\(\int x^{3/2} (A+B x) (b x+c x^2)^{3/2} \, dx\) [183]

Optimal result

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

2/5*b^3*(-A*c+B*b)*(c*x^2+b*x)^(5/2)/c^5/x^(5/2)-2/7*b^2*(-3*A*c+4*B*b)*(c 
*x^2+b*x)^(7/2)/c^5/x^(7/2)+2/3*b*(-A*c+2*B*b)*(c*x^2+b*x)^(9/2)/c^5/x^(9/ 
2)-2/11*(-A*c+4*B*b)*(c*x^2+b*x)^(11/2)/c^5/x^(11/2)+2/13*B*(c*x^2+b*x)^(1 
3/2)/c^5/x^(13/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 (x (b+c x))^{5/2} \left (128 b^4 B+105 c^4 x^3 (13 A+11 B x)-70 b c^3 x^2 (13 A+12 B x)+40 b^2 c^2 x (13 A+14 B x)-16 b^3 c (13 A+20 B x)\right )}{15015 c^5 x^{5/2}} \] Input:

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

Output:

(2*(x*(b + c*x))^(5/2)*(128*b^4*B + 105*c^4*x^3*(13*A + 11*B*x) - 70*b*c^3 
*x^2*(13*A + 12*B*x) + 40*b^2*c^2*x*(13*A + 14*B*x) - 16*b^3*c*(13*A + 20* 
B*x)))/(15015*c^5*x^(5/2))
 

Rubi [A] (verified)

Time = 0.54 (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.

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

\(\Big \downarrow \) 1221

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

\(\Big \downarrow \) 1128

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

\(\Big \downarrow \) 1128

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

\(\Big \downarrow \) 1128

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

\(\Big \downarrow \) 1122

\(\displaystyle \frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {\left (\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{9 c \sqrt {x}}-\frac {4 b \left (\frac {2 \left (b x+c x^2\right )^{5/2}}{7 c x^{3/2}}-\frac {4 b \left (b x+c x^2\right )^{5/2}}{35 c^2 x^{5/2}}\right )}{9 c}\right )}{11 c}\right ) (8 b B-13 A c)}{13 c}\)

Input:

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

Output:

(2*B*x^(3/2)*(b*x + c*x^2)^(5/2))/(13*c) - ((8*b*B - 13*A*c)*((2*Sqrt[x]*( 
b*x + c*x^2)^(5/2))/(11*c) - (6*b*((2*(b*x + c*x^2)^(5/2))/(9*c*Sqrt[x]) - 
 (4*b*((-4*b*(b*x + c*x^2)^(5/2))/(35*c^2*x^(5/2)) + (2*(b*x + c*x^2)^(5/2 
))/(7*c*x^(3/2))))/(9*c)))/(11*c)))/(13*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 = 0.96 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.63

method	result	size
gosper	\(-\frac {2 \left (c x +b \right ) \left (-1155 B \,c^{4} x^{4}-1365 A \,c^{4} x^{3}+840 B \,c^{3} x^{3} b +910 A b \,c^{3} x^{2}-560 c^{2} x^{2} B \,b^{2}-520 A \,b^{2} c^{2} x +320 B \,b^{3} c x +208 A \,b^{3} c -128 B \,b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 c^{5} x^{\frac {3}{2}}}\)	\(107\)
default	\(-\frac {2 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right )^{2} \left (-1155 B \,c^{4} x^{4}-1365 A \,c^{4} x^{3}+840 B \,c^{3} x^{3} b +910 A b \,c^{3} x^{2}-560 c^{2} x^{2} B \,b^{2}-520 A \,b^{2} c^{2} x +320 B \,b^{3} c x +208 A \,b^{3} c -128 B \,b^{4}\right )}{15015 \sqrt {x}\, c^{5}}\)	\(107\)
orering	\(-\frac {2 \left (c x +b \right ) \left (-1155 B \,c^{4} x^{4}-1365 A \,c^{4} x^{3}+840 B \,c^{3} x^{3} b +910 A b \,c^{3} x^{2}-560 c^{2} x^{2} B \,b^{2}-520 A \,b^{2} c^{2} x +320 B \,b^{3} c x +208 A \,b^{3} c -128 B \,b^{4}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 c^{5} x^{\frac {3}{2}}}\)	\(107\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-1155 B \,c^{6} x^{6}-1365 A \,c^{6} x^{5}-1470 B b \,c^{5} x^{5}-1820 A b \,c^{5} x^{4}-35 B \,b^{2} c^{4} x^{4}-65 A \,b^{2} c^{4} x^{3}+40 B \,b^{3} c^{3} x^{3}+78 A \,b^{3} c^{3} x^{2}-48 B \,b^{4} c^{2} x^{2}-104 A \,b^{4} c^{2} x +64 B \,b^{5} c x +208 A \,b^{5} c -128 B \,b^{6}\right )}{15015 \sqrt {x \left (c x +b \right )}\, c^{5}}\)	\(153\)

Input:

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

Output:

-2/15015*(c*x+b)*(-1155*B*c^4*x^4-1365*A*c^4*x^3+840*B*b*c^3*x^3+910*A*b*c 
^3*x^2-560*B*b^2*c^2*x^2-520*A*b^2*c^2*x+320*B*b^3*c*x+208*A*b^3*c-128*B*b 
^4)*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.89 \[ \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (1155 \, B c^{6} x^{6} + 128 \, B b^{6} - 208 \, A b^{5} c + 105 \, {\left (14 \, B b c^{5} + 13 \, A c^{6}\right )} x^{5} + 35 \, {\left (B b^{2} c^{4} + 52 \, A b c^{5}\right )} x^{4} - 5 \, {\left (8 \, B b^{3} c^{3} - 13 \, A b^{2} c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{15015 \, c^{5} \sqrt {x}} \] Input:

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

Output:

2/15015*(1155*B*c^6*x^6 + 128*B*b^6 - 208*A*b^5*c + 105*(14*B*b*c^5 + 13*A 
*c^6)*x^5 + 35*(B*b^2*c^4 + 52*A*b*c^5)*x^4 - 5*(8*B*b^3*c^3 - 13*A*b^2*c^ 
4)*x^3 + 6*(8*B*b^4*c^2 - 13*A*b^3*c^3)*x^2 - 8*(8*B*b^5*c - 13*A*b^4*c^2) 
*x)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.62 \[ \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt {c x + b} A}{3465 \, c^{4} x^{4}} + \frac {2 \, {\left (5 \, {\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 13 \, {\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4}\right )} \sqrt {c x + b} B}{45045 \, c^{5} x^{5}} \] Input:

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

Output:

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)*x^4 + 11*(35*b*c^4*x^5 + 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 
 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)*A/(c^4*x^4) + 2/45045*(5*(69 
3*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 + 13*(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)*sqrt(c*x + b) 
*B/(c^5*x^5)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.49 \[ \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \, {\left (35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}\right )} A b}{315 \, c^{4}} + \frac {2 \, {\left (315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}\right )} B b}{3465 \, c^{5}} + \frac {2 \, {\left (315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}\right )} A}{3465 \, c^{4}} + \frac {2 \, {\left (693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}\right )} B}{9009 \, c^{5}} \] Input:

integrate(x^(3/2)*(B*x+A)*(c*x^2+b*x)^(3/2),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/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 + 
 1155*(c*x + b)^(3/2)*b^4)*B*b/c^5 + 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)*A/c^4 + 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 + 90 
09*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)*B/c^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.81 \[ \int x^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {2 \sqrt {c x +b}\, \left (1155 b \,c^{6} x^{6}+1365 a \,c^{6} x^{5}+1470 b^{2} c^{5} x^{5}+1820 a b \,c^{5} x^{4}+35 b^{3} c^{4} x^{4}+65 a \,b^{2} c^{4} x^{3}-40 b^{4} c^{3} x^{3}-78 a \,b^{3} c^{3} x^{2}+48 b^{5} c^{2} x^{2}+104 a \,b^{4} c^{2} x -64 b^{6} c x -208 a \,b^{5} c +128 b^{7}\right )}{15015 c^{5}} \] Input:

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

Output:

(2*sqrt(b + c*x)*( - 208*a*b**5*c + 104*a*b**4*c**2*x - 78*a*b**3*c**3*x** 
2 + 65*a*b**2*c**4*x**3 + 1820*a*b*c**5*x**4 + 1365*a*c**6*x**5 + 128*b**7 
 - 64*b**6*c*x + 48*b**5*c**2*x**2 - 40*b**4*c**3*x**3 + 35*b**3*c**4*x**4 
 + 1470*b**2*c**5*x**5 + 1155*b*c**6*x**6))/(15015*c**5)