\(\int \frac {x^{9/2} (A+B x)}{(b x+c x^2)^{3/2}} \, dx\) [212]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 b^3 (b B-A c) \sqrt {x}}{c^5 \sqrt {b x+c x^2}}-\frac {2 b^2 (4 b B-3 A c) \sqrt {b x+c x^2}}{c^5 \sqrt {x}}+\frac {2 b (2 b B-A c) \left (b x+c x^2\right )^{3/2}}{c^5 x^{3/2}}-\frac {2 (4 b B-A c) \left (b x+c x^2\right )^{5/2}}{5 c^5 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{7 c^5 x^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.57 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {x} \left (-128 b^4 B+16 b^3 c (7 A-4 B x)+8 b^2 c^2 x (7 A+2 B x)-2 b c^3 x^2 (7 A+4 B x)+c^4 x^3 (7 A+5 B x)\right )}{35 c^5 \sqrt {x (b+c x)}} \] Input:

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

Output:

(2*Sqrt[x]*(-128*b^4*B + 16*b^3*c*(7*A - 4*B*x) + 8*b^2*c^2*x*(7*A + 2*B*x 
) - 2*b*c^3*x^2*(7*A + 4*B*x) + c^4*x^3*(7*A + 5*B*x)))/(35*c^5*Sqrt[x*(b 
+ c*x)])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1218, 1128, 1128, 1128, 1122}

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.

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

\(\Big \downarrow \) 1218

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

\(\Big \downarrow \) 1128

\(\displaystyle -\left (\frac {7 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \int \frac {x^{5/2}}{\sqrt {c x^2+b x}}dx}{7 c}\right )-\frac {2 x^{9/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1128

\(\displaystyle -\left (\frac {7 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \int \frac {x^{3/2}}{\sqrt {c x^2+b x}}dx}{5 c}\right )}{7 c}\right )-\frac {2 x^{9/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1128

\(\displaystyle -\left (\frac {7 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {2 b \int \frac {\sqrt {x}}{\sqrt {c x^2+b x}}dx}{3 c}\right )}{5 c}\right )}{7 c}\right )-\frac {2 x^{9/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1122

\(\displaystyle -\left (\frac {2 x^{5/2} \sqrt {b x+c x^2}}{7 c}-\frac {6 b \left (\frac {2 x^{3/2} \sqrt {b x+c x^2}}{5 c}-\frac {4 b \left (\frac {2 \sqrt {x} \sqrt {b x+c x^2}}{3 c}-\frac {4 b \sqrt {b x+c x^2}}{3 c^2 \sqrt {x}}\right )}{5 c}\right )}{7 c}\right ) \left (\frac {7 A}{b}-\frac {8 B}{c}\right )-\frac {2 x^{9/2} (b B-A c)}{b c \sqrt {b x+c x^2}}\)

Input:

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

Output:

(-2*(b*B - A*c)*x^(9/2))/(b*c*Sqrt[b*x + c*x^2]) - ((7*A)/b - (8*B)/c)*((2 
*x^(5/2)*Sqrt[b*x + c*x^2])/(7*c) - (6*b*((2*x^(3/2)*Sqrt[b*x + c*x^2])/(5 
*c) - (4*b*((-4*b*Sqrt[b*x + c*x^2])/(3*c^2*Sqrt[x]) + (2*Sqrt[x]*Sqrt[b*x 
 + c*x^2])/(3*c)))/(5*c)))/(7*c))
 

Defintions of rubi rules used

rule 1122

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

rule 1128

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1)))   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] && IGtQ[Simplify[m + p], 0]
 

rule 1218

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

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.66

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (5 B \,c^{4} x^{4}+7 A \,c^{4} x^{3}-8 B \,c^{3} x^{3} b -14 A b \,c^{3} x^{2}+16 c^{2} x^{2} B \,b^{2}+56 A \,b^{2} c^{2} x -64 B \,b^{3} c x +112 A \,b^{3} c -128 B \,b^{4}\right ) x^{\frac {3}{2}}}{35 c^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(107\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (5 B \,c^{4} x^{4}+7 A \,c^{4} x^{3}-8 B \,c^{3} x^{3} b -14 A b \,c^{3} x^{2}+16 c^{2} x^{2} B \,b^{2}+56 A \,b^{2} c^{2} x -64 B \,b^{3} c x +112 A \,b^{3} c -128 B \,b^{4}\right )}{35 \sqrt {x}\, \left (c x +b \right ) c^{5}}\)	\(107\)
orering	\(\frac {2 \left (c x +b \right ) \left (5 B \,c^{4} x^{4}+7 A \,c^{4} x^{3}-8 B \,c^{3} x^{3} b -14 A b \,c^{3} x^{2}+16 c^{2} x^{2} B \,b^{2}+56 A \,b^{2} c^{2} x -64 B \,b^{3} c x +112 A \,b^{3} c -128 B \,b^{4}\right ) x^{\frac {3}{2}}}{35 c^{5} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(107\)
risch	\(\frac {2 \left (5 B \,c^{3} x^{3}+7 A \,c^{3} x^{2}-13 B b \,c^{2} x^{2}-21 A b \,c^{2} x +29 B \,b^{2} c x +77 A \,b^{2} c -93 B \,b^{3}\right ) \left (c x +b \right ) \sqrt {x}}{35 c^{5} \sqrt {x \left (c x +b \right )}}+\frac {2 b^{3} \left (A c -B b \right ) \sqrt {x}}{c^{5} \sqrt {x \left (c x +b \right )}}\)	\(110\)

Input:

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

Output:

2/35*(c*x+b)*(5*B*c^4*x^4+7*A*c^4*x^3-8*B*b*c^3*x^3-14*A*b*c^3*x^2+16*B*b^ 
2*c^2*x^2+56*A*b^2*c^2*x-64*B*b^3*c*x+112*A*b^3*c-128*B*b^4)*x^(3/2)/c^5/( 
c*x^2+b*x)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (5 \, B c^{4} x^{4} - 128 \, B b^{4} + 112 \, A b^{3} c - {\left (8 \, B b c^{3} - 7 \, A c^{4}\right )} x^{3} + 2 \, {\left (8 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{35 \, {\left (c^{6} x^{2} + b c^{5} x\right )}} \] Input:

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

Output:

2/35*(5*B*c^4*x^4 - 128*B*b^4 + 112*A*b^3*c - (8*B*b*c^3 - 7*A*c^4)*x^3 + 
2*(8*B*b^2*c^2 - 7*A*b*c^3)*x^2 - 8*(8*B*b^3*c - 7*A*b^2*c^2)*x)*sqrt(c*x^ 
2 + b*x)*sqrt(x)/(c^6*x^2 + b*c^5*x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

2/105*((15*B*c^5*x^3 + 3*B*b*c^4*x^2 - 4*B*b^2*c^3*x + 8*B*b^3*c^2)*x^4 + 
(16*B*b^4*c - 3*(4*B*b*c^4 - 7*A*c^5)*x^3 - (8*B*b^2*c^3 - 7*A*b*c^4)*x^2 
+ 2*(10*B*b^3*c^2 - 7*A*b^2*c^3)*x)*x^3 + 4*(2*B*b^5 + (9*B*b^2*c^3 - 7*A* 
b*c^4)*x^3 + 2*(10*B*b^3*c^2 - 7*A*b^2*c^3)*x^2 + (13*B*b^4*c - 7*A*b^3*c^ 
2)*x)*x^2)*sqrt(c*x + b)/(c^7*x^4 + 2*b*c^6*x^3 + b^2*c^5*x^2) + integrate 
(-4/15*(4*B*b^5 - 2*A*b^4*c + (9*B*b^3*c^2 - 7*A*b^2*c^3)*x^2 + (13*B*b^4* 
c - 9*A*b^3*c^2)*x)*sqrt(c*x + b)*x^2/(c^7*x^5 + 3*b*c^6*x^4 + 3*b^2*c^5*x 
^3 + b^3*c^4*x^2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.82 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (B b^{4} - A b^{3} c\right )}}{\sqrt {c x + b} c^{5}} + \frac {2 \, {\left (5 \, {\left (c x + b\right )}^{\frac {7}{2}} B c^{30} - 28 \, {\left (c x + b\right )}^{\frac {5}{2}} B b c^{30} + 70 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{2} c^{30} - 140 \, \sqrt {c x + b} B b^{3} c^{30} + 7 \, {\left (c x + b\right )}^{\frac {5}{2}} A c^{31} - 35 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c^{31} + 105 \, \sqrt {c x + b} A b^{2} c^{31}\right )}}{35 \, c^{35}} \] Input:

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

Output:

-2*(B*b^4 - A*b^3*c)/(sqrt(c*x + b)*c^5) + 2/35*(5*(c*x + b)^(7/2)*B*c^30 
- 28*(c*x + b)^(5/2)*B*b*c^30 + 70*(c*x + b)^(3/2)*B*b^2*c^30 - 140*sqrt(c 
*x + b)*B*b^3*c^30 + 7*(c*x + b)^(5/2)*A*c^31 - 35*(c*x + b)^(3/2)*A*b*c^3 
1 + 105*sqrt(c*x + b)*A*b^2*c^31)/c^35
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.57 \[ \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {2}{7} b \,c^{4} x^{4}+\frac {2}{5} a \,c^{4} x^{3}-\frac {16}{35} b^{2} c^{3} x^{3}-\frac {4}{5} a b \,c^{3} x^{2}+\frac {32}{35} b^{3} c^{2} x^{2}+\frac {16}{5} a \,b^{2} c^{2} x -\frac {128}{35} b^{4} c x +\frac {32}{5} a \,b^{3} c -\frac {256}{35} b^{5}}{\sqrt {c x +b}\, c^{5}} \] Input:

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

Output:

(2*(112*a*b**3*c + 56*a*b**2*c**2*x - 14*a*b*c**3*x**2 + 7*a*c**4*x**3 - 1 
28*b**5 - 64*b**4*c*x + 16*b**3*c**2*x**2 - 8*b**2*c**3*x**3 + 5*b*c**4*x* 
*4))/(35*sqrt(b + c*x)*c**5)