\(\int \frac {1}{(a+b x^2)^{3/2} (c+d x^2)^2} \, dx\) [101]

Optimal result

Integrand size = 21, antiderivative size = 143 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx=\frac {b (2 b c+a d) x}{2 a c (b c-a d)^2 \sqrt {a+b x^2}}-\frac {d x}{2 c (b c-a d) \sqrt {a+b x^2} \left (c+d x^2\right )}-\frac {d (4 b c-a d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx=\frac {\frac {\sqrt {c} x \left (a^2 d^2+a b d^2 x^2+2 b^2 c \left (c+d x^2\right )\right )}{a (b c-a d)^2 \sqrt {a+b x^2} \left (c+d x^2\right )}+\frac {d (4 b c-a d) \arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}}{2 c^{3/2}} \] Input:

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

Output:

((Sqrt[c]*x*(a^2*d^2 + a*b*d^2*x^2 + 2*b^2*c*(c + d*x^2)))/(a*(b*c - a*d)^ 
2*Sqrt[a + b*x^2]*(c + d*x^2)) + (d*(4*b*c - a*d)*ArcTan[(-(d*x*Sqrt[a + b 
*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[-(b*c) + a*d])])/(-(b*c) + a*d 
)^(5/2))/(2*c^(3/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {316, 402, 27, 291, 221}

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.

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {b x (a d+2 b c)}{a \sqrt {a+b x^2} (b c-a d)}-\frac {d (4 b c-a d) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{3/2}}}{2 c (b c-a d)}-\frac {d x}{2 c \sqrt {a+b x^2} \left (c+d x^2\right ) (b c-a d)}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

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 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 = 0.71 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(\frac {-a d \sqrt {b \,x^{2}+a}\, \left (x^{2} d +c \right ) \left (a d -4 b c \right ) \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )+\left (2 b^{2} c^{2}+2 x^{2} b^{2} c d +a \,d^{2} \left (b \,x^{2}+a \right )\right ) \sqrt {\left (a d -b c \right ) c}\, x}{2 \sqrt {\left (a d -b c \right ) c}\, \sqrt {b \,x^{2}+a}\, c \left (x^{2} d +c \right ) \left (a d -b c \right )^{2} a}\)	\(150\)
default	\(\text {Expression too large to display}\)	\(1906\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (123) = 246\).

Time = 0.33 (sec) , antiderivative size = 864, normalized size of antiderivative = 6.04 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/8*((4*a^2*b*c^2*d - a^3*c*d^2 + (4*a*b^2*c*d^2 - a^2*b*d^3)*x^4 + (4*a 
*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2 
*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 
+ 4*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2* 
x^4 + 2*c*d*x^2 + c^2)) - 4*((2*b^3*c^3*d - a*b^2*c^2*d^2 - a^2*b*c*d^3)*x 
^3 + (2*b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2 - a^3*c*d^3)*x)*sqrt(b*x^2 
 + a))/(a^2*b^3*c^6 - 3*a^3*b^2*c^5*d + 3*a^4*b*c^4*d^2 - a^5*c^3*d^3 + (a 
*b^4*c^5*d - 3*a^2*b^3*c^4*d^2 + 3*a^3*b^2*c^3*d^3 - a^4*b*c^2*d^4)*x^4 + 
(a*b^4*c^6 - 2*a^2*b^3*c^5*d + 2*a^4*b*c^3*d^3 - a^5*c^2*d^4)*x^2), 1/4*(( 
4*a^2*b*c^2*d - a^3*c*d^2 + (4*a*b^2*c*d^2 - a^2*b*d^3)*x^4 + (4*a*b^2*c^2 
*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqrt(-b 
*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c* 
d)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*((2*b^3*c^3*d - a*b^2*c^2*d^2 - a^2*b 
*c*d^3)*x^3 + (2*b^3*c^4 - 2*a*b^2*c^3*d + a^2*b*c^2*d^2 - a^3*c*d^3)*x)*s 
qrt(b*x^2 + a))/(a^2*b^3*c^6 - 3*a^3*b^2*c^5*d + 3*a^4*b*c^4*d^2 - a^5*c^3 
*d^3 + (a*b^4*c^5*d - 3*a^2*b^3*c^4*d^2 + 3*a^3*b^2*c^3*d^3 - a^4*b*c^2*d^ 
4)*x^4 + (a*b^4*c^6 - 2*a^2*b^3*c^5*d + 2*a^4*b*c^3*d^3 - a^5*c^2*d^4)*x^2 
)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (123) = 246\).

Time = 0.43 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.22 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx=\frac {b^{2} x}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {b x^{2} + a}} + \frac {{\left (4 \, b^{\frac {3}{2}} c d - a \sqrt {b} d^{2}\right )} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{2 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {-b^{2} c^{2} + a b c d}} + \frac {2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} c d - {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d^{2} + a^{2} \sqrt {b} d^{2}}{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 1275, normalized size of antiderivative = 8.92 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 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*c*d**2 - 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*d**3*x**2 + 4*sqrt(c)*sqrt(a*d - b*c)*atan((sqr 
t(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt 
(b)))*a**2*b*c**2*d + 3*sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sq 
rt(d)*sqrt(a + b*x**2) - sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**2*b*c*d* 
*2*x**2 - 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*d**3*x**4 + 4*sqrt 
(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b*x**2) - sqr 
t(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a*b**2*c**2*d*x**2 + 4*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*b**2*c*d**2*x**4 - 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**3*c*d**2 - sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqr 
t(d)*sqrt(a + b*x**2) + sqrt(d)*sqrt(b)*x)/(sqrt(c)*sqrt(b)))*a**3*d**3*x* 
*2 + 4*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*c**2*d + 3*sqrt(c)*sq 
rt(a*d - b*c)*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x**2) + sqrt(d)*s 
qrt(b)*x)/(sqrt(c)*sqrt(b)))*a**2*b*c*d**2*x**2 - sqrt(c)*sqrt(a*d - b*...