\(\int \frac {A+B x^2+C x^4+B x^6}{x^4 (a+b x^2)^{9/2}} \, dx\) [245]

Optimal result

Integrand size = 32, antiderivative size = 246 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {\left (A b^3-a \left (a^2 B+b^2 B-a b C\right )\right ) x}{7 a^3 b \left (a+b x^2\right )^{7/2}}+\frac {\left (20 A b^3+a \left (a^2 B-13 b^2 B+6 a b C\right )\right ) x}{35 a^4 b \left (a+b x^2\right )^{5/2}}+\frac {\left (185 A b^3+a \left (4 a^2 B-87 b^2 B+24 a b C\right )\right ) x}{105 a^5 b \left (a+b x^2\right )^{3/2}}+\frac {\left (790 A b^3+a \left (8 a^2 B-279 b^2 B+48 a b C\right )\right ) x}{105 a^6 b \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{3 a^5 x^3}+\frac {(14 A b-3 a B) \sqrt {a+b x^2}}{3 a^6 x} \] Output:

1/7*(A*b^3-a*(B*a^2+B*b^2-C*a*b))*x/a^3/b/(b*x^2+a)^(7/2)+1/35*(20*A*b^3+a 
*(B*a^2-13*B*b^2+6*C*a*b))*x/a^4/b/(b*x^2+a)^(5/2)+1/105*(185*A*b^3+a*(4*B 
*a^2-87*B*b^2+24*C*a*b))*x/a^5/b/(b*x^2+a)^(3/2)+1/105*(790*A*b^3+a*(8*B*a 
^2-279*B*b^2+48*C*a*b))*x/a^6/b/(b*x^2+a)^(1/2)-1/3*A*(b*x^2+a)^(1/2)/a^5/ 
x^3+1/3*(14*A*b-3*B*a)*(b*x^2+a)^(1/2)/a^6/x
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {1280 A b^5 x^{10}+128 a b^4 x^8 \left (35 A-3 B x^2\right )+16 a^2 b^3 x^6 \left (350 A-84 B x^2+3 C x^4\right )+8 a^3 b^2 x^4 \left (350 A+21 C x^4+B x^2 \left (-210+x^4\right )\right )+14 a^4 b x^2 \left (25 A+15 C x^4+2 B x^2 \left (-30+x^4\right )\right )-35 a^5 \left (A-3 C x^4-B x^2 \left (-3+x^4\right )\right )}{105 a^6 x^3 \left (a+b x^2\right )^{7/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + B*x^6)/(x^4*(a + b*x^2)^(9/2)),x]
 

Output:

(1280*A*b^5*x^10 + 128*a*b^4*x^8*(35*A - 3*B*x^2) + 16*a^2*b^3*x^6*(350*A 
- 84*B*x^2 + 3*C*x^4) + 8*a^3*b^2*x^4*(350*A + 21*C*x^4 + B*x^2*(-210 + x^ 
4)) + 14*a^4*b*x^2*(25*A + 15*C*x^4 + 2*B*x^2*(-30 + x^4)) - 35*a^5*(A - 3 
*C*x^4 - B*x^2*(-3 + x^4)))/(105*a^6*x^3*(a + b*x^2)^(7/2))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2334, 2089, 1588, 298, 209, 209, 208}

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.

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

\(\Big \downarrow \) 2334

\(\displaystyle -\frac {\int \frac {10 A b-3 a \left (B x^4+C x^2+B\right )}{x^2 \left (b x^2+a\right )^{9/2}}dx}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 2089

\(\displaystyle -\frac {\int \frac {-3 a B x^4-3 a C x^2+10 A b-3 a B}{x^2 \left (b x^2+a\right )^{9/2}}dx}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 1588

\(\displaystyle -\frac {-\frac {\int \frac {80 A b^2+3 a^2 B x^2-3 a (8 b B-a C)}{\left (b x^2+a\right )^{9/2}}dx}{a}-\frac {10 A b-3 a B}{a x \left (a+b x^2\right )^{7/2}}}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 298

\(\displaystyle -\frac {-\frac {\frac {3 \left (a \left (a^2 B+6 a b C-48 b^2 B\right )+160 A b^3\right ) \int \frac {1}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}+\frac {x \left (80 A b^3-3 a \left (a^2 B-a b C+8 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}}{a}-\frac {10 A b-3 a B}{a x \left (a+b x^2\right )^{7/2}}}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 209

\(\displaystyle -\frac {-\frac {\frac {3 \left (a \left (a^2 B+6 a b C-48 b^2 B\right )+160 A b^3\right ) \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}+\frac {x \left (80 A b^3-3 a \left (a^2 B-a b C+8 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}}{a}-\frac {10 A b-3 a B}{a x \left (a+b x^2\right )^{7/2}}}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 209

\(\displaystyle -\frac {-\frac {\frac {3 \left (a \left (a^2 B+6 a b C-48 b^2 B\right )+160 A b^3\right ) \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 a b}+\frac {x \left (80 A b^3-3 a \left (a^2 B-a b C+8 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}}{a}-\frac {10 A b-3 a B}{a x \left (a+b x^2\right )^{7/2}}}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

\(\Big \downarrow \) 208

\(\displaystyle -\frac {-\frac {\frac {x \left (80 A b^3-3 a \left (a^2 B-a b C+8 b^2 B\right )\right )}{7 a b \left (a+b x^2\right )^{7/2}}+\frac {3 \left (\frac {4 \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right ) \left (a \left (a^2 B+6 a b C-48 b^2 B\right )+160 A b^3\right )}{7 a b}}{a}-\frac {10 A b-3 a B}{a x \left (a+b x^2\right )^{7/2}}}{3 a}-\frac {A}{3 a x^3 \left (a+b x^2\right )^{7/2}}\)

Input:

Int[(A + B*x^2 + C*x^4 + B*x^6)/(x^4*(a + b*x^2)^(9/2)),x]
 

Output:

-1/3*A/(a*x^3*(a + b*x^2)^(7/2)) - (-((10*A*b - 3*a*B)/(a*x*(a + b*x^2)^(7 
/2))) - (((80*A*b^3 - 3*a*(a^2*B + 8*b^2*B - a*b*C))*x)/(7*a*b*(a + b*x^2) 
^(7/2)) + (3*(160*A*b^3 + a*(a^2*B - 48*b^2*B + 6*a*b*C))*(x/(5*a*(a + b*x 
^2)^(5/2)) + (4*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2]) 
))/(5*a)))/(7*a*b))/a)/(3*a)
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 298

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

rule 1588

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c 
_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + 
c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, 
 Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f 
^2*(m + 1))   Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x 
) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne 
Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]
 

rule 2089

Int[(u_)^(p_.)*((f_.)*(x_))^(m_.)*(z_)^(q_.), x_Symbol] :> Int[(f*x)^m*Expa 
ndToSum[z, x]^q*ExpandToSum[u, x]^p, x] /; FreeQ[{f, m, p, q}, x] && Binomi 
alQ[z, x] && TrinomialQ[u, x] &&  !(BinomialMatchQ[z, x] && TrinomialMatchQ 
[u, 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]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.63

method	result	size
pseudoelliptic	\(-\frac {\left (-B \,x^{6}-3 C \,x^{4}+3 x^{2} B +A \right ) a^{5}-10 \left (\frac {2}{25} B \,x^{6}+\frac {3}{5} C \,x^{4}-\frac {12}{5} x^{2} B +A \right ) x^{2} b \,a^{4}-80 \left (\frac {1}{350} B \,x^{6}+\frac {3}{50} C \,x^{4}-\frac {3}{5} x^{2} B +A \right ) x^{4} b^{2} a^{3}-160 \left (\frac {3}{350} C \,x^{4}-\frac {6}{25} x^{2} B +A \right ) x^{6} b^{3} a^{2}-128 \left (-\frac {3 x^{2} B}{35}+A \right ) x^{8} b^{4} a -\frac {256 A \,b^{5} x^{10}}{7}}{3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} x^{3} a^{6}}\)	\(154\)
gosper	\(-\frac {-1280 A \,b^{5} x^{10}-8 B \,a^{3} b^{2} x^{10}+384 B a \,b^{4} x^{10}-48 C \,a^{2} b^{3} x^{10}-4480 a A \,b^{4} x^{8}-28 B \,a^{4} b \,x^{8}+1344 B \,a^{2} b^{3} x^{8}-168 C \,a^{3} b^{2} x^{8}-5600 a^{2} A \,b^{3} x^{6}-35 B \,a^{5} x^{6}+1680 B \,a^{3} b^{2} x^{6}-210 C \,a^{4} b \,x^{6}-2800 a^{3} A \,b^{2} x^{4}+840 B \,a^{4} b \,x^{4}-105 C \,a^{5} x^{4}-350 a^{4} A b \,x^{2}+105 B \,a^{5} x^{2}+35 a^{5} A}{105 x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{6}}\)	\(205\)
trager	\(-\frac {-1280 A \,b^{5} x^{10}-8 B \,a^{3} b^{2} x^{10}+384 B a \,b^{4} x^{10}-48 C \,a^{2} b^{3} x^{10}-4480 a A \,b^{4} x^{8}-28 B \,a^{4} b \,x^{8}+1344 B \,a^{2} b^{3} x^{8}-168 C \,a^{3} b^{2} x^{8}-5600 a^{2} A \,b^{3} x^{6}-35 B \,a^{5} x^{6}+1680 B \,a^{3} b^{2} x^{6}-210 C \,a^{4} b \,x^{6}-2800 a^{3} A \,b^{2} x^{4}+840 B \,a^{4} b \,x^{4}-105 C \,a^{5} x^{4}-350 a^{4} A b \,x^{2}+105 B \,a^{5} x^{2}+35 a^{5} A}{105 x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{6}}\)	\(205\)
orering	\(-\frac {-1280 A \,b^{5} x^{10}-8 B \,a^{3} b^{2} x^{10}+384 B a \,b^{4} x^{10}-48 C \,a^{2} b^{3} x^{10}-4480 a A \,b^{4} x^{8}-28 B \,a^{4} b \,x^{8}+1344 B \,a^{2} b^{3} x^{8}-168 C \,a^{3} b^{2} x^{8}-5600 a^{2} A \,b^{3} x^{6}-35 B \,a^{5} x^{6}+1680 B \,a^{3} b^{2} x^{6}-210 C \,a^{4} b \,x^{6}-2800 a^{3} A \,b^{2} x^{4}+840 B \,a^{4} b \,x^{4}-105 C \,a^{5} x^{4}-350 a^{4} A b \,x^{2}+105 B \,a^{5} x^{2}+35 a^{5} A}{105 x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{6}}\)	\(205\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-14 A b \,x^{2}+3 B a \,x^{2}+A a \right )}{3 a^{6} x^{3}}+\frac {\sqrt {b \,x^{2}+a}\, x \left (790 b^{5} A \,x^{6}+8 B \,a^{3} b^{2} x^{6}-279 B a \,b^{4} x^{6}+48 C \,a^{2} b^{3} x^{6}+2555 a A \,b^{4} x^{4}+28 B \,a^{4} b \,x^{4}-924 B \,a^{2} b^{3} x^{4}+168 a^{3} C \,b^{2} x^{4}+2800 A \,a^{2} b^{3} x^{2}+35 B \,a^{5} x^{2}-1050 B \,a^{3} b^{2} x^{2}+210 a^{4} b \,x^{2} C +1050 a^{3} b^{2} A -420 a^{4} b B +105 a^{5} C \right )}{105 a^{6} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )}\)	\(247\)
default	\(C \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )+A \left (-\frac {1}{3 a \,x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {10 b \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {8 b \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{a}\right )}{3 a}\right )+B \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {8 b \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{a}\right )+B \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )\)	\(396\)

Input:

int((B*x^6+C*x^4+B*x^2+A)/x^4/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3*((-B*x^6-3*C*x^4+3*B*x^2+A)*a^5-10*(2/25*B*x^6+3/5*C*x^4-12/5*x^2*B+A 
)*x^2*b*a^4-80*(1/350*B*x^6+3/50*C*x^4-3/5*x^2*B+A)*x^4*b^2*a^3-160*(3/350 
*C*x^4-6/25*x^2*B+A)*x^6*b^3*a^2-128*(-3/35*x^2*B+A)*x^8*b^4*a-256/7*A*b^5 
*x^10)/(b*x^2+a)^(7/2)/x^3/a^6
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, {\left (B a^{3} b^{2} + 6 \, C a^{2} b^{3} - 48 \, B a b^{4} + 160 \, A b^{5}\right )} x^{10} + 28 \, {\left (B a^{4} b + 6 \, C a^{3} b^{2} - 48 \, B a^{2} b^{3} + 160 \, A a b^{4}\right )} x^{8} + 35 \, {\left (B a^{5} + 6 \, C a^{4} b - 48 \, B a^{3} b^{2} + 160 \, A a^{2} b^{3}\right )} x^{6} - 35 \, A a^{5} + 35 \, {\left (3 \, C a^{5} - 24 \, B a^{4} b + 80 \, A a^{3} b^{2}\right )} x^{4} - 35 \, {\left (3 \, B a^{5} - 10 \, A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{6} b^{4} x^{11} + 4 \, a^{7} b^{3} x^{9} + 6 \, a^{8} b^{2} x^{7} + 4 \, a^{9} b x^{5} + a^{10} x^{3}\right )}} \] Input:

integrate((B*x^6+C*x^4+B*x^2+A)/x^4/(b*x^2+a)^(9/2),x, algorithm="fricas")
 

Output:

1/105*(8*(B*a^3*b^2 + 6*C*a^2*b^3 - 48*B*a*b^4 + 160*A*b^5)*x^10 + 28*(B*a 
^4*b + 6*C*a^3*b^2 - 48*B*a^2*b^3 + 160*A*a*b^4)*x^8 + 35*(B*a^5 + 6*C*a^4 
*b - 48*B*a^3*b^2 + 160*A*a^2*b^3)*x^6 - 35*A*a^5 + 35*(3*C*a^5 - 24*B*a^4 
*b + 80*A*a^3*b^2)*x^4 - 35*(3*B*a^5 - 10*A*a^4*b)*x^2)*sqrt(b*x^2 + a)/(a 
^6*b^4*x^11 + 4*a^7*b^3*x^9 + 6*a^8*b^2*x^7 + 4*a^9*b*x^5 + a^10*x^3)
 

Sympy [F(-1)]

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

integrate((B*x**6+C*x**4+B*x**2+A)/x**4/(b*x**2+a)**(9/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {16 \, C x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {C x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} - \frac {B x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, B x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, B x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {128 \, B b x}{35 \, \sqrt {b x^{2} + a} a^{5}} - \frac {64 \, B b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {48 \, B b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3}} - \frac {8 \, B b x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2}} + \frac {256 \, A b^{2} x}{21 \, \sqrt {b x^{2} + a} a^{6}} + \frac {128 \, A b^{2} x}{21 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{5}} + \frac {32 \, A b^{2} x}{7 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{4}} + \frac {80 \, A b^{2} x}{21 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{3}} - \frac {B}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} a x} + \frac {10 \, A b}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2} x} - \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a x^{3}} \] Input:

integrate((B*x^6+C*x^4+B*x^2+A)/x^4/(b*x^2+a)^(9/2),x, algorithm="maxima")
 

Output:

16/35*C*x/(sqrt(b*x^2 + a)*a^4) + 8/35*C*x/((b*x^2 + a)^(3/2)*a^3) + 6/35* 
C*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*C*x/((b*x^2 + a)^(7/2)*a) - 1/7*B*x/((b* 
x^2 + a)^(7/2)*b) + 8/105*B*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*B*x/((b*x^2 
+ a)^(3/2)*a^2*b) + 1/35*B*x/((b*x^2 + a)^(5/2)*a*b) - 128/35*B*b*x/(sqrt( 
b*x^2 + a)*a^5) - 64/35*B*b*x/((b*x^2 + a)^(3/2)*a^4) - 48/35*B*b*x/((b*x^ 
2 + a)^(5/2)*a^3) - 8/7*B*b*x/((b*x^2 + a)^(7/2)*a^2) + 256/21*A*b^2*x/(sq 
rt(b*x^2 + a)*a^6) + 128/21*A*b^2*x/((b*x^2 + a)^(3/2)*a^5) + 32/7*A*b^2*x 
/((b*x^2 + a)^(5/2)*a^4) + 80/21*A*b^2*x/((b*x^2 + a)^(7/2)*a^3) - B/((b*x 
^2 + a)^(7/2)*a*x) + 10/3*A*b/((b*x^2 + a)^(7/2)*a^2*x) - 1/3*A/((b*x^2 + 
a)^(7/2)*a*x^3)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.42 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (x^{2} {\left (\frac {{\left (8 \, B a^{15} b^{5} + 48 \, C a^{14} b^{6} - 279 \, B a^{13} b^{7} + 790 \, A a^{12} b^{8}\right )} x^{2}}{a^{18} b^{3}} + \frac {7 \, {\left (4 \, B a^{16} b^{4} + 24 \, C a^{15} b^{5} - 132 \, B a^{14} b^{6} + 365 \, A a^{13} b^{7}\right )}}{a^{18} b^{3}}\right )} + \frac {35 \, {\left (B a^{17} b^{3} + 6 \, C a^{16} b^{4} - 30 \, B a^{15} b^{5} + 80 \, A a^{14} b^{6}\right )}}{a^{18} b^{3}}\right )} x^{2} + \frac {105 \, {\left (C a^{17} b^{3} - 4 \, B a^{16} b^{4} + 10 \, A a^{15} b^{5}\right )}}{a^{18} b^{3}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a \sqrt {b} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a b^{\frac {3}{2}} + 3 \, B a^{3} \sqrt {b} - 14 \, A a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{5}} \] Input:

integrate((B*x^6+C*x^4+B*x^2+A)/x^4/(b*x^2+a)^(9/2),x, algorithm="giac")
 

Output:

1/105*((x^2*((8*B*a^15*b^5 + 48*C*a^14*b^6 - 279*B*a^13*b^7 + 790*A*a^12*b 
^8)*x^2/(a^18*b^3) + 7*(4*B*a^16*b^4 + 24*C*a^15*b^5 - 132*B*a^14*b^6 + 36 
5*A*a^13*b^7)/(a^18*b^3)) + 35*(B*a^17*b^3 + 6*C*a^16*b^4 - 30*B*a^15*b^5 
+ 80*A*a^14*b^6)/(a^18*b^3))*x^2 + 105*(C*a^17*b^3 - 4*B*a^16*b^4 + 10*A*a 
^15*b^5)/(a^18*b^3))*x/(b*x^2 + a)^(7/2) + 2/3*(3*(sqrt(b)*x - sqrt(b*x^2 
+ a))^4*B*a*sqrt(b) - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*b^(3/2) - 6*(sq 
rt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*sqrt(b) + 30*(sqrt(b)*x - sqrt(b*x^2 + 
a))^2*A*a*b^(3/2) + 3*B*a^3*sqrt(b) - 14*A*a^2*b^(3/2))/(((sqrt(b)*x - sqr 
t(b*x^2 + a))^2 - a)^3*a^5)
 

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {\frac {4\,B\,a^3+24\,C\,a^2\,b-192\,B\,a\,b^2+640\,A\,b^3}{105\,a^5\,b^2}+\frac {2\,x^2\,\left (4\,B\,a^3+24\,C\,a^2\,b-192\,B\,a\,b^2+640\,A\,b^3\right )}{105\,a^6\,b}}{x\,\sqrt {b\,x^2+a}}+\frac {x\,\left (\frac {A\,b^2}{7\,a^3}-\frac {a\,\left (\frac {B\,b^2}{7\,a^3}-\frac {a\,\left (\frac {C\,b^2}{7\,a^3}-\frac {B\,b}{7\,a^2}\right )}{b}\right )}{b}\right )}{{\left (b\,x^2+a\right )}^{7/2}}-\frac {x^2\,\left (\frac {a\,\left (\frac {B\,a^3+6\,C\,a^2\,b-6\,B\,a\,b^2+6\,A\,b^3}{35\,a^4\,b}+\frac {8\,b\,\left (A\,b-B\,a\right )}{15\,a^4}\right )}{b}-\frac {A\,b-B\,a}{3\,a^3}+\frac {8\,A\,b}{15\,a^3}\right )+\frac {A}{3\,a^2}}{x^3\,{\left (b\,x^2+a\right )}^{5/2}}-\frac {B\,a^3+6\,C\,a^2\,b-48\,B\,a\,b^2+160\,A\,b^3}{105\,a^4\,b^2\,x\,{\left (b\,x^2+a\right )}^{3/2}} \] Input:

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

Output:

((640*A*b^3 + 4*B*a^3 - 192*B*a*b^2 + 24*C*a^2*b)/(105*a^5*b^2) + (2*x^2*( 
640*A*b^3 + 4*B*a^3 - 192*B*a*b^2 + 24*C*a^2*b))/(105*a^6*b))/(x*(a + b*x^ 
2)^(1/2)) + (x*((A*b^2)/(7*a^3) - (a*((B*b^2)/(7*a^3) - (a*((C*b^2)/(7*a^3 
) - (B*b)/(7*a^2)))/b))/b))/(a + b*x^2)^(7/2) - (x^2*((a*((6*A*b^3 + B*a^3 
 - 6*B*a*b^2 + 6*C*a^2*b)/(35*a^4*b) + (8*b*(A*b - B*a))/(15*a^4)))/b - (A 
*b - B*a)/(3*a^3) + (8*A*b)/(15*a^3)) + A/(3*a^2))/(x^3*(a + b*x^2)^(5/2)) 
 - (160*A*b^3 + B*a^3 - 48*B*a*b^2 + 6*C*a^2*b)/(105*a^4*b^2*x*(a + b*x^2) 
^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.93 \[ \int \frac {A+B x^2+C x^4+B x^6}{x^4 \left (a+b x^2\right )^{9/2}} \, dx=\frac {-35 \sqrt {b \,x^{2}+a}\, a^{5} b -896 \sqrt {b}\, b^{6} x^{11}+896 \sqrt {b \,x^{2}+a}\, b^{6} x^{10}-8 \sqrt {b}\, a^{6} x^{3}+35 \sqrt {b \,x^{2}+a}\, a^{4} b^{2} x^{6}+28 \sqrt {b \,x^{2}+a}\, a^{3} b^{3} x^{8}+8 \sqrt {b \,x^{2}+a}\, a^{2} b^{4} x^{10}-32 \sqrt {b}\, a^{5} b \,x^{5}-48 \sqrt {b}\, a^{5} c \,x^{3}-48 \sqrt {b}\, a^{4} b^{2} x^{7}-32 \sqrt {b}\, a^{3} b^{3} x^{9}-8 \sqrt {b}\, a^{2} b^{4} x^{11}+245 \sqrt {b \,x^{2}+a}\, a^{4} b^{2} x^{2}+1960 \sqrt {b \,x^{2}+a}\, a^{3} b^{3} x^{4}+3920 \sqrt {b \,x^{2}+a}\, a^{2} b^{4} x^{6}+3136 \sqrt {b \,x^{2}+a}\, a \,b^{5} x^{8}-896 \sqrt {b}\, a^{4} b^{2} x^{3}-3584 \sqrt {b}\, a^{3} b^{3} x^{5}-5376 \sqrt {b}\, a^{2} b^{4} x^{7}-3584 \sqrt {b}\, a \,b^{5} x^{9}+105 \sqrt {b \,x^{2}+a}\, a^{4} b c \,x^{4}+210 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} c \,x^{6}+168 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} c \,x^{8}+48 \sqrt {b \,x^{2}+a}\, a \,b^{4} c \,x^{10}-192 \sqrt {b}\, a^{4} b c \,x^{5}-288 \sqrt {b}\, a^{3} b^{2} c \,x^{7}-192 \sqrt {b}\, a^{2} b^{3} c \,x^{9}-48 \sqrt {b}\, a \,b^{4} c \,x^{11}}{105 a^{5} b \,x^{3} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

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

Output:

( - 35*sqrt(a + b*x**2)*a**5*b + 35*sqrt(a + b*x**2)*a**4*b**2*x**6 + 245* 
sqrt(a + b*x**2)*a**4*b**2*x**2 + 105*sqrt(a + b*x**2)*a**4*b*c*x**4 + 28* 
sqrt(a + b*x**2)*a**3*b**3*x**8 + 1960*sqrt(a + b*x**2)*a**3*b**3*x**4 + 2 
10*sqrt(a + b*x**2)*a**3*b**2*c*x**6 + 8*sqrt(a + b*x**2)*a**2*b**4*x**10 
+ 3920*sqrt(a + b*x**2)*a**2*b**4*x**6 + 168*sqrt(a + b*x**2)*a**2*b**3*c* 
x**8 + 3136*sqrt(a + b*x**2)*a*b**5*x**8 + 48*sqrt(a + b*x**2)*a*b**4*c*x* 
*10 + 896*sqrt(a + b*x**2)*b**6*x**10 - 8*sqrt(b)*a**6*x**3 - 32*sqrt(b)*a 
**5*b*x**5 - 48*sqrt(b)*a**5*c*x**3 - 48*sqrt(b)*a**4*b**2*x**7 - 896*sqrt 
(b)*a**4*b**2*x**3 - 192*sqrt(b)*a**4*b*c*x**5 - 32*sqrt(b)*a**3*b**3*x**9 
 - 3584*sqrt(b)*a**3*b**3*x**5 - 288*sqrt(b)*a**3*b**2*c*x**7 - 8*sqrt(b)* 
a**2*b**4*x**11 - 5376*sqrt(b)*a**2*b**4*x**7 - 192*sqrt(b)*a**2*b**3*c*x* 
*9 - 3584*sqrt(b)*a*b**5*x**9 - 48*sqrt(b)*a*b**4*c*x**11 - 896*sqrt(b)*b* 
*6*x**11)/(105*a**5*b*x**3*(a**4 + 4*a**3*b*x**2 + 6*a**2*b**2*x**4 + 4*a* 
b**3*x**6 + b**4*x**8))