\(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{17/2}} \, dx\) [203]

Optimal result

Integrand size = 24, antiderivative size = 206 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=-\frac {c (22 b B+3 A c) \sqrt {b x+c x^2}}{48 x^{7/2}}-\frac {c^2 (118 b B+3 A c) \sqrt {b x+c x^2}}{192 b x^{5/2}}-\frac {c^3 (10 b B-3 A c) \sqrt {b x+c x^2}}{128 b^2 x^{3/2}}-\frac {(2 b B+A c) \left (b x+c x^2\right )^{3/2}}{8 x^{11/2}}-\frac {A \left (b x+c x^2\right )^{5/2}}{5 x^{15/2}}+\frac {c^4 (10 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{128 b^{5/2}} \] Output:

-1/48*c*(3*A*c+22*B*b)*(c*x^2+b*x)^(1/2)/x^(7/2)-1/192*c^2*(3*A*c+118*B*b) 
*(c*x^2+b*x)^(1/2)/b/x^(5/2)-1/128*c^3*(-3*A*c+10*B*b)*(c*x^2+b*x)^(1/2)/b 
^2/x^(3/2)-1/8*(A*c+2*B*b)*(c*x^2+b*x)^(3/2)/x^(11/2)-1/5*A*(c*x^2+b*x)^(5 
/2)/x^(15/2)+1/128*c^4*(-3*A*c+10*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x 
^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\frac {-\sqrt {b} (b+c x) \left (10 b B x \left (48 b^3+136 b^2 c x+118 b c^2 x^2+15 c^3 x^3\right )+3 A \left (128 b^4+336 b^3 c x+248 b^2 c^2 x^2+10 b c^3 x^3-15 c^4 x^4\right )\right )+15 c^4 (10 b B-3 A c) x^5 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{1920 b^{5/2} x^{9/2} \sqrt {x (b+c x)}} \] Input:

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]
 

Output:

(-(Sqrt[b]*(b + c*x)*(10*b*B*x*(48*b^3 + 136*b^2*c*x + 118*b*c^2*x^2 + 15* 
c^3*x^3) + 3*A*(128*b^4 + 336*b^3*c*x + 248*b^2*c^2*x^2 + 10*b*c^3*x^3 - 1 
5*c^4*x^4))) + 15*c^4*(10*b*B - 3*A*c)*x^5*Sqrt[b + c*x]*ArcTanh[Sqrt[b + 
c*x]/Sqrt[b]])/(1920*b^(5/2)*x^(9/2)*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.89, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1220, 1130, 1130, 1130, 1135, 1136, 220}

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.

\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(10 b B-3 A c) \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{15/2}}dx}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^{11/2}}dx-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \int \frac {\sqrt {c x^2+b x}}{x^{7/2}}dx-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(10 b B-3 A c) \left (\frac {5}{8} c \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{3 x^{9/2}}\right )-\frac {\left (b x+c x^2\right )^{5/2}}{4 x^{13/2}}\right )}{10 b}-\frac {A \left (b x+c x^2\right )^{7/2}}{5 b x^{17/2}}\)

Input:

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(17/2),x]
 

Output:

-1/5*(A*(b*x + c*x^2)^(7/2))/(b*x^(17/2)) + ((10*b*B - 3*A*c)*(-1/4*(b*x + 
 c*x^2)^(5/2)/x^(13/2) + (5*c*(-1/3*(b*x + c*x^2)^(3/2)/x^(9/2) + (c*(-1/2 
*Sqrt[b*x + c*x^2]/x^(5/2) + (c*(-(Sqrt[b*x + c*x^2]/(b*x^(3/2))) + (c*Arc 
Tanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/4))/2))/8))/(10*b)
 

Defintions of rubi rules used

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 1130

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] 
- Simp[c*(p/(e^2*(m + p + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p 
 - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & 
& IntegerQ[2*p]
 

rule 1135

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e)))   Int 
[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I 
ntegerQ[2*p]
 

rule 1136

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-45 A \,c^{4} x^{4}+150 B b \,c^{3} x^{4}+30 A b \,c^{3} x^{3}+1180 B \,b^{2} c^{2} x^{3}+744 A \,b^{2} c^{2} x^{2}+1360 B \,b^{3} c \,x^{2}+1008 A \,b^{3} c x +480 B \,b^{4} x +384 A \,b^{4}\right )}{1920 x^{\frac {9}{2}} b^{2} \sqrt {x \left (c x +b \right )}}-\frac {c^{4} \left (3 A c -10 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{128 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\)	\(157\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (45 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{5} x^{5}-150 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{4} x^{5}-45 A \,c^{4} x^{4} \sqrt {c x +b}\, \sqrt {b}+150 B \,b^{\frac {3}{2}} c^{3} x^{4} \sqrt {c x +b}+30 A \,b^{\frac {3}{2}} c^{3} x^{3} \sqrt {c x +b}+1180 B \,b^{\frac {5}{2}} c^{2} x^{3} \sqrt {c x +b}+744 A \,b^{\frac {5}{2}} c^{2} x^{2} \sqrt {c x +b}+1360 B \,b^{\frac {7}{2}} c \,x^{2} \sqrt {c x +b}+1008 A \,b^{\frac {7}{2}} c x \sqrt {c x +b}+480 B \,b^{\frac {9}{2}} x \sqrt {c x +b}+384 A \,b^{\frac {9}{2}} \sqrt {c x +b}\right )}{1920 b^{\frac {5}{2}} x^{\frac {11}{2}} \sqrt {c x +b}}\)	\(223\)

Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(17/2),x,method=_RETURNVERBOSE)
 

Output:

-1/1920*(c*x+b)*(-45*A*c^4*x^4+150*B*b*c^3*x^4+30*A*b*c^3*x^3+1180*B*b^2*c 
^2*x^3+744*A*b^2*c^2*x^2+1360*B*b^3*c*x^2+1008*A*b^3*c*x+480*B*b^4*x+384*A 
*b^4)/x^(9/2)/b^2/(x*(c*x+b))^(1/2)-1/128*c^4*(3*A*c-10*B*b)/b^(5/2)*arcta 
nh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.65 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\left [-\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {b} x^{6} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{3840 \, b^{3} x^{6}}, -\frac {15 \, {\left (10 \, B b c^{4} - 3 \, A c^{5}\right )} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (384 \, A b^{5} + 15 \, {\left (10 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} + 10 \, {\left (118 \, B b^{3} c^{2} + 3 \, A b^{2} c^{3}\right )} x^{3} + 8 \, {\left (170 \, B b^{4} c + 93 \, A b^{3} c^{2}\right )} x^{2} + 48 \, {\left (10 \, B b^{5} + 21 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{1920 \, b^{3} x^{6}}\right ] \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(17/2),x, algorithm="fricas")
 

Output:

[-1/3840*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(b)*x^6*log(-(c*x^2 + 2*b*x - 2*sq 
rt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(384*A*b^5 + 15*(10*B*b^2*c^3 - 
3*A*b*c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 9 
3*A*b^3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x) 
)/(b^3*x^6), -1/1920*(15*(10*B*b*c^4 - 3*A*c^5)*sqrt(-b)*x^6*arctan(sqrt(c 
*x^2 + b*x)*sqrt(-b)/(b*sqrt(x))) + (384*A*b^5 + 15*(10*B*b^2*c^3 - 3*A*b* 
c^4)*x^4 + 10*(118*B*b^3*c^2 + 3*A*b^2*c^3)*x^3 + 8*(170*B*b^4*c + 93*A*b^ 
3*c^2)*x^2 + 48*(10*B*b^5 + 21*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3 
*x^6)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(17/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{x^{\frac {17}{2}}} \,d x } \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(17/2),x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x)^(5/2)*(B*x + A)/x^(17/2), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=-\frac {1}{1920} \, c^{5} {\left (\frac {15 \, {\left (10 \, B b - 3 \, A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2} c} + \frac {150 \, {\left (c x + b\right )}^{\frac {9}{2}} B b + 580 \, {\left (c x + b\right )}^{\frac {7}{2}} B b^{2} - 1280 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{3} + 700 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{4} - 150 \, \sqrt {c x + b} B b^{5} - 45 \, {\left (c x + b\right )}^{\frac {9}{2}} A c + 210 \, {\left (c x + b\right )}^{\frac {7}{2}} A b c + 384 \, {\left (c x + b\right )}^{\frac {5}{2}} A b^{2} c - 210 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{3} c + 45 \, \sqrt {c x + b} A b^{4} c}{b^{2} c^{6} x^{5}}\right )} \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^(17/2),x, algorithm="giac")
 

Output:

-1/1920*c^5*(15*(10*B*b - 3*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)* 
b^2*c) + (150*(c*x + b)^(9/2)*B*b + 580*(c*x + b)^(7/2)*B*b^2 - 1280*(c*x 
+ b)^(5/2)*B*b^3 + 700*(c*x + b)^(3/2)*B*b^4 - 150*sqrt(c*x + b)*B*b^5 - 4 
5*(c*x + b)^(9/2)*A*c + 210*(c*x + b)^(7/2)*A*b*c + 384*(c*x + b)^(5/2)*A* 
b^2*c - 210*(c*x + b)^(3/2)*A*b^3*c + 45*sqrt(c*x + b)*A*b^4*c)/(b^2*c^6*x 
^5))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{17/2}} \,d x \] Input:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(17/2),x)
 

Output:

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(17/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{17/2}} \, dx=\frac {-768 \sqrt {c x +b}\, a \,b^{5}-2016 \sqrt {c x +b}\, a \,b^{4} c x -1488 \sqrt {c x +b}\, a \,b^{3} c^{2} x^{2}-60 \sqrt {c x +b}\, a \,b^{2} c^{3} x^{3}+90 \sqrt {c x +b}\, a b \,c^{4} x^{4}-960 \sqrt {c x +b}\, b^{6} x -2720 \sqrt {c x +b}\, b^{5} c \,x^{2}-2360 \sqrt {c x +b}\, b^{4} c^{2} x^{3}-300 \sqrt {c x +b}\, b^{3} c^{3} x^{4}+45 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{5} x^{5}-150 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{4} x^{5}-45 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{5} x^{5}+150 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{4} x^{5}}{3840 b^{3} x^{5}} \] Input:

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(17/2),x)
 

Output:

( - 768*sqrt(b + c*x)*a*b**5 - 2016*sqrt(b + c*x)*a*b**4*c*x - 1488*sqrt(b 
 + c*x)*a*b**3*c**2*x**2 - 60*sqrt(b + c*x)*a*b**2*c**3*x**3 + 90*sqrt(b + 
 c*x)*a*b*c**4*x**4 - 960*sqrt(b + c*x)*b**6*x - 2720*sqrt(b + c*x)*b**5*c 
*x**2 - 2360*sqrt(b + c*x)*b**4*c**2*x**3 - 300*sqrt(b + c*x)*b**3*c**3*x* 
*4 + 45*sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c**5*x**5 - 150*sqrt(b)*log 
(sqrt(b + c*x) - sqrt(b))*b**2*c**4*x**5 - 45*sqrt(b)*log(sqrt(b + c*x) + 
sqrt(b))*a*c**5*x**5 + 150*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*b**2*c**4* 
x**5)/(3840*b**3*x**5)