Integrand size = 25, antiderivative size = 149 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {x^4 (a B-(A b-a C) x)}{7 a b \left (a+b x^2\right )^{7/2}}-\frac {x^2 (4 a B+(2 A b+5 a C) x)}{35 a b^2 \left (a+b x^2\right )^{5/2}}-\frac {8 a B+3 (2 A b+5 a C) x}{105 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 A b+5 a C) x}{35 a^2 b^3 \sqrt {a+b x^2}} \]
-1/7*x^4*(B*a-(A*b-C*a)*x)/a/b/(b*x^2+a)^(7/2)-1/35*x^2*(4*B*a+(2*A*b+5*C* a)*x)/a/b^2/(b*x^2+a)^(5/2)+1/105*(-8*B*a-3*(2*A*b+5*C*a)*x)/a/b^3/(b*x^2+ a)^(3/2)+1/35*(2*A*b+5*C*a)*x/a^2/b^3/(b*x^2+a)^(1/2)
Time = 0.71 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.53 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-8 a^4 B-28 a^3 b B x^2-35 a^2 b^2 B x^4+21 a A b^3 x^5+6 A b^4 x^7+15 a b^3 C x^7}{105 a^2 b^3 \left (a+b x^2\right )^{7/2}} \]
(-8*a^4*B - 28*a^3*b*B*x^2 - 35*a^2*b^2*B*x^4 + 21*a*A*b^3*x^5 + 6*A*b^4*x ^7 + 15*a*b^3*C*x^7)/(105*a^2*b^3*(a + b*x^2)^(7/2))
Time = 0.45 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2335, 25, 530, 2345, 27, 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.
| \(\displaystyle \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx\) |
| \(\Big \downarrow \) 2335 |
| \(\displaystyle -\frac {\int -\frac {x^3 (4 a B+(2 A b+5 a C) x)}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {x^3 (4 a B+(2 A b+5 a C) x)}{\left (b x^2+a\right )^{7/2}}dx}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 530 |
| \(\displaystyle \frac {\frac {a \left (b x \left (\frac {5 a C}{b}+2 A\right )+4 a B\right )}{5 b^2 \left (a+b x^2\right )^{5/2}}-\frac {\int \frac {\frac {(2 A b+5 a C) a^2}{b^2}-\frac {20 B x a^2}{b}-5 \left (2 A+\frac {5 a C}{b}\right ) x^2 a}{\left (b x^2+a\right )^{5/2}}dx}{5 a}}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {\frac {a \left (b x \left (\frac {5 a C}{b}+2 A\right )+4 a B\right )}{5 b^2 \left (a+b x^2\right )^{5/2}}-\frac {\frac {2 a \left (3 b x \left (\frac {5 a C}{b}+2 A\right )+10 a B\right )}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {3 a^2 (2 A b+5 a C)}{b^2 \left (b x^2+a\right )^{3/2}}dx}{3 a}}{5 a}}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a \left (b x \left (\frac {5 a C}{b}+2 A\right )+4 a B\right )}{5 b^2 \left (a+b x^2\right )^{5/2}}-\frac {\frac {2 a \left (3 b x \left (\frac {5 a C}{b}+2 A\right )+10 a B\right )}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {a (5 a C+2 A b) \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{b^2}}{5 a}}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {a \left (b x \left (\frac {5 a C}{b}+2 A\right )+4 a B\right )}{5 b^2 \left (a+b x^2\right )^{5/2}}-\frac {\frac {2 a \left (3 b x \left (\frac {5 a C}{b}+2 A\right )+10 a B\right )}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {x (5 a C+2 A b)}{b^2 \sqrt {a+b x^2}}}{5 a}}{7 a b}-\frac {x^4 (a B-x (A b-a C))}{7 a b \left (a+b x^2\right )^{7/2}}\) |
-1/7*(x^4*(a*B - (A*b - a*C)*x))/(a*b*(a + b*x^2)^(7/2)) + ((a*(4*a*B + b* (2*A + (5*a*C)/b)*x))/(5*b^2*(a + b*x^2)^(5/2)) - ((2*a*(10*a*B + 3*b*(2*A + (5*a*C)/b)*x))/(3*b^2*(a + b*x^2)^(3/2)) - ((2*A*b + 5*a*C)*x)/(b^2*Sqr t[a + b*x^2]))/(5*a))/(7*a*b)
3.1.50.3.1 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[(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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[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[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 3.49 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.51
| method | result | size |
| gosper | \(\frac {6 A \,b^{4} x^{7}+15 C a \,x^{7} b^{3}+21 A a \,b^{3} x^{5}-35 a^{2} B \,b^{2} x^{4}-28 B \,a^{3} b \,x^{2}-8 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}}\) | \(76\) |
| trager | \(\frac {6 A \,b^{4} x^{7}+15 C a \,x^{7} b^{3}+21 A a \,b^{3} x^{5}-35 a^{2} B \,b^{2} x^{4}-28 B \,a^{3} b \,x^{2}-8 B \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{2} b^{3}}\) | \(76\) |
| default | \(C \left (-\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {5 a \left (-\frac {x^{3}}{4 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {3 a \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right )+B \left (-\frac {x^{4}}{3 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {4 a \left (-\frac {x^{2}}{5 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}-\frac {2 a}{35 b^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}\right )}{3 b}\right )+A \left (-\frac {x^{3}}{4 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {3 a \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )}{4 b}\right )\) | \(327\) |
1/105*(6*A*b^4*x^7+15*C*a*b^3*x^7+21*A*a*b^3*x^5-35*B*a^2*b^2*x^4-28*B*a^3 *b*x^2-8*B*a^4)/(b*x^2+a)^(7/2)/a^2/b^3
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (21 \, A a b^{3} x^{5} - 35 \, B a^{2} b^{2} x^{4} + 3 \, {\left (5 \, C a b^{3} + 2 \, A b^{4}\right )} x^{7} - 28 \, B a^{3} b x^{2} - 8 \, B a^{4}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{2} b^{7} x^{8} + 4 \, a^{3} b^{6} x^{6} + 6 \, a^{4} b^{5} x^{4} + 4 \, a^{5} b^{4} x^{2} + a^{6} b^{3}\right )}} \]
1/105*(21*A*a*b^3*x^5 - 35*B*a^2*b^2*x^4 + 3*(5*C*a*b^3 + 2*A*b^4)*x^7 - 2 8*B*a^3*b*x^2 - 8*B*a^4)*sqrt(b*x^2 + a)/(a^2*b^7*x^8 + 4*a^3*b^6*x^6 + 6* a^4*b^5*x^4 + 4*a^5*b^4*x^2 + a^6*b^3)
Time = 31.27 (sec) , antiderivative size = 575, normalized size of antiderivative = 3.86 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=A \left (\frac {7 a x^{5}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{7}}{35 a^{\frac {11}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {9}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {7}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 35 a^{\frac {5}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {8 a^{2}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} - \frac {28 a b x^{2}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} - \frac {35 b^{2} x^{4}}{105 a^{3} b^{3} \sqrt {a + b x^{2}} + 315 a^{2} b^{4} x^{2} \sqrt {a + b x^{2}} + 315 a b^{5} x^{4} \sqrt {a + b x^{2}} + 105 b^{6} x^{6} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{6}}{6 a^{\frac {9}{2}}} & \text {otherwise} \end {cases}\right ) + \frac {C x^{7}}{7 a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 21 a^{\frac {7}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 21 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 7 a^{\frac {3}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} \]
A*(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*x**2/a) + 10 5*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))) + B*Piecewise((-8*a**2 /(105*a**3*b**3*sqrt(a + b*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2) + 3 15*a*b**5*x**4*sqrt(a + b*x**2) + 105*b**6*x**6*sqrt(a + b*x**2)) - 28*a*b *x**2/(105*a**3*b**3*sqrt(a + b*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2 ) + 315*a*b**5*x**4*sqrt(a + b*x**2) + 105*b**6*x**6*sqrt(a + b*x**2)) - 3 5*b**2*x**4/(105*a**3*b**3*sqrt(a + b*x**2) + 315*a**2*b**4*x**2*sqrt(a + b*x**2) + 315*a*b**5*x**4*sqrt(a + b*x**2) + 105*b**6*x**6*sqrt(a + b*x**2 )), Ne(b, 0)), (x**6/(6*a**(9/2)), True)) + C*x**7/(7*a**(9/2)*sqrt(1 + b* x**2/a) + 21*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 21*a**(5/2)*b**2*x**4*sq rt(1 + b*x**2/a) + 7*a**(3/2)*b**3*x**6*sqrt(1 + b*x**2/a))
Time = 0.21 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.70 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {C x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {B x^{4}}{3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {5 \, C a x^{3}}{8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {A x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {4 \, B a x^{2}}{15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} + \frac {C x}{14 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {C x}{7 \, \sqrt {b x^{2} + a} a b^{3}} + \frac {3 \, C a x}{56 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}} - \frac {15 \, C a^{2} x}{56 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} + \frac {3 \, A x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, A x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, A a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {8 \, B a^{2}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} \]
-1/2*C*x^5/((b*x^2 + a)^(7/2)*b) - 1/3*B*x^4/((b*x^2 + a)^(7/2)*b) - 5/8*C *a*x^3/((b*x^2 + a)^(7/2)*b^2) - 1/4*A*x^3/((b*x^2 + a)^(7/2)*b) - 4/15*B* a*x^2/((b*x^2 + a)^(7/2)*b^2) + 1/14*C*x/((b*x^2 + a)^(3/2)*b^3) + 1/7*C*x /(sqrt(b*x^2 + a)*a*b^3) + 3/56*C*a*x/((b*x^2 + a)^(5/2)*b^3) - 15/56*C*a^ 2*x/((b*x^2 + a)^(7/2)*b^3) + 3/140*A*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*A*x /(sqrt(b*x^2 + a)*a^2*b^2) + 1/35*A*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*A*a *x/((b*x^2 + a)^(7/2)*b^2) - 8/105*B*a^2/((b*x^2 + a)^(7/2)*b^3)
Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.54 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left ({\left (3 \, x {\left (\frac {7 \, A}{a} + \frac {{\left (5 \, C a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{2}}{a^{3} b^{3}}\right )} - \frac {35 \, B}{b}\right )} x^{2} - \frac {28 \, B a}{b^{2}}\right )} x^{2} - \frac {8 \, B a^{2}}{b^{3}}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \]
1/105*(((3*x*(7*A/a + (5*C*a^2*b^3 + 2*A*a*b^4)*x^2/(a^3*b^3)) - 35*B/b)*x ^2 - 28*B*a/b^2)*x^2 - 8*B*a^2/b^3)/(b*x^2 + a)^(7/2)
Time = 5.57 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 \left (A+B x+C x^2\right )}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {x\,\left (\frac {C\,a^2-A\,a\,b}{35\,a\,b^3}+\frac {a\,\left (\frac {C}{5\,b^2}-\frac {7\,A\,b^2-7\,C\,a\,b}{35\,a\,b^3}\right )}{b}\right )+\frac {2\,B\,a}{5\,b^3}}{{\left (b\,x^2+a\right )}^{5/2}}-\frac {\frac {B}{3\,b^3}+x\,\left (\frac {C}{3\,b^3}-\frac {3\,A\,b-10\,C\,a}{105\,a\,b^3}\right )}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {\frac {B\,a^2}{7\,b^3}-\frac {a\,x\,\left (\frac {A}{7\,b}-\frac {C\,a}{7\,b^2}\right )}{b}}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {x\,\left (2\,A\,b+5\,C\,a\right )}{35\,a^2\,b^3\,\sqrt {b\,x^2+a}} \]
(x*((C*a^2 - A*a*b)/(35*a*b^3) + (a*(C/(5*b^2) - (7*A*b^2 - 7*C*a*b)/(35*a *b^3)))/b) + (2*B*a)/(5*b^3))/(a + b*x^2)^(5/2) - (B/(3*b^3) + x*(C/(3*b^3 ) - (3*A*b - 10*C*a)/(105*a*b^3)))/(a + b*x^2)^(3/2) - ((B*a^2)/(7*b^3) - (a*x*(A/(7*b) - (C*a)/(7*b^2)))/b)/(a + b*x^2)^(7/2) + (x*(2*A*b + 5*C*a)) /(35*a^2*b^3*(a + b*x^2)^(1/2))