\(\int \frac {x^{13/2} (A+B x)}{(b x+c x^2)^3} \, dx\) [96]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {7 b (9 b B-5 A c) \sqrt {x}}{4 c^5}-\frac {7 (9 b B-5 A c) x^{3/2}}{12 c^4}+\frac {2 B x^{5/2}}{5 c^3}+\frac {(b B-A c) x^{7/2}}{2 c^2 (b+c x)^2}+\frac {(11 b B-7 A c) x^{5/2}}{4 c^3 (b+c x)}-\frac {7 b^{3/2} (9 b B-5 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{11/2}} \] Output:

7/4*b*(-5*A*c+9*B*b)*x^(1/2)/c^5-7/12*(-5*A*c+9*B*b)*x^(3/2)/c^4+2/5*B*x^( 
5/2)/c^3+1/2*(-A*c+B*b)*x^(7/2)/c^2/(c*x+b)^2+1/4*(-7*A*c+11*B*b)*x^(5/2)/ 
c^3/(c*x+b)-7/4*b^(3/2)*(-5*A*c+9*B*b)*arctan(c^(1/2)*x^(1/2)/b^(1/2))/c^( 
11/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.85 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {\sqrt {x} \left (945 b^4 B-525 b^3 c (A-3 B x)+8 c^4 x^3 (5 A+3 B x)-8 b c^3 x^2 (35 A+9 B x)+7 b^2 c^2 x (-125 A+72 B x)\right )}{60 c^5 (b+c x)^2}-\frac {7 b^{3/2} (9 b B-5 A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 c^{11/2}} \] Input:

Integrate[(x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

(Sqrt[x]*(945*b^4*B - 525*b^3*c*(A - 3*B*x) + 8*c^4*x^3*(5*A + 3*B*x) - 8* 
b*c^3*x^2*(35*A + 9*B*x) + 7*b^2*c^2*x*(-125*A + 72*B*x)))/(60*c^5*(b + c* 
x)^2) - (7*b^(3/2)*(9*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*c 
^(11/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {9, 87, 51, 60, 60, 60, 73, 218}

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.

\(\displaystyle \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

Int[(x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

-1/2*((b*B - A*c)*x^(9/2))/(b*c*(b + c*x)^2) + ((9*b*B - 5*A*c)*(-(x^(7/2) 
/(c*(b + c*x))) + (7*((2*x^(5/2))/(5*c) - (b*((2*x^(3/2))/(3*c) - (b*((2*S 
qrt[x])/c - (2*Sqrt[b]*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(3/2)))/c))/c) 
)/(2*c)))/(4*b*c)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {2 \left (-3 B \,c^{2} x^{2}-5 A \,c^{2} x +15 B b c x +45 A b c -90 B \,b^{2}\right ) \sqrt {x}}{15 c^{5}}+\frac {b^{2} \left (\frac {2 \left (-\frac {13}{8} A \,c^{2}+\frac {17}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (11 A c -15 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {7 \left (5 A c -9 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}\right )}{c^{5}}\)	\(120\)
derivativedivides	\(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {5}{2}}}{5}-\frac {A \,c^{2} x^{\frac {3}{2}}}{3}+B b c \,x^{\frac {3}{2}}+3 A b c \sqrt {x}-6 B \,b^{2} \sqrt {x}\right )}{c^{5}}+\frac {2 b^{2} \left (\frac {\left (-\frac {13}{8} A \,c^{2}+\frac {17}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (11 A c -15 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (5 A c -9 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{5}}\)	\(126\)
default	\(-\frac {2 \left (-\frac {B \,c^{2} x^{\frac {5}{2}}}{5}-\frac {A \,c^{2} x^{\frac {3}{2}}}{3}+B b c \,x^{\frac {3}{2}}+3 A b c \sqrt {x}-6 B \,b^{2} \sqrt {x}\right )}{c^{5}}+\frac {2 b^{2} \left (\frac {\left (-\frac {13}{8} A \,c^{2}+\frac {17}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (11 A c -15 B b \right ) \sqrt {x}}{8}}{\left (c x +b \right )^{2}}+\frac {7 \left (5 A c -9 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{c^{5}}\)	\(126\)

Input:

int(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)
 

Output:

-2/15*(-3*B*c^2*x^2-5*A*c^2*x+15*B*b*c*x+45*A*b*c-90*B*b^2)*x^(1/2)/c^5+b^ 
2/c^5*(2*((-13/8*A*c^2+17/8*B*b*c)*x^(3/2)-1/8*b*(11*A*c-15*B*b)*x^(1/2))/ 
(c*x+b)^2+7/4*(5*A*c-9*B*b)/(b*c)^(1/2)*arctan(c*x^(1/2)/(b*c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.68 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\left [-\frac {105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x + 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (24 \, B c^{4} x^{4} + 945 \, B b^{4} - 525 \, A b^{3} c - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{120 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}, -\frac {105 \, {\left (9 \, B b^{4} - 5 \, A b^{3} c + {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 2 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) - {\left (24 \, B c^{4} x^{4} + 945 \, B b^{4} - 525 \, A b^{3} c - 8 \, {\left (9 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 56 \, {\left (9 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 175 \, {\left (9 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {x}}{60 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}}\right ] \] Input:

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")
 

Output:

[-1/120*(105*(9*B*b^4 - 5*A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 2*(9*B 
*b^3*c - 5*A*b^2*c^2)*x)*sqrt(-b/c)*log((c*x + 2*c*sqrt(x)*sqrt(-b/c) - b) 
/(c*x + b)) - 2*(24*B*c^4*x^4 + 945*B*b^4 - 525*A*b^3*c - 8*(9*B*b*c^3 - 5 
*A*c^4)*x^3 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 175*(9*B*b^3*c - 5*A*b^2* 
c^2)*x)*sqrt(x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5), -1/60*(105*(9*B*b^4 - 5* 
A*b^3*c + (9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 2*(9*B*b^3*c - 5*A*b^2*c^2)*x)*s 
qrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) - (24*B*c^4*x^4 + 945*B*b^4 - 525*A 
*b^3*c - 8*(9*B*b*c^3 - 5*A*c^4)*x^3 + 56*(9*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 
175*(9*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(x))/(c^7*x^2 + 2*b*c^6*x + b^2*c^5)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(13/2)*(B*x+A)/(c*x**2+b*x)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.99 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {{\left (17 \, B b^{3} c - 13 \, A b^{2} c^{2}\right )} x^{\frac {3}{2}} + {\left (15 \, B b^{4} - 11 \, A b^{3} c\right )} \sqrt {x}}{4 \, {\left (c^{7} x^{2} + 2 \, b c^{6} x + b^{2} c^{5}\right )}} - \frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{5}} + \frac {2 \, {\left (3 \, B c^{2} x^{\frac {5}{2}} - 5 \, {\left (3 \, B b c - A c^{2}\right )} x^{\frac {3}{2}} + 45 \, {\left (2 \, B b^{2} - A b c\right )} \sqrt {x}\right )}}{15 \, c^{5}} \] Input:

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")
 

Output:

1/4*((17*B*b^3*c - 13*A*b^2*c^2)*x^(3/2) + (15*B*b^4 - 11*A*b^3*c)*sqrt(x) 
)/(c^7*x^2 + 2*b*c^6*x + b^2*c^5) - 7/4*(9*B*b^3 - 5*A*b^2*c)*arctan(c*sqr 
t(x)/sqrt(b*c))/(sqrt(b*c)*c^5) + 2/15*(3*B*c^2*x^(5/2) - 5*(3*B*b*c - A*c 
^2)*x^(3/2) + 45*(2*B*b^2 - A*b*c)*sqrt(x))/c^5
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.96 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=-\frac {7 \, {\left (9 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{5}} + \frac {17 \, B b^{3} c x^{\frac {3}{2}} - 13 \, A b^{2} c^{2} x^{\frac {3}{2}} + 15 \, B b^{4} \sqrt {x} - 11 \, A b^{3} c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} c^{5}} + \frac {2 \, {\left (3 \, B c^{12} x^{\frac {5}{2}} - 15 \, B b c^{11} x^{\frac {3}{2}} + 5 \, A c^{12} x^{\frac {3}{2}} + 90 \, B b^{2} c^{10} \sqrt {x} - 45 \, A b c^{11} \sqrt {x}\right )}}{15 \, c^{15}} \] Input:

integrate(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")
 

Output:

-7/4*(9*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^5) + 1 
/4*(17*B*b^3*c*x^(3/2) - 13*A*b^2*c^2*x^(3/2) + 15*B*b^4*sqrt(x) - 11*A*b^ 
3*c*sqrt(x))/((c*x + b)^2*c^5) + 2/15*(3*B*c^12*x^(5/2) - 15*B*b*c^11*x^(3 
/2) + 5*A*c^12*x^(3/2) + 90*B*b^2*c^10*sqrt(x) - 45*A*b*c^11*sqrt(x))/c^15
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.20 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=x^{3/2}\,\left (\frac {2\,A}{3\,c^3}-\frac {2\,B\,b}{c^4}\right )-\frac {x^{3/2}\,\left (\frac {13\,A\,b^2\,c^2}{4}-\frac {17\,B\,b^3\,c}{4}\right )-\sqrt {x}\,\left (\frac {15\,B\,b^4}{4}-\frac {11\,A\,b^3\,c}{4}\right )}{b^2\,c^5+2\,b\,c^6\,x+c^7\,x^2}-\sqrt {x}\,\left (\frac {3\,b\,\left (\frac {2\,A}{c^3}-\frac {6\,B\,b}{c^4}\right )}{c}+\frac {6\,B\,b^2}{c^5}\right )+\frac {2\,B\,x^{5/2}}{5\,c^3}-\frac {7\,b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,\sqrt {x}\,\left (5\,A\,c-9\,B\,b\right )}{9\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-9\,B\,b\right )}{4\,c^{11/2}} \] Input:

int((x^(13/2)*(A + B*x))/(b*x + c*x^2)^3,x)
 

Output:

x^(3/2)*((2*A)/(3*c^3) - (2*B*b)/c^4) - (x^(3/2)*((13*A*b^2*c^2)/4 - (17*B 
*b^3*c)/4) - x^(1/2)*((15*B*b^4)/4 - (11*A*b^3*c)/4))/(b^2*c^5 + c^7*x^2 + 
 2*b*c^6*x) - x^(1/2)*((3*b*((2*A)/c^3 - (6*B*b)/c^4))/c + (6*B*b^2)/c^5) 
+ (2*B*x^(5/2))/(5*c^3) - (7*b^(3/2)*atan((b^(3/2)*c^(1/2)*x^(1/2)*(5*A*c 
- 9*B*b))/(9*B*b^3 - 5*A*b^2*c))*(5*A*c - 9*B*b))/(4*c^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.83 \[ \int \frac {x^{13/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{3} c +1050 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c^{2} x +525 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{3} x^{2}-945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{5}-1890 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{4} c x -945 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c^{2} x^{2}-525 \sqrt {x}\, a \,b^{3} c^{2}-875 \sqrt {x}\, a \,b^{2} c^{3} x -280 \sqrt {x}\, a b \,c^{4} x^{2}+40 \sqrt {x}\, a \,c^{5} x^{3}+945 \sqrt {x}\, b^{5} c +1575 \sqrt {x}\, b^{4} c^{2} x +504 \sqrt {x}\, b^{3} c^{3} x^{2}-72 \sqrt {x}\, b^{2} c^{4} x^{3}+24 \sqrt {x}\, b \,c^{5} x^{4}}{60 c^{6} \left (c^{2} x^{2}+2 b c x +b^{2}\right )} \] Input:

int(x^(13/2)*(B*x+A)/(c*x^2+b*x)^3,x)
 

Output:

(525*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b**3*c + 1050*s 
qrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b**2*c**2*x + 525*sqr 
t(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b*c**3*x**2 - 945*sqrt( 
c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**5 - 1890*sqrt(c)*sqrt(b) 
*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**4*c*x - 945*sqrt(c)*sqrt(b)*atan(( 
sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**3*c**2*x**2 - 525*sqrt(x)*a*b**3*c**2 - 8 
75*sqrt(x)*a*b**2*c**3*x - 280*sqrt(x)*a*b*c**4*x**2 + 40*sqrt(x)*a*c**5*x 
**3 + 945*sqrt(x)*b**5*c + 1575*sqrt(x)*b**4*c**2*x + 504*sqrt(x)*b**3*c** 
3*x**2 - 72*sqrt(x)*b**2*c**4*x**3 + 24*sqrt(x)*b*c**5*x**4)/(60*c**6*(b** 
2 + 2*b*c*x + c**2*x**2))