3.2.64 \(\int \frac {A+B x^2+C x^4+D x^6}{x^2 (a+b x^2)^{9/2}} \, dx\) [164]

3.2.64.1 Optimal result
3.2.64.2 Mathematica [A] (verified)
3.2.64.3 Rubi [A] (verified)
3.2.64.4 Maple [A] (verified)
3.2.64.5 Fricas [A] (verification not implemented)
3.2.64.6 Sympy [B] (verification not implemented)
3.2.64.7 Maxima [A] (verification not implemented)
3.2.64.8 Giac [A] (verification not implemented)
3.2.64.9 Mupad [F(-1)]

3.2.64.1 Optimal result

Integrand size = 32, antiderivative size = 185 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\left (48 A b^2-a (6 b B+a C)\right ) x^3}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {\left (4 b \left (48 A b^2-a (6 b B+a C)\right )-3 a^3 D\right ) x^5}{15 a^4 \left (a+b x^2\right )^{7/2}}-\frac {2 b \left (4 b \left (48 A b^2-a (6 b B+a C)\right )-3 a^3 D\right ) x^7}{105 a^5 \left (a+b x^2\right )^{7/2}} \]

output
-A/a/x/(b*x^2+a)^(7/2)-(8*A*b-B*a)*x/a^2/(b*x^2+a)^(7/2)-1/3*(48*A*b^2-a*( 
6*B*b+C*a))*x^3/a^3/(b*x^2+a)^(7/2)-1/15*(4*b*(48*A*b^2-a*(6*B*b+C*a))-3*D 
*a^3)*x^5/a^4/(b*x^2+a)^(7/2)-2/105*b*(4*b*(48*A*b^2-a*(6*B*b+C*a))-3*D*a^ 
3)*x^7/a^5/(b*x^2+a)^(7/2)
 
3.2.64.2 Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=\frac {-384 A b^4 x^8+48 a b^3 x^6 \left (-28 A+B x^2\right )+8 a^2 b^2 x^4 \left (-210 A+21 B x^2+C x^4\right )-7 a^4 \left (15 A-15 B x^2-5 C x^4-3 D x^6\right )+2 a^3 b x^2 \left (-420 A+105 B x^2+14 C x^4+3 D x^6\right )}{105 a^5 x \left (a+b x^2\right )^{7/2}} \]

input
Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(x^2*(a + b*x^2)^(9/2)),x]
 
output
(-384*A*b^4*x^8 + 48*a*b^3*x^6*(-28*A + B*x^2) + 8*a^2*b^2*x^4*(-210*A + 2 
1*B*x^2 + C*x^4) - 7*a^4*(15*A - 15*B*x^2 - 5*C*x^4 - 3*D*x^6) + 2*a^3*b*x 
^2*(-420*A + 105*B*x^2 + 14*C*x^4 + 3*D*x^6))/(105*a^5*x*(a + b*x^2)^(7/2) 
)
 
3.2.64.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2334, 2087, 1469, 362, 245, 242}

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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx\)

\(\Big \downarrow \) 2334

\(\displaystyle -\frac {\int \frac {8 A b-a \left (D x^4+C x^2+B\right )}{\left (b x^2+a\right )^{9/2}}dx}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 2087

\(\displaystyle -\frac {\int \frac {-a D x^4-a C x^2+8 A b-a B}{\left (b x^2+a\right )^{9/2}}dx}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 1469

\(\displaystyle -\frac {\frac {\int \frac {x^2 \left (-D x^2 a^2-C a^2+6 b (8 A b-a B)\right )}{\left (b x^2+a\right )^{9/2}}dx}{a}+\frac {x (8 A b-a B)}{a \left (a+b x^2\right )^{7/2}}}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 362

\(\displaystyle -\frac {\frac {\frac {\left (-3 a^3 D-4 a b (a C+6 b B)+192 A b^3\right ) \int \frac {x^2}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}+\frac {x^3 \left (48 A b^3-a \left (a^2 (-D)+a b C+6 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}}{a}+\frac {x (8 A b-a B)}{a \left (a+b x^2\right )^{7/2}}}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 245

\(\displaystyle -\frac {\frac {\frac {\left (-3 a^3 D-4 a b (a C+6 b B)+192 A b^3\right ) \left (\frac {2 b \int \frac {x^4}{\left (b x^2+a\right )^{7/2}}dx}{3 a}+\frac {x^3}{3 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}+\frac {x^3 \left (48 A b^3-a \left (a^2 (-D)+a b C+6 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}}{a}+\frac {x (8 A b-a B)}{a \left (a+b x^2\right )^{7/2}}}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 242

\(\displaystyle -\frac {\frac {\frac {x^3 \left (48 A b^3-a \left (a^2 (-D)+a b C+6 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}+\frac {\left (\frac {2 b x^5}{15 a^2 \left (a+b x^2\right )^{5/2}}+\frac {x^3}{3 a \left (a+b x^2\right )^{5/2}}\right ) \left (-3 a^3 D-4 a b (a C+6 b B)+192 A b^3\right )}{7 a b}}{a}+\frac {x (8 A b-a B)}{a \left (a+b x^2\right )^{7/2}}}{a}-\frac {A}{a x \left (a+b x^2\right )^{7/2}}\)

input
Int[(A + B*x^2 + C*x^4 + D*x^6)/(x^2*(a + b*x^2)^(9/2)),x]
 
output
-(A/(a*x*(a + b*x^2)^(7/2))) - (((8*A*b - a*B)*x)/(a*(a + b*x^2)^(7/2)) + 
(((48*A*b^3 - a*(6*b^2*B + a*b*C - a^2*D))*x^3)/(7*a*b*(a + b*x^2)^(7/2)) 
+ ((192*A*b^3 - 4*a*b*(6*b*B + a*C) - 3*a^3*D)*(x^3/(3*a*(a + b*x^2)^(5/2) 
) + (2*b*x^5)/(15*a^2*(a + b*x^2)^(5/2))))/(7*a*b))/a)/a
 

3.2.64.3.1 Defintions of rubi rules used

rule 242
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 245
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + 
 b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) 
   Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si 
mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
 

rule 362
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 1469
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[a^p*x*((d + e*x^2)^(q + 1)/d), x] + Simp[1/d   Int[x^2*(d 
 + e*x^2)^q*(d*PolynomialQuotient[(a + b*x^2 + c*x^4)^p - a^p, x^2, x] - e* 
a^p*(2*q + 3)), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4 
*p + 2*q + 1, 0]
 

rule 2087
Int[(u_)^(q_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*ExpandToSum 
[v, x]^p, x] /; FreeQ[{p, q}, x] && BinomialQ[u, x] && TrinomialQ[v, x] && 
 !(BinomialMatchQ[u, x] && TrinomialMatchQ[v, x])
 

rule 2334
Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{A = Coef 
f[Pq, x, 0], Q = PolynomialQuotient[Pq - Coeff[Pq, x, 0], x^2, x]}, Simp[A* 
x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1))   Int[ 
x^(m + 2)*(a + b*x^2)^p*(a*(m + 1)*Q - A*b*(m + 2*(p + 1) + 1)), x], x]] /; 
 FreeQ[{a, b}, x] && PolyQ[Pq, x^2] && IntegerQ[m/2] && ILtQ[(m + 1)/2 + p, 
 0] && LtQ[m + Expon[Pq, x] + 2*p + 1, 0]
 
3.2.64.4 Maple [A] (verified)

Time = 3.55 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.66

method result size
pseudoelliptic \(-\frac {\left (-\frac {1}{5} D x^{6}-\frac {1}{3} C \,x^{4}-x^{2} B +A \right ) a^{4}+8 b \left (-\frac {1}{140} D x^{6}-\frac {1}{30} C \,x^{4}-\frac {1}{4} x^{2} B +A \right ) x^{2} a^{3}+16 \left (-\frac {1}{210} C \,x^{4}-\frac {1}{10} x^{2} B +A \right ) b^{2} x^{4} a^{2}+\frac {64 b^{3} x^{6} \left (-\frac {x^{2} B}{28}+A \right ) a}{5}+\frac {128 A \,b^{4} x^{8}}{35}}{\left (b \,x^{2}+a \right )^{\frac {7}{2}} x \,a^{5}}\) \(123\)
gosper \(-\frac {384 A \,b^{4} x^{8}-48 B a \,b^{3} x^{8}-8 C \,a^{2} b^{2} x^{8}-6 D a^{3} b \,x^{8}+1344 A a \,b^{3} x^{6}-168 B \,a^{2} b^{2} x^{6}-28 C \,a^{3} b \,x^{6}-21 D a^{4} x^{6}+1680 A \,a^{2} b^{2} x^{4}-210 B \,a^{3} b \,x^{4}-35 C \,a^{4} x^{4}+840 A \,a^{3} b \,x^{2}-105 B \,a^{4} x^{2}+105 A \,a^{4}}{105 x \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{5}}\) \(157\)
trager \(-\frac {384 A \,b^{4} x^{8}-48 B a \,b^{3} x^{8}-8 C \,a^{2} b^{2} x^{8}-6 D a^{3} b \,x^{8}+1344 A a \,b^{3} x^{6}-168 B \,a^{2} b^{2} x^{6}-28 C \,a^{3} b \,x^{6}-21 D a^{4} x^{6}+1680 A \,a^{2} b^{2} x^{4}-210 B \,a^{3} b \,x^{4}-35 C \,a^{4} x^{4}+840 A \,a^{3} b \,x^{2}-105 B \,a^{4} x^{2}+105 A \,a^{4}}{105 x \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{5}}\) \(157\)
default \(B \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )+D \left (-\frac {x^{3}}{4 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {3 a \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )}{4 b}\right )+C \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )+A \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {8 b \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{a}\right )\) \(394\)

input
int((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
 
output
-((-1/5*D*x^6-1/3*C*x^4-x^2*B+A)*a^4+8*b*(-1/140*D*x^6-1/30*C*x^4-1/4*x^2* 
B+A)*x^2*a^3+16*(-1/210*C*x^4-1/10*x^2*B+A)*b^2*x^4*a^2+64/5*b^3*x^6*(-1/2 
8*x^2*B+A)*a+128/35*A*b^4*x^8)/(b*x^2+a)^(7/2)/x/a^5
 
3.2.64.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (2 \, {\left (3 \, D a^{3} b + 4 \, C a^{2} b^{2} + 24 \, B a b^{3} - 192 \, A b^{4}\right )} x^{8} + 7 \, {\left (3 \, D a^{4} + 4 \, C a^{3} b + 24 \, B a^{2} b^{2} - 192 \, A a b^{3}\right )} x^{6} - 105 \, A a^{4} + 35 \, {\left (C a^{4} + 6 \, B a^{3} b - 48 \, A a^{2} b^{2}\right )} x^{4} + 105 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}} \]

input
integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="fricas")
 
output
1/105*(2*(3*D*a^3*b + 4*C*a^2*b^2 + 24*B*a*b^3 - 192*A*b^4)*x^8 + 7*(3*D*a 
^4 + 4*C*a^3*b + 24*B*a^2*b^2 - 192*A*a*b^3)*x^6 - 105*A*a^4 + 35*(C*a^4 + 
 6*B*a^3*b - 48*A*a^2*b^2)*x^4 + 105*(B*a^4 - 8*A*a^3*b)*x^2)*sqrt(b*x^2 + 
 a)/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x)
 
3.2.64.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2392 vs. \(2 (170) = 340\).

Time = 77.57 (sec) , antiderivative size = 2392, normalized size of antiderivative = 12.93 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=\text {Too large to display} \]

input
integrate((D*x**6+C*x**4+B*x**2+A)/x**2/(b*x**2+a)**(9/2),x)
 
output
A*(-35*a**4*b**(33/2)*sqrt(a/(b*x**2) + 1)/(35*a**9*b**16 + 140*a**8*b**17 
*x**2 + 210*a**7*b**18*x**4 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**8) - 
280*a**3*b**(35/2)*x**2*sqrt(a/(b*x**2) + 1)/(35*a**9*b**16 + 140*a**8*b** 
17*x**2 + 210*a**7*b**18*x**4 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**8) 
- 560*a**2*b**(37/2)*x**4*sqrt(a/(b*x**2) + 1)/(35*a**9*b**16 + 140*a**8*b 
**17*x**2 + 210*a**7*b**18*x**4 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**8 
) - 448*a*b**(39/2)*x**6*sqrt(a/(b*x**2) + 1)/(35*a**9*b**16 + 140*a**8*b* 
*17*x**2 + 210*a**7*b**18*x**4 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**8) 
 - 128*b**(41/2)*x**8*sqrt(a/(b*x**2) + 1)/(35*a**9*b**16 + 140*a**8*b**17 
*x**2 + 210*a**7*b**18*x**4 + 140*a**6*b**19*x**6 + 35*a**5*b**20*x**8)) + 
 B*(35*a**14*x/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqr 
t(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 700*a**(31/ 
2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b**4*x**8*sqrt(1 + b*x**2/ 
a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) + 35*a**(25/2)*b**6*x**12 
*sqrt(1 + b*x**2/a)) + 175*a**13*b*x**3/(35*a**(37/2)*sqrt(1 + b*x**2/a) + 
 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + 525*a**(33/2)*b**2*x**4*sqrt(1 
+ b*x**2/a) + 700*a**(31/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 525*a**(29/2)*b 
**4*x**8*sqrt(1 + b*x**2/a) + 210*a**(27/2)*b**5*x**10*sqrt(1 + b*x**2/a) 
+ 35*a**(25/2)*b**6*x**12*sqrt(1 + b*x**2/a)) + 371*a**12*b**2*x**5/(35*a* 
*(37/2)*sqrt(1 + b*x**2/a) + 210*a**(35/2)*b*x**2*sqrt(1 + b*x**2/a) + ...
 
3.2.64.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=-\frac {D x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {16 \, B x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {B x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} + \frac {3 \, D x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, D x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {D x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, D a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {C x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, C x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, C x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {128 \, A b x}{35 \, \sqrt {b x^{2} + a} a^{5}} - \frac {64 \, A b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {48 \, A b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3}} - \frac {8 \, A b x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2}} - \frac {A}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} a x} \]

input
integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="maxima")
 
output
-1/4*D*x^3/((b*x^2 + a)^(7/2)*b) + 16/35*B*x/(sqrt(b*x^2 + a)*a^4) + 8/35* 
B*x/((b*x^2 + a)^(3/2)*a^3) + 6/35*B*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*B*x/( 
(b*x^2 + a)^(7/2)*a) + 3/140*D*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*D*x/(sqrt( 
b*x^2 + a)*a^2*b^2) + 1/35*D*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*D*a*x/((b* 
x^2 + a)^(7/2)*b^2) - 1/7*C*x/((b*x^2 + a)^(7/2)*b) + 8/105*C*x/(sqrt(b*x^ 
2 + a)*a^3*b) + 4/105*C*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*C*x/((b*x^2 + a 
)^(5/2)*a*b) - 128/35*A*b*x/(sqrt(b*x^2 + a)*a^5) - 64/35*A*b*x/((b*x^2 + 
a)^(3/2)*a^4) - 48/35*A*b*x/((b*x^2 + a)^(5/2)*a^3) - 8/7*A*b*x/((b*x^2 + 
a)^(7/2)*a^2) - A/((b*x^2 + a)^(7/2)*a*x)
 
3.2.64.8 Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (x^{2} {\left (\frac {{\left (6 \, D a^{12} b^{4} + 8 \, C a^{11} b^{5} + 48 \, B a^{10} b^{6} - 279 \, A a^{9} b^{7}\right )} x^{2}}{a^{14} b^{3}} + \frac {7 \, {\left (3 \, D a^{13} b^{3} + 4 \, C a^{12} b^{4} + 24 \, B a^{11} b^{5} - 132 \, A a^{10} b^{6}\right )}}{a^{14} b^{3}}\right )} + \frac {35 \, {\left (C a^{13} b^{3} + 6 \, B a^{12} b^{4} - 30 \, A a^{11} b^{5}\right )}}{a^{14} b^{3}}\right )} x^{2} + \frac {105 \, {\left (B a^{13} b^{3} - 4 \, A a^{12} b^{4}\right )}}{a^{14} b^{3}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{4}} \]

input
integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="giac")
 
output
1/105*((x^2*((6*D*a^12*b^4 + 8*C*a^11*b^5 + 48*B*a^10*b^6 - 279*A*a^9*b^7) 
*x^2/(a^14*b^3) + 7*(3*D*a^13*b^3 + 4*C*a^12*b^4 + 24*B*a^11*b^5 - 132*A*a 
^10*b^6)/(a^14*b^3)) + 35*(C*a^13*b^3 + 6*B*a^12*b^4 - 30*A*a^11*b^5)/(a^1 
4*b^3))*x^2 + 105*(B*a^13*b^3 - 4*A*a^12*b^4)/(a^14*b^3))*x/(b*x^2 + a)^(7 
/2) + 2*A*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a^4)
 
3.2.64.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{x^2\,{\left (b\,x^2+a\right )}^{9/2}} \,d x \]

input
int((A + B*x^2 + C*x^4 + x^6*D)/(x^2*(a + b*x^2)^(9/2)),x)
 
output
int((A + B*x^2 + C*x^4 + x^6*D)/(x^2*(a + b*x^2)^(9/2)), x)