\(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^5} \, dx\) [134]

Optimal result

Integrand size = 22, antiderivative size = 140 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\frac {5}{4} c (3 b B+4 A c) \sqrt {b x+c x^2}-\frac {2 (3 b B+5 A c) \left (b x+c x^2\right )^{3/2}}{3 x^2}+\frac {B c \left (b x+c x^2\right )^{3/2}}{2 x}-\frac {2 A \left (b x+c x^2\right )^{5/2}}{3 x^4}+\frac {5}{4} b \sqrt {c} (3 b B+4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \] Output:

5/4*c*(4*A*c+3*B*b)*(c*x^2+b*x)^(1/2)-2/3*(5*A*c+3*B*b)*(c*x^2+b*x)^(3/2)/ 
x^2+1/2*B*c*(c*x^2+b*x)^(3/2)/x-2/3*A*(c*x^2+b*x)^(5/2)/x^4+5/4*b*c^(1/2)* 
(4*A*c+3*B*b)*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {b+c x} \left (-4 A \left (2 b^2+14 b c x-3 c^2 x^2\right )+3 B x \left (-8 b^2+9 b c x+2 c^2 x^2\right )\right )+30 b \sqrt {c} (3 b B+4 A c) x^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{12 x^2 \sqrt {b+c x}} \] Input:

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

Output:

(Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(-4*A*(2*b^2 + 14*b*c*x - 3*c^2*x^2) + 3 
*B*x*(-8*b^2 + 9*b*c*x + 2*c^2*x^2)) + 30*b*Sqrt[c]*(3*b*B + 4*A*c)*x^(3/2 
)*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(12*x^2*Sqrt[b + 
 c*x])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1220, 1125, 25, 2192, 27, 1160, 1091, 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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1125

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2192

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1125

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 
2)*(d + e*x))), x] - Simp[e^(2*m + 2)   Int[(1/Sqrt[a + b*x + c*x^2])*Expan 
dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x 
), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && ILtQ[m, 0] && EqQ[m + p, -3/2]
 

rule 1160

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

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 
]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {15 c \,x^{2} \left (A c +\frac {3 B b}{4}\right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{2}+\left (7 \left (-\frac {27 B x}{56}+A \right ) x b \,c^{\frac {3}{2}}-\frac {3 \left (\frac {B x}{2}+A \right ) x^{2} c^{\frac {5}{2}}}{2}+b^{2} \sqrt {c}\, \left (3 B x +A \right )\right ) \sqrt {x \left (c x +b \right )}\right )}{3 \sqrt {c}\, x^{2}}\)	\(93\)
risch	\(-\frac {\left (c x +b \right ) \left (-6 B \,c^{2} x^{3}-12 A \,c^{2} x^{2}-27 x^{2} B b c +56 A b c x +24 x B \,b^{2}+8 b^{2} A \right )}{12 x \sqrt {x \left (c x +b \right )}}+\frac {5 \left (4 A c +3 B b \right ) \sqrt {c}\, b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8}\)	\(106\)
default	\(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{5}}+\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{4}}+\frac {6 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )}{b}\right )}{3 b}\right )+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{4}}+\frac {6 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )}{b}\right )\)	\(394\)

Input:

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

Output:

-2/3*(-15/2*c*x^2*(A*c+3/4*B*b)*b*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(7* 
(-27/56*B*x+A)*x*b*c^(3/2)-3/2*(1/2*B*x+A)*x^2*c^(5/2)+b^2*c^(1/2)*(3*B*x+ 
A))*(x*(c*x+b))^(1/2))/c^(1/2)/x^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.61 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\left [\frac {15 \, {\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt {c} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \, {\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \, {\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, x^{2}}, -\frac {15 \, {\left (3 \, B b^{2} + 4 \, A b c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (6 \, B c^{2} x^{3} - 8 \, A b^{2} + 3 \, {\left (9 \, B b c + 4 \, A c^{2}\right )} x^{2} - 8 \, {\left (3 \, B b^{2} + 7 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{12 \, x^{2}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.36 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\frac {15}{8} \, B b^{2} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \frac {5}{2} \, A b c^{\frac {3}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {15 \, \sqrt {c x^{2} + b x} B b^{2}}{4 \, x} - \frac {35 \, \sqrt {c x^{2} + b x} A b c}{6 \, x} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{4 \, x^{2}} - \frac {5 \, \sqrt {c x^{2} + b x} A b^{2}}{6 \, x^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{2 \, x^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{6 \, x^{3}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{x^{4}} \] Input:

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

Output:

15/8*B*b^2*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 5/2*A*b* 
c^(3/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 15/4*sqrt(c*x^2 + b 
*x)*B*b^2/x - 35/6*sqrt(c*x^2 + b*x)*A*b*c/x + 5/4*(c*x^2 + b*x)^(3/2)*B*b 
/x^2 - 5/6*sqrt(c*x^2 + b*x)*A*b^2/x^2 + 1/2*(c*x^2 + b*x)^(5/2)*B/x^3 - 5 
/6*(c*x^2 + b*x)^(3/2)*A*b/x^3 + (c*x^2 + b*x)^(5/2)*A/x^4
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\frac {1}{4} \, {\left (2 \, B c^{2} x + \frac {9 \, B b c^{2} + 4 \, A c^{3}}{c}\right )} \sqrt {c x^{2} + b x} - \frac {5 \, {\left (3 \, B b^{2} c + 4 \, A b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, \sqrt {c}} + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{3} + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{2} c + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{3} \sqrt {c} + A b^{4}\right )}}{3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3}} \] Input:

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

Output:

1/4*(2*B*c^2*x + (9*B*b*c^2 + 4*A*c^3)/c)*sqrt(c*x^2 + b*x) - 5/8*(3*B*b^2 
*c + 4*A*b*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/sq 
rt(c) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^3 + 9*(sqrt(c)*x - sq 
rt(c*x^2 + b*x))^2*A*b^2*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^3*sqrt( 
c) + A*b^4)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^3
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^5} \, dx=\frac {-64 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{2}-448 \sqrt {x}\, \sqrt {c x +b}\, a b c x +96 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{2} x^{2}-192 \sqrt {x}\, \sqrt {c x +b}\, b^{3} x +216 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c \,x^{2}+48 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} x^{3}+480 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a b c \,x^{2}+360 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} x^{2}+80 \sqrt {c}\, a b c \,x^{2}+95 \sqrt {c}\, b^{3} x^{2}}{96 x^{2}} \] Input:

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

Output:

( - 64*sqrt(x)*sqrt(b + c*x)*a*b**2 - 448*sqrt(x)*sqrt(b + c*x)*a*b*c*x + 
96*sqrt(x)*sqrt(b + c*x)*a*c**2*x**2 - 192*sqrt(x)*sqrt(b + c*x)*b**3*x + 
216*sqrt(x)*sqrt(b + c*x)*b**2*c*x**2 + 48*sqrt(x)*sqrt(b + c*x)*b*c**2*x* 
*3 + 480*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*a*b*c*x**2 
 + 360*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*b**3*x**2 + 
80*sqrt(c)*a*b*c*x**2 + 95*sqrt(c)*b**3*x**2)/(96*x**2)