\(\int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx\) [227]

Optimal result

Integrand size = 17, antiderivative size = 136 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=-\frac {d \sqrt {c+d x}}{4 b^2 (a+b x)^2}-\frac {d^2 \sqrt {c+d x}}{8 b^2 (b c-a d) (a+b x)}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}+\frac {d^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}} \] Output:

-1/4*d*(d*x+c)^(1/2)/b^2/(b*x+a)^2-1/8*d^2*(d*x+c)^(1/2)/b^2/(-a*d+b*c)/(b 
*x+a)-1/3*(d*x+c)^(3/2)/b/(b*x+a)^3+1/8*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/ 
(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(3/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.95 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\frac {\sqrt {c+d x} \left (-3 a^2 d^2-2 a b d (c+4 d x)+b^2 \left (8 c^2+14 c d x+3 d^2 x^2\right )\right )}{24 b^2 (-b c+a d) (a+b x)^3}+\frac {d^3 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{8 b^{5/2} (-b c+a d)^{3/2}} \] Input:

Integrate[(c + d*x)^(3/2)/(a + b*x)^4,x]
 

Output:

(Sqrt[c + d*x]*(-3*a^2*d^2 - 2*a*b*d*(c + 4*d*x) + b^2*(8*c^2 + 14*c*d*x + 
 3*d^2*x^2)))/(24*b^2*(-(b*c) + a*d)*(a + b*x)^3) + (d^3*ArcTan[(Sqrt[b]*S 
qrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(8*b^(5/2)*(-(b*c) + a*d)^(3/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {51, 51, 52, 73, 221}

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.

\(\displaystyle \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {d \int \frac {\sqrt {c+d x}}{(a+b x)^3}dx}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {d \left (\frac {d \int \frac {1}{(a+b x)^2 \sqrt {c+d x}}dx}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{2 (b c-a d)}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {\int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{b c-a d}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {d \left (\frac {d \left (\frac {d \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x}}{(a+b x) (b c-a d)}\right )}{4 b}-\frac {\sqrt {c+d x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(c+d x)^{3/2}}{3 b (a+b x)^3}\)

Input:

Int[(c + d*x)^(3/2)/(a + b*x)^4,x]
 

Output:

-1/3*(c + d*x)^(3/2)/(b*(a + b*x)^3) + (d*(-1/2*Sqrt[c + d*x]/(b*(a + b*x) 
^2) + (d*(-(Sqrt[c + d*x]/((b*c - a*d)*(a + b*x))) + (d*ArcTanh[(Sqrt[b]*S 
qrt[c + d*x])/Sqrt[b*c - a*d]])/(Sqrt[b]*(b*c - a*d)^(3/2))))/(4*b)))/(2*b 
)
 

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

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*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {d^{3} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )-\left (\frac {\left (-x d -4 c \right ) b}{3}+a d \right ) \left (\left (3 x d +2 c \right ) b +a d \right ) \sqrt {\left (a d -b c \right ) b}\, \sqrt {x d +c}}{8 \sqrt {\left (a d -b c \right ) b}\, \left (a d -b c \right ) b^{2} \left (b x +a \right )^{3}}\)	\(119\)
derivativedivides	\(2 d^{3} \left (\frac {\frac {\left (x d +c \right )^{\frac {5}{2}}}{16 a d -16 b c}-\frac {\left (x d +c \right )^{\frac {3}{2}}}{6 b}-\frac {\left (a d -b c \right ) \sqrt {x d +c}}{16 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(126\)
default	\(2 d^{3} \left (\frac {\frac {\left (x d +c \right )^{\frac {5}{2}}}{16 a d -16 b c}-\frac {\left (x d +c \right )^{\frac {3}{2}}}{6 b}-\frac {\left (a d -b c \right ) \sqrt {x d +c}}{16 b^{2}}}{\left (\left (x d +c \right ) b +a d -b c \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {x d +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \left (a d -b c \right ) b^{2} \sqrt {\left (a d -b c \right ) b}}\right )\)	\(126\)

Input:

int((d*x+c)^(3/2)/(b*x+a)^4,x,method=_RETURNVERBOSE)
 

Output:

1/8*(d^3*(b*x+a)^3*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-(1/3*(-d*x- 
4*c)*b+a*d)*((3*d*x+2*c)*b+a*d)*((a*d-b*c)*b)^(1/2)*(d*x+c)^(1/2))/((a*d-b 
*c)*b)^(1/2)/(a*d-b*c)/b^2/(b*x+a)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (112) = 224\).

Time = 0.11 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.90 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\left [-\frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {b^{2} c - a b d} \sqrt {d x + c}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} c^{3} - 10 \, a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + 3 \, a^{3} b d^{3} + 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} c^{2} d - 11 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{48 \, {\left (a^{3} b^{5} c^{2} - 2 \, a^{4} b^{4} c d + a^{5} b^{3} d^{2} + {\left (b^{8} c^{2} - 2 \, a b^{7} c d + a^{2} b^{6} d^{2}\right )} x^{3} + 3 \, {\left (a b^{7} c^{2} - 2 \, a^{2} b^{6} c d + a^{3} b^{5} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{2} - 2 \, a^{3} b^{5} c d + a^{4} b^{4} d^{2}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x + a^{3} d^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {-b^{2} c + a b d} \sqrt {d x + c}}{b d x + b c}\right ) + {\left (8 \, b^{4} c^{3} - 10 \, a b^{3} c^{2} d - a^{2} b^{2} c d^{2} + 3 \, a^{3} b d^{3} + 3 \, {\left (b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} c^{2} d - 11 \, a b^{3} c d^{2} + 4 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {d x + c}}{24 \, {\left (a^{3} b^{5} c^{2} - 2 \, a^{4} b^{4} c d + a^{5} b^{3} d^{2} + {\left (b^{8} c^{2} - 2 \, a b^{7} c d + a^{2} b^{6} d^{2}\right )} x^{3} + 3 \, {\left (a b^{7} c^{2} - 2 \, a^{2} b^{6} c d + a^{3} b^{5} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} c^{2} - 2 \, a^{3} b^{5} c d + a^{4} b^{4} d^{2}\right )} x\right )}}\right ] \] Input:

integrate((d*x+c)^(3/2)/(b*x+a)^4,x, algorithm="fricas")
 

Output:

[-1/48*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(b 
^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + 
c))/(b*x + a)) + 2*(8*b^4*c^3 - 10*a*b^3*c^2*d - a^2*b^2*c*d^2 + 3*a^3*b*d 
^3 + 3*(b^4*c*d^2 - a*b^3*d^3)*x^2 + 2*(7*b^4*c^2*d - 11*a*b^3*c*d^2 + 4*a 
^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^3*b^5*c^2 - 2*a^4*b^4*c*d + a^5*b^3*d^2 + 
 (b^8*c^2 - 2*a*b^7*c*d + a^2*b^6*d^2)*x^3 + 3*(a*b^7*c^2 - 2*a^2*b^6*c*d 
+ a^3*b^5*d^2)*x^2 + 3*(a^2*b^6*c^2 - 2*a^3*b^5*c*d + a^4*b^4*d^2)*x), -1/ 
24*(3*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(-b^2* 
c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (8*b 
^4*c^3 - 10*a*b^3*c^2*d - a^2*b^2*c*d^2 + 3*a^3*b*d^3 + 3*(b^4*c*d^2 - a*b 
^3*d^3)*x^2 + 2*(7*b^4*c^2*d - 11*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x 
 + c))/(a^3*b^5*c^2 - 2*a^4*b^4*c*d + a^5*b^3*d^2 + (b^8*c^2 - 2*a*b^7*c*d 
 + a^2*b^6*d^2)*x^3 + 3*(a*b^7*c^2 - 2*a^2*b^6*c*d + a^3*b^5*d^2)*x^2 + 3* 
(a^2*b^6*c^2 - 2*a^3*b^5*c*d + a^4*b^4*d^2)*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(3/2)/(b*x+a)**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x+c)^(3/2)/(b*x+a)^4,x, algorithm="maxima")
 

Output:

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
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.36 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=-\frac {d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{3} + 8 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c d^{3} - 3 \, \sqrt {d x + c} b^{2} c^{2} d^{3} - 8 \, {\left (d x + c\right )}^{\frac {3}{2}} a b d^{4} + 6 \, \sqrt {d x + c} a b c d^{4} - 3 \, \sqrt {d x + c} a^{2} d^{5}}{24 \, {\left (b^{3} c - a b^{2} d\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3}} \] Input:

integrate((d*x+c)^(3/2)/(b*x+a)^4,x, algorithm="giac")
 

Output:

-1/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c - a*b^2*d)*s 
qrt(-b^2*c + a*b*d)) - 1/24*(3*(d*x + c)^(5/2)*b^2*d^3 + 8*(d*x + c)^(3/2) 
*b^2*c*d^3 - 3*sqrt(d*x + c)*b^2*c^2*d^3 - 8*(d*x + c)^(3/2)*a*b*d^4 + 6*s 
qrt(d*x + c)*a*b*c*d^4 - 3*sqrt(d*x + c)*a^2*d^5)/((b^3*c - a*b^2*d)*((d*x 
 + c)*b - b*c + a*d)^3)
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\frac {d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}}{\sqrt {a\,d-b\,c}}\right )}{8\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{3/2}}-\frac {\frac {d^3\,{\left (c+d\,x\right )}^{3/2}}{3\,b}-\frac {d^3\,{\left (c+d\,x\right )}^{5/2}}{8\,\left (a\,d-b\,c\right )}+\frac {d^3\,\left (a\,d-b\,c\right )\,\sqrt {c+d\,x}}{8\,b^2}}{\left (c+d\,x\right )\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )+b^3\,{\left (c+d\,x\right )}^3-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^2+a^3\,d^3-b^3\,c^3+3\,a\,b^2\,c^2\,d-3\,a^2\,b\,c\,d^2} \] Input:

int((c + d*x)^(3/2)/(a + b*x)^4,x)
 

Output:

(d^3*atan((b^(1/2)*(c + d*x)^(1/2))/(a*d - b*c)^(1/2)))/(8*b^(5/2)*(a*d - 
b*c)^(3/2)) - ((d^3*(c + d*x)^(3/2))/(3*b) - (d^3*(c + d*x)^(5/2))/(8*(a*d 
 - b*c)) + (d^3*(a*d - b*c)*(c + d*x)^(1/2))/(8*b^2))/((c + d*x)*(3*b^3*c^ 
2 + 3*a^2*b*d^2 - 6*a*b^2*c*d) + b^3*(c + d*x)^3 - (3*b^3*c - 3*a*b^2*d)*( 
c + d*x)^2 + a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.46 \[ \int \frac {(c+d x)^{3/2}}{(a+b x)^4} \, dx=\frac {3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} d^{3}+9 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b \,d^{3} x +9 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{2} d^{3} x^{2}+3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{3} d^{3} x^{3}-3 \sqrt {d x +c}\, a^{3} b \,d^{3}+\sqrt {d x +c}\, a^{2} b^{2} c \,d^{2}-8 \sqrt {d x +c}\, a^{2} b^{2} d^{3} x +10 \sqrt {d x +c}\, a \,b^{3} c^{2} d +22 \sqrt {d x +c}\, a \,b^{3} c \,d^{2} x +3 \sqrt {d x +c}\, a \,b^{3} d^{3} x^{2}-8 \sqrt {d x +c}\, b^{4} c^{3}-14 \sqrt {d x +c}\, b^{4} c^{2} d x -3 \sqrt {d x +c}\, b^{4} c \,d^{2} x^{2}}{24 b^{3} \left (a^{2} b^{3} d^{2} x^{3}-2 a \,b^{4} c d \,x^{3}+b^{5} c^{2} x^{3}+3 a^{3} b^{2} d^{2} x^{2}-6 a^{2} b^{3} c d \,x^{2}+3 a \,b^{4} c^{2} x^{2}+3 a^{4} b \,d^{2} x -6 a^{3} b^{2} c d x +3 a^{2} b^{3} c^{2} x +a^{5} d^{2}-2 a^{4} b c d +a^{3} b^{2} c^{2}\right )} \] Input:

int((d*x+c)^(3/2)/(b*x+a)^4,x)
 

Output:

(3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c) 
))*a**3*d**3 + 9*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*s 
qrt(a*d - b*c)))*a**2*b*d**3*x + 9*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(c + 
d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a*b**2*d**3*x**2 + 3*sqrt(b)*sqrt(a*d - 
 b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**3*d**3*x**3 - 3 
*sqrt(c + d*x)*a**3*b*d**3 + sqrt(c + d*x)*a**2*b**2*c*d**2 - 8*sqrt(c + d 
*x)*a**2*b**2*d**3*x + 10*sqrt(c + d*x)*a*b**3*c**2*d + 22*sqrt(c + d*x)*a 
*b**3*c*d**2*x + 3*sqrt(c + d*x)*a*b**3*d**3*x**2 - 8*sqrt(c + d*x)*b**4*c 
**3 - 14*sqrt(c + d*x)*b**4*c**2*d*x - 3*sqrt(c + d*x)*b**4*c*d**2*x**2)/( 
24*b**3*(a**5*d**2 - 2*a**4*b*c*d + 3*a**4*b*d**2*x + a**3*b**2*c**2 - 6*a 
**3*b**2*c*d*x + 3*a**3*b**2*d**2*x**2 + 3*a**2*b**3*c**2*x - 6*a**2*b**3* 
c*d*x**2 + a**2*b**3*d**2*x**3 + 3*a*b**4*c**2*x**2 - 2*a*b**4*c*d*x**3 + 
b**5*c**2*x**3))