\(\int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\) [325]

Optimal result

Integrand size = 20, antiderivative size = 171 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=-\frac {(b c-a d) (b c+5 a d) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d}-\frac {(b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d}+\frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(b c-a d)^2 (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}} \] Output:

-1/8*(-a*d+b*c)*(5*a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d-1/12*(5*a*d+ 
b*c)*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b^2/d+1/3*(b*x+a)^(1/2)*(d*x+c)^(5/2)/b/d 
-1/8*(-a*d+b*c)^2*(5*a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c 
)^(1/2))/b^(7/2)/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-2 a b d (11 c+5 d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{24 b^3 d}-\frac {(b c-a d)^2 (b c+5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 - 2*a*b*d*(11*c + 5*d*x) + b^2*(3 
*c^2 + 14*c*d*x + 8*d^2*x^2)))/(24*b^3*d) - ((b*c - a*d)^2*(b*c + 5*a*d)*A 
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(7/2)*d^(3/2 
))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 60, 60, 66, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}}dx}{6 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {a+b x} (c+d x)^{5/2}}{3 b d}-\frac {(5 a d+b c) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}\right )}{6 b d}\)

Input:

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

Output:

(Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b*d) - ((b*c + 5*a*d)*((Sqrt[a + b*x]*( 
c + d*x)^(3/2))/(2*b) + (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + 
((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^ 
(3/2)*Sqrt[d])))/(4*b)))/(6*b*d)
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(139)=278\).

Time = 0.22 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.31

method	result	size
default	\(-\frac {\sqrt {x d +c}\, \sqrt {b x +a}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} d^{3}-27 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{3}+20 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b \,d^{2} x -28 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c d x -30 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} d^{2}+44 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b c d -6 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c^{2}\right )}{48 d \,b^{3} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}}\)	\(395\)

Input:

int(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2) 
*(d*b)^(1/2)+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+ 
b*c)/(d*b)^(1/2))*a^3*d^3-27*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d* 
b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b*c*d^2+9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d 
*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^2*c^2*d+3*ln(1/2*(2*b*d 
*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^3*c^3+20* 
(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*d^2*x-28*(d*b)^(1/2)*((b*x+a)*(d*x 
+c))^(1/2)*b^2*c*d*x-30*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+44*(d* 
b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d-6*(d*b)^(1/2)*((b*x+a)*(d*x+c))^( 
1/2)*b^2*c^2)/d/b^3/((b*x+a)*(d*x+c))^(1/2)/(d*b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.41 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\left [\frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{2}}, \frac {3 \, {\left (b^{3} c^{3} + 3 \, a b^{2} c^{2} d - 9 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{2}}\right ] \] Input:

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/96*(3*(b^3*c^3 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(b*d)*l 
og(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d) 
*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8*b 
^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 
- 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^2), 1/48*(3*(b^3*c^3 
 + 3*a*b^2*c^2*d - 9*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*b*d 
*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b* 
c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(8*b^3*d^3*x^2 + 3*b^3*c^2*d - 22*a*b^2* 
c*d^2 + 15*a^2*b*d^3 + 2*(7*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt 
(d*x + c))/(b^4*d^2)]
 

Sympy [F]

\[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\int \frac {x \left (c + d x\right )^{\frac {3}{2}}}{\sqrt {a + b x}}\, dx \] Input:

integrate(x*(d*x+c)**(3/2)/(b*x+a)**(1/2),x)
 

Output:

Integral(x*(c + d*x)**(3/2)/sqrt(a + b*x), x)
 

Maxima [F(-2)]

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

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (139) = 278\).

Time = 0.17 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.91 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {\frac {6 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} c {\left | b \right |}}{b^{3}} + \frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, {\left (4 \, b x + 4 \, a + \frac {b c d^{3} - 13 \, a d^{4}}{d^{4}}\right )} {\left (b x + a\right )} - \frac {3 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b c d^{3} - 11 \, a^{2} d^{4}\right )}}{d^{4}}\right )} \sqrt {b x + a} - \frac {3 \, {\left (b^{4} c^{3} + a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - 5 \, a^{3} b d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} d {\left | b \right |}}{b^{4}}}{24 \, b} \] Input:

integrate(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/24*(6*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d 
^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sq 
rt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d 
))*c*abs(b)/b^3 + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*b*x + 4*a + ( 
b*c*d^3 - 13*a*d^4)/d^4)*(b*x + a) - 3*(b^2*c^2*d^2 + 2*a*b*c*d^3 - 11*a^2 
*d^4)/d^4)*sqrt(b*x + a) - 3*(b^4*c^3 + a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - 5* 
a^3*b*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - 
 a*b*d)))/(sqrt(b*d)*d^2))*d*abs(b)/b^4)/b
 

Mupad [F(-1)]

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

int((x*(c + d*x)^(3/2))/(a + b*x)^(1/2),x)
 

Output:

int((x*(c + d*x)^(3/2))/(a + b*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.87 \[ \int \frac {x (c+d x)^{3/2}}{\sqrt {a+b x}} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}-22 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}-10 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x +3 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d +14 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}+27 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}-9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d -3 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b^{4} d^{2}} \] Input:

int(x*(d*x+c)^(3/2)/(b*x+a)^(1/2),x)
 

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**3 - 22*sqrt(c + d*x)*sqrt(a + b* 
x)*a*b**2*c*d**2 - 10*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x + 3*sqrt(c 
 + d*x)*sqrt(a + b*x)*b**3*c**2*d + 14*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c* 
d**2*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**3*x**2 - 15*sqrt(d)*sqrt(b) 
*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3 
*d**3 + 27*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d 
*x))/sqrt(a*d - b*c))*a**2*b*c*d**2 - 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt( 
a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*d - 3*sqrt( 
d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - 
b*c))*b**3*c**3)/(24*b**4*d**2)