\(\int \frac {x^{11/2} (A+B x)}{(b x+c x^2)^{5/2}} \, dx\) [221]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 b^3 (b B-A c) x^{3/2}}{3 c^5 \left (b x+c x^2\right )^{3/2}}+\frac {2 b^2 (4 b B-3 A c) \sqrt {x}}{c^5 \sqrt {b x+c x^2}}+\frac {6 b (2 b B-A c) \sqrt {b x+c x^2}}{c^5 \sqrt {x}}-\frac {2 (4 b B-A c) \left (b x+c x^2\right )^{3/2}}{3 c^5 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{5/2}}{5 c^5 x^{5/2}} \] Output:

-2/3*b^3*(-A*c+B*b)*x^(3/2)/c^5/(c*x^2+b*x)^(3/2)+2*b^2*(-3*A*c+4*B*b)*x^( 
1/2)/c^5/(c*x^2+b*x)^(1/2)+6*b*(-A*c+2*B*b)*(c*x^2+b*x)^(1/2)/c^5/x^(1/2)- 
2/3*(-A*c+4*B*b)*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)+2/5*B*(c*x^2+b*x)^(5/2)/c^5 
/x^(5/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.56 \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 x^{3/2} \left (128 b^4 B+24 b^2 c^2 x (-5 A+2 B x)+c^4 x^3 (5 A+3 B x)-2 b c^3 x^2 (15 A+4 B x)+b^3 (-80 A c+192 B c x)\right )}{15 c^5 (x (b+c x))^{3/2}} \] Input:

Integrate[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
 

Output:

(2*x^(3/2)*(128*b^4*B + 24*b^2*c^2*x*(-5*A + 2*B*x) + c^4*x^3*(5*A + 3*B*x 
) - 2*b*c^3*x^2*(15*A + 4*B*x) + b^3*(-80*A*c + 192*B*c*x)))/(15*c^5*(x*(b 
 + c*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.03, 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.

\(\displaystyle \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1218

\(\displaystyle -\frac {1}{3} \left (\frac {5 A}{b}-\frac {8 B}{c}\right ) \int \frac {x^{9/2}}{\left (c x^2+b x\right )^{3/2}}dx-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1128

\(\displaystyle -\frac {1}{3} \left (\frac {5 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {6 b \int \frac {x^{7/2}}{\left (c x^2+b x\right )^{3/2}}dx}{5 c}\right )-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1128

\(\displaystyle -\frac {1}{3} \left (\frac {5 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {6 b \left (\frac {2 x^{5/2}}{3 c \sqrt {b x+c x^2}}-\frac {4 b \int \frac {x^{5/2}}{\left (c x^2+b x\right )^{3/2}}dx}{3 c}\right )}{5 c}\right )-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1128

\(\displaystyle -\frac {1}{3} \left (\frac {5 A}{b}-\frac {8 B}{c}\right ) \left (\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {6 b \left (\frac {2 x^{5/2}}{3 c \sqrt {b x+c x^2}}-\frac {4 b \left (\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}-\frac {2 b \int \frac {x^{3/2}}{\left (c x^2+b x\right )^{3/2}}dx}{c}\right )}{3 c}\right )}{5 c}\right )-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1122

\(\displaystyle -\frac {1}{3} \left (\frac {2 x^{7/2}}{5 c \sqrt {b x+c x^2}}-\frac {6 b \left (\frac {2 x^{5/2}}{3 c \sqrt {b x+c x^2}}-\frac {4 b \left (\frac {4 b \sqrt {x}}{c^2 \sqrt {b x+c x^2}}+\frac {2 x^{3/2}}{c \sqrt {b x+c x^2}}\right )}{3 c}\right )}{5 c}\right ) \left (\frac {5 A}{b}-\frac {8 B}{c}\right )-\frac {2 x^{11/2} (b B-A c)}{3 b c \left (b x+c x^2\right )^{3/2}}\)

Input:

Int[(x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x]
 

Output:

(-2*(b*B - A*c)*x^(11/2))/(3*b*c*(b*x + c*x^2)^(3/2)) - (((5*A)/b - (8*B)/ 
c)*((2*x^(7/2))/(5*c*Sqrt[b*x + c*x^2]) - (6*b*((2*x^(5/2))/(3*c*Sqrt[b*x 
+ c*x^2]) - (4*b*((4*b*Sqrt[x])/(c^2*Sqrt[b*x + c*x^2]) + (2*x^(3/2))/(c*S 
qrt[b*x + c*x^2])))/(3*c)))/(5*c)))/3
 

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 = 0.65

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-3 B \,c^{4} x^{4}-5 A \,c^{4} x^{3}+8 B \,c^{3} x^{3} b +30 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}+120 A \,b^{2} c^{2} x -192 B \,b^{3} c x +80 A \,b^{3} c -128 B \,b^{4}\right ) x^{\frac {5}{2}}}{15 c^{5} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(107\)
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (-3 B \,c^{4} x^{4}-5 A \,c^{4} x^{3}+8 B \,c^{3} x^{3} b +30 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}+120 A \,b^{2} c^{2} x -192 B \,b^{3} c x +80 A \,b^{3} c -128 B \,b^{4}\right )}{15 \sqrt {x}\, \left (c x +b \right )^{2} c^{5}}\)	\(107\)
orering	\(-\frac {2 \left (c x +b \right ) \left (-3 B \,c^{4} x^{4}-5 A \,c^{4} x^{3}+8 B \,c^{3} x^{3} b +30 A b \,c^{3} x^{2}-48 c^{2} x^{2} B \,b^{2}+120 A \,b^{2} c^{2} x -192 B \,b^{3} c x +80 A \,b^{3} c -128 B \,b^{4}\right ) x^{\frac {5}{2}}}{15 c^{5} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(107\)
risch	\(-\frac {2 \left (-3 B \,c^{2} x^{2}-5 A \,c^{2} x +14 B b c x +40 A b c -73 B \,b^{2}\right ) \left (c x +b \right ) \sqrt {x}}{15 c^{5} \sqrt {x \left (c x +b \right )}}-\frac {2 b^{2} \left (9 A \,c^{2} x -12 B b c x +8 A b c -11 B \,b^{2}\right ) \sqrt {x}}{3 c^{5} \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}\)	\(110\)

Input:

int(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/15*(c*x+b)*(-3*B*c^4*x^4-5*A*c^4*x^3+8*B*b*c^3*x^3+30*A*b*c^3*x^2-48*B* 
b^2*c^2*x^2+120*A*b^2*c^2*x-192*B*b^3*c*x+80*A*b^3*c-128*B*b^4)*x^(5/2)/c^ 
5/(c*x^2+b*x)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.77 \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, B c^{4} x^{4} + 128 \, B b^{4} - 80 \, A b^{3} c - {\left (8 \, B b c^{3} - 5 \, A c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{2} c^{2} - 5 \, A b c^{3}\right )} x^{2} + 24 \, {\left (8 \, B b^{3} c - 5 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{15 \, {\left (c^{7} x^{3} + 2 \, b c^{6} x^{2} + b^{2} c^{5} x\right )}} \] Input:

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
 

Output:

2/15*(3*B*c^4*x^4 + 128*B*b^4 - 80*A*b^3*c - (8*B*b*c^3 - 5*A*c^4)*x^3 + 6 
*(8*B*b^2*c^2 - 5*A*b*c^3)*x^2 + 24*(8*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x^ 
2 + b*x)*sqrt(x)/(c^7*x^3 + 2*b*c^6*x^2 + b^2*c^5*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(11/2)*(B*x+A)/(c*x**2+b*x)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int { \frac {{\left (B x + A\right )} x^{\frac {11}{2}}}{{\left (c x^{2} + b x\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")
 

Output:

2/15*((3*B*c^3*x^2 + B*b*c^2*x - 2*B*b^2*c)*x^4 - (6*B*b^3 + (6*B*b*c^2 - 
5*A*c^3)*x^2 + (12*B*b^2*c - 5*A*b*c^2)*x)*x^3)*sqrt(c*x + b)/(c^6*x^4 + 3 
*b*c^5*x^3 + 3*b^2*c^4*x^2 + b^3*c^3*x) - integrate(-2/5*(4*B*b^4 + (9*B*b 
^2*c^2 - 5*A*b*c^3)*x^2 + (13*B*b^3*c - 5*A*b^2*c^2)*x)*sqrt(c*x + b)*x^3/ 
(c^7*x^6 + 4*b*c^6*x^5 + 6*b^2*c^5*x^4 + 4*b^3*c^4*x^3 + b^4*c^3*x^2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.76 \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (c x + b\right )} B b^{3} - B b^{4} - 9 \, {\left (c x + b\right )} A b^{2} c + A b^{3} c\right )}}{3 \, {\left (c x + b\right )}^{\frac {3}{2}} c^{5}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} B c^{20} - 20 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c^{20} + 90 \, \sqrt {c x + b} B b^{2} c^{20} + 5 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{21} - 45 \, \sqrt {c x + b} A b c^{21}\right )}}{15 \, c^{25}} \] Input:

integrate(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
 

Output:

2/3*(12*(c*x + b)*B*b^3 - B*b^4 - 9*(c*x + b)*A*b^2*c + A*b^3*c)/((c*x + b 
)^(3/2)*c^5) + 2/15*(3*(c*x + b)^(5/2)*B*c^20 - 20*(c*x + b)^(3/2)*B*b*c^2 
0 + 90*sqrt(c*x + b)*B*b^2*c^20 + 5*(c*x + b)^(3/2)*A*c^21 - 45*sqrt(c*x + 
 b)*A*b*c^21)/c^25
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{11/2}\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \] Input:

int((x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2),x)
 

Output:

int((x^(11/2)*(A + B*x))/(b*x + c*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61 \[ \int \frac {x^{11/2} (A+B x)}{\left (b x+c x^2\right )^{5/2}} \, dx=\frac {\frac {2}{5} b \,c^{4} x^{4}+\frac {2}{3} a \,c^{4} x^{3}-\frac {16}{15} b^{2} c^{3} x^{3}-4 a b \,c^{3} x^{2}+\frac {32}{5} b^{3} c^{2} x^{2}-16 a \,b^{2} c^{2} x +\frac {128}{5} b^{4} c x -\frac {32}{3} a \,b^{3} c +\frac {256}{15} b^{5}}{\sqrt {c x +b}\, c^{5} \left (c x +b \right )} \] Input:

int(x^(11/2)*(B*x+A)/(c*x^2+b*x)^(5/2),x)
 

Output:

(2*( - 80*a*b**3*c - 120*a*b**2*c**2*x - 30*a*b*c**3*x**2 + 5*a*c**4*x**3 
+ 128*b**5 + 192*b**4*c*x + 48*b**3*c**2*x**2 - 8*b**2*c**3*x**3 + 3*b*c** 
4*x**4))/(15*sqrt(b + c*x)*c**5*(b + c*x))