\(\int \frac {1}{x^4 (a+b x^2) (c+d x^2)^{5/2}} \, dx\) [986]

Optimal result

Integrand size = 24, antiderivative size = 245 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {d}{3 c (b c-a d) x^3 \left (c+d x^2\right )^{3/2}}-\frac {d (3 b c-2 a d)}{c^2 (b c-a d)^2 x^3 \sqrt {c+d x^2}}-\frac {\left (b^2 c^2-12 a b c d+8 a^2 d^2\right ) \sqrt {c+d x^2}}{3 a c^3 (b c-a d)^2 x^3}+\frac {(b c-2 a d) \left (3 b^2 c^2+8 a b c d-8 a^2 d^2\right ) \sqrt {c+d x^2}}{3 a^2 c^4 (b c-a d)^2 x}+\frac {b^4 \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{5/2} (b c-a d)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {3 b^3 c^3 x^2 \left (c+d x^2\right )^2-a b^2 c^2 \left (c-2 d x^2\right ) \left (c+d x^2\right )^2+a^2 b c d \left (2 c^3-9 c^2 d x^2-36 c d^2 x^4-24 d^3 x^6\right )+a^3 d^2 \left (-c^3+6 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )}{3 a^2 c^4 (b c-a d)^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {b^4 \arctan \left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{5/2} (b c-a d)^{5/2}} \] Input:

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

(3*b^3*c^3*x^2*(c + d*x^2)^2 - a*b^2*c^2*(c - 2*d*x^2)*(c + d*x^2)^2 + a^2 
*b*c*d*(2*c^3 - 9*c^2*d*x^2 - 36*c*d^2*x^4 - 24*d^3*x^6) + a^3*d^2*(-c^3 + 
 6*c^2*d*x^2 + 24*c*d^2*x^4 + 16*d^3*x^6))/(3*a^2*c^4*(b*c - a*d)^2*x^3*(c 
 + d*x^2)^(3/2)) - (b^4*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d]*x - Sqrt[c + d*x^ 
2]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(a^(5/2)*(b*c - a*d)^(5/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {374, 27, 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.

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {\frac {\int \frac {b^2 c^2-12 a b d c+8 a^2 d^2-4 b d (3 b c-2 a d) x^2}{x^4 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {\int \frac {2 b d \left (b^2 c^2-12 a b d c+8 a^2 d^2\right ) x^2+(b c-2 a d) \left (3 b^2 c^2+8 a b d c-8 a^2 d^2\right )}{x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {b^2 c}{a}+\frac {8 a d^2}{c}-12 b d\right )}{3 x^3}}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 445

\(\displaystyle \frac {\frac {-\frac {-\frac {\int \frac {3 b^4 c^4}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{a c}-\frac {\sqrt {c+d x^2} (b c-2 a d) \left (-8 a^2 d^2+8 a b c d+3 b^2 c^2\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {b^2 c}{a}+\frac {8 a d^2}{c}-12 b d\right )}{3 x^3}}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^4 c^3 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{a}-\frac {\sqrt {c+d x^2} (b c-2 a d) \left (-8 a^2 d^2+8 a b c d+3 b^2 c^2\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {b^2 c}{a}+\frac {8 a d^2}{c}-12 b d\right )}{3 x^3}}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^4 c^3 \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{a}-\frac {\sqrt {c+d x^2} (b c-2 a d) \left (-8 a^2 d^2+8 a b c d+3 b^2 c^2\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {b^2 c}{a}+\frac {8 a d^2}{c}-12 b d\right )}{3 x^3}}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {-\frac {-\frac {3 b^4 c^3 \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{a^{3/2} \sqrt {b c-a d}}-\frac {\sqrt {c+d x^2} (b c-2 a d) \left (-8 a^2 d^2+8 a b c d+3 b^2 c^2\right )}{a c x}}{3 a c}-\frac {\sqrt {c+d x^2} \left (\frac {b^2 c}{a}+\frac {8 a d^2}{c}-12 b d\right )}{3 x^3}}{c (b c-a d)}-\frac {d (3 b c-2 a d)}{c x^3 \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}-\frac {d}{3 c x^3 \left (c+d x^2\right )^{3/2} (b c-a d)}\)

Input:

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*d/(c*(b*c - a*d)*x^3*(c + d*x^2)^(3/2)) + (-((d*(3*b*c - 2*a*d))/(c*( 
b*c - a*d)*x^3*Sqrt[c + d*x^2])) + (-1/3*(((b^2*c)/a - 12*b*d + (8*a*d^2)/ 
c)*Sqrt[c + d*x^2])/x^3 - (-(((b*c - 2*a*d)*(3*b^2*c^2 + 8*a*b*c*d - 8*a^2 
*d^2)*Sqrt[c + d*x^2])/(a*c*x)) - (3*b^4*c^3*ArcTan[(Sqrt[b*c - a*d]*x)/(S 
qrt[a]*Sqrt[c + d*x^2])])/(a^(3/2)*Sqrt[b*c - a*d]))/(3*a*c))/(c*(b*c - a* 
d)))/(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 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 = 0.83 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(-\frac {-3 c^{4} b^{4} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right ) x^{3} \left (x^{2} d +c \right )^{\frac {3}{2}}+\left (b^{2} \left (-3 b \,x^{2}+a \right ) c^{5}-2 b d \left (3 b^{2} x^{4}+a^{2}\right ) c^{4}+d^{2} \left (-3 b^{3} x^{6}-3 a \,b^{2} x^{4}+9 a^{2} b \,x^{2}+a^{3}\right ) c^{3}-6 a \,d^{3} \left (\frac {1}{3} b^{2} x^{4}-6 a b \,x^{2}+a^{2}\right ) x^{2} c^{2}-24 a^{2} d^{4} x^{4} \left (-b \,x^{2}+a \right ) c -16 a^{3} d^{5} x^{6}\right ) \sqrt {\left (a d -b c \right ) a}}{3 \sqrt {\left (a d -b c \right ) a}\, \left (x^{2} d +c \right )^{\frac {3}{2}} a^{2} x^{3} \left (a d -b c \right )^{2} c^{4}}\)	\(238\)
risch	\(\text {Expression too large to display}\)	\(1023\)
default	\(\text {Expression too large to display}\)	\(1552\)

Input:

int(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*(-3*c^4*b^4*arctanh((d*x^2+c)^(1/2)/x*a/((a*d-b*c)*a)^(1/2))*x^3*(d*x 
^2+c)^(3/2)+(b^2*(-3*b*x^2+a)*c^5-2*b*d*(3*b^2*x^4+a^2)*c^4+d^2*(-3*b^3*x^ 
6-3*a*b^2*x^4+9*a^2*b*x^2+a^3)*c^3-6*a*d^3*(1/3*b^2*x^4-6*a*b*x^2+a^2)*x^2 
*c^2-24*a^2*d^4*x^4*(-b*x^2+a)*c-16*a^3*d^5*x^6)*((a*d-b*c)*a)^(1/2))/((a* 
d-b*c)*a)^(1/2)/(d*x^2+c)^(3/2)/a^2/x^3/(a*d-b*c)^2/c^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 544 vs. \(2 (221) = 442\).

Time = 0.50 (sec) , antiderivative size = 1128, normalized size of antiderivative = 4.60 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(b^4*c^4*d^2*x^7 + 2*b^4*c^5*d*x^5 + b^4*c^6*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)*sqr 
t(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(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 - (3*a*b^4*c^4*d^2 - a^2*b^3*c^3*d^3 - 
 26*a^3*b^2*c^2*d^4 + 40*a^4*b*c*d^5 - 16*a^5*d^6)*x^6 - 3*(2*a*b^4*c^5*d 
- a^2*b^3*c^4*d^2 - 13*a^3*b^2*c^3*d^3 + 20*a^4*b*c^2*d^4 - 8*a^5*c*d^5)*x 
^4 - 3*(a*b^4*c^6 - a^2*b^3*c^5*d - 3*a^3*b^2*c^4*d^2 + 5*a^4*b*c^3*d^3 - 
2*a^5*c^2*d^4)*x^2)*sqrt(d*x^2 + c))/((a^3*b^3*c^7*d^2 - 3*a^4*b^2*c^6*d^3 
 + 3*a^5*b*c^5*d^4 - a^6*c^4*d^5)*x^7 + 2*(a^3*b^3*c^8*d - 3*a^4*b^2*c^7*d 
^2 + 3*a^5*b*c^6*d^3 - a^6*c^5*d^4)*x^5 + (a^3*b^3*c^9 - 3*a^4*b^2*c^8*d + 
 3*a^5*b*c^7*d^2 - a^6*c^6*d^3)*x^3), 1/6*(3*(b^4*c^4*d^2*x^7 + 2*b^4*c^5* 
d*x^5 + b^4*c^6*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*(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 - (3*a*b^4*c^4*d^2 - a^2*b^3*c^3*d^3 - 26*a^3*b^2*c^2*d^4 + 40*a 
^4*b*c*d^5 - 16*a^5*d^6)*x^6 - 3*(2*a*b^4*c^5*d - a^2*b^3*c^4*d^2 - 13*a^3 
*b^2*c^3*d^3 + 20*a^4*b*c^2*d^4 - 8*a^5*c*d^5)*x^4 - 3*(a*b^4*c^6 - a^2*b^ 
3*c^5*d - 3*a^3*b^2*c^4*d^2 + 5*a^4*b*c^3*d^3 - 2*a^5*c^2*d^4)*x^2)*sqrt(d 
*x^2 + c))/((a^3*b^3*c^7*d^2 - 3*a^4*b^2*c^6*d^3 + 3*a^5*b*c^5*d^4 - a^...
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x**4/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(1/(x**4*(a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{4}} \,d x } \] Input:

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(5/2)*x^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (221) = 442\).

Time = 0.38 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.00 \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {b^{4} \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\frac {{\left (11 \, b^{3} c^{6} d^{5} - 30 \, a b^{2} c^{5} d^{6} + 27 \, a^{2} b c^{4} d^{7} - 8 \, a^{3} c^{3} d^{8}\right )} x^{2}}{b^{4} c^{11} d - 4 \, a b^{3} c^{10} d^{2} + 6 \, a^{2} b^{2} c^{9} d^{3} - 4 \, a^{3} b c^{8} d^{4} + a^{4} c^{7} d^{5}} + \frac {3 \, {\left (4 \, b^{3} c^{7} d^{4} - 11 \, a b^{2} c^{6} d^{5} + 10 \, a^{2} b c^{5} d^{6} - 3 \, a^{3} c^{4} d^{7}\right )}}{b^{4} c^{11} d - 4 \, a b^{3} c^{10} d^{2} + 6 \, a^{2} b^{2} c^{9} d^{3} - 4 \, a^{3} b c^{8} d^{4} + a^{4} c^{7} d^{5}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b c \sqrt {d} + 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a d^{\frac {3}{2}} - 6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c^{2} \sqrt {d} - 18 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a c d^{\frac {3}{2}} + 3 \, b c^{3} \sqrt {d} + 8 \, a c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2} c^{3}} \] Input:

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

-b^4*sqrt(d)*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^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*sqrt(a*b*c 
*d - a^2*d^2)) - 1/3*((11*b^3*c^6*d^5 - 30*a*b^2*c^5*d^6 + 27*a^2*b*c^4*d^ 
7 - 8*a^3*c^3*d^8)*x^2/(b^4*c^11*d - 4*a*b^3*c^10*d^2 + 6*a^2*b^2*c^9*d^3 
- 4*a^3*b*c^8*d^4 + a^4*c^7*d^5) + 3*(4*b^3*c^7*d^4 - 11*a*b^2*c^6*d^5 + 1 
0*a^2*b*c^5*d^6 - 3*a^3*c^4*d^7)/(b^4*c^11*d - 4*a*b^3*c^10*d^2 + 6*a^2*b^ 
2*c^9*d^3 - 4*a^3*b*c^8*d^4 + a^4*c^7*d^5))*x/(d*x^2 + c)^(3/2) - 2/3*(3*( 
sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c*sqrt(d) + 6*(sqrt(d)*x - sqrt(d*x^2 + c 
))^4*a*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2*sqrt(d) - 18*(sqr 
t(d)*x - sqrt(d*x^2 + c))^2*a*c*d^(3/2) + 3*b*c^3*sqrt(d) + 8*a*c^2*d^(3/2 
))/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^3*a^2*c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{x^4\,\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(1/(x^4*(a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{x^{4} \left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )^{\frac {5}{2}}}d x \] Input:

int(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x)
 

Output:

int(1/x^4/(b*x^2+a)/(d*x^2+c)^(5/2),x)