3.670 \(\int \frac {(a+b x^2)^2}{x^6 (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {8 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^5 \sqrt {c+d x^2}}-\frac {4 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^4 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2 c^2-4 a d (5 b c-4 a d)}{5 c^3 x \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}} \]

[Out]

-1/5*a^2/c/x^5/(d*x^2+c)^(3/2)-2/15*a*(-4*a*d+5*b*c)/c^2/x^3/(d*x^2+c)^(3/2)+1/5*(-5*b^2*c^2+4*a*d*(-4*a*d+5*b
*c))/c^3/x/(d*x^2+c)^(3/2)-4/15*d*(5*b^2*c^2-4*a*d*(-4*a*d+5*b*c))*x/c^4/(d*x^2+c)^(3/2)-8/15*d*(5*b^2*c^2-4*a
*d*(-4*a*d+5*b*c))*x/c^5/(d*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {462, 453, 271, 192, 191} \[ -\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {8 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^5 \sqrt {c+d x^2}}-\frac {4 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^4 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}}{5 c x \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^6*(c + d*x^2)^(5/2)),x]

[Out]

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx &=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {2 a (5 b c-4 a d)+5 b^2 c x^2}{x^4 \left (c+d x^2\right )^{5/2}} \, dx}{5 c}\\ &=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {1}{5} \left (-5 b^2+\frac {4 a d (5 b c-4 a d)}{c^2}\right ) \int \frac {1}{x^2 \left (c+d x^2\right )^{5/2}} \, dx\\ &=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}}{5 c x \left (c+d x^2\right )^{3/2}}-\frac {\left (4 d \left (5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}\right )\right ) \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx}{5 c}\\ &=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}}{5 c x \left (c+d x^2\right )^{3/2}}-\frac {4 d \left (5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}\right ) x}{15 c^2 \left (c+d x^2\right )^{3/2}}-\frac {\left (8 d \left (5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}\right )\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{15 c^2}\\ &=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}}{5 c x \left (c+d x^2\right )^{3/2}}-\frac {4 d \left (5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}\right ) x}{15 c^2 \left (c+d x^2\right )^{3/2}}-\frac {8 d \left (5 b^2-\frac {4 a d (5 b c-4 a d)}{c^2}\right ) x}{15 c^3 \sqrt {c+d x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 142, normalized size = 0.78 \[ \frac {-a^2 \left (3 c^4-8 c^3 d x^2+48 c^2 d^2 x^4+192 c d^3 x^6+128 d^4 x^8\right )+10 a b c x^2 \left (-c^3+6 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )-5 b^2 c^2 x^4 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )}{15 c^5 x^5 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^6*(c + d*x^2)^(5/2)),x]

[Out]

(-5*b^2*c^2*x^4*(3*c^2 + 12*c*d*x^2 + 8*d^2*x^4) + 10*a*b*c*x^2*(-c^3 + 6*c^2*d*x^2 + 24*c*d^2*x^4 + 16*d^3*x^
6) - a^2*(3*c^4 - 8*c^3*d*x^2 + 48*c^2*d^2*x^4 + 192*c*d^3*x^6 + 128*d^4*x^8))/(15*c^5*x^5*(c + d*x^2)^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 171, normalized size = 0.93 \[ -\frac {{\left (8 \, {\left (5 \, b^{2} c^{2} d^{2} - 20 \, a b c d^{3} + 16 \, a^{2} d^{4}\right )} x^{8} + 12 \, {\left (5 \, b^{2} c^{3} d - 20 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x^{6} + 3 \, a^{2} c^{4} + 3 \, {\left (5 \, b^{2} c^{4} - 20 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} x^{4} + 2 \, {\left (5 \, a b c^{4} - 4 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (c^{5} d^{2} x^{9} + 2 \, c^{6} d x^{7} + c^{7} x^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(8*(5*b^2*c^2*d^2 - 20*a*b*c*d^3 + 16*a^2*d^4)*x^8 + 12*(5*b^2*c^3*d - 20*a*b*c^2*d^2 + 16*a^2*c*d^3)*x^
6 + 3*a^2*c^4 + 3*(5*b^2*c^4 - 20*a*b*c^3*d + 16*a^2*c^2*d^2)*x^4 + 2*(5*a*b*c^4 - 4*a^2*c^3*d)*x^2)*sqrt(d*x^
2 + c)/(c^5*d^2*x^9 + 2*c^6*d*x^7 + c^7*x^5)

________________________________________________________________________________________

giac [B]  time = 0.54, size = 509, normalized size = 2.78 \[ -\frac {x {\left (\frac {{\left (5 \, b^{2} c^{6} d^{3} - 16 \, a b c^{5} d^{4} + 11 \, a^{2} c^{4} d^{5}\right )} x^{2}}{c^{9} d} + \frac {6 \, {\left (b^{2} c^{7} d^{2} - 3 \, a b c^{6} d^{3} + 2 \, a^{2} c^{5} d^{4}\right )}}{c^{9} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt {d} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c d^{\frac {3}{2}} + 45 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{3} \sqrt {d} + 300 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac {3}{2}} - 240 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{4} \sqrt {d} - 500 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac {3}{2}} + 490 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{5} \sqrt {d} + 340 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac {3}{2}} - 320 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac {5}{2}} + 15 \, b^{2} c^{6} \sqrt {d} - 80 \, a b c^{5} d^{\frac {3}{2}} + 73 \, a^{2} c^{4} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

-1/3*x*((5*b^2*c^6*d^3 - 16*a*b*c^5*d^4 + 11*a^2*c^4*d^5)*x^2/(c^9*d) + 6*(b^2*c^7*d^2 - 3*a*b*c^6*d^3 + 2*a^2
*c^5*d^4)/(c^9*d))/(d*x^2 + c)^(3/2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^2*sqrt(d) - 60*(sqrt(d)*
x - sqrt(d*x^2 + c))^8*a*b*c*d^(3/2) + 45*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(d
*x^2 + c))^6*b^2*c^3*sqrt(d) + 300*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(3/2) - 240*(sqrt(d)*x - sqrt(d*x
^2 + c))^6*a^2*c*d^(5/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^4*sqrt(d) - 500*(sqrt(d)*x - sqrt(d*x^2 +
c))^4*a*b*c^3*d^(3/2) + 490*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))
^2*b^2*c^5*sqrt(d) + 340*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^(3/2) - 320*(sqrt(d)*x - sqrt(d*x^2 + c))^2
*a^2*c^3*d^(5/2) + 15*b^2*c^6*sqrt(d) - 80*a*b*c^5*d^(3/2) + 73*a^2*c^4*d^(5/2))/(((sqrt(d)*x - sqrt(d*x^2 + c
))^2 - c)^5*c^4)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 158, normalized size = 0.86 \[ -\frac {128 a^{2} d^{4} x^{8}-160 a b c \,d^{3} x^{8}+40 b^{2} c^{2} d^{2} x^{8}+192 a^{2} c \,d^{3} x^{6}-240 a b \,c^{2} d^{2} x^{6}+60 b^{2} c^{3} d \,x^{6}+48 a^{2} c^{2} d^{2} x^{4}-60 a b \,c^{3} d \,x^{4}+15 b^{2} c^{4} x^{4}-8 a^{2} c^{3} d \,x^{2}+10 a b \,c^{4} x^{2}+3 a^{2} c^{4}}{15 \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{5} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x)

[Out]

-1/15*(128*a^2*d^4*x^8-160*a*b*c*d^3*x^8+40*b^2*c^2*d^2*x^8+192*a^2*c*d^3*x^6-240*a*b*c^2*d^2*x^6+60*b^2*c^3*d
*x^6+48*a^2*c^2*d^2*x^4-60*a*b*c^3*d*x^4+15*b^2*c^4*x^4-8*a^2*c^3*d*x^2+10*a*b*c^4*x^2+3*a^2*c^4)/(d*x^2+c)^(3
/2)/x^5/c^5

________________________________________________________________________________________

maxima [A]  time = 0.96, size = 244, normalized size = 1.33 \[ -\frac {8 \, b^{2} d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {4 \, b^{2} d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {32 \, a b d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{4}} + \frac {16 \, a b d^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {128 \, a^{2} d^{3} x}{15 \, \sqrt {d x^{2} + c} c^{5}} - \frac {64 \, a^{2} d^{3} x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{4}} - \frac {b^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} + \frac {4 \, a b d}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x} - \frac {16 \, a^{2} d^{2}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3} x} - \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{3}} + \frac {8 \, a^{2} d}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x^{3}} - \frac {a^{2}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-8/3*b^2*d*x/(sqrt(d*x^2 + c)*c^3) - 4/3*b^2*d*x/((d*x^2 + c)^(3/2)*c^2) + 32/3*a*b*d^2*x/(sqrt(d*x^2 + c)*c^4
) + 16/3*a*b*d^2*x/((d*x^2 + c)^(3/2)*c^3) - 128/15*a^2*d^3*x/(sqrt(d*x^2 + c)*c^5) - 64/15*a^2*d^3*x/((d*x^2
+ c)^(3/2)*c^4) - b^2/((d*x^2 + c)^(3/2)*c*x) + 4*a*b*d/((d*x^2 + c)^(3/2)*c^2*x) - 16/5*a^2*d^2/((d*x^2 + c)^
(3/2)*c^3*x) - 2/3*a*b/((d*x^2 + c)^(3/2)*c*x^3) + 8/15*a^2*d/((d*x^2 + c)^(3/2)*c^2*x^3) - 1/5*a^2/((d*x^2 +
c)^(3/2)*c*x^5)

________________________________________________________________________________________

mupad [B]  time = 0.98, size = 298, normalized size = 1.63 \[ \frac {2\,a\,\sqrt {d\,x^2+c}\,\left (7\,a\,d-5\,b\,c\right )}{15\,c^4\,x^3}-\frac {\frac {73\,a^2\,c^2\,d^2-80\,a\,b\,c^3\,d+15\,b^2\,c^4}{30\,c^5}-\frac {c\,\left (\frac {d\,\left (73\,a^2\,c^2\,d^2-80\,a\,b\,c^3\,d+15\,b^2\,c^4\right )}{18\,c^6}+\frac {c\,\left (\frac {4\,a\,d^3\,\left (7\,a\,d-5\,b\,c\right )}{45\,c^5}-\frac {a\,d^3\,\left (43\,a\,d-35\,b\,c\right )}{9\,c^5}\right )}{d}+\frac {a\,d^2\,\left (43\,a\,d-35\,b\,c\right )}{15\,c^4}\right )}{d}}{x\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a^2\,\sqrt {d\,x^2+c}}{5\,c^3\,x^5}-\frac {x^2\,\left (\frac {2\,d\,\left (78\,a^2\,c\,d^2-90\,a\,b\,c^2\,d+20\,b^2\,c^3\right )}{15\,c^6}-\frac {4\,a\,d^2\,\left (7\,a\,d-5\,b\,c\right )}{15\,c^5}\right )+\frac {78\,a^2\,c\,d^2-90\,a\,b\,c^2\,d+20\,b^2\,c^3}{15\,c^5}}{x\,\sqrt {d\,x^2+c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/(x^6*(c + d*x^2)^(5/2)),x)

[Out]

(2*a*(c + d*x^2)^(1/2)*(7*a*d - 5*b*c))/(15*c^4*x^3) - ((15*b^2*c^4 + 73*a^2*c^2*d^2 - 80*a*b*c^3*d)/(30*c^5)
- (c*((d*(15*b^2*c^4 + 73*a^2*c^2*d^2 - 80*a*b*c^3*d))/(18*c^6) + (c*((4*a*d^3*(7*a*d - 5*b*c))/(45*c^5) - (a*
d^3*(43*a*d - 35*b*c))/(9*c^5)))/d + (a*d^2*(43*a*d - 35*b*c))/(15*c^4)))/d)/(x*(c + d*x^2)^(3/2)) - (a^2*(c +
d*x^2)^(1/2))/(5*c^3*x^5) - (x^2*((2*d*(20*b^2*c^3 + 78*a^2*c*d^2 - 90*a*b*c^2*d))/(15*c^6) - (4*a*d^2*(7*a*d
- 5*b*c))/(15*c^5)) + (20*b^2*c^3 + 78*a^2*c*d^2 - 90*a*b*c^2*d)/(15*c^5))/(x*(c + d*x^2)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right )^{2}}{x^{6} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**6/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(x**6*(c + d*x**2)**(5/2)), x)

________________________________________________________________________________________