Integrand size = 25, antiderivative size = 139 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {x^2 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {2 a B+(4 A b+3 a C) x}{35 a b^2 \left (a+b x^2\right )^{5/2}}+\frac {(4 A b+3 a C) x}{105 a^2 b^2 \left (a+b x^2\right )^{3/2}}+\frac {2 (4 A b+3 a C) x}{105 a^3 b^2 \sqrt {a+b x^2}} \]
-1/7*x^2*(B*a-(A*b-C*a)*x)/a/b/(b*x^2+a)^(7/2)+1/35*(-2*B*a-(4*A*b+3*C*a)* x)/a/b^2/(b*x^2+a)^(5/2)+1/105*(4*A*b+3*C*a)*x/a^2/b^2/(b*x^2+a)^(3/2)+2/1 05*(4*A*b+3*C*a)*x/a^3/b^2/(b*x^2+a)^(1/2)
Time = 0.67 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.63 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-6 a^4 B-21 a^3 b B x^2+8 A b^4 x^7+7 a^2 b^2 x^3 \left (5 A+3 C x^2\right )+2 a b^3 x^5 \left (14 A+3 C x^2\right )}{105 a^3 b^2 \left (a+b x^2\right )^{7/2}} \]
(-6*a^4*B - 21*a^3*b*B*x^2 + 8*A*b^4*x^7 + 7*a^2*b^2*x^3*(5*A + 3*C*x^2) + 2*a*b^3*x^5*(14*A + 3*C*x^2))/(105*a^3*b^2*(a + b*x^2)^(7/2))
Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2335, 25, 530, 25, 27, 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.
| \(\displaystyle \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle -\frac {\int -\frac {x (2 a B+(4 A b+3 a C) x)}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x (2 a B+(4 A b+3 a C) x)}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {-\frac {\int -\frac {a \left (4 A+\frac {3 a C}{b}\right )}{\left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {x (3 a C+4 A b)+2 a B}{5 b \left (a+b x^2\right )^{5/2}}}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (4 A+\frac {3 a C}{b}\right )}{\left (b x^2+a\right )^{5/2}}dx}{5 a}-\frac {x (3 a C+4 A b)+2 a B}{5 b \left (a+b x^2\right )^{5/2}}}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {3 a C}{b}+4 A\right ) \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx-\frac {x (3 a C+4 A b)+2 a B}{5 b \left (a+b x^2\right )^{5/2}}}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {3 a C}{b}+4 A\right ) \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )-\frac {x (3 a C+4 A b)+2 a B}{5 b \left (a+b x^2\right )^{5/2}}}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {1}{5} \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right ) \left (\frac {3 a C}{b}+4 A\right )-\frac {x (3 a C+4 A b)+2 a B}{5 b \left (a+b x^2\right )^{5/2}}}{7 a b}-\frac {x^2 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
-1/7*(x^2*(a*B - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(7/2)) + (-1/5*(2*a*B + (4*A*b + 3*a*C)*x)/(b*(a + b*x^2)^(5/2)) + ((4*A + (3*a*C)/b)*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/5)/(7*a*b)
3.1.52.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
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]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq , a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(c*x)^m*(a + b*x^2)^(p + 1)*((a*g - b*f*x)/(2*a*b*(p + 1))), x] + Simp[c/(2*a*b*(p + 1)) Int[(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSu m[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]
Time = 3.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.63
| method | result | size |
| gosper | \(\frac {8 A \,b^{4} x^{7}+6 C a \,x^{7} b^{3}+28 A a \,b^{3} x^{5}+21 C \,a^{2} x^{5} b^{2}+35 A \,a^{2} b^{2} x^{3}-21 B \,a^{3} b \,x^{2}-6 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3} b^{2}}\) | \(88\) |
| trager | \(\frac {8 A \,b^{4} x^{7}+6 C a \,x^{7} b^{3}+28 A a \,b^{3} x^{5}+21 C \,a^{2} x^{5} b^{2}+35 A \,a^{2} b^{2} x^{3}-21 B \,a^{3} b \,x^{2}-6 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3} b^{2}}\) | \(88\) |
| default | \(C \left (-\frac {x^{3}}{4 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {3 a \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )}{4 b}\right )+B \left (-\frac {x^{2}}{5 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {2 a}{35 b^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\right )+A \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )\) | \(255\) |
1/105*(8*A*b^4*x^7+6*C*a*b^3*x^7+28*A*a*b^3*x^5+21*C*a^2*b^2*x^5+35*A*a^2* b^2*x^3-21*B*a^3*b*x^2-6*B*a^4)/(b*x^2+a)^(7/2)/a^3/b^2
Time = 0.36 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (35 \, A a^{2} b^{2} x^{3} + 2 \, {\left (3 \, C a b^{3} + 4 \, A b^{4}\right )} x^{7} - 21 \, B a^{3} b x^{2} + 7 \, {\left (3 \, C a^{2} b^{2} + 4 \, A a b^{3}\right )} x^{5} - 6 \, B a^{4}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{3} b^{6} x^{8} + 4 \, a^{4} b^{5} x^{6} + 6 \, a^{5} b^{4} x^{4} + 4 \, a^{6} b^{3} x^{2} + a^{7} b^{2}\right )}} \]
1/105*(35*A*a^2*b^2*x^3 + 2*(3*C*a*b^3 + 4*A*b^4)*x^7 - 21*B*a^3*b*x^2 + 7 *(3*C*a^2*b^2 + 4*A*a*b^3)*x^5 - 6*B*a^4)*sqrt(b*x^2 + a)/(a^3*b^6*x^8 + 4 *a^4*b^5*x^6 + 6*a^5*b^4*x^4 + 4*a^6*b^3*x^2 + a^7*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (129) = 258\).
Time = 30.65 (sec) , antiderivative size = 904, normalized size of antiderivative = 6.50 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=A \left (\frac {35 a^{5} x^{3}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {63 a^{4} b x^{5}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {36 a^{3} b^{2} x^{7}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {8 a^{2} b^{3} x^{9}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {2 a}{35 a^{3} b^{2} \sqrt {a + b x^{2}} + 105 a^{2} b^{3} x^{2} \sqrt {a + b x^{2}} + 105 a b^{4} x^{4} \sqrt {a + b x^{2}} + 35 b^{5} x^{6} \sqrt {a + b x^{2}}} - \frac {7 b x^{2}}{35 a^{3} b^{2} \sqrt {a + b x^{2}} + 105 a^{2} b^{3} x^{2} \sqrt {a + b x^{2}} + 105 a b^{4} x^{4} \sqrt {a + b x^{2}} + 35 b^{5} x^{6} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {9}{2}}} & \text {otherwise} \end {cases}\right ) + C \left (\frac {7 a x^{5}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{7}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]
A*(35*a**5*x**3/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*s qrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(1 3/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x** 2/a)) + 63*a**4*b*x**5/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b *x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 42 0*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a)) + 36*a**3*b**2*x**7/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a **(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x* *2/a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x* *8*sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**9/(105*a**(19/2)*sqrt(1 + b*x**2/a ) + 420*a**(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt (1 + b*x**2/a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2 )*b**4*x**8*sqrt(1 + b*x**2/a))) + B*Piecewise((-2*a/(35*a**3*b**2*sqrt(a + b*x**2) + 105*a**2*b**3*x**2*sqrt(a + b*x**2) + 105*a*b**4*x**4*sqrt(a + b*x**2) + 35*b**5*x**6*sqrt(a + b*x**2)) - 7*b*x**2/(35*a**3*b**2*sqrt(a + b*x**2) + 105*a**2*b**3*x**2*sqrt(a + b*x**2) + 105*a*b**4*x**4*sqrt(a + b*x**2) + 35*b**5*x**6*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(9/2)), True)) + C*(7*a*x**5/(35*a**(11/2)*sqrt(1 + b*x**2/a) + 105*a**(9/2)*b*x** 2*sqrt(1 + b*x**2/a) + 105*a**(7/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 35*a**( 5/2)*b**3*x**6*sqrt(1 + b*x**2/a)) + 2*b*x**7/(35*a**(11/2)*sqrt(1 + b*...
Time = 0.20 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.42 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {C x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {B x^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {3 \, C x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, C x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, C a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, A x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, A x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {2 \, B a}{35 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} \]
-1/4*C*x^3/((b*x^2 + a)^(7/2)*b) - 1/5*B*x^2/((b*x^2 + a)^(7/2)*b) + 3/140 *C*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*C*x/(sqrt(b*x^2 + a)*a^2*b^2) + 1/35*C *x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*C*a*x/((b*x^2 + a)^(7/2)*b^2) - 1/7*A* x/((b*x^2 + a)^(7/2)*b) + 8/105*A*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*A*x/(( b*x^2 + a)^(3/2)*a^2*b) + 1/35*A*x/((b*x^2 + a)^(5/2)*a*b) - 2/35*B*a/((b* x^2 + a)^(7/2)*b^2)
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.68 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (x^{2} {\left (\frac {2 \, {\left (3 \, C a b^{4} + 4 \, A b^{5}\right )} x^{2}}{a^{3} b^{3}} + \frac {7 \, {\left (3 \, C a^{2} b^{3} + 4 \, A a b^{4}\right )}}{a^{3} b^{3}}\right )} + \frac {35 \, A}{a}\right )} x - \frac {21 \, B}{b}\right )} x^{2} - \frac {6 \, B a}{b^{2}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \]
1/105*(((x^2*(2*(3*C*a*b^4 + 4*A*b^5)*x^2/(a^3*b^3) + 7*(3*C*a^2*b^3 + 4*A *a*b^4)/(a^3*b^3)) + 35*A/a)*x - 21*B/b)*x^2 - 6*B*a/b^2)/(b*x^2 + a)^(7/2 )
Time = 5.48 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {x\,\left (4\,A\,b+3\,C\,a\right )}{105\,a^2\,b^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {\frac {B}{5\,b^2}+x\,\left (\frac {C}{5\,b^2}-\frac {A\,b-C\,a}{35\,a\,b^2}\right )}{{\left (b\,x^2+a\right )}^{5/2}}-\frac {x\,\left (\frac {A}{7\,b}-\frac {C\,a}{7\,b^2}\right )-\frac {B\,a}{7\,b^2}}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {x\,\left (8\,A\,b+6\,C\,a\right )}{105\,a^3\,b^2\,\sqrt {b\,x^2+a}} \]