Integrand size = 24, antiderivative size = 183 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac {2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}}-\frac {5 b^2 c^2-4 a d (5 b c-4 a d)}{5 c^3 x \left (c+d x^2\right )^{3/2}}-\frac {4 d \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right ) x}{15 c^4 \left (c+d x^2\right )^{3/2}}-\frac {8 d \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right ) x}{15 c^5 \sqrt {c+d x^2}} \]
-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)
Time = 0.21 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=\frac {-5 b^2 c^2 x^4 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )+10 a b c x^2 \left (-c^3+6 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )-a^2 \left (3 c^4-8 c^3 d x^2+48 c^2 d^2 x^4+192 c d^3 x^6+128 d^4 x^8\right )}{15 c^5 x^5 \left (c+d x^2\right )^{3/2}} \]
(-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))
Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {365, 359, 245, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\int \frac {5 b^2 c x^2+2 a (5 b c-4 a d)}{x^4 \left (d x^2+c\right )^{5/2}}dx}{5 c}-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {\left (5 b^2 c^2-4 a d (5 b c-4 a d)\right ) \int \frac {1}{x^2 \left (d x^2+c\right )^{5/2}}dx}{c}-\frac {2 a (5 b c-4 a d)}{3 c x^3 \left (c+d x^2\right )^{3/2}}}{5 c}-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {\frac {\left (5 b^2 c^2-4 a d (5 b c-4 a d)\right ) \left (-\frac {4 d \int \frac {1}{\left (d x^2+c\right )^{5/2}}dx}{c}-\frac {1}{c x \left (c+d x^2\right )^{3/2}}\right )}{c}-\frac {2 a (5 b c-4 a d)}{3 c x^3 \left (c+d x^2\right )^{3/2}}}{5 c}-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {\frac {\left (5 b^2 c^2-4 a d (5 b c-4 a d)\right ) \left (-\frac {4 d \left (\frac {2 \int \frac {1}{\left (d x^2+c\right )^{3/2}}dx}{3 c}+\frac {x}{3 c \left (c+d x^2\right )^{3/2}}\right )}{c}-\frac {1}{c x \left (c+d x^2\right )^{3/2}}\right )}{c}-\frac {2 a (5 b c-4 a d)}{3 c x^3 \left (c+d x^2\right )^{3/2}}}{5 c}-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {\left (-\frac {4 d \left (\frac {2 x}{3 c^2 \sqrt {c+d x^2}}+\frac {x}{3 c \left (c+d x^2\right )^{3/2}}\right )}{c}-\frac {1}{c x \left (c+d x^2\right )^{3/2}}\right ) \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{c}-\frac {2 a (5 b c-4 a d)}{3 c x^3 \left (c+d x^2\right )^{3/2}}}{5 c}-\frac {a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}\) |
-1/5*a^2/(c*x^5*(c + d*x^2)^(3/2)) + ((-2*a*(5*b*c - 4*a*d))/(3*c*x^3*(c + d*x^2)^(3/2)) + ((5*b^2*c^2 - 4*a*d*(5*b*c - 4*a*d))*(-(1/(c*x*(c + d*x^2 )^(3/2))) - (4*d*(x/(3*c*(c + d*x^2)^(3/2)) + (2*x)/(3*c^2*Sqrt[c + d*x^2] )))/c))/c)/(5*c)
3.7.70.3.1 Defintions of rubi rules used
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_)*((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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 2.98 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.72
| method | result | size |
| pseudoelliptic | \(\frac {\left (-15 b^{2} x^{4}-10 a b \,x^{2}-3 a^{2}\right ) c^{4}+8 x^{2} d \left (-\frac {15}{2} b^{2} x^{4}+\frac {15}{2} a b \,x^{2}+a^{2}\right ) c^{3}-48 x^{4} d^{2} \left (\frac {5}{6} b^{2} x^{4}-5 a b \,x^{2}+a^{2}\right ) c^{2}-192 x^{6} d^{3} \left (-\frac {5 b \,x^{2}}{6}+a \right ) a c -128 a^{2} d^{4} x^{8}}{15 \left (d \,x^{2}+c \right )^{\frac {3}{2}} x^{5} c^{5}}\) | \(131\) |
| risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (73 a^{2} d^{2} x^{4}-80 x^{4} a b c d +15 b^{2} c^{2} x^{4}-14 a^{2} c d \,x^{2}+10 a b \,c^{2} x^{2}+3 a^{2} c^{2}\right )}{15 c^{5} x^{5}}-\frac {d \left (a d -b c \right ) \left (11 a \,d^{2} x^{2}-5 b c d \,x^{2}+12 a c d -6 b \,c^{2}\right ) x \sqrt {d \,x^{2}+c}}{3 \left (d^{2} x^{4}+2 c d \,x^{2}+c^{2}\right ) c^{5}}\) | \(152\) |
| gosper | \(-\frac {128 a^{2} d^{4} x^{8}-160 a b c \,d^{3} x^{8}+40 b^{2} c^{2} d^{2} x^{8}+192 a^{2} c \,d^{3} x^{6}-240 a b \,c^{2} d^{2} x^{6}+60 b^{2} c^{3} d \,x^{6}+48 a^{2} c^{2} d^{2} x^{4}-60 a b \,c^{3} d \,x^{4}+15 b^{2} c^{4} x^{4}-8 a^{2} c^{3} d \,x^{2}+10 a b \,c^{4} x^{2}+3 a^{2} c^{4}}{15 x^{5} \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{5}}\) | \(158\) |
| trager | \(-\frac {128 a^{2} d^{4} x^{8}-160 a b c \,d^{3} x^{8}+40 b^{2} c^{2} d^{2} x^{8}+192 a^{2} c \,d^{3} x^{6}-240 a b \,c^{2} d^{2} x^{6}+60 b^{2} c^{3} d \,x^{6}+48 a^{2} c^{2} d^{2} x^{4}-60 a b \,c^{3} d \,x^{4}+15 b^{2} c^{4} x^{4}-8 a^{2} c^{3} d \,x^{2}+10 a b \,c^{4} x^{2}+3 a^{2} c^{4}}{15 x^{5} \left (d \,x^{2}+c \right )^{\frac {3}{2}} c^{5}}\) | \(158\) |
| default | \(b^{2} \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )+a^{2} \left (-\frac {1}{5 c \,x^{5} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {8 d \left (-\frac {1}{3 c \,x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 d \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )}{c}\right )}{5 c}\right )+2 a b \left (-\frac {1}{3 c \,x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 d \left (-\frac {1}{c x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {4 d \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{c}\right )}{c}\right )\) | \(251\) |
1/15*((-15*b^2*x^4-10*a*b*x^2-3*a^2)*c^4+8*x^2*d*(-15/2*b^2*x^4+15/2*a*b*x ^2+a^2)*c^3-48*x^4*d^2*(5/6*b^2*x^4-5*a*b*x^2+a^2)*c^2-192*x^6*d^3*(-5/6*b *x^2+a)*a*c-128*a^2*d^4*x^8)/(d*x^2+c)^(3/2)/x^5/c^5
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {{\left (8 \, {\left (5 \, b^{2} c^{2} d^{2} - 20 \, a b c d^{3} + 16 \, a^{2} d^{4}\right )} x^{8} + 12 \, {\left (5 \, b^{2} c^{3} d - 20 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x^{6} + 3 \, a^{2} c^{4} + 3 \, {\left (5 \, b^{2} c^{4} - 20 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} x^{4} + 2 \, {\left (5 \, a b c^{4} - 4 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (c^{5} d^{2} x^{9} + 2 \, c^{6} d x^{7} + c^{7} x^{5}\right )}} \]
-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)
\[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{x^{6} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {8 \, b^{2} d x}{3 \, \sqrt {d x^{2} + c} c^{3}} - \frac {4 \, b^{2} d x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {32 \, a b d^{2} x}{3 \, \sqrt {d x^{2} + c} c^{4}} + \frac {16 \, a b d^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} - \frac {128 \, a^{2} d^{3} x}{15 \, \sqrt {d x^{2} + c} c^{5}} - \frac {64 \, a^{2} d^{3} x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{4}} - \frac {b^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c x} + \frac {4 \, a b d}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x} - \frac {16 \, a^{2} d^{2}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3} x} - \frac {2 \, a b}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{3}} + \frac {8 \, a^{2} d}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} x^{3}} - \frac {a^{2}}{5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c x^{5}} \]
-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)
Leaf count of result is larger than twice the leaf count of optimal. 509 vs. \(2 (163) = 326\).
Time = 0.31 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.78 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=-\frac {x {\left (\frac {{\left (5 \, b^{2} c^{6} d^{3} - 16 \, a b c^{5} d^{4} + 11 \, a^{2} c^{4} d^{5}\right )} x^{2}}{c^{9} d} + \frac {6 \, {\left (b^{2} c^{7} d^{2} - 3 \, a b c^{6} d^{3} + 2 \, a^{2} c^{5} d^{4}\right )}}{c^{9} d}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (15 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt {d} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c d^{\frac {3}{2}} + 45 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{3} \sqrt {d} + 300 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac {3}{2}} - 240 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c d^{\frac {5}{2}} + 90 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{4} \sqrt {d} - 500 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac {3}{2}} + 490 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac {5}{2}} - 60 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{5} \sqrt {d} + 340 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac {3}{2}} - 320 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac {5}{2}} + 15 \, b^{2} c^{6} \sqrt {d} - 80 \, a b c^{5} d^{\frac {3}{2}} + 73 \, a^{2} c^{4} d^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{5} c^{4}} \]
-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 - s qrt(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)
Time = 5.88 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a+b x^2\right )^2}{x^6 \left (c+d x^2\right )^{5/2}} \, dx=\frac {2\,a\,\sqrt {d\,x^2+c}\,\left (7\,a\,d-5\,b\,c\right )}{15\,c^4\,x^3}-\frac {\frac {73\,a^2\,c^2\,d^2-80\,a\,b\,c^3\,d+15\,b^2\,c^4}{30\,c^5}-\frac {c\,\left (\frac {d\,\left (73\,a^2\,c^2\,d^2-80\,a\,b\,c^3\,d+15\,b^2\,c^4\right )}{18\,c^6}+\frac {c\,\left (\frac {4\,a\,d^3\,\left (7\,a\,d-5\,b\,c\right )}{45\,c^5}-\frac {a\,d^3\,\left (43\,a\,d-35\,b\,c\right )}{9\,c^5}\right )}{d}+\frac {a\,d^2\,\left (43\,a\,d-35\,b\,c\right )}{15\,c^4}\right )}{d}}{x\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a^2\,\sqrt {d\,x^2+c}}{5\,c^3\,x^5}-\frac {x^2\,\left (\frac {2\,d\,\left (78\,a^2\,c\,d^2-90\,a\,b\,c^2\,d+20\,b^2\,c^3\right )}{15\,c^6}-\frac {4\,a\,d^2\,\left (7\,a\,d-5\,b\,c\right )}{15\,c^5}\right )+\frac {78\,a^2\,c\,d^2-90\,a\,b\,c^2\,d+20\,b^2\,c^3}{15\,c^5}}{x\,\sqrt {d\,x^2+c}} \]
(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))