\(\int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx\) [172]

Optimal result

Integrand size = 24, antiderivative size = 169 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 b^3 (b B-A c) \left (b x+c x^2\right )^{3/2}}{3 c^5 x^{3/2}}-\frac {2 b^2 (4 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{5 c^5 x^{5/2}}+\frac {6 b (2 b B-A c) \left (b x+c x^2\right )^{7/2}}{7 c^5 x^{7/2}}-\frac {2 (4 b B-A c) \left (b x+c x^2\right )^{9/2}}{9 c^5 x^{9/2}}+\frac {2 B \left (b x+c x^2\right )^{11/2}}{11 c^5 x^{11/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 (x (b+c x))^{3/2} \left (128 b^4 B+35 c^4 x^3 (11 A+9 B x)+24 b^2 c^2 x (11 A+10 B x)-16 b^3 c (11 A+12 B x)-10 b c^3 x^2 (33 A+28 B x)\right )}{3465 c^5 x^{3/2}} \] Input:

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

Output:

(2*(x*(b + c*x))^(3/2)*(128*b^4*B + 35*c^4*x^3*(11*A + 9*B*x) + 24*b^2*c^2 
*x*(11*A + 10*B*x) - 16*b^3*c*(11*A + 12*B*x) - 10*b*c^3*x^2*(33*A + 28*B* 
x)))/(3465*c^5*x^(3/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1221, 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.

\(\displaystyle \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx\)

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1128

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

\(\Big \downarrow \) 1128

\(\displaystyle \frac {2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac {(8 b B-11 A c) \left (\frac {2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {2 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \int \sqrt {x} \sqrt {c x^2+b x}dx}{7 c}\right )}{3 c}\right )}{11 c}\)

\(\Big \downarrow \) 1128

\(\displaystyle \frac {2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac {(8 b B-11 A c) \left (\frac {2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {2 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {2 b \int \frac {\sqrt {c x^2+b x}}{\sqrt {x}}dx}{5 c}\right )}{7 c}\right )}{3 c}\right )}{11 c}\)

\(\Big \downarrow \) 1122

\(\displaystyle \frac {2 B x^{5/2} \left (b x+c x^2\right )^{3/2}}{11 c}-\frac {\left (\frac {2 x^{3/2} \left (b x+c x^2\right )^{3/2}}{9 c}-\frac {2 b \left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}{7 c}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}\right )}{7 c}\right )}{3 c}\right ) (8 b B-11 A c)}{11 c}\)

Input:

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

Output:

(2*B*x^(5/2)*(b*x + c*x^2)^(3/2))/(11*c) - ((8*b*B - 11*A*c)*((2*x^(3/2)*( 
b*x + c*x^2)^(3/2))/(9*c) - (2*b*((2*Sqrt[x]*(b*x + c*x^2)^(3/2))/(7*c) - 
(4*b*((-4*b*(b*x + c*x^2)^(3/2))/(15*c^2*x^(3/2)) + (2*(b*x + c*x^2)^(3/2) 
)/(5*c*Sqrt[x])))/(7*c)))/(3*c)))/(11*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 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(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] 
 && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {2 \left (c x +b \right ) \left (-315 B \,c^{4} x^{4}-385 A \,c^{4} x^{3}+280 B \,c^{3} x^{3} b +330 A b \,c^{3} x^{2}-240 c^{2} x^{2} B \,b^{2}-264 A \,b^{2} c^{2} x +192 B \,b^{3} c x +176 A \,b^{3} c -128 B \,b^{4}\right ) \sqrt {x \left (c x +b \right )}}{3465 c^{5} \sqrt {x}}\)	\(105\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-315 B \,c^{4} x^{4}-385 A \,c^{4} x^{3}+280 B \,c^{3} x^{3} b +330 A b \,c^{3} x^{2}-240 c^{2} x^{2} B \,b^{2}-264 A \,b^{2} c^{2} x +192 B \,b^{3} c x +176 A \,b^{3} c -128 B \,b^{4}\right ) \sqrt {c \,x^{2}+b x}}{3465 c^{5} \sqrt {x}}\)	\(107\)
orering	\(-\frac {2 \left (c x +b \right ) \left (-315 B \,c^{4} x^{4}-385 A \,c^{4} x^{3}+280 B \,c^{3} x^{3} b +330 A b \,c^{3} x^{2}-240 c^{2} x^{2} B \,b^{2}-264 A \,b^{2} c^{2} x +192 B \,b^{3} c x +176 A \,b^{3} c -128 B \,b^{4}\right ) \sqrt {c \,x^{2}+b x}}{3465 c^{5} \sqrt {x}}\)	\(107\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-315 B \,c^{5} x^{5}-385 A \,c^{5} x^{4}-35 B b \,c^{4} x^{4}-55 A b \,c^{4} x^{3}+40 B \,b^{2} c^{3} x^{3}+66 A \,b^{2} c^{3} x^{2}-48 B \,b^{3} c^{2} x^{2}-88 A \,b^{3} c^{2} x +64 B \,b^{4} c x +176 A \,b^{4} c -128 b^{5} B \right )}{3465 \sqrt {x \left (c x +b \right )}\, c^{5}}\)	\(129\)

Input:

int(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3465*(c*x+b)*(-315*B*c^4*x^4-385*A*c^4*x^3+280*B*b*c^3*x^3+330*A*b*c^3* 
x^2-240*B*b^2*c^2*x^2-264*A*b^2*c^2*x+192*B*b^3*c*x+176*A*b^3*c-128*B*b^4) 
*(x*(c*x+b))^(1/2)/c^5/x^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.75 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (315 \, B c^{5} x^{5} + 128 \, B b^{5} - 176 \, A b^{4} c + 35 \, {\left (B b c^{4} + 11 \, A c^{5}\right )} x^{4} - 5 \, {\left (8 \, B b^{2} c^{3} - 11 \, A b c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{3} c^{2} - 11 \, A b^{2} c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{3465 \, c^{5} \sqrt {x}} \] Input:

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="fricas")
 

Output:

2/3465*(315*B*c^5*x^5 + 128*B*b^5 - 176*A*b^4*c + 35*(B*b*c^4 + 11*A*c^5)* 
x^4 - 5*(8*B*b^2*c^3 - 11*A*b*c^4)*x^3 + 6*(8*B*b^3*c^2 - 11*A*b^2*c^3)*x^ 
2 - 8*(8*B*b^4*c - 11*A*b^3*c^2)*x)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x))
 

Sympy [F]

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

integrate(x**(5/2)*(B*x+A)*(c*x**2+b*x)**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x + b} A}{315 \, c^{4}} + \frac {2 \, {\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt {c x + b} B}{3465 \, c^{5}} \] Input:

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="maxima")
 

Output:

2/315*(35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*sqrt 
(c*x + b)*A/c^4 + 2/3465*(315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48 
*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*sqrt(c*x + b)*B/c^5
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.67 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}\right )} A}{315 \, c^{4}} + \frac {2 \, {\left (315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}\right )} B}{3465 \, c^{5}} \] Input:

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x, algorithm="giac")
 

Output:

2/315*(35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^ 
2 - 105*(c*x + b)^(3/2)*b^3)*A/c^4 + 2/3465*(315*(c*x + b)^(11/2) - 1540*( 
c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 + 1 
155*(c*x + b)^(3/2)*b^4)*B/c^5
 

Mupad [F(-1)]

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

int(x^(5/2)*(b*x + c*x^2)^(1/2)*(A + B*x),x)
 

Output:

int(x^(5/2)*(b*x + c*x^2)^(1/2)*(A + B*x), x)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.67 \[ \int x^{5/2} (A+B x) \sqrt {b x+c x^2} \, dx=\frac {2 \sqrt {c x +b}\, \left (315 b \,c^{5} x^{5}+385 a \,c^{5} x^{4}+35 b^{2} c^{4} x^{4}+55 a b \,c^{4} x^{3}-40 b^{3} c^{3} x^{3}-66 a \,b^{2} c^{3} x^{2}+48 b^{4} c^{2} x^{2}+88 a \,b^{3} c^{2} x -64 b^{5} c x -176 a \,b^{4} c +128 b^{6}\right )}{3465 c^{5}} \] Input:

int(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(1/2),x)
 

Output:

(2*sqrt(b + c*x)*( - 176*a*b**4*c + 88*a*b**3*c**2*x - 66*a*b**2*c**3*x**2 
 + 55*a*b*c**4*x**3 + 385*a*c**5*x**4 + 128*b**6 - 64*b**5*c*x + 48*b**4*c 
**2*x**2 - 40*b**3*c**3*x**3 + 35*b**2*c**4*x**4 + 315*b*c**5*x**5))/(3465 
*c**5)