3.5.79 \(\int x^{5/2} \sqrt {a+b x} (A+B x) \, dx\) [479]

3.5.79.1 Optimal result
3.5.79.2 Mathematica [A] (verified)
3.5.79.3 Rubi [A] (verified)
3.5.79.4 Maple [A] (verified)
3.5.79.5 Fricas [A] (verification not implemented)
3.5.79.6 Sympy [A] (verification not implemented)
3.5.79.7 Maxima [A] (verification not implemented)
3.5.79.8 Giac [F(-1)]
3.5.79.9 Mupad [F(-1)]

3.5.79.1 Optimal result

Integrand size = 20, antiderivative size = 192 \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\frac {a^3 (10 A b-7 a B) \sqrt {x} \sqrt {a+b x}}{128 b^4}-\frac {a^2 (10 A b-7 a B) x^{3/2} \sqrt {a+b x}}{192 b^3}+\frac {a (10 A b-7 a B) x^{5/2} \sqrt {a+b x}}{240 b^2}+\frac {(10 A b-7 a B) x^{7/2} \sqrt {a+b x}}{40 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}-\frac {a^4 (10 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{128 b^{9/2}} \]

output
1/5*B*x^(7/2)*(b*x+a)^(3/2)/b-1/128*a^4*(10*A*b-7*B*a)*arctanh(b^(1/2)*x^( 
1/2)/(b*x+a)^(1/2))/b^(9/2)-1/192*a^2*(10*A*b-7*B*a)*x^(3/2)*(b*x+a)^(1/2) 
/b^3+1/240*a*(10*A*b-7*B*a)*x^(5/2)*(b*x+a)^(1/2)/b^2+1/40*(10*A*b-7*B*a)* 
x^(7/2)*(b*x+a)^(1/2)/b+1/128*a^3*(10*A*b-7*B*a)*x^(1/2)*(b*x+a)^(1/2)/b^4
 
3.5.79.2 Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.92 \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\frac {\sqrt {b} \sqrt {x} \sqrt {a+b x} \left (-105 a^4 B+16 a b^3 x^2 (5 A+3 B x)+96 b^4 x^3 (5 A+4 B x)+10 a^3 b (15 A+7 B x)-4 a^2 b^2 x (25 A+14 B x)\right )+300 a^4 A b \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )+210 a^5 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{1920 b^{9/2}} \]

input
Integrate[x^(5/2)*Sqrt[a + b*x]*(A + B*x),x]
 
output
(Sqrt[b]*Sqrt[x]*Sqrt[a + b*x]*(-105*a^4*B + 16*a*b^3*x^2*(5*A + 3*B*x) + 
96*b^4*x^3*(5*A + 4*B*x) + 10*a^3*b*(15*A + 7*B*x) - 4*a^2*b^2*x*(25*A + 1 
4*B*x)) + 300*a^4*A*b*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])] 
 + 210*a^5*B*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])])/(1920* 
b^(9/2))
 
3.5.79.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {90, 60, 60, 60, 60, 65, 219}

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.

\(\displaystyle \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {(10 A b-7 a B) \int x^{5/2} \sqrt {a+b x}dx}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \int \frac {x^{5/2}}{\sqrt {a+b x}}dx+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \int \frac {x^{3/2}}{\sqrt {a+b x}}dx}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \int \frac {\sqrt {x}}{\sqrt {a+b x}}dx}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{\sqrt {x} \sqrt {a+b x}}dx}{2 b}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \int \frac {1}{1-\frac {b x}{a+b x}}d\frac {\sqrt {x}}{\sqrt {a+b x}}}{b}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(10 A b-7 a B) \left (\frac {1}{8} a \left (\frac {x^{5/2} \sqrt {a+b x}}{3 b}-\frac {5 a \left (\frac {x^{3/2} \sqrt {a+b x}}{2 b}-\frac {3 a \left (\frac {\sqrt {x} \sqrt {a+b x}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{b^{3/2}}\right )}{4 b}\right )}{6 b}\right )+\frac {1}{4} x^{7/2} \sqrt {a+b x}\right )}{10 b}+\frac {B x^{7/2} (a+b x)^{3/2}}{5 b}\)

input
Int[x^(5/2)*Sqrt[a + b*x]*(A + B*x),x]
 
output
(B*x^(7/2)*(a + b*x)^(3/2))/(5*b) + ((10*A*b - 7*a*B)*((x^(7/2)*Sqrt[a + b 
*x])/4 + (a*((x^(5/2)*Sqrt[a + b*x])/(3*b) - (5*a*((x^(3/2)*Sqrt[a + b*x]) 
/(2*b) - (3*a*((Sqrt[x]*Sqrt[a + b*x])/b - (a*ArcTanh[(Sqrt[b]*Sqrt[x])/Sq 
rt[a + b*x]])/b^(3/2)))/(4*b)))/(6*b)))/8))/(10*b)
 

3.5.79.3.1 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 65
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 
3.5.79.4 Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.83

method result size
risch \(\frac {\left (384 B \,x^{4} b^{4}+480 A \,x^{3} b^{4}+48 B \,x^{3} a \,b^{3}+80 A \,x^{2} a \,b^{3}-56 B \,x^{2} a^{2} b^{2}-100 A x \,a^{2} b^{2}+70 B x \,a^{3} b +150 A \,a^{3} b -105 B \,a^{4}\right ) \sqrt {x}\, \sqrt {b x +a}}{1920 b^{4}}-\frac {a^{4} \left (10 A b -7 B a \right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) \sqrt {x \left (b x +a \right )}}{256 b^{\frac {9}{2}} \sqrt {x}\, \sqrt {b x +a}}\) \(159\)
default \(-\frac {\sqrt {x}\, \sqrt {b x +a}\, \left (-768 B \,b^{\frac {9}{2}} x^{4} \sqrt {x \left (b x +a \right )}-960 A \,b^{\frac {9}{2}} x^{3} \sqrt {x \left (b x +a \right )}-96 B a \,b^{\frac {7}{2}} x^{3} \sqrt {x \left (b x +a \right )}-160 A a \,b^{\frac {7}{2}} x^{2} \sqrt {x \left (b x +a \right )}+112 B \,a^{2} b^{\frac {5}{2}} x^{2} \sqrt {x \left (b x +a \right )}+200 A \,b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, a^{2} x -140 B \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3} x +150 A \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{4} b -300 A \,b^{\frac {3}{2}} \sqrt {x \left (b x +a \right )}\, a^{3}-105 B \ln \left (\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{5}+210 B \sqrt {b}\, \sqrt {x \left (b x +a \right )}\, a^{4}\right )}{3840 b^{\frac {9}{2}} \sqrt {x \left (b x +a \right )}}\) \(260\)

input
int(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 
output
1/1920*(384*B*b^4*x^4+480*A*b^4*x^3+48*B*a*b^3*x^3+80*A*a*b^3*x^2-56*B*a^2 
*b^2*x^2-100*A*a^2*b^2*x+70*B*a^3*b*x+150*A*a^3*b-105*B*a^4)*x^(1/2)*(b*x+ 
a)^(1/2)/b^4-1/256*a^4*(10*A*b-7*B*a)/b^(9/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^ 
2+a*x)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)
 
3.5.79.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.54 \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\left [-\frac {15 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {b} \log \left (2 \, b x - 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (384 \, B b^{5} x^{4} - 105 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (B a b^{4} + 10 \, A b^{5}\right )} x^{3} - 8 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{3840 \, b^{5}}, -\frac {15 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (384 \, B b^{5} x^{4} - 105 \, B a^{4} b + 150 \, A a^{3} b^{2} + 48 \, {\left (B a b^{4} + 10 \, A b^{5}\right )} x^{3} - 8 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{2} + 10 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{1920 \, b^{5}}\right ] \]

input
integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="fricas")
 
output
[-1/3840*(15*(7*B*a^5 - 10*A*a^4*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sq 
rt(b)*sqrt(x) + a) - 2*(384*B*b^5*x^4 - 105*B*a^4*b + 150*A*a^3*b^2 + 48*( 
B*a*b^4 + 10*A*b^5)*x^3 - 8*(7*B*a^2*b^3 - 10*A*a*b^4)*x^2 + 10*(7*B*a^3*b 
^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^5, -1/1920*(15*(7*B*a^5 - 1 
0*A*a^4*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (384*B*b^ 
5*x^4 - 105*B*a^4*b + 150*A*a^3*b^2 + 48*(B*a*b^4 + 10*A*b^5)*x^3 - 8*(7*B 
*a^2*b^3 - 10*A*a*b^4)*x^2 + 10*(7*B*a^3*b^2 - 10*A*a^2*b^3)*x)*sqrt(b*x + 
 a)*sqrt(x))/b^5]
 
3.5.79.6 Sympy [A] (verification not implemented)

Time = 127.16 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.86 \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\frac {5 A a^{\frac {7}{2}} \sqrt {x}}{64 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {5 A a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {A a^{\frac {3}{2}} x^{\frac {5}{2}}}{96 b \sqrt {1 + \frac {b x}{a}}} + \frac {7 A \sqrt {a} x^{\frac {7}{2}}}{24 \sqrt {1 + \frac {b x}{a}}} - \frac {5 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {7}{2}}} + \frac {A b x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} - \frac {7 B a^{\frac {9}{2}} \sqrt {x}}{128 b^{4} \sqrt {1 + \frac {b x}{a}}} - \frac {7 B a^{\frac {7}{2}} x^{\frac {3}{2}}}{384 b^{3} \sqrt {1 + \frac {b x}{a}}} + \frac {7 B a^{\frac {5}{2}} x^{\frac {5}{2}}}{960 b^{2} \sqrt {1 + \frac {b x}{a}}} - \frac {B a^{\frac {3}{2}} x^{\frac {7}{2}}}{240 b \sqrt {1 + \frac {b x}{a}}} + \frac {9 B \sqrt {a} x^{\frac {9}{2}}}{40 \sqrt {1 + \frac {b x}{a}}} + \frac {7 B a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{128 b^{\frac {9}{2}}} + \frac {B b x^{\frac {11}{2}}}{5 \sqrt {a} \sqrt {1 + \frac {b x}{a}}} \]

input
integrate(x**(5/2)*(B*x+A)*(b*x+a)**(1/2),x)
 
output
5*A*a**(7/2)*sqrt(x)/(64*b**3*sqrt(1 + b*x/a)) + 5*A*a**(5/2)*x**(3/2)/(19 
2*b**2*sqrt(1 + b*x/a)) - A*a**(3/2)*x**(5/2)/(96*b*sqrt(1 + b*x/a)) + 7*A 
*sqrt(a)*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*A*a**4*asinh(sqrt(b)*sqrt(x)/sq 
rt(a))/(64*b**(7/2)) + A*b*x**(9/2)/(4*sqrt(a)*sqrt(1 + b*x/a)) - 7*B*a**( 
9/2)*sqrt(x)/(128*b**4*sqrt(1 + b*x/a)) - 7*B*a**(7/2)*x**(3/2)/(384*b**3* 
sqrt(1 + b*x/a)) + 7*B*a**(5/2)*x**(5/2)/(960*b**2*sqrt(1 + b*x/a)) - B*a* 
*(3/2)*x**(7/2)/(240*b*sqrt(1 + b*x/a)) + 9*B*sqrt(a)*x**(9/2)/(40*sqrt(1 
+ b*x/a)) + 7*B*a**5*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(128*b**(9/2)) + B*b*x 
**(11/2)/(5*sqrt(a)*sqrt(1 + b*x/a))
 
3.5.79.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.26 \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B x^{2}}{5 \, b} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{3} x}{64 \, b^{3}} - \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a x}{40 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a x} A a^{2} x}{32 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A x}{4 \, b} + \frac {7 \, B a^{5} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{256 \, b^{\frac {9}{2}}} - \frac {5 \, A a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} - \frac {7 \, \sqrt {b x^{2} + a x} B a^{4}}{128 \, b^{4}} + \frac {7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a^{2}}{48 \, b^{3}} + \frac {5 \, \sqrt {b x^{2} + a x} A a^{3}}{64 \, b^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} A a}{24 \, b^{2}} \]

input
integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="maxima")
 
output
1/5*(b*x^2 + a*x)^(3/2)*B*x^2/b - 7/64*sqrt(b*x^2 + a*x)*B*a^3*x/b^3 - 7/4 
0*(b*x^2 + a*x)^(3/2)*B*a*x/b^2 + 5/32*sqrt(b*x^2 + a*x)*A*a^2*x/b^2 + 1/4 
*(b*x^2 + a*x)^(3/2)*A*x/b + 7/256*B*a^5*log(2*b*x + a + 2*sqrt(b*x^2 + a* 
x)*sqrt(b))/b^(9/2) - 5/128*A*a^4*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt 
(b))/b^(7/2) - 7/128*sqrt(b*x^2 + a*x)*B*a^4/b^4 + 7/48*(b*x^2 + a*x)^(3/2 
)*B*a^2/b^3 + 5/64*sqrt(b*x^2 + a*x)*A*a^3/b^3 - 5/24*(b*x^2 + a*x)^(3/2)* 
A*a/b^2
 
3.5.79.8 Giac [F(-1)]

Timed out. \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\text {Timed out} \]

input
integrate(x^(5/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="giac")
 
output
Timed out
 
3.5.79.9 Mupad [F(-1)]

Timed out. \[ \int x^{5/2} \sqrt {a+b x} (A+B x) \, dx=\int x^{5/2}\,\left (A+B\,x\right )\,\sqrt {a+b\,x} \,d x \]

input
int(x^(5/2)*(A + B*x)*(a + b*x)^(1/2),x)
 
output
int(x^(5/2)*(A + B*x)*(a + b*x)^(1/2), x)