\(\int \frac {1}{x^4 (a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [1031]

Optimal result

Integrand size = 24, antiderivative size = 277 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {d (b c+2 a d)}{2 a c (b c-a d)^2 x^3 \sqrt {c+d x^2}}+\frac {b}{2 a (b c-a d) x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {\left (5 b^2 c^2-4 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{6 a^2 c^2 (b c-a d)^2 x^3}+\frac {\left (15 b^3 c^3-14 a b^2 c^2 d-8 a^2 b c d^2+16 a^3 d^3\right ) \sqrt {c+d x^2}}{6 a^3 c^3 (b c-a d)^2 x}+\frac {b^3 (5 b c-8 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{7/2} (b c-a d)^{5/2}} \] Output:

1/2*d*(2*a*d+b*c)/a/c/(-a*d+b*c)^2/x^3/(d*x^2+c)^(1/2)+1/2*b/a/(-a*d+b*c)/ 
x^3/(b*x^2+a)/(d*x^2+c)^(1/2)-1/6*(8*a^2*d^2-4*a*b*c*d+5*b^2*c^2)*(d*x^2+c 
)^(1/2)/a^2/c^2/(-a*d+b*c)^2/x^3+1/6*(16*a^3*d^3-8*a^2*b*c*d^2-14*a*b^2*c^ 
2*d+15*b^3*c^3)*(d*x^2+c)^(1/2)/a^3/c^3/(-a*d+b*c)^2/x+1/2*b^3*(-8*a*d+5*b 
*c)*arctan((-a*d+b*c)^(1/2)*x/a^(1/2)/(d*x^2+c)^(1/2))/a^(7/2)/(-a*d+b*c)^ 
(5/2)
 

Mathematica [A] (verified)

Time = 1.29 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {15 b^4 c^3 x^4 \left (c+d x^2\right )-2 a^2 b^2 c \left (c+d x^2\right )^2 \left (c+4 d x^2\right )+2 a b^3 c^2 x^2 \left (5 c^2-2 c d x^2-7 d^2 x^4\right )+2 a^4 d^2 \left (-c^2+4 c d x^2+8 d^2 x^4\right )+2 a^3 b d \left (2 c^3-3 c^2 d x^2+8 d^3 x^6\right )}{6 a^3 c^3 (b c-a d)^2 x^3 \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {b^3 (5 b c-8 a d) \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{2 a^{7/2} (b c-a d)^{5/2}} \] Input:

Integrate[1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)),x]
 

Output:

(15*b^4*c^3*x^4*(c + d*x^2) - 2*a^2*b^2*c*(c + d*x^2)^2*(c + 4*d*x^2) + 2* 
a*b^3*c^2*x^2*(5*c^2 - 2*c*d*x^2 - 7*d^2*x^4) + 2*a^4*d^2*(-c^2 + 4*c*d*x^ 
2 + 8*d^2*x^4) + 2*a^3*b*d*(2*c^3 - 3*c^2*d*x^2 + 8*d^3*x^6))/(6*a^3*c^3*( 
b*c - a*d)^2*x^3*(a + b*x^2)*Sqrt[c + d*x^2]) - (b^3*(5*b*c - 8*a*d)*ArcTa 
n[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d] 
)])/(2*a^(7/2)*(b*c - a*d)^(5/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {374, 25, 441, 445, 445, 27, 291, 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.

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

\(\Big \downarrow \) 374

\(\displaystyle \frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}-\frac {\int -\frac {6 b d x^2+5 b c-2 a d}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx}{2 a (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {6 b d x^2+5 b c-2 a d}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 441

\(\displaystyle \frac {\frac {\int \frac {5 b^2 c^2-4 a b d c+8 a^2 d^2+4 b d (b c+2 a d) x^2}{x^4 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int \frac {15 b^3 c^3-14 a b^2 d c^2-8 a^2 b d^2 c+16 a^3 d^3+2 b d \left (5 b^2 c^2-4 a b d c+8 a^2 d^2\right ) x^2}{x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {5 b^2 c}{a}+\frac {8 a d^2}{c}-4 b d\right )}{3 x^3}}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {3 b^3 c^3 (5 b c-8 a d)}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{a c}-\frac {\sqrt {c+d x^2} \left (16 a^3 d^3-8 a^2 b c d^2-14 a b^2 c^2 d+15 b^3 c^3\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {5 b^2 c}{a}+\frac {8 a d^2}{c}-4 b d\right )}{3 x^3}}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^3 c^2 (5 b c-8 a d) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{a}-\frac {\sqrt {c+d x^2} \left (16 a^3 d^3-8 a^2 b c d^2-14 a b^2 c^2 d+15 b^3 c^3\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {5 b^2 c}{a}+\frac {8 a d^2}{c}-4 b d\right )}{3 x^3}}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^3 c^2 (5 b c-8 a d) \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{a}-\frac {\sqrt {c+d x^2} \left (16 a^3 d^3-8 a^2 b c d^2-14 a b^2 c^2 d+15 b^3 c^3\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {5 b^2 c}{a}+\frac {8 a d^2}{c}-4 b d\right )}{3 x^3}}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^3 c^2 (5 b c-8 a d) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2} \left (16 a^3 d^3-8 a^2 b c d^2-14 a b^2 c^2 d+15 b^3 c^3\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {5 b^2 c}{a}+\frac {8 a d^2}{c}-4 b d\right )}{3 x^3}}{c (b c-a d)}+\frac {d (2 a d+b c)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{2 a (b c-a d)}+\frac {b}{2 a x^3 \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)}\)

Input:

Int[1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)),x]
 

Output:

b/(2*a*(b*c - a*d)*x^3*(a + b*x^2)*Sqrt[c + d*x^2]) + ((d*(b*c + 2*a*d))/( 
c*(b*c - a*d)*x^3*Sqrt[c + d*x^2]) + (-1/3*(((5*b^2*c)/a - 4*b*d + (8*a*d^ 
2)/c)*Sqrt[c + d*x^2])/x^3 - (-(((15*b^3*c^3 - 14*a*b^2*c^2*d - 8*a^2*b*c* 
d^2 + 16*a^3*d^3)*Sqrt[c + d*x^2])/(a*c*x)) - (3*b^3*c^2*(5*b*c - 8*a*d)*A 
rcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(a^(3/2)*Sqrt[b*c - 
a*d]))/(3*a*c))/(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 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 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 441

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

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 = 0.92 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.57

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {x^{2} d +c}\, \left (-5 a d \,x^{2}-6 x^{2} b c +a c \right )}{3 x^{3} a^{3}}+\frac {b^{3} c^{3} \left (\frac {\sqrt {x^{2} d +c}\, b x}{b \,x^{2}+a}-\frac {\left (8 a d -5 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{2 \left (a d -b c \right )^{2} a^{3}}+\frac {d^{4} x}{\left (a d -b c \right )^{2} \sqrt {x^{2} d +c}}}{c^{3}}\)	\(158\)
risch	\(\text {Expression too large to display}\)	\(1271\)
default	\(\text {Expression too large to display}\)	\(2032\)

Input:

int(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(-1/3*(d*x^2+c)^(1/2)*(-5*a*d*x^2-6*b*c*x^2+a*c)/x^3/a^3+1/2*b^3*c^3*((d*x 
^2+c)^(1/2)*b*x/(b*x^2+a)-(8*a*d-5*b*c)/((a*d-b*c)*a)^(1/2)*arctanh((d*x^2 
+c)^(1/2)/x*a/((a*d-b*c)*a)^(1/2)))/(a*d-b*c)^2/a^3+d^4/(a*d-b*c)^2/(d*x^2 
+c)^(1/2)*x)/c^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (249) = 498\).

Time = 0.59 (sec) , antiderivative size = 1252, normalized size of antiderivative = 4.52 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/24*(3*((5*b^5*c^4*d - 8*a*b^4*c^3*d^2)*x^7 + (5*b^5*c^5 - 3*a*b^4*c^4*d 
 - 8*a^2*b^3*c^3*d^2)*x^5 + (5*a*b^4*c^5 - 8*a^2*b^3*c^4*d)*x^3)*sqrt(-a*b 
*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a* 
b*c^2 - 4*a^2*c*d)*x^2 + 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d 
)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(2*a^3*b^3*c^5 - 6*a^4 
*b^2*c^4*d + 6*a^5*b*c^3*d^2 - 2*a^6*c^2*d^3 - (15*a*b^5*c^4*d - 29*a^2*b^ 
4*c^3*d^2 + 6*a^3*b^3*c^2*d^3 + 24*a^4*b^2*c*d^4 - 16*a^5*b*d^5)*x^6 - (15 
*a*b^5*c^5 - 19*a^2*b^4*c^4*d - 14*a^3*b^3*c^3*d^2 + 18*a^4*b^2*c^2*d^3 + 
16*a^5*b*c*d^4 - 16*a^6*d^5)*x^4 - 2*(5*a^2*b^4*c^5 - 11*a^3*b^3*c^4*d + 3 
*a^4*b^2*c^3*d^2 + 7*a^5*b*c^2*d^3 - 4*a^6*c*d^4)*x^2)*sqrt(d*x^2 + c))/(( 
a^4*b^4*c^6*d - 3*a^5*b^3*c^5*d^2 + 3*a^6*b^2*c^4*d^3 - a^7*b*c^3*d^4)*x^7 
 + (a^4*b^4*c^7 - 2*a^5*b^3*c^6*d + 2*a^7*b*c^4*d^3 - a^8*c^3*d^4)*x^5 + ( 
a^5*b^3*c^7 - 3*a^6*b^2*c^6*d + 3*a^7*b*c^5*d^2 - a^8*c^4*d^3)*x^3), 1/12* 
(3*((5*b^5*c^4*d - 8*a*b^4*c^3*d^2)*x^7 + (5*b^5*c^5 - 3*a*b^4*c^4*d - 8*a 
^2*b^3*c^3*d^2)*x^5 + (5*a*b^4*c^5 - 8*a^2*b^3*c^4*d)*x^3)*sqrt(a*b*c - a^ 
2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + 
 c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*(2*a^3*b^3*c^5 
- 6*a^4*b^2*c^4*d + 6*a^5*b*c^3*d^2 - 2*a^6*c^2*d^3 - (15*a*b^5*c^4*d - 29 
*a^2*b^4*c^3*d^2 + 6*a^3*b^3*c^2*d^3 + 24*a^4*b^2*c*d^4 - 16*a^5*b*d^5)*x^ 
6 - (15*a*b^5*c^5 - 19*a^2*b^4*c^4*d - 14*a^3*b^3*c^3*d^2 + 18*a^4*b^2*...
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**4/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)
 

Output:

Integral(1/(x**4*(a + b*x**2)**2*(c + d*x**2)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}} \,d x } \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^2*(d*x^2 + c)^(3/2)*x^4), x)
 

Giac [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.75 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {d^{4} x}{{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (5 \, b^{4} c \sqrt {d} - 8 \, a b^{3} d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{4} c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{3} d^{\frac {3}{2}} - b^{4} c^{2} \sqrt {d}}{{\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )} {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} - \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d^{\frac {3}{2}} + 6 \, b c^{3} \sqrt {d} + 5 \, a c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{3} c^{2}} \] Input:

integrate(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")
 

Output:

d^4*x/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2)*sqrt(d*x^2 + c)) - 1/2*(5*b^4 
*c*sqrt(d) - 8*a*b^3*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2* 
b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a^3*b^2*c^2 - 2*a^4*b*c*d + a^ 
5*d^2)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^4*c*s 
qrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b^3*d^(3/2) - b^4*c^2*sqrt(d) 
)/((a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2)*((sqrt(d)*x - sqrt(d*x^2 + c))^4* 
b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c) 
)^2*a*d + b*c^2)) - 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c*sqrt(d) + 3 
*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + 
c))^2*b*c^2*sqrt(d) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d^(3/2) + 6*b 
*c^3*sqrt(d) + 5*a*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a 
^3*c^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \] Input:

int(1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)),x)
 

Output:

int(1/(x^4*(a + b*x^2)^2*(c + d*x^2)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}d x \] Input:

int(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x)
 

Output:

int(1/x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x)