a^4*b^5*c^3*d^4 + 32*a^5*b 
^4*c^2*d^5))/(8*(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d)*(a^5*d^2 + a^3*b^2*c 
^2 - 2*a^4*b*c*d)))*(-a^3*b)^(1/2)*(3*a*d - b*c))/(4*(a^5*d^2 + a^3*b^2*c^ 
2 - 2*a^4*b*c*d)))*1i)/(4*(a^5*d^2 + a^3*b^2*c^2 - 2*a^4*b*c*d)))/(((3*a*b 
^3*d^5)/2 - (b^4*c*d^4)/2)/(a^5*d^3 - a^2*b^3*c^3 + 3*a^3*b^2*c^2*d - 3*a^ 
4*b*c*d^2) - ((-a^3*b)^(1/2)*(3*a*d - b*c)*((x*(13*a^2*b^3*d^5 + b^5*c^...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.12 \[ \int \frac {1}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} c d +\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,c^{2}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b c d \,x^{2}+\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} c^{2} x^{2}+2 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{3} d +2 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {d x}{\sqrt {d}\, \sqrt {c}}\right ) a^{2} b d \,x^{2}-a^{2} b c d x +a \,b^{2} c^{2} x}{2 a^{2} c \left (a^{2} b \,d^{2} x^{2}-2 a \,b^{2} c d \,x^{2}+b^{3} c^{2} x^{2}+a^{3} d^{2}-2 a^{2} b c d +a \,b^{2} c^{2}\right )} \] Input:

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

Output:

( - 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*c*d + sqrt(b)*sqr 
t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*c**2 - 3*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a*b*c*d*x**2 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqr 
t(a)))*b**2*c**2*x**2 + 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a* 
*3*d + 2*sqrt(d)*sqrt(c)*atan((d*x)/(sqrt(d)*sqrt(c)))*a**2*b*d*x**2 - a** 
2*b*c*d*x + a*b**2*c**2*x)/(2*a**2*c*(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))