Integrand size = 24, antiderivative size = 187 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {4 c (4 b c-5 a d)}{15 a^2 \left (a x+b x^2\right )^{3/2}}-\frac {2 c^2}{5 a x \left (a x+b x^2\right )^{3/2}}+\frac {2 \left (5 d^2+\frac {4 b c (4 b c-5 a d)}{a^2}\right ) x}{15 a \left (a x+b x^2\right )^{3/2}}+\frac {8 \left (5 a^2 d^2+4 b c (4 b c-5 a d)\right )}{15 a^4 \sqrt {a x+b x^2}}-\frac {16 \left (5 a^2 d^2+4 b c (4 b c-5 a d)\right ) \sqrt {a x+b x^2}}{15 a^5 x} \] Output:
4/15*c*(-5*a*d+4*b*c)/a^2/(b*x^2+a*x)^(3/2)-2/5*c^2/a/x/(b*x^2+a*x)^(3/2)+ 2/15*(5*d^2+4*b*c*(-5*a*d+4*b*c)/a^2)*x/a/(b*x^2+a*x)^(3/2)+8/15*(5*a^2*d^ 2+4*b*c*(-5*a*d+4*b*c))/a^4/(b*x^2+a*x)^(1/2)-16/15*(5*a^2*d^2+4*b*c*(-5*a *d+4*b*c))*(b*x^2+a*x)^(1/2)/a^5/x
Time = 0.34 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.70 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 \left (128 b^4 c^2 x^4+32 a b^3 c x^3 (6 c-5 d x)+8 a^2 b^2 x^2 \left (6 c^2-30 c d x+5 d^2 x^2\right )+4 a^3 b x \left (-2 c^2-15 c d x+15 d^2 x^2\right )+a^4 \left (3 c^2+10 c d x+15 d^2 x^2\right )\right )}{15 a^5 x (x (a+b x))^{3/2}} \] Input:
Integrate[(c + d*x)^2/(x*(a*x + b*x^2)^(5/2)),x]
Output:
(-2*(128*b^4*c^2*x^4 + 32*a*b^3*c*x^3*(6*c - 5*d*x) + 8*a^2*b^2*x^2*(6*c^2 - 30*c*d*x + 5*d^2*x^2) + 4*a^3*b*x*(-2*c^2 - 15*c*d*x + 15*d^2*x^2) + a^ 4*(3*c^2 + 10*c*d*x + 15*d^2*x^2)))/(15*a^5*x*(x*(a + b*x))^(3/2))
Time = 0.55 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.76, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1262, 27, 1220, 1089, 1088}
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.
| \(\displaystyle \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1262 |
| \(\displaystyle -\frac {\int -\frac {3 \left (2 b c^2+d (4 b c-a d) x\right )}{2 x \left (b x^2+a x\right )^{5/2}}dx}{3 b}-\frac {d^2}{3 b \left (a x+b x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 b c^2+d (4 b c-a d) x}{x \left (b x^2+a x\right )^{5/2}}dx}{2 b}-\frac {d^2}{3 b \left (a x+b x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {-\frac {\left (5 a^2 d^2-20 a b c d+16 b^2 c^2\right ) \int \frac {1}{\left (b x^2+a x\right )^{5/2}}dx}{5 a}-\frac {4 b c^2}{5 a x \left (a x+b x^2\right )^{3/2}}}{2 b}-\frac {d^2}{3 b \left (a x+b x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1089 |
| \(\displaystyle \frac {-\frac {\left (5 a^2 d^2-20 a b c d+16 b^2 c^2\right ) \left (-\frac {8 b \int \frac {1}{\left (b x^2+a x\right )^{3/2}}dx}{3 a^2}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}\right )}{5 a}-\frac {4 b c^2}{5 a x \left (a x+b x^2\right )^{3/2}}}{2 b}-\frac {d^2}{3 b \left (a x+b x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1088 |
| \(\displaystyle \frac {-\frac {\left (\frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}}\right ) \left (5 a^2 d^2-20 a b c d+16 b^2 c^2\right )}{5 a}-\frac {4 b c^2}{5 a x \left (a x+b x^2\right )^{3/2}}}{2 b}-\frac {d^2}{3 b \left (a x+b x^2\right )^{3/2}}\) |
Input:
Int[(c + d*x)^2/(x*(a*x + b*x^2)^(5/2)),x]
Output:
-1/3*d^2/(b*(a*x + b*x^2)^(3/2)) + ((-4*b*c^2)/(5*a*x*(a*x + b*x^2)^(3/2)) - ((16*b^2*c^2 - 20*a*b*c*d + 5*a^2*d^2)*((-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*Sqrt[a*x + b*x^2])))/(5*a))/(2*b )
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_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b *x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n + e*g^n*( m + p + n)*(d + e*x)^(n - 2)*(b*d - 2*a*e + (2*c*d - b*e)*x), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
Time = 0.52 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {\left (-30 d^{2} x^{2}-20 c d x -6 c^{2}\right ) a^{4}+16 x b \left (-\frac {15}{2} d^{2} x^{2}+\frac {15}{2} c d x +c^{2}\right ) a^{3}-96 \left (\frac {5}{6} d^{2} x^{2}-5 c d x +c^{2}\right ) x^{2} b^{2} a^{2}-384 x^{3} b^{3} c \left (-\frac {5 d x}{6}+c \right ) a -256 b^{4} c^{2} x^{4}}{15 \sqrt {x \left (b x +a \right )}\, x^{2} \left (b x +a \right ) a^{5}}\) | \(128\) |
| risch | \(-\frac {2 \left (b x +a \right ) \left (15 a^{2} d^{2} x^{2}-80 a b c d \,x^{2}+73 b^{2} c^{2} x^{2}+10 a^{2} c d x -14 a b \,c^{2} x +3 a^{2} c^{2}\right )}{15 a^{5} x^{2} \sqrt {x \left (b x +a \right )}}-\frac {2 x \left (5 a b d x -11 b^{2} c x +6 a^{2} d -12 a b c \right ) \left (a d -b c \right ) b}{3 \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) a^{5}}\) | \(136\) |
| gosper | \(-\frac {2 \left (b x +a \right ) \left (40 a^{2} b^{2} d^{2} x^{4}-160 a \,b^{3} c d \,x^{4}+128 b^{4} c^{2} x^{4}+60 a^{3} b \,d^{2} x^{3}-240 a^{2} b^{2} c d \,x^{3}+192 a \,b^{3} c^{2} x^{3}+15 a^{4} d^{2} x^{2}-60 a^{3} d c b \,x^{2}+48 a^{2} b^{2} c^{2} x^{2}+10 a^{4} c d x -8 a^{3} b \,c^{2} x +3 a^{4} c^{2}\right )}{15 a^{5} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\) | \(158\) |
| orering | \(-\frac {2 \left (b x +a \right ) \left (40 a^{2} b^{2} d^{2} x^{4}-160 a \,b^{3} c d \,x^{4}+128 b^{4} c^{2} x^{4}+60 a^{3} b \,d^{2} x^{3}-240 a^{2} b^{2} c d \,x^{3}+192 a \,b^{3} c^{2} x^{3}+15 a^{4} d^{2} x^{2}-60 a^{3} d c b \,x^{2}+48 a^{2} b^{2} c^{2} x^{2}+10 a^{4} c d x -8 a^{3} b \,c^{2} x +3 a^{4} c^{2}\right )}{15 a^{5} \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\) | \(158\) |
| trager | \(-\frac {2 \left (40 a^{2} b^{2} d^{2} x^{4}-160 a \,b^{3} c d \,x^{4}+128 b^{4} c^{2} x^{4}+60 a^{3} b \,d^{2} x^{3}-240 a^{2} b^{2} c d \,x^{3}+192 a \,b^{3} c^{2} x^{3}+15 a^{4} d^{2} x^{2}-60 a^{3} d c b \,x^{2}+48 a^{2} b^{2} c^{2} x^{2}+10 a^{4} c d x -8 a^{3} b \,c^{2} x +3 a^{4} c^{2}\right ) \sqrt {b \,x^{2}+a x}}{15 a^{5} x^{3} \left (b x +a \right )^{2}}\) | \(163\) |
| default | \(c^{2} \left (-\frac {2}{5 a x \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {8 b \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{5 a}\right )+d^{2} \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )+2 c d \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )\) | \(201\) |
Input:
int((d*x+c)^2/x/(b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/15*((-30*d^2*x^2-20*c*d*x-6*c^2)*a^4+16*x*b*(-15/2*d^2*x^2+15/2*c*d*x+c^ 2)*a^3-96*(5/6*d^2*x^2-5*c*d*x+c^2)*x^2*b^2*a^2-384*x^3*b^3*c*(-5/6*d*x+c) *a-256*b^4*c^2*x^4)/(x*(b*x+a))^(1/2)/x^2/(b*x+a)/a^5
Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, a^{4} c^{2} + 8 \, {\left (16 \, b^{4} c^{2} - 20 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{4} + 12 \, {\left (16 \, a b^{3} c^{2} - 20 \, a^{2} b^{2} c d + 5 \, a^{3} b d^{2}\right )} x^{3} + 3 \, {\left (16 \, a^{2} b^{2} c^{2} - 20 \, a^{3} b c d + 5 \, a^{4} d^{2}\right )} x^{2} - 2 \, {\left (4 \, a^{3} b c^{2} - 5 \, a^{4} c d\right )} x\right )} \sqrt {b x^{2} + a x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \] Input:
integrate((d*x+c)^2/x/(b*x^2+a*x)^(5/2),x, algorithm="fricas")
Output:
-2/15*(3*a^4*c^2 + 8*(16*b^4*c^2 - 20*a*b^3*c*d + 5*a^2*b^2*d^2)*x^4 + 12* (16*a*b^3*c^2 - 20*a^2*b^2*c*d + 5*a^3*b*d^2)*x^3 + 3*(16*a^2*b^2*c^2 - 20 *a^3*b*c*d + 5*a^4*d^2)*x^2 - 2*(4*a^3*b*c^2 - 5*a^4*c*d)*x)*sqrt(b*x^2 + a*x)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3)
\[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {\left (c + d x\right )^{2}}{x \left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((d*x+c)**2/x/(b*x**2+a*x)**(5/2),x)
Output:
Integral((c + d*x)**2/(x*(x*(a + b*x))**(5/2)), x)
Time = 0.04 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.34 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {32 \, b^{2} c^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, b^{3} c^{2} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {8 \, b c d x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} + \frac {64 \, b^{2} c d x}{3 \, \sqrt {b x^{2} + a x} a^{4}} + \frac {2 \, d^{2} x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} - \frac {16 \, b d^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{3}} + \frac {16 \, b c^{2}}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, b^{2} c^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {4 \, c d}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} + \frac {32 \, b c d}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {8 \, d^{2}}{3 \, \sqrt {b x^{2} + a x} a^{2}} - \frac {2 \, c^{2}}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} \] Input:
integrate((d*x+c)^2/x/(b*x^2+a*x)^(5/2),x, algorithm="maxima")
Output:
32/15*b^2*c^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*b^3*c^2*x/(sqrt(b*x^2 + a*x)*a^5) - 8/3*b*c*d*x/((b*x^2 + a*x)^(3/2)*a^2) + 64/3*b^2*c*d*x/(sqrt( b*x^2 + a*x)*a^4) + 2/3*d^2*x/((b*x^2 + a*x)^(3/2)*a) - 16/3*b*d^2*x/(sqrt (b*x^2 + a*x)*a^3) + 16/15*b*c^2/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*b^2*c^ 2/(sqrt(b*x^2 + a*x)*a^4) - 4/3*c*d/((b*x^2 + a*x)^(3/2)*a) + 32/3*b*c*d/( sqrt(b*x^2 + a*x)*a^3) - 8/3*d^2/(sqrt(b*x^2 + a*x)*a^2) - 2/5*c^2/((b*x^2 + a*x)^(3/2)*a*x)
\[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (b x^{2} + a x\right )}^{\frac {5}{2}} x} \,d x } \] Input:
integrate((d*x+c)^2/x/(b*x^2+a*x)^(5/2),x, algorithm="giac")
Output:
integrate((d*x + c)^2/((b*x^2 + a*x)^(5/2)*x), x)
Time = 9.75 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^4\,c^2+10\,a^4\,c\,d\,x+15\,a^4\,d^2\,x^2-8\,a^3\,b\,c^2\,x-60\,a^3\,b\,c\,d\,x^2+60\,a^3\,b\,d^2\,x^3+48\,a^2\,b^2\,c^2\,x^2-240\,a^2\,b^2\,c\,d\,x^3+40\,a^2\,b^2\,d^2\,x^4+192\,a\,b^3\,c^2\,x^3-160\,a\,b^3\,c\,d\,x^4+128\,b^4\,c^2\,x^4\right )}{15\,a^5\,x^3\,{\left (a+b\,x\right )}^2} \] Input:
int((c + d*x)^2/(x*(a*x + b*x^2)^(5/2)),x)
Output:
-(2*(a*x + b*x^2)^(1/2)*(3*a^4*c^2 + 15*a^4*d^2*x^2 + 128*b^4*c^2*x^4 + 19 2*a*b^3*c^2*x^3 + 60*a^3*b*d^2*x^3 + 10*a^4*c*d*x + 48*a^2*b^2*c^2*x^2 + 4 0*a^2*b^2*d^2*x^4 - 8*a^3*b*c^2*x - 60*a^3*b*c*d*x^2 - 160*a*b^3*c*d*x^4 - 240*a^2*b^2*c*d*x^3))/(15*a^5*x^3*(a + b*x)^2)
Time = 0.22 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.60 \[ \int \frac {(c+d x)^2}{x \left (a x+b x^2\right )^{5/2}} \, dx=\frac {\frac {16 \sqrt {b}\, \sqrt {b x +a}\, a^{3} d^{2} x^{3}}{3}-\frac {64 \sqrt {b}\, \sqrt {b x +a}\, a^{2} b c d \,x^{3}}{3}+\frac {16 \sqrt {b}\, \sqrt {b x +a}\, a^{2} b \,d^{2} x^{4}}{3}+\frac {256 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} c^{2} x^{3}}{15}-\frac {64 \sqrt {b}\, \sqrt {b x +a}\, a \,b^{2} c d \,x^{4}}{3}+\frac {256 \sqrt {b}\, \sqrt {b x +a}\, b^{3} c^{2} x^{4}}{15}-\frac {2 \sqrt {x}\, a^{4} c^{2}}{5}-\frac {4 \sqrt {x}\, a^{4} c d x}{3}-2 \sqrt {x}\, a^{4} d^{2} x^{2}+\frac {16 \sqrt {x}\, a^{3} b \,c^{2} x}{15}+8 \sqrt {x}\, a^{3} b c d \,x^{2}-8 \sqrt {x}\, a^{3} b \,d^{2} x^{3}-\frac {32 \sqrt {x}\, a^{2} b^{2} c^{2} x^{2}}{5}+32 \sqrt {x}\, a^{2} b^{2} c d \,x^{3}-\frac {16 \sqrt {x}\, a^{2} b^{2} d^{2} x^{4}}{3}-\frac {128 \sqrt {x}\, a \,b^{3} c^{2} x^{3}}{5}+\frac {64 \sqrt {x}\, a \,b^{3} c d \,x^{4}}{3}-\frac {256 \sqrt {x}\, b^{4} c^{2} x^{4}}{15}}{\sqrt {b x +a}\, a^{5} x^{3} \left (b x +a \right )} \] Input:
int((d*x+c)^2/x/(b*x^2+a*x)^(5/2),x)
Output:
(2*(40*sqrt(b)*sqrt(a + b*x)*a**3*d**2*x**3 - 160*sqrt(b)*sqrt(a + b*x)*a* *2*b*c*d*x**3 + 40*sqrt(b)*sqrt(a + b*x)*a**2*b*d**2*x**4 + 128*sqrt(b)*sq rt(a + b*x)*a*b**2*c**2*x**3 - 160*sqrt(b)*sqrt(a + b*x)*a*b**2*c*d*x**4 + 128*sqrt(b)*sqrt(a + b*x)*b**3*c**2*x**4 - 3*sqrt(x)*a**4*c**2 - 10*sqrt( x)*a**4*c*d*x - 15*sqrt(x)*a**4*d**2*x**2 + 8*sqrt(x)*a**3*b*c**2*x + 60*s qrt(x)*a**3*b*c*d*x**2 - 60*sqrt(x)*a**3*b*d**2*x**3 - 48*sqrt(x)*a**2*b** 2*c**2*x**2 + 240*sqrt(x)*a**2*b**2*c*d*x**3 - 40*sqrt(x)*a**2*b**2*d**2*x **4 - 192*sqrt(x)*a*b**3*c**2*x**3 + 160*sqrt(x)*a*b**3*c*d*x**4 - 128*sqr t(x)*b**4*c**2*x**4))/(15*sqrt(a + b*x)*a**5*x**3*(a + b*x))