\(\int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx\) [343]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 177 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=-\frac {2 A}{7 a x^{7/2} (a+b x)^{3/2}}-\frac {2 (10 A b-7 a B)}{21 a^2 x^{5/2} (a+b x)^{3/2}}-\frac {16 (10 A b-7 a B)}{21 a^3 x^{5/2} \sqrt {a+b x}}+\frac {32 (10 A b-7 a B) \sqrt {a+b x}}{35 a^4 x^{5/2}}-\frac {128 b (10 A b-7 a B) \sqrt {a+b x}}{105 a^5 x^{3/2}}+\frac {256 b^2 (10 A b-7 a B) \sqrt {a+b x}}{105 a^6 \sqrt {x}} \] Output:

-2/7*A/a/x^(7/2)/(b*x+a)^(3/2)-2/21*(10*A*b-7*B*a)/a^2/x^(5/2)/(b*x+a)^(3/ 
2)-16/21*(10*A*b-7*B*a)/a^3/x^(5/2)/(b*x+a)^(1/2)+32/35*(10*A*b-7*B*a)*(b* 
x+a)^(1/2)/a^4/x^(5/2)-128/105*b*(10*A*b-7*B*a)*(b*x+a)^(1/2)/a^5/x^(3/2)+ 
256/105*b^2*(10*A*b-7*B*a)*(b*x+a)^(1/2)/a^6/x^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.64 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=-\frac {2 \left (-1280 A b^5 x^5+128 a b^4 x^4 (-15 A+7 B x)+3 a^5 (5 A+7 B x)+96 a^2 b^3 x^3 (-5 A+14 B x)+16 a^3 b^2 x^2 (5 A+21 B x)-2 a^4 b x (15 A+28 B x)\right )}{105 a^6 x^{7/2} (a+b x)^{3/2}} \] Input:

Integrate[(A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x]
 

Output:

(-2*(-1280*A*b^5*x^5 + 128*a*b^4*x^4*(-15*A + 7*B*x) + 3*a^5*(5*A + 7*B*x) 
 + 96*a^2*b^3*x^3*(-5*A + 14*B*x) + 16*a^3*b^2*x^2*(5*A + 21*B*x) - 2*a^4* 
b*x*(15*A + 28*B*x)))/(105*a^6*x^(7/2)*(a + b*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {87, 55, 55, 55, 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.

\(\displaystyle \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

Input:

Int[(A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x]
 

Output:

(-2*A)/(7*a*x^(7/2)*(a + b*x)^(3/2)) - ((10*A*b - 7*a*B)*(2/(3*a*x^(5/2)*( 
a + b*x)^(3/2)) + (8*(2/(a*x^(5/2)*Sqrt[a + b*x]) + (6*((-2*Sqrt[a + b*x]) 
/(5*a*x^(5/2)) - (4*b*((-2*Sqrt[a + b*x])/(3*a*x^(3/2)) + (4*b*Sqrt[a + b* 
x])/(3*a^2*Sqrt[x])))/(5*a)))/a))/(3*a)))/(7*a)
 

Defintions of rubi rules used

rule 48
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]
 

rule 55
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])
 

rule 87
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]))))
 
Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.68

method result size
risch \(-\frac {2 \sqrt {b x +a}\, \left (-790 A \,b^{3} x^{3}+511 B a \,b^{2} x^{3}+185 a A \,b^{2} x^{2}-98 B \,a^{2} b \,x^{2}-60 a^{2} A b x +21 B \,a^{3} x +15 a^{3} A \right )}{105 a^{6} x^{\frac {7}{2}}}+\frac {2 b^{3} \left (14 A \,b^{2} x -11 B a b x +15 a b A -12 a^{2} B \right ) \sqrt {x}}{3 \left (b x +a \right )^{\frac {3}{2}} a^{6}}\) \(121\)
gosper \(-\frac {2 \left (-1280 A \,b^{5} x^{5}+896 B a \,b^{4} x^{5}-1920 a A \,b^{4} x^{4}+1344 B \,a^{2} b^{3} x^{4}-480 a^{2} A \,b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+80 a^{3} A \,b^{2} x^{2}-56 B \,a^{4} b \,x^{2}-30 a^{4} A b x +21 B \,a^{5} x +15 a^{5} A \right )}{105 x^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}} a^{6}}\) \(125\)
default \(-\frac {2 \left (-1280 A \,b^{5} x^{5}+896 B a \,b^{4} x^{5}-1920 a A \,b^{4} x^{4}+1344 B \,a^{2} b^{3} x^{4}-480 a^{2} A \,b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+80 a^{3} A \,b^{2} x^{2}-56 B \,a^{4} b \,x^{2}-30 a^{4} A b x +21 B \,a^{5} x +15 a^{5} A \right )}{105 x^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}} a^{6}}\) \(125\)
orering \(-\frac {2 \left (-1280 A \,b^{5} x^{5}+896 B a \,b^{4} x^{5}-1920 a A \,b^{4} x^{4}+1344 B \,a^{2} b^{3} x^{4}-480 a^{2} A \,b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+80 a^{3} A \,b^{2} x^{2}-56 B \,a^{4} b \,x^{2}-30 a^{4} A b x +21 B \,a^{5} x +15 a^{5} A \right )}{105 x^{\frac {7}{2}} \left (b x +a \right )^{\frac {3}{2}} a^{6}}\) \(125\)

Input:

int((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/105*(b*x+a)^(1/2)*(-790*A*b^3*x^3+511*B*a*b^2*x^3+185*A*a*b^2*x^2-98*B* 
a^2*b*x^2-60*A*a^2*b*x+21*B*a^3*x+15*A*a^3)/a^6/x^(7/2)+2/3*b^3*(14*A*b^2* 
x-11*B*a*b*x+15*A*a*b-12*B*a^2)*x^(1/2)/(b*x+a)^(3/2)/a^6
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=-\frac {2 \, {\left (15 \, A a^{5} + 128 \, {\left (7 \, B a b^{4} - 10 \, A b^{5}\right )} x^{5} + 192 \, {\left (7 \, B a^{2} b^{3} - 10 \, A a b^{4}\right )} x^{4} + 48 \, {\left (7 \, B a^{3} b^{2} - 10 \, A a^{2} b^{3}\right )} x^{3} - 8 \, {\left (7 \, B a^{4} b - 10 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (7 \, B a^{5} - 10 \, A a^{4} b\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{105 \, {\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}} \] Input:

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

-2/105*(15*A*a^5 + 128*(7*B*a*b^4 - 10*A*b^5)*x^5 + 192*(7*B*a^2*b^3 - 10* 
A*a*b^4)*x^4 + 48*(7*B*a^3*b^2 - 10*A*a^2*b^3)*x^3 - 8*(7*B*a^4*b - 10*A*a 
^3*b^2)*x^2 + 3*(7*B*a^5 - 10*A*a^4*b)*x)*sqrt(b*x + a)*sqrt(x)/(a^6*b^2*x 
^6 + 2*a^7*b*x^5 + a^8*x^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)/x**(9/2)/(b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=\frac {32 \, B b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, B b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {64 \, A b^{3} x}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{4}} + \frac {512 \, A b^{4} x}{21 \, \sqrt {b x^{2} + a x} a^{6}} + \frac {16 \, B b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, B b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {32 \, A b^{2}}{21 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} + \frac {256 \, A b^{3}}{21 \, \sqrt {b x^{2} + a x} a^{5}} - \frac {2 \, B}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} + \frac {4 \, A b}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2} x} - \frac {2 \, A}{7 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x^{2}} \] Input:

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

32/15*B*b^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*B*b^3*x/(sqrt(b*x^2 + a*x 
)*a^5) - 64/21*A*b^3*x/((b*x^2 + a*x)^(3/2)*a^4) + 512/21*A*b^4*x/(sqrt(b* 
x^2 + a*x)*a^6) + 16/15*B*b/((b*x^2 + a*x)^(3/2)*a^2) - 128/15*B*b^2/(sqrt 
(b*x^2 + a*x)*a^4) - 32/21*A*b^2/((b*x^2 + a*x)^(3/2)*a^3) + 256/21*A*b^3/ 
(sqrt(b*x^2 + a*x)*a^5) - 2/5*B/((b*x^2 + a*x)^(3/2)*a*x) + 4/7*A*b/((b*x^ 
2 + a*x)^(3/2)*a^2*x) - 2/7*A/((b*x^2 + a*x)^(3/2)*a*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (141) = 282\).

Time = 0.24 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.15 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=-\frac {2 \, {\left ({\left (b x + a\right )} {\left ({\left (b x + a\right )} {\left (\frac {{\left (511 \, B a^{13} b^{9} {\left | b \right |} - 790 \, A a^{12} b^{10} {\left | b \right |}\right )} {\left (b x + a\right )}}{a^{18} b^{4}} - \frac {7 \, {\left (233 \, B a^{14} b^{9} {\left | b \right |} - 365 \, A a^{13} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} + \frac {350 \, {\left (5 \, B a^{15} b^{9} {\left | b \right |} - 8 \, A a^{14} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} - \frac {210 \, {\left (3 \, B a^{16} b^{9} {\left | b \right |} - 5 \, A a^{15} b^{10} {\left | b \right |}\right )}}{a^{18} b^{4}}\right )} \sqrt {b x + a}}{105 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {7}{2}}} - \frac {4 \, {\left (9 \, B a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {9}{2}} + 24 \, B a^{2} {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {11}{2}} - 12 \, A {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac {11}{2}} + 11 \, B a^{3} b^{\frac {13}{2}} - 30 \, A a {\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac {13}{2}} - 14 \, A a^{2} b^{\frac {15}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} a^{5} {\left | b \right |}} \] Input:

integrate((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x, algorithm="giac")
 

Output:

-2/105*((b*x + a)*((b*x + a)*((511*B*a^13*b^9*abs(b) - 790*A*a^12*b^10*abs 
(b))*(b*x + a)/(a^18*b^4) - 7*(233*B*a^14*b^9*abs(b) - 365*A*a^13*b^10*abs 
(b))/(a^18*b^4)) + 350*(5*B*a^15*b^9*abs(b) - 8*A*a^14*b^10*abs(b))/(a^18* 
b^4)) - 210*(3*B*a^16*b^9*abs(b) - 5*A*a^15*b^10*abs(b))/(a^18*b^4))*sqrt( 
b*x + a)/((b*x + a)*b - a*b)^(7/2) - 4/3*(9*B*a*(sqrt(b*x + a)*sqrt(b) - s 
qrt((b*x + a)*b - a*b))^4*b^(9/2) + 24*B*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt 
((b*x + a)*b - a*b))^2*b^(11/2) - 12*A*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x 
+ a)*b - a*b))^4*b^(11/2) + 11*B*a^3*b^(13/2) - 30*A*a*(sqrt(b*x + a)*sqrt 
(b) - sqrt((b*x + a)*b - a*b))^2*b^(13/2) - 14*A*a^2*b^(15/2))/(((sqrt(b*x 
 + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*a^5*abs(b))
 

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=-\frac {\sqrt {a+b\,x}\,\left (\frac {2\,A}{7\,a\,b^2}-\frac {32\,x^3\,\left (10\,A\,b-7\,B\,a\right )}{35\,a^4}+\frac {16\,x^2\,\left (10\,A\,b-7\,B\,a\right )}{105\,a^3\,b}-\frac {x^5\,\left (2560\,A\,b^5-1792\,B\,a\,b^4\right )}{105\,a^6\,b^2}-\frac {128\,b\,x^4\,\left (10\,A\,b-7\,B\,a\right )}{35\,a^5}+\frac {x\,\left (42\,B\,a^5-60\,A\,a^4\,b\right )}{105\,a^6\,b^2}\right )}{x^{11/2}+\frac {2\,a\,x^{9/2}}{b}+\frac {a^2\,x^{7/2}}{b^2}} \] Input:

int((A + B*x)/(x^(9/2)*(a + b*x)^(5/2)),x)
 

Output:

-((a + b*x)^(1/2)*((2*A)/(7*a*b^2) - (32*x^3*(10*A*b - 7*B*a))/(35*a^4) + 
(16*x^2*(10*A*b - 7*B*a))/(105*a^3*b) - (x^5*(2560*A*b^5 - 1792*B*a*b^4))/ 
(105*a^6*b^2) - (128*b*x^4*(10*A*b - 7*B*a))/(35*a^5) + (x*(42*B*a^5 - 60* 
A*a^4*b))/(105*a^6*b^2)))/(x^(11/2) + (2*a*x^(9/2))/b + (a^2*x^(7/2))/b^2)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.47 \[ \int \frac {A+B x}{x^{9/2} (a+b x)^{5/2}} \, dx=\frac {-\frac {256 \sqrt {b}\, \sqrt {b x +a}\, b^{3} x^{4}}{35}-\frac {2 \sqrt {x}\, a^{4}}{7}+\frac {16 \sqrt {x}\, a^{3} b x}{35}-\frac {32 \sqrt {x}\, a^{2} b^{2} x^{2}}{35}+\frac {128 \sqrt {x}\, a \,b^{3} x^{3}}{35}+\frac {256 \sqrt {x}\, b^{4} x^{4}}{35}}{\sqrt {b x +a}\, a^{5} x^{4}} \] Input:

int((B*x+A)/x^(9/2)/(b*x+a)^(5/2),x)
 

Output:

(2*( - 128*sqrt(b)*sqrt(a + b*x)*b**3*x**4 - 5*sqrt(x)*a**4 + 8*sqrt(x)*a* 
*3*b*x - 16*sqrt(x)*a**2*b**2*x**2 + 64*sqrt(x)*a*b**3*x**3 + 128*sqrt(x)* 
b**4*x**4))/(35*sqrt(a + b*x)*a**5*x**4)