\(\int \frac {x^{9/2} (A+B x)}{(b x+c x^2)^3} \, dx\) [98]

Optimal result

Integrand size = 22, antiderivative size = 107 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {2 B \sqrt {x}}{c^3}+\frac {(b B-A c) x^{3/2}}{2 c^2 (b+c x)^2}+\frac {(7 b B-3 A c) \sqrt {x}}{4 c^3 (b+c x)}-\frac {3 (5 b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {\sqrt {x} \left (15 b^2 B-3 A b c+25 b B c x-5 A c^2 x+8 B c^2 x^2\right )}{4 c^3 (b+c x)^2}-\frac {3 (5 b B-A c) \arctan \left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{4 \sqrt {b} c^{7/2}} \] Input:

Integrate[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

(Sqrt[x]*(15*b^2*B - 3*A*b*c + 25*b*B*c*x - 5*A*c^2*x + 8*B*c^2*x^2))/(4*c 
^3*(b + c*x)^2) - (3*(5*b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*S 
qrt[b]*c^(7/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {9, 87, 51, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

Int[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x]
 

Output:

-1/2*((b*B - A*c)*x^(5/2))/(b*c*(b + c*x)^2) + ((5*b*B - A*c)*(-(x^(3/2)/( 
c*(b + c*x))) + (3*((2*Sqrt[x])/c - (2*Sqrt[b]*ArcTan[(Sqrt[c]*Sqrt[x])/Sq 
rt[b]])/c^(3/2)))/(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 = 0.82 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} A \,c^{2}+\frac {9}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (3 A c -7 B b \right ) \sqrt {x}}{8}\right )}{\left (c x +b \right )^{2}}+\frac {3 \left (A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}}{c^{3}}\)	\(83\)
default	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (\left (-\frac {5}{8} A \,c^{2}+\frac {9}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (3 A c -7 B b \right ) \sqrt {x}}{8}\right )}{\left (c x +b \right )^{2}}+\frac {3 \left (A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}}{c^{3}}\)	\(83\)
risch	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (-\frac {5}{8} A \,c^{2}+\frac {9}{8} B b c \right ) x^{\frac {3}{2}}-\frac {b \left (3 A c -7 B b \right ) \sqrt {x}}{4}}{\left (c x +b \right )^{2}}+\frac {3 \left (A c -5 B b \right ) \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}}}{c^{3}}\)	\(83\)

Input:

int(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x,method=_RETURNVERBOSE)
 

Output:

2*B*x^(1/2)/c^3+2/c^3*(((-5/8*A*c^2+9/8*B*b*c)*x^(3/2)-1/8*b*(3*A*c-7*B*b) 
*x^(1/2))/(c*x+b)^2+3/8*(A*c-5*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 = 319, normalized size of antiderivative = 2.98 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\left [\frac {3 \, {\left (5 \, B b^{3} - A b^{2} c + {\left (5 \, B b c^{2} - A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x - b - 2 \, \sqrt {-b c} \sqrt {x}}{c x + b}\right ) + 2 \, {\left (8 \, B b c^{3} x^{2} + 15 \, B b^{3} c - 3 \, A b^{2} c^{2} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (b c^{6} x^{2} + 2 \, b^{2} c^{5} x + b^{3} c^{4}\right )}}, \frac {3 \, {\left (5 \, B b^{3} - A b^{2} c + {\left (5 \, B b c^{2} - A c^{3}\right )} x^{2} + 2 \, {\left (5 \, B b^{2} c - A b c^{2}\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c}}{c \sqrt {x}}\right ) + {\left (8 \, B b c^{3} x^{2} + 15 \, B b^{3} c - 3 \, A b^{2} c^{2} + 5 \, {\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (b c^{6} x^{2} + 2 \, b^{2} c^{5} x + b^{3} c^{4}\right )}}\right ] \] Input:

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(9/2)*(B*x+A)/(c*x**2+b*x)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {{\left (9 \, B b c - 5 \, A c^{2}\right )} x^{\frac {3}{2}} + {\left (7 \, B b^{2} - 3 \, A b c\right )} \sqrt {x}}{4 \, {\left (c^{5} x^{2} + 2 \, b c^{4} x + b^{2} c^{3}\right )}} + \frac {2 \, B \sqrt {x}}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{3}} \] Input:

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {2 \, B \sqrt {x}}{c^{3}} - \frac {3 \, {\left (5 \, B b - A c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \, \sqrt {b c} c^{3}} + \frac {9 \, B b c x^{\frac {3}{2}} - 5 \, A c^{2} x^{\frac {3}{2}} + 7 \, B b^{2} \sqrt {x} - 3 \, A b c \sqrt {x}}{4 \, {\left (c x + b\right )}^{2} c^{3}} \] Input:

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {\sqrt {x}\,\left (\frac {7\,B\,b^2}{4}-\frac {3\,A\,b\,c}{4}\right )-x^{3/2}\,\left (\frac {5\,A\,c^2}{4}-\frac {9\,B\,b\,c}{4}\right )}{b^2\,c^3+2\,b\,c^4\,x+c^5\,x^2}+\frac {2\,B\,\sqrt {x}}{c^3}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c-5\,B\,b\right )}{4\,\sqrt {b}\,c^{7/2}} \] Input:

int((x^(9/2)*(A + B*x))/(b*x + c*x^2)^3,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.14 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^3} \, dx=\frac {3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c +6 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x +3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{2}-15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{4}-30 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c x -15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {\sqrt {x}\, c}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{2}-3 \sqrt {x}\, a \,b^{2} c^{2}-5 \sqrt {x}\, a b \,c^{3} x +15 \sqrt {x}\, b^{4} c +25 \sqrt {x}\, b^{3} c^{2} x +8 \sqrt {x}\, b^{2} c^{3} x^{2}}{4 b \,c^{4} \left (c^{2} x^{2}+2 b c x +b^{2}\right )} \] Input:

int(x^(9/2)*(B*x+A)/(c*x^2+b*x)^3,x)
 

Output:

(3*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b**2*c + 6*sqrt(c 
)*sqrt(b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*b*c**2*x + 3*sqrt(c)*sqrt( 
b)*atan((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*a*c**3*x**2 - 15*sqrt(c)*sqrt(b)*at 
an((sqrt(x)*c)/(sqrt(c)*sqrt(b)))*b**4 - 30*sqrt(c)*sqrt(b)*atan((sqrt(x)* 
c)/(sqrt(c)*sqrt(b)))*b**3*c*x - 15*sqrt(c)*sqrt(b)*atan((sqrt(x)*c)/(sqrt 
(c)*sqrt(b)))*b**2*c**2*x**2 - 3*sqrt(x)*a*b**2*c**2 - 5*sqrt(x)*a*b*c**3* 
x + 15*sqrt(x)*b**4*c + 25*sqrt(x)*b**3*c**2*x + 8*sqrt(x)*b**2*c**3*x**2) 
/(4*b*c**4*(b**2 + 2*b*c*x + c**2*x**2))