\(\int \frac {A+B x}{x^3 (b x+c x^2)^{5/2}} \, dx\) [170]

Optimal result

Integrand size = 22, antiderivative size = 224 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}+\frac {2 (3 b B-4 A c)}{9 b^2 x^2 \left (b x+c x^2\right )^{3/2}}+\frac {20 (3 b B-4 A c)}{9 b^3 x^3 \sqrt {b x+c x^2}}-\frac {160 (3 b B-4 A c) \sqrt {b x+c x^2}}{63 b^4 x^4}+\frac {64 c (3 b B-4 A c) \sqrt {b x+c x^2}}{21 b^5 x^3}-\frac {256 c^2 (3 b B-4 A c) \sqrt {b x+c x^2}}{63 b^6 x^2}+\frac {512 c^3 (3 b B-4 A c) \sqrt {b x+c x^2}}{63 b^7 x} \] Output:

-2/9*A/b/x^3/(c*x^2+b*x)^(3/2)+2/9*(-4*A*c+3*B*b)/b^2/x^2/(c*x^2+b*x)^(3/2 
)+20/9*(-4*A*c+3*B*b)/b^3/x^3/(c*x^2+b*x)^(1/2)-160/63*(-4*A*c+3*B*b)*(c*x 
^2+b*x)^(1/2)/b^4/x^4+64/21*c*(-4*A*c+3*B*b)*(c*x^2+b*x)^(1/2)/b^5/x^3-256 
/63*c^2*(-4*A*c+3*B*b)*(c*x^2+b*x)^(1/2)/b^6/x^2+512/63*c^3*(-4*A*c+3*B*b) 
*(c*x^2+b*x)^(1/2)/b^7/x
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=\frac {6 b B x \left (-3 b^5+6 b^4 c x-16 b^3 c^2 x^2+96 b^2 c^3 x^3+384 b c^4 x^4+256 c^5 x^5\right )-2 A \left (7 b^6-12 b^5 c x+24 b^4 c^2 x^2-64 b^3 c^3 x^3+384 b^2 c^4 x^4+1536 b c^5 x^5+1024 c^6 x^6\right )}{63 b^7 x^3 (x (b+c x))^{3/2}} \] Input:

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

Output:

(6*b*B*x*(-3*b^5 + 6*b^4*c*x - 16*b^3*c^2*x^2 + 96*b^2*c^3*x^3 + 384*b*c^4 
*x^4 + 256*c^5*x^5) - 2*A*(7*b^6 - 12*b^5*c*x + 24*b^4*c^2*x^2 - 64*b^3*c^ 
3*x^3 + 384*b^2*c^4*x^4 + 1536*b*c^5*x^5 + 1024*c^6*x^6))/(63*b^7*x^3*(x*( 
b + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.71, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1220, 1129, 1129, 1089, 1088}

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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1089

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

\(\Big \downarrow \) 1088

\(\displaystyle \frac {\left (-\frac {10 c \left (-\frac {8 c \left (\frac {16 c (b+2 c x)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (b+2 c x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}\right )}{5 b}-\frac {2}{5 b x \left (b x+c x^2\right )^{3/2}}\right )}{7 b}-\frac {2}{7 b x^2 \left (b x+c x^2\right )^{3/2}}\right ) (3 b B-4 A c)}{3 b}-\frac {2 A}{9 b x^3 \left (b x+c x^2\right )^{3/2}}\)

Input:

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

Output:

(-2*A)/(9*b*x^3*(b*x + c*x^2)^(3/2)) + ((3*b*B - 4*A*c)*(-2/(7*b*x^2*(b*x 
+ c*x^2)^(3/2)) - (10*c*(-2/(5*b*x*(b*x + c*x^2)^(3/2)) - (8*c*((-2*(b + 2 
*c*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (16*c*(b + 2*c*x))/(3*b^4*Sqrt[b*x + 
c*x^2])))/(5*b)))/(7*b)))/(3*b)
 

Defintions of rubi rules used

rule 1088

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 
2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0]
 

rule 1089

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1129

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*(Simplify[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, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 
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 = 1.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.56

method	result	size
pseudoelliptic	\(\frac {\left (-18 B x -14 A \right ) b^{6}+24 c x \left (\frac {3 B x}{2}+A \right ) b^{5}-48 c^{2} x^{2} \left (2 B x +A \right ) b^{4}+128 c^{3} x^{3} \left (\frac {9 B x}{2}+A \right ) b^{3}-768 c^{4} x^{4} \left (-3 B x +A \right ) b^{2}-3072 c^{5} \left (-\frac {B x}{2}+A \right ) x^{5} b -2048 A \,c^{6} x^{6}}{63 \sqrt {x \left (c x +b \right )}\, x^{4} \left (c x +b \right ) b^{7}}\)	\(126\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (1024 A \,c^{6} x^{6}-768 B b \,c^{5} x^{6}+1536 A b \,c^{5} x^{5}-1152 B \,b^{2} c^{4} x^{5}+384 A \,b^{2} c^{4} x^{4}-288 B \,b^{3} c^{3} x^{4}-64 A \,b^{3} c^{3} x^{3}+48 B \,b^{4} c^{2} x^{3}+24 A \,b^{4} c^{2} x^{2}-18 B \,b^{5} c \,x^{2}-12 A \,b^{5} c x +9 B \,b^{6} x +7 A \,b^{6}\right )}{63 x^{2} b^{7} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(158\)
orering	\(-\frac {2 \left (c x +b \right ) \left (1024 A \,c^{6} x^{6}-768 B b \,c^{5} x^{6}+1536 A b \,c^{5} x^{5}-1152 B \,b^{2} c^{4} x^{5}+384 A \,b^{2} c^{4} x^{4}-288 B \,b^{3} c^{3} x^{4}-64 A \,b^{3} c^{3} x^{3}+48 B \,b^{4} c^{2} x^{3}+24 A \,b^{4} c^{2} x^{2}-18 B \,b^{5} c \,x^{2}-12 A \,b^{5} c x +9 B \,b^{6} x +7 A \,b^{6}\right )}{63 x^{2} b^{7} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(158\)
risch	\(-\frac {2 \left (c x +b \right ) \left (667 A \,c^{4} x^{4}-474 B b \,c^{3} x^{4}-176 A b \,c^{3} x^{3}+111 B \,b^{2} c^{2} x^{3}+69 A \,b^{2} c^{2} x^{2}-36 B \,b^{3} c \,x^{2}-26 A \,b^{3} c x +9 B \,b^{4} x +7 A \,b^{4}\right )}{63 b^{7} x^{4} \sqrt {x \left (c x +b \right )}}-\frac {2 c^{4} \left (17 A \,c^{2} x -14 B b c x +18 A b c -15 B \,b^{2}\right ) x}{3 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{7}}\)	\(159\)
trager	\(-\frac {2 \left (1024 A \,c^{6} x^{6}-768 B b \,c^{5} x^{6}+1536 A b \,c^{5} x^{5}-1152 B \,b^{2} c^{4} x^{5}+384 A \,b^{2} c^{4} x^{4}-288 B \,b^{3} c^{3} x^{4}-64 A \,b^{3} c^{3} x^{3}+48 B \,b^{4} c^{2} x^{3}+24 A \,b^{4} c^{2} x^{2}-18 B \,b^{5} c \,x^{2}-12 A \,b^{5} c x +9 B \,b^{6} x +7 A \,b^{6}\right ) \sqrt {c \,x^{2}+b x}}{63 b^{7} x^{5} \left (c x +b \right )^{2}}\)	\(160\)
default	\(A \left (-\frac {2}{9 b \,x^{3} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {4 c \left (-\frac {2}{7 b \,x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {10 c \left (-\frac {2}{5 b x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {8 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\right )}{3 b}\right )+B \left (-\frac {2}{7 b \,x^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {10 c \left (-\frac {2}{5 b x \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {8 c \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{5 b}\right )}{7 b}\right )\)	\(228\)

Input:

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

Output:

1/63*((-18*B*x-14*A)*b^6+24*c*x*(3/2*B*x+A)*b^5-48*c^2*x^2*(2*B*x+A)*b^4+1 
28*c^3*x^3*(9/2*B*x+A)*b^3-768*c^4*x^4*(-3*B*x+A)*b^2-3072*c^5*(-1/2*B*x+A 
)*x^5*b-2048*A*c^6*x^6)/(x*(c*x+b))^(1/2)/x^4/(c*x+b)/b^7
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (7 \, A b^{6} - 256 \, {\left (3 \, B b c^{5} - 4 \, A c^{6}\right )} x^{6} - 384 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{5} - 96 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} + 16 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{3} - 6 \, {\left (3 \, B b^{5} c - 4 \, A b^{4} c^{2}\right )} x^{2} + 3 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} x\right )} \sqrt {c x^{2} + b x}}{63 \, {\left (b^{7} c^{2} x^{7} + 2 \, b^{8} c x^{6} + b^{9} x^{5}\right )}} \] Input:

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

Output:

-2/63*(7*A*b^6 - 256*(3*B*b*c^5 - 4*A*c^6)*x^6 - 384*(3*B*b^2*c^4 - 4*A*b* 
c^5)*x^5 - 96*(3*B*b^3*c^3 - 4*A*b^2*c^4)*x^4 + 16*(3*B*b^4*c^2 - 4*A*b^3* 
c^3)*x^3 - 6*(3*B*b^5*c - 4*A*b^4*c^2)*x^2 + 3*(3*B*b^6 - 4*A*b^5*c)*x)*sq 
rt(c*x^2 + b*x)/(b^7*c^2*x^7 + 2*b^8*c*x^6 + b^9*x^5)
 

Sympy [F]

\[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {A + B x}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((A + B*x)/(x**3*(x*(b + c*x))**(5/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {64 \, B c^{3} x}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} + \frac {512 \, B c^{4} x}{21 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {256 \, A c^{4} x}{63 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{5}} - \frac {2048 \, A c^{5} x}{63 \, \sqrt {c x^{2} + b x} b^{7}} - \frac {32 \, B c^{2}}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}} + \frac {256 \, B c^{3}}{21 \, \sqrt {c x^{2} + b x} b^{5}} + \frac {128 \, A c^{3}}{63 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4}} - \frac {1024 \, A c^{4}}{63 \, \sqrt {c x^{2} + b x} b^{6}} + \frac {4 \, B c}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x} - \frac {16 \, A c^{2}}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3} x} - \frac {2 \, B}{7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x^{2}} + \frac {8 \, A c}{21 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x^{2}} - \frac {2 \, A}{9 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b x^{3}} \] Input:

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

Output:

-64/21*B*c^3*x/((c*x^2 + b*x)^(3/2)*b^4) + 512/21*B*c^4*x/(sqrt(c*x^2 + b* 
x)*b^6) + 256/63*A*c^4*x/((c*x^2 + b*x)^(3/2)*b^5) - 2048/63*A*c^5*x/(sqrt 
(c*x^2 + b*x)*b^7) - 32/21*B*c^2/((c*x^2 + b*x)^(3/2)*b^3) + 256/21*B*c^3/ 
(sqrt(c*x^2 + b*x)*b^5) + 128/63*A*c^3/((c*x^2 + b*x)^(3/2)*b^4) - 1024/63 
*A*c^4/(sqrt(c*x^2 + b*x)*b^6) + 4/7*B*c/((c*x^2 + b*x)^(3/2)*b^2*x) - 16/ 
21*A*c^2/((c*x^2 + b*x)^(3/2)*b^3*x) - 2/7*B/((c*x^2 + b*x)^(3/2)*b*x^2) + 
 8/21*A*c/((c*x^2 + b*x)^(3/2)*b^2*x^2) - 2/9*A/((c*x^2 + b*x)^(3/2)*b*x^3 
)
 

Giac [F]

\[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}} x^{3}} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {4\,c^3\,\left (176\,A\,c-111\,B\,b\right )}{63\,b^5}+\frac {2\,c^3\,\left (247\,A\,c-138\,B\,b\right )}{63\,b^5}+\frac {b\,\left (\frac {184\,A\,c^5-96\,B\,b\,c^4}{63\,b^6}-\frac {4\,c^4\,\left (247\,A\,c-138\,B\,b\right )}{63\,b^6}\right )}{c}\right )+\frac {2\,c^2\,\left (176\,A\,c-111\,B\,b\right )}{63\,b^4}\right )}{x^2\,{\left (b+c\,x\right )}^2}-\frac {\sqrt {c\,x^2+b\,x}\,\left (18\,B\,b^3-52\,A\,b^2\,c\right )}{63\,b^6\,x^4}-\frac {\sqrt {c\,x^2+b\,x}\,\left (\frac {1024\,A\,c^4-768\,B\,b\,c^3}{63\,b^6}+\frac {2\,c\,x\,\left (1024\,A\,c^4-768\,B\,b\,c^3\right )}{63\,b^7}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{9\,b^3\,x^5}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (23\,A\,c-12\,B\,b\right )}{21\,b^5\,x^3} \] Input:

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

Output:

((b*x + c*x^2)^(1/2)*(x*((4*c^3*(176*A*c - 111*B*b))/(63*b^5) + (2*c^3*(24 
7*A*c - 138*B*b))/(63*b^5) + (b*((184*A*c^5 - 96*B*b*c^4)/(63*b^6) - (4*c^ 
4*(247*A*c - 138*B*b))/(63*b^6)))/c) + (2*c^2*(176*A*c - 111*B*b))/(63*b^4 
)))/(x^2*(b + c*x)^2) - ((b*x + c*x^2)^(1/2)*(18*B*b^3 - 52*A*b^2*c))/(63* 
b^6*x^4) - ((b*x + c*x^2)^(1/2)*((1024*A*c^4 - 768*B*b*c^3)/(63*b^6) + (2* 
c*x*(1024*A*c^4 - 768*B*b*c^3))/(63*b^7)))/(x*(b + c*x)) - (2*A*(b*x + c*x 
^2)^(1/2))/(9*b^3*x^5) - (2*c*(b*x + c*x^2)^(1/2)*(23*A*c - 12*B*b))/(21*b 
^5*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x}{x^3 \left (b x+c x^2\right )^{5/2}} \, dx=\frac {\frac {2048 \sqrt {c}\, \sqrt {c x +b}\, a b \,c^{4} x^{5}}{63}+\frac {2048 \sqrt {c}\, \sqrt {c x +b}\, a \,c^{5} x^{6}}{63}-\frac {512 \sqrt {c}\, \sqrt {c x +b}\, b^{3} c^{3} x^{5}}{21}-\frac {512 \sqrt {c}\, \sqrt {c x +b}\, b^{2} c^{4} x^{6}}{21}-\frac {2 \sqrt {x}\, a \,b^{6}}{9}+\frac {8 \sqrt {x}\, a \,b^{5} c x}{21}-\frac {16 \sqrt {x}\, a \,b^{4} c^{2} x^{2}}{21}+\frac {128 \sqrt {x}\, a \,b^{3} c^{3} x^{3}}{63}-\frac {256 \sqrt {x}\, a \,b^{2} c^{4} x^{4}}{21}-\frac {1024 \sqrt {x}\, a b \,c^{5} x^{5}}{21}-\frac {2048 \sqrt {x}\, a \,c^{6} x^{6}}{63}-\frac {2 \sqrt {x}\, b^{7} x}{7}+\frac {4 \sqrt {x}\, b^{6} c \,x^{2}}{7}-\frac {32 \sqrt {x}\, b^{5} c^{2} x^{3}}{21}+\frac {64 \sqrt {x}\, b^{4} c^{3} x^{4}}{7}+\frac {256 \sqrt {x}\, b^{3} c^{4} x^{5}}{7}+\frac {512 \sqrt {x}\, b^{2} c^{5} x^{6}}{21}}{\sqrt {c x +b}\, b^{7} x^{5} \left (c x +b \right )} \] Input:

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

Output:

(2*(1024*sqrt(c)*sqrt(b + c*x)*a*b*c**4*x**5 + 1024*sqrt(c)*sqrt(b + c*x)* 
a*c**5*x**6 - 768*sqrt(c)*sqrt(b + c*x)*b**3*c**3*x**5 - 768*sqrt(c)*sqrt( 
b + c*x)*b**2*c**4*x**6 - 7*sqrt(x)*a*b**6 + 12*sqrt(x)*a*b**5*c*x - 24*sq 
rt(x)*a*b**4*c**2*x**2 + 64*sqrt(x)*a*b**3*c**3*x**3 - 384*sqrt(x)*a*b**2* 
c**4*x**4 - 1536*sqrt(x)*a*b*c**5*x**5 - 1024*sqrt(x)*a*c**6*x**6 - 9*sqrt 
(x)*b**7*x + 18*sqrt(x)*b**6*c*x**2 - 48*sqrt(x)*b**5*c**2*x**3 + 288*sqrt 
(x)*b**4*c**3*x**4 + 1152*sqrt(x)*b**3*c**4*x**5 + 768*sqrt(x)*b**2*c**5*x 
**6))/(63*sqrt(b + c*x)*b**7*x**5*(b + c*x))