\(\int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx\) [252]

Optimal result

Integrand size = 18, antiderivative size = 117 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\frac {1}{2} b (5 A b+8 a B) \sqrt {a+b x}-\frac {a (5 A b+4 a B) \sqrt {a+b x}}{4 x}+\frac {2}{3} b B (a+b x)^{3/2}-\frac {A (a+b x)^{5/2}}{2 x^2}-\frac {5}{4} \sqrt {a} b (3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \] Output:

1/2*b*(5*A*b+8*B*a)*(b*x+a)^(1/2)-1/4*a*(5*A*b+4*B*a)*(b*x+a)^(1/2)/x+2/3* 
b*B*(b*x+a)^(3/2)-1/2*A*(b*x+a)^(5/2)/x^2-5/4*a^(1/2)*b*(3*A*b+4*B*a)*arct 
anh((b*x+a)^(1/2)/a^(1/2))
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\frac {\sqrt {a+b x} \left (8 b^2 x^2 (3 A+B x)-6 a^2 (A+2 B x)+a b x (-27 A+56 B x)\right )}{12 x^2}-\frac {5}{4} \sqrt {a} b (3 A b+4 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \] Input:

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^3,x]
 

Output:

(Sqrt[a + b*x]*(8*b^2*x^2*(3*A + B*x) - 6*a^2*(A + 2*B*x) + a*b*x*(-27*A + 
 56*B*x)))/(12*x^2) - (5*Sqrt[a]*b*(3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/S 
qrt[a]])/4
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {87, 51, 60, 60, 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.

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {(4 a B+3 A b) \int \frac {(a+b x)^{5/2}}{x^2}dx}{4 a}-\frac {A (a+b x)^{7/2}}{2 a x^2}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x)^(5/2)*(A + B*x))/x^3,x]
 

Output:

-1/2*(A*(a + b*x)^(7/2))/(a*x^2) + ((3*A*b + 4*a*B)*(-((a + b*x)^(5/2)/x) 
+ (5*b*((2*(a + b*x)^(3/2))/3 + a*(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqr 
t[a + b*x]/Sqrt[a]])))/2))/(4*a)
 

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 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 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 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]))))
 

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 = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {\frac {15 a b \left (A b +\frac {4 B a}{3}\right ) x^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2}+\sqrt {b x +a}\, \left (-4 b^{2} \left (\frac {B x}{3}+A \right ) x^{2} \sqrt {a}+\left (-\frac {28 b B \,x^{2}}{3}+\left (2 B a +\frac {9 A b}{2}\right ) x +A a \right ) a^{\frac {3}{2}}\right )}{2 \sqrt {a}\, x^{2}}\)	\(89\)
risch	\(-\frac {a \sqrt {b x +a}\, \left (9 A b x +4 B a x +2 A a \right )}{4 x^{2}}+\frac {b \left (\frac {16 B \left (b x +a \right )^{\frac {3}{2}}}{3}+16 A b \sqrt {b x +a}+32 B a \sqrt {b x +a}-10 \sqrt {a}\, \left (3 A b +4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{8}\)	\(92\)
derivativedivides	\(2 b \left (\frac {B \left (b x +a \right )^{\frac {3}{2}}}{3}+A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-a \left (\frac {\left (\frac {9 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (3 A b +4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(111\)
default	\(2 b \left (\frac {B \left (b x +a \right )^{\frac {3}{2}}}{3}+A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-a \left (\frac {\left (\frac {9 A b}{8}+\frac {B a}{2}\right ) \left (b x +a \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a b A -\frac {1}{2} a^{2} B \right ) \sqrt {b x +a}}{b^{2} x^{2}}+\frac {5 \left (3 A b +4 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(111\)

Input:

int((b*x+a)^(5/2)*(B*x+A)/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*(15/2*a*b*(A*b+4/3*B*a)*x^2*arctanh((b*x+a)^(1/2)/a^(1/2))+(b*x+a)^(1 
/2)*(-4*b^2*(1/3*B*x+A)*x^2*a^(1/2)+(-28/3*b*B*x^2+(2*B*a+9/2*A*b)*x+A*a)* 
a^(3/2)))/a^(1/2)/x^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\left [\frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {a} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt {b x + a}}{24 \, x^{2}}, \frac {15 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \, {\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \, {\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt {b x + a}}{12 \, x^{2}}\right ] \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="fricas")
 

Output:

[1/24*(15*(4*B*a*b + 3*A*b^2)*sqrt(a)*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt( 
a) + 2*a)/x) + 2*(8*B*b^2*x^3 - 6*A*a^2 + 8*(7*B*a*b + 3*A*b^2)*x^2 - 3*(4 
*B*a^2 + 9*A*a*b)*x)*sqrt(b*x + a))/x^2, 1/12*(15*(4*B*a*b + 3*A*b^2)*sqrt 
(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x + a)) + (8*B*b^2*x^3 - 6*A*a^2 + 8*(7*B* 
a*b + 3*A*b^2)*x^2 - 3*(4*B*a^2 + 9*A*a*b)*x)*sqrt(b*x + a))/x^2]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 41.48 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.70 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=- \frac {7 A \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {A a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {3 A a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {2 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} - \frac {A a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + A b^{2} \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) - B a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} + 2 B a b \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} \frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x+a)**(5/2)*(B*x+A)/x**3,x)
 

Output:

-7*A*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - A*a**3/(2*sqrt(b)*x 
**(5/2)*sqrt(a/(b*x) + 1)) - 3*A*a**2*sqrt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1 
)) - 2*A*a*b**(3/2)*sqrt(a/(b*x) + 1)/sqrt(x) - A*a*b**(3/2)/(4*sqrt(x)*sq 
rt(a/(b*x) + 1)) + A*b**2*Piecewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt 
(-a) + 2*sqrt(a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True)) - B*a**(3/2)*b* 
asinh(sqrt(a)/(sqrt(b)*sqrt(x))) - B*a**2*sqrt(b)*sqrt(a/(b*x) + 1)/sqrt(x 
) + 2*B*a*b*Piecewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*sqrt( 
a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True)) + B*b**2*Piecewise((2*(a + b* 
x)**(3/2)/(3*b), Ne(b, 0)), (sqrt(a)*x, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\frac {1}{24} \, {\left (\frac {15 \, {\left (4 \, B a + 3 \, A b\right )} \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{b} - \frac {6 \, {\left ({\left (4 \, B a^{2} + 9 \, A a b\right )} {\left (b x + a\right )}^{\frac {3}{2}} - {\left (4 \, B a^{3} + 7 \, A a^{2} b\right )} \sqrt {b x + a}\right )}}{{\left (b x + a\right )}^{2} b - 2 \, {\left (b x + a\right )} a b + a^{2} b} + \frac {16 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, {\left (2 \, B a + A b\right )} \sqrt {b x + a}\right )}}{b}\right )} b^{2} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="maxima")
 

Output:

1/24*(15*(4*B*a + 3*A*b)*sqrt(a)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + 
 a) + sqrt(a)))/b - 6*((4*B*a^2 + 9*A*a*b)*(b*x + a)^(3/2) - (4*B*a^3 + 7* 
A*a^2*b)*sqrt(b*x + a))/((b*x + a)^2*b - 2*(b*x + a)*a*b + a^2*b) + 16*((b 
*x + a)^(3/2)*B + 3*(2*B*a + A*b)*sqrt(b*x + a))/b)*b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\frac {8 \, {\left (b x + a\right )}^{\frac {3}{2}} B b^{2} + 48 \, \sqrt {b x + a} B a b^{2} + 24 \, \sqrt {b x + a} A b^{3} + \frac {15 \, {\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3 \, {\left (4 \, {\left (b x + a\right )}^{\frac {3}{2}} B a^{2} b^{2} - 4 \, \sqrt {b x + a} B a^{3} b^{2} + 9 \, {\left (b x + a\right )}^{\frac {3}{2}} A a b^{3} - 7 \, \sqrt {b x + a} A a^{2} b^{3}\right )}}{b^{2} x^{2}}}{12 \, b} \] Input:

integrate((b*x+a)^(5/2)*(B*x+A)/x^3,x, algorithm="giac")
 

Output:

1/12*(8*(b*x + a)^(3/2)*B*b^2 + 48*sqrt(b*x + a)*B*a*b^2 + 24*sqrt(b*x + a 
)*A*b^3 + 15*(4*B*a^2*b^2 + 3*A*a*b^3)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt 
(-a) - 3*(4*(b*x + a)^(3/2)*B*a^2*b^2 - 4*sqrt(b*x + a)*B*a^3*b^2 + 9*(b*x 
 + a)^(3/2)*A*a*b^3 - 7*sqrt(b*x + a)*A*a^2*b^3)/(b^2*x^2))/b
 

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.38 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\left (2\,A\,b^2+4\,B\,a\,b\right )\,\sqrt {a+b\,x}-\frac {\left (B\,a^2\,b+\frac {9\,A\,a\,b^2}{4}\right )\,{\left (a+b\,x\right )}^{3/2}-\left (B\,a^3\,b+\frac {7\,A\,a^2\,b^2}{4}\right )\,\sqrt {a+b\,x}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+\frac {2\,B\,b\,{\left (a+b\,x\right )}^{3/2}}{3}+2\,b\,\mathrm {atan}\left (\frac {2\,b\,\left (3\,A\,b+4\,B\,a\right )\,\sqrt {-\frac {25\,a}{64}}\,\sqrt {a+b\,x}}{5\,B\,a^2\,b+\frac {15\,A\,a\,b^2}{4}}\right )\,\left (3\,A\,b+4\,B\,a\right )\,\sqrt {-\frac {25\,a}{64}} \] Input:

int(((A + B*x)*(a + b*x)^(5/2))/x^3,x)
 

Output:

(2*A*b^2 + 4*B*a*b)*(a + b*x)^(1/2) - (((9*A*a*b^2)/4 + B*a^2*b)*(a + b*x) 
^(3/2) - ((7*A*a^2*b^2)/4 + B*a^3*b)*(a + b*x)^(1/2))/((a + b*x)^2 - 2*a*( 
a + b*x) + a^2) + (2*B*b*(a + b*x)^(3/2))/3 + 2*b*atan((2*b*(3*A*b + 4*B*a 
)*(-(25*a)/64)^(1/2)*(a + b*x)^(1/2))/((15*A*a*b^2)/4 + 5*B*a^2*b))*(3*A*b 
 + 4*B*a)*(-(25*a)/64)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{5/2} (A+B x)}{x^3} \, dx=\frac {-12 \sqrt {b x +a}\, a^{3}-78 \sqrt {b x +a}\, a^{2} b x +160 \sqrt {b x +a}\, a \,b^{2} x^{2}+16 \sqrt {b x +a}\, b^{3} x^{3}+105 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} x^{2}-105 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} x^{2}}{24 x^{2}} \] Input:

int((b*x+a)^(5/2)*(B*x+A)/x^3,x)
 

Output:

( - 12*sqrt(a + b*x)*a**3 - 78*sqrt(a + b*x)*a**2*b*x + 160*sqrt(a + b*x)* 
a*b**2*x**2 + 16*sqrt(a + b*x)*b**3*x**3 + 105*sqrt(a)*log(sqrt(a + b*x) - 
 sqrt(a))*a*b**2*x**2 - 105*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*x* 
*2)/(24*x**2)