Integrand size = 22, antiderivative size = 186 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 a^2}{3 b^2 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {4 a c}{b (b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/2}}+\frac {2 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x}}{3 b^2 (b c-a d)^3 (c+d x)^{3/2}}+\frac {4 \left (b^2 c^2+6 a b c d+a^2 d^2\right ) \sqrt {a+b x}}{3 b (b c-a d)^4 \sqrt {c+d x}} \]
-2/3*a^2/b^2/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/2)+4*a*c/b/(-a*d+b*c)^2/( d*x+c)^(3/2)/(b*x+a)^(1/2)+2/3*(a^2*d^2+6*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)/b ^2/(-a*d+b*c)^3/(d*x+c)^(3/2)+4/3*(a^2*d^2+6*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2 )/b/(-a*d+b*c)^4/(d*x+c)^(1/2)
Time = 0.17 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 (a+b x)^{3/2} \left (-c^2 d+\frac {3 b c^2 (c+d x)}{a+b x}+\frac {6 a c d (c+d x)}{a+b x}+\frac {6 a b c (c+d x)^2}{(a+b x)^2}+\frac {3 a^2 d (c+d x)^2}{(a+b x)^2}-\frac {a^2 b (c+d x)^3}{(a+b x)^3}\right )}{3 (b c-a d)^4 (c+d x)^{3/2}} \]
(2*(a + b*x)^(3/2)*(-(c^2*d) + (3*b*c^2*(c + d*x))/(a + b*x) + (6*a*c*d*(c + d*x))/(a + b*x) + (6*a*b*c*(c + d*x)^2)/(a + b*x)^2 + (3*a^2*d*(c + d*x )^2)/(a + b*x)^2 - (a^2*b*(c + d*x)^3)/(a + b*x)^3))/(3*(b*c - a*d)^4*(c + d*x)^(3/2))
Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {100, 27, 87, 55, 48}
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.
| \(\displaystyle \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {2 \int -\frac {3 (a (b c+a d)-b (b c-a d) x)}{2 (a+b x)^{3/2} (c+d x)^{5/2}}dx}{3 b^2 (b c-a d)}-\frac {2 a^2}{3 b^2 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a (b c+a d)-b (b c-a d) x}{(a+b x)^{3/2} (c+d x)^{5/2}}dx}{b^2 (b c-a d)}-\frac {2 a^2}{3 b^2 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {-\frac {\left (a^2 d^2+6 a b c d+b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}}dx}{b c-a d}-\frac {4 a b c}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b^2 (b c-a d)}-\frac {2 a^2}{3 b^2 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle -\frac {-\frac {\left (a^2 d^2+6 a b c d+b^2 c^2\right ) \left (\frac {2 b \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 (b c-a d)}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {4 a b c}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b^2 (b c-a d)}-\frac {2 a^2}{3 b^2 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {-\frac {\left (a^2 d^2+6 a b c d+b^2 c^2\right ) \left (\frac {4 b \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {4 a b c}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b^2 (b c-a d)}-\frac {2 a^2}{3 b^2 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\) |
(-2*a^2)/(3*b^2*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) - ((-4*a*b*c) /((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((b^2*c^2 + 6*a*b*c*d + a^2 *d^2)*((2*Sqrt[a + b*x])/(3*(b*c - a*d)*(c + d*x)^(3/2)) + (4*b*Sqrt[a + b *x])/(3*(b*c - a*d)^2*Sqrt[c + d*x])))/(b*c - a*d))/(b^2*(b*c - a*d))
3.9.8.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Time = 1.74 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.87
| method | result | size |
| default | \(\frac {\frac {4}{3} a^{2} b \,d^{3} x^{3}+8 a \,b^{2} c \,d^{2} x^{3}+\frac {4}{3} b^{3} c^{2} d \,x^{3}+2 a^{3} d^{3} x^{2}+14 a^{2} b c \,d^{2} x^{2}+14 a \,b^{2} c^{2} d \,x^{2}+2 b^{3} c^{3} x^{2}+8 x \,a^{3} c \,d^{2}+16 x \,a^{2} b \,c^{2} d +8 x a \,b^{2} c^{3}+\frac {16}{3} a^{3} c^{2} d +\frac {16}{3} a^{2} b \,c^{3}}{\left (a d -b c \right )^{4} \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}}}\) | \(162\) |
| gosper | \(\frac {\frac {4}{3} a^{2} b \,d^{3} x^{3}+8 a \,b^{2} c \,d^{2} x^{3}+\frac {4}{3} b^{3} c^{2} d \,x^{3}+2 a^{3} d^{3} x^{2}+14 a^{2} b c \,d^{2} x^{2}+14 a \,b^{2} c^{2} d \,x^{2}+2 b^{3} c^{3} x^{2}+8 x \,a^{3} c \,d^{2}+16 x \,a^{2} b \,c^{2} d +8 x a \,b^{2} c^{3}+\frac {16}{3} a^{3} c^{2} d +\frac {16}{3} a^{2} b \,c^{3}}{\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\) | \(203\) |
2/3*(2*a^2*b*d^3*x^3+12*a*b^2*c*d^2*x^3+2*b^3*c^2*d*x^3+3*a^3*d^3*x^2+21*a ^2*b*c*d^2*x^2+21*a*b^2*c^2*d*x^2+3*b^3*c^3*x^2+12*a^3*c*d^2*x+24*a^2*b*c^ 2*d*x+12*a*b^2*c^3*x+8*a^3*c^2*d+8*a^2*b*c^3)/(a*d-b*c)^4/(b*x+a)^(3/2)/(d *x+c)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 468 vs. \(2 (164) = 328\).
Time = 0.60 (sec) , antiderivative size = 468, normalized size of antiderivative = 2.52 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 \, {\left (8 \, a^{2} b c^{3} + 8 \, a^{3} c^{2} d + 2 \, {\left (b^{3} c^{2} d + 6 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{3} + 3 \, {\left (b^{3} c^{3} + 7 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2} + 12 \, {\left (a b^{2} c^{3} + 2 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{3 \, {\left (a^{2} b^{4} c^{6} - 4 \, a^{3} b^{3} c^{5} d + 6 \, a^{4} b^{2} c^{4} d^{2} - 4 \, a^{5} b c^{3} d^{3} + a^{6} c^{2} d^{4} + {\left (b^{6} c^{4} d^{2} - 4 \, a b^{5} c^{3} d^{3} + 6 \, a^{2} b^{4} c^{2} d^{4} - 4 \, a^{3} b^{3} c d^{5} + a^{4} b^{2} d^{6}\right )} x^{4} + 2 \, {\left (b^{6} c^{5} d - 3 \, a b^{5} c^{4} d^{2} + 2 \, a^{2} b^{4} c^{3} d^{3} + 2 \, a^{3} b^{3} c^{2} d^{4} - 3 \, a^{4} b^{2} c d^{5} + a^{5} b d^{6}\right )} x^{3} + {\left (b^{6} c^{6} - 9 \, a^{2} b^{4} c^{4} d^{2} + 16 \, a^{3} b^{3} c^{3} d^{3} - 9 \, a^{4} b^{2} c^{2} d^{4} + a^{6} d^{6}\right )} x^{2} + 2 \, {\left (a b^{5} c^{6} - 3 \, a^{2} b^{4} c^{5} d + 2 \, a^{3} b^{3} c^{4} d^{2} + 2 \, a^{4} b^{2} c^{3} d^{3} - 3 \, a^{5} b c^{2} d^{4} + a^{6} c d^{5}\right )} x\right )}} \]
2/3*(8*a^2*b*c^3 + 8*a^3*c^2*d + 2*(b^3*c^2*d + 6*a*b^2*c*d^2 + a^2*b*d^3) *x^3 + 3*(b^3*c^3 + 7*a*b^2*c^2*d + 7*a^2*b*c*d^2 + a^3*d^3)*x^2 + 12*(a*b ^2*c^3 + 2*a^2*b*c^2*d + a^3*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a^2*b^ 4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d^3 + a^6*c^2*d^ 4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2*b^4*c^3*d^3 + 2 *a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6*c^6 - 9*a^2*b^4 *c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6)*x^2 + 2*(a*b^ 5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3*d^3 - 3*a^5*b* c^2*d^4 + a^6*c*d^5)*x)
\[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{2}}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 769 vs. \(2 (164) = 328\).
Time = 0.55 (sec) , antiderivative size = 769, normalized size of antiderivative = 4.13 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {2 \, \sqrt {b x + a} {\left (\frac {2 \, {\left (b^{7} c^{5} d^{2} {\left | b \right |} - 6 \, a^{2} b^{5} c^{3} d^{4} {\left | b \right |} + 8 \, a^{3} b^{4} c^{2} d^{5} {\left | b \right |} - 3 \, a^{4} b^{3} c d^{6} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}} + \frac {3 \, {\left (b^{8} c^{6} d {\left | b \right |} - 2 \, a b^{7} c^{5} d^{2} {\left | b \right |} - 2 \, a^{2} b^{6} c^{4} d^{3} {\left | b \right |} + 8 \, a^{3} b^{5} c^{3} d^{4} {\left | b \right |} - 7 \, a^{4} b^{4} c^{2} d^{5} {\left | b \right |} + 2 \, a^{5} b^{3} c d^{6} {\left | b \right |}\right )}}{b^{9} c^{7} d - 7 \, a b^{8} c^{6} d^{2} + 21 \, a^{2} b^{7} c^{5} d^{3} - 35 \, a^{3} b^{6} c^{4} d^{4} + 35 \, a^{4} b^{5} c^{3} d^{5} - 21 \, a^{5} b^{4} c^{2} d^{6} + 7 \, a^{6} b^{3} c d^{7} - a^{7} b^{2} d^{8}}\right )}}{3 \, {\left (b^{2} c + {\left (b x + a\right )} b d - a b d\right )}^{\frac {3}{2}}} + \frac {8 \, {\left (3 \, \sqrt {b d} a b^{6} c^{3} - 5 \, \sqrt {b d} a^{2} b^{5} c^{2} d + \sqrt {b d} a^{3} b^{4} c d^{2} + \sqrt {b d} a^{4} b^{3} d^{3} - 6 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b^{4} c^{2} + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} b^{3} c d + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{3} b^{2} d^{2} + 3 \, \sqrt {b d} {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4} a b^{2} c\right )}}{3 \, {\left (b^{3} c^{3} {\left | b \right |} - 3 \, a b^{2} c^{2} d {\left | b \right |} + 3 \, a^{2} b c d^{2} {\left | b \right |} - a^{3} d^{3} {\left | b \right |}\right )} {\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}^{3}} \]
2/3*sqrt(b*x + a)*(2*(b^7*c^5*d^2*abs(b) - 6*a^2*b^5*c^3*d^4*abs(b) + 8*a^ 3*b^4*c^2*d^5*abs(b) - 3*a^4*b^3*c*d^6*abs(b))*(b*x + a)/(b^9*c^7*d - 7*a* b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3*d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8) + 3*(b^8*c^6*d*abs( b) - 2*a*b^7*c^5*d^2*abs(b) - 2*a^2*b^6*c^4*d^3*abs(b) + 8*a^3*b^5*c^3*d^4 *abs(b) - 7*a^4*b^4*c^2*d^5*abs(b) + 2*a^5*b^3*c*d^6*abs(b))/(b^9*c^7*d - 7*a*b^8*c^6*d^2 + 21*a^2*b^7*c^5*d^3 - 35*a^3*b^6*c^4*d^4 + 35*a^4*b^5*c^3 *d^5 - 21*a^5*b^4*c^2*d^6 + 7*a^6*b^3*c*d^7 - a^7*b^2*d^8))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 8/3*(3*sqrt(b*d)*a*b^6*c^3 - 5*sqrt(b*d)*a^2*b^5 *c^2*d + sqrt(b*d)*a^3*b^4*c*d^2 + sqrt(b*d)*a^4*b^3*d^3 - 6*sqrt(b*d)*(sq rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^4*c^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d ))^2*a^2*b^3*c*d + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b* x + a)*b*d - a*b*d))^2*a^3*b^2*d^2 + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^2*c)/((b^3*c^3*abs(b) - 3*a*b ^2*c^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^3)
Time = 2.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.33 \[ \int \frac {x^2}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {4\,x^3\,\left (a^2\,d^2+6\,a\,b\,c\,d+b^2\,c^2\right )}{3\,d\,{\left (a\,d-b\,c\right )}^4}+\frac {x^2\,\left (6\,a^3\,d^3+42\,a^2\,b\,c\,d^2+42\,a\,b^2\,c^2\,d+6\,b^3\,c^3\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {16\,a^2\,c^2\,\left (a\,d+b\,c\right )}{3\,b\,d^2\,{\left (a\,d-b\,c\right )}^4}+\frac {8\,a\,c\,x\,{\left (a\,d+b\,c\right )}^2}{b\,d^2\,{\left (a\,d-b\,c\right )}^4}\right )}{x^3\,\sqrt {a+b\,x}+\frac {a\,c^2\,\sqrt {a+b\,x}}{b\,d^2}+\frac {x^2\,\left (a\,d+2\,b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {c\,x\,\left (2\,a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d^2}} \]
((c + d*x)^(1/2)*((4*x^3*(a^2*d^2 + b^2*c^2 + 6*a*b*c*d))/(3*d*(a*d - b*c) ^4) + (x^2*(6*a^3*d^3 + 6*b^3*c^3 + 42*a*b^2*c^2*d + 42*a^2*b*c*d^2))/(3*b *d^2*(a*d - b*c)^4) + (16*a^2*c^2*(a*d + b*c))/(3*b*d^2*(a*d - b*c)^4) + ( 8*a*c*x*(a*d + b*c)^2)/(b*d^2*(a*d - b*c)^4)))/(x^3*(a + b*x)^(1/2) + (a*c ^2*(a + b*x)^(1/2))/(b*d^2) + (x^2*(a*d + 2*b*c)*(a + b*x)^(1/2))/(b*d) + (c*x*(2*a*d + b*c)*(a + b*x)^(1/2))/(b*d^2))