\(\int \sqrt {x} (A+B x) (b x+c x^2)^{5/2} \, dx\) [194]

Optimal result

Integrand size = 24, antiderivative size = 169 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 b^3 (b B-A c) \left (b x+c x^2\right )^{7/2}}{7 c^5 x^{7/2}}-\frac {2 b^2 (4 b B-3 A c) \left (b x+c x^2\right )^{9/2}}{9 c^5 x^{9/2}}+\frac {6 b (2 b B-A c) \left (b x+c x^2\right )^{11/2}}{11 c^5 x^{11/2}}-\frac {2 (4 b B-A c) \left (b x+c x^2\right )^{13/2}}{13 c^5 x^{13/2}}+\frac {2 B \left (b x+c x^2\right )^{15/2}}{15 c^5 x^{15/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (128 b^4 B+168 b^2 c^2 x (5 A+6 B x)+231 c^4 x^3 (15 A+13 B x)-16 b^3 c (15 A+28 B x)-42 b c^3 x^2 (45 A+44 B x)\right )}{45045 c^5 \sqrt {x}} \] Input:

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

Output:

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(128*b^4*B + 168*b^2*c^2*x*(5*A + 6*B*x) 
+ 231*c^4*x^3*(15*A + 13*B*x) - 16*b^3*c*(15*A + 28*B*x) - 42*b*c^3*x^2*(4 
5*A + 44*B*x)))/(45045*c^5*Sqrt[x])
 

Rubi [A] (verified)

Time = 0.51 (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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1128

\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}-\frac {(8 b B-15 A c) \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}-\frac {6 b \int \frac {\left (c x^2+b x\right )^{5/2}}{\sqrt {x}}dx}{13 c}\right )}{15 c}\)

\(\Big \downarrow \) 1128

\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}-\frac {(8 b B-15 A c) \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}-\frac {4 b \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{3/2}}dx}{11 c}\right )}{13 c}\right )}{15 c}\)

\(\Big \downarrow \) 1128

\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}-\frac {(8 b B-15 A c) \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac {2 b \int \frac {\left (c x^2+b x\right )^{5/2}}{x^{5/2}}dx}{9 c}\right )}{11 c}\right )}{13 c}\right )}{15 c}\)

\(\Big \downarrow \) 1122

\(\displaystyle \frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}-\frac {\left (\frac {2 \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac {4 b \left (b x+c x^2\right )^{7/2}}{63 c^2 x^{7/2}}\right )}{11 c}\right )}{13 c}\right ) (8 b B-15 A c)}{15 c}\)

Input:

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

Output:

(2*B*Sqrt[x]*(b*x + c*x^2)^(7/2))/(15*c) - ((8*b*B - 15*A*c)*((2*(b*x + c* 
x^2)^(7/2))/(13*c*Sqrt[x]) - (6*b*((2*(b*x + c*x^2)^(7/2))/(11*c*x^(3/2)) 
- (4*b*((-4*b*(b*x + c*x^2)^(7/2))/(63*c^2*x^(7/2)) + (2*(b*x + c*x^2)^(7/ 
2))/(9*c*x^(5/2))))/(11*c)))/(13*c)))/(15*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 = 1.00 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-3003 B \,c^{4} x^{4}-3465 A \,c^{4} x^{3}+1848 B \,c^{3} x^{3} b +1890 A b \,c^{3} x^{2}-1008 c^{2} x^{2} B \,b^{2}-840 A \,b^{2} c^{2} x +448 B \,b^{3} c x +240 A \,b^{3} c -128 B \,b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{45045 c^{5} x^{\frac {5}{2}}}\)	\(107\)
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{3} \left (-3003 B \,c^{4} x^{4}-3465 A \,c^{4} x^{3}+1848 B \,c^{3} x^{3} b +1890 A b \,c^{3} x^{2}-1008 c^{2} x^{2} B \,b^{2}-840 A \,b^{2} c^{2} x +448 B \,b^{3} c x +240 A \,b^{3} c -128 B \,b^{4}\right )}{45045 \sqrt {x}\, c^{5}}\)	\(107\)
orering	\(-\frac {2 \left (c x +b \right ) \left (-3003 B \,c^{4} x^{4}-3465 A \,c^{4} x^{3}+1848 B \,c^{3} x^{3} b +1890 A b \,c^{3} x^{2}-1008 c^{2} x^{2} B \,b^{2}-840 A \,b^{2} c^{2} x +448 B \,b^{3} c x +240 A \,b^{3} c -128 B \,b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{45045 c^{5} x^{\frac {5}{2}}}\)	\(107\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-3003 B \,c^{7} x^{7}-3465 A \,c^{7} x^{6}-7161 B b \,c^{6} x^{6}-8505 A b \,c^{6} x^{5}-4473 B \,b^{2} c^{5} x^{5}-5565 A \,b^{2} c^{5} x^{4}-35 B \,b^{3} c^{4} x^{4}-75 A \,b^{3} c^{4} x^{3}+40 B \,b^{4} c^{3} x^{3}+90 A \,b^{4} c^{3} x^{2}-48 B \,b^{5} c^{2} x^{2}-120 A \,b^{5} c^{2} x +64 B \,b^{6} c x +240 A \,b^{6} c -128 B \,b^{7}\right )}{45045 \sqrt {x \left (c x +b \right )}\, c^{5}}\)	\(177\)

Input:

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

Output:

-2/45045*(c*x+b)*(-3003*B*c^4*x^4-3465*A*c^4*x^3+1848*B*b*c^3*x^3+1890*A*b 
*c^3*x^2-1008*B*b^2*c^2*x^2-840*A*b^2*c^2*x+448*B*b^3*c*x+240*A*b^3*c-128* 
B*b^4)*(c*x^2+b*x)^(5/2)/c^5/x^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.03 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 \, {\left (3003 \, B c^{7} x^{7} + 128 \, B b^{7} - 240 \, A b^{6} c + 231 \, {\left (31 \, B b c^{6} + 15 \, A c^{7}\right )} x^{6} + 63 \, {\left (71 \, B b^{2} c^{5} + 135 \, A b c^{6}\right )} x^{5} + 35 \, {\left (B b^{3} c^{4} + 159 \, A b^{2} c^{5}\right )} x^{4} - 5 \, {\left (8 \, B b^{4} c^{3} - 15 \, A b^{3} c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{5} c^{2} - 15 \, A b^{4} c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{6} c - 15 \, A b^{5} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{45045 \, c^{5} \sqrt {x}} \] Input:

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

Output:

2/45045*(3003*B*c^7*x^7 + 128*B*b^7 - 240*A*b^6*c + 231*(31*B*b*c^6 + 15*A 
*c^7)*x^6 + 63*(71*B*b^2*c^5 + 135*A*b*c^6)*x^5 + 35*(B*b^3*c^4 + 159*A*b^ 
2*c^5)*x^4 - 5*(8*B*b^4*c^3 - 15*A*b^3*c^4)*x^3 + 6*(8*B*b^5*c^2 - 15*A*b^ 
4*c^3)*x^2 - 8*(8*B*b^6*c - 15*A*b^5*c^2)*x)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x 
))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (139) = 278\).

Time = 0.06 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.61 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 \, {\left (5 \, {\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 26 \, {\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4} + 143 \, {\left (35 \, b^{2} c^{4} x^{6} + 5 \, b^{3} c^{3} x^{5} - 6 \, b^{4} c^{2} x^{4} + 8 \, b^{5} c x^{3} - 16 \, b^{6} x^{2}\right )} x^{3}\right )} \sqrt {c x + b} A}{45045 \, c^{4} x^{5}} + \frac {2 \, {\left ({\left (3003 \, c^{7} x^{7} + 231 \, b c^{6} x^{6} - 252 \, b^{2} c^{5} x^{5} + 280 \, b^{3} c^{4} x^{4} - 320 \, b^{4} c^{3} x^{3} + 384 \, b^{5} c^{2} x^{2} - 512 \, b^{6} c x + 1024 \, b^{7}\right )} x^{6} + 10 \, {\left (693 \, b c^{6} x^{7} + 63 \, b^{2} c^{5} x^{6} - 70 \, b^{3} c^{4} x^{5} + 80 \, b^{4} c^{3} x^{4} - 96 \, b^{5} c^{2} x^{3} + 128 \, b^{6} c x^{2} - 256 \, b^{7} x\right )} x^{5} + 13 \, {\left (315 \, b^{2} c^{5} x^{7} + 35 \, b^{3} c^{4} x^{6} - 40 \, b^{4} c^{3} x^{5} + 48 \, b^{5} c^{2} x^{4} - 64 \, b^{6} c x^{3} + 128 \, b^{7} x^{2}\right )} x^{4}\right )} \sqrt {c x + b} B}{45045 \, c^{5} x^{6}} \] Input:

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

Output:

2/45045*(5*(693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^4 + 80*b^3*c^3*x^3 - 
 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*b^6)*x^5 + 26*(315*b*c^5*x^6 + 35*b^2* 
c^4*x^5 - 40*b^3*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^4 
+ 143*(35*b^2*c^4*x^6 + 5*b^3*c^3*x^5 - 6*b^4*c^2*x^4 + 8*b^5*c*x^3 - 16*b 
^6*x^2)*x^3)*sqrt(c*x + b)*A/(c^4*x^5) + 2/45045*((3003*c^7*x^7 + 231*b*c^ 
6*x^6 - 252*b^2*c^5*x^5 + 280*b^3*c^4*x^4 - 320*b^4*c^3*x^3 + 384*b^5*c^2* 
x^2 - 512*b^6*c*x + 1024*b^7)*x^6 + 10*(693*b*c^6*x^7 + 63*b^2*c^5*x^6 - 7 
0*b^3*c^4*x^5 + 80*b^4*c^3*x^4 - 96*b^5*c^2*x^3 + 128*b^6*c*x^2 - 256*b^7* 
x)*x^5 + 13*(315*b^2*c^5*x^7 + 35*b^3*c^4*x^6 - 40*b^4*c^3*x^5 + 48*b^5*c^ 
2*x^4 - 64*b^6*c*x^3 + 128*b^7*x^2)*x^4)*sqrt(c*x + b)*B/(c^5*x^6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 417 vs. \(2 (139) = 278\).

Time = 0.13 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.47 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, 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 b^{2}}{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 b^{2}}{3465 \, c^{5}} + \frac {4 \, {\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 )} A b}{3465 \, c^{4}} + \frac {4 \, {\left (693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}\right )} B b}{9009 \, c^{5}} + \frac {2 \, {\left (693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}\right )} A}{9009 \, c^{4}} + \frac {2 \, {\left (3003 \, {\left (c x + b\right )}^{\frac {15}{2}} - 20790 \, {\left (c x + b\right )}^{\frac {13}{2}} b + 61425 \, {\left (c x + b\right )}^{\frac {11}{2}} b^{2} - 100100 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{3} + 96525 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{4} - 54054 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{5} + 15015 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{6}\right )} B}{45045 \, c^{5}} \] Input:

integrate(x^(1/2)*(B*x+A)*(c*x^2+b*x)^(5/2),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*b^2/c^4 + 2/3465*(315*(c*x + b)^(11/2) - 15 
40*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 
 + 1155*(c*x + b)^(3/2)*b^4)*B*b^2/c^5 + 4/3465*(315*(c*x + b)^(11/2) - 15 
40*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 
 + 1155*(c*x + b)^(3/2)*b^4)*A*b/c^4 + 4/9009*(693*(c*x + b)^(13/2) - 4095 
*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^2 - 12870*(c*x + b)^(7/2)*b^ 
3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)*B*b/c^5 + 2/9009* 
(693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^ 
2 - 12870*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^ 
(3/2)*b^5)*A/c^4 + 2/45045*(3003*(c*x + b)^(15/2) - 20790*(c*x + b)^(13/2) 
*b + 61425*(c*x + b)^(11/2)*b^2 - 100100*(c*x + b)^(9/2)*b^3 + 96525*(c*x 
+ b)^(7/2)*b^4 - 54054*(c*x + b)^(5/2)*b^5 + 15015*(c*x + b)^(3/2)*b^6)*B/ 
c^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95 \[ \int \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx=\frac {2 \sqrt {c x +b}\, \left (3003 b \,c^{7} x^{7}+3465 a \,c^{7} x^{6}+7161 b^{2} c^{6} x^{6}+8505 a b \,c^{6} x^{5}+4473 b^{3} c^{5} x^{5}+5565 a \,b^{2} c^{5} x^{4}+35 b^{4} c^{4} x^{4}+75 a \,b^{3} c^{4} x^{3}-40 b^{5} c^{3} x^{3}-90 a \,b^{4} c^{3} x^{2}+48 b^{6} c^{2} x^{2}+120 a \,b^{5} c^{2} x -64 b^{7} c x -240 a \,b^{6} c +128 b^{8}\right )}{45045 c^{5}} \] Input:

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

Output:

(2*sqrt(b + c*x)*( - 240*a*b**6*c + 120*a*b**5*c**2*x - 90*a*b**4*c**3*x** 
2 + 75*a*b**3*c**4*x**3 + 5565*a*b**2*c**5*x**4 + 8505*a*b*c**6*x**5 + 346 
5*a*c**7*x**6 + 128*b**8 - 64*b**7*c*x + 48*b**6*c**2*x**2 - 40*b**5*c**3* 
x**3 + 35*b**4*c**4*x**4 + 4473*b**3*c**5*x**5 + 7161*b**2*c**6*x**6 + 300 
3*b*c**7*x**7))/(45045*c**5)