\(\int x (A+B x) (b x+c x^2)^{3/2} \, dx\) [115]

Optimal result

Integrand size = 20, antiderivative size = 227 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {b^4 (7 b B-12 A c) \sqrt {b x+c x^2}}{512 c^4}+\frac {b^3 (7 b B-12 A c) x \sqrt {b x+c x^2}}{768 c^3}-\frac {b^2 (7 b B-12 A c) x^2 \sqrt {b x+c x^2}}{960 c^2}-\frac {11 b (7 b B-12 A c) x^3 \sqrt {b x+c x^2}}{480 c}-\frac {1}{60} (7 b B-12 A c) x^4 \sqrt {b x+c x^2}+\frac {B x \left (b x+c x^2\right )^{5/2}}{6 c}+\frac {b^5 (7 b B-12 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{9/2}} \] Output:

-1/512*b^4*(-12*A*c+7*B*b)*(c*x^2+b*x)^(1/2)/c^4+1/768*b^3*(-12*A*c+7*B*b) 
*x*(c*x^2+b*x)^(1/2)/c^3-1/960*b^2*(-12*A*c+7*B*b)*x^2*(c*x^2+b*x)^(1/2)/c 
^2-11/480*b*(-12*A*c+7*B*b)*x^3*(c*x^2+b*x)^(1/2)/c-1/60*(-12*A*c+7*B*b)*x 
^4*(c*x^2+b*x)^(1/2)+1/6*B*x*(c*x^2+b*x)^(5/2)/c+1/512*b^5*(-12*A*c+7*B*b) 
*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.96 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {x} \sqrt {b+c x} \left (\sqrt {c} \sqrt {x} \sqrt {b+c x} \left (-105 b^5 B+48 b^2 c^3 x^2 (2 A+B x)+256 c^5 x^4 (6 A+5 B x)-8 b^3 c^2 x (15 A+7 B x)+10 b^4 c (18 A+7 B x)+64 b c^4 x^3 (33 A+26 B x)\right )+360 A b^5 c \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+210 b^6 B \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )\right )}{7680 c^{9/2} \sqrt {x (b+c x)}} \] Input:

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

Output:

(Sqrt[x]*Sqrt[b + c*x]*(Sqrt[c]*Sqrt[x]*Sqrt[b + c*x]*(-105*b^5*B + 48*b^2 
*c^3*x^2*(2*A + B*x) + 256*c^5*x^4*(6*A + 5*B*x) - 8*b^3*c^2*x*(15*A + 7*B 
*x) + 10*b^4*c*(18*A + 7*B*x) + 64*b*c^4*x^3*(33*A + 26*B*x)) + 360*A*b^5* 
c*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 210*b^6*B*ArcTanh 
[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])]))/(7680*c^(9/2)*Sqrt[x*(b + 
 c*x)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1225, 1087, 1087, 1091, 219}

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.

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {b (7 b B-12 A c) \int \left (c x^2+b x\right )^{3/2}dx}{24 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {b (7 b B-12 A c) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \int \sqrt {c x^2+b x}dx}{16 c}\right )}{24 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {b (7 b B-12 A c) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {b (7 b B-12 A c) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {b (7 b B-12 A c) \left (\frac {(b+2 c x) \left (b x+c x^2\right )^{3/2}}{8 c}-\frac {3 b^2 \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )}{16 c}\right )}{24 c^2}-\frac {\left (b x+c x^2\right )^{5/2} (-12 A c+7 b B-10 B c x)}{60 c^2}\)

Input:

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

Output:

-1/60*((7*b*B - 12*A*c - 10*B*c*x)*(b*x + c*x^2)^(5/2))/c^2 + (b*(7*b*B - 
12*A*c)*(((b + 2*c*x)*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2*(((b + 2*c*x)*Sq 
rt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c 
^(3/2))))/(16*c)))/(24*c^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.59

method	result	size
pseudoelliptic	\(-\frac {\left (\frac {3}{2} A \,b^{5} c -\frac {7}{8} B \,b^{6}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (-\frac {64 x^{4} \left (\frac {5 B x}{6}+A \right ) c^{\frac {11}{2}}}{5}+\left (-\frac {3 \left (\frac {7 B x}{18}+A \right ) b^{3} c^{\frac {3}{2}}}{2}+b^{2} x \left (\frac {7 B x}{15}+A \right ) c^{\frac {5}{2}}-\frac {4 \left (\frac {B x}{2}+A \right ) x^{2} b \,c^{\frac {7}{2}}}{5}-\frac {88 \left (\frac {26 B x}{33}+A \right ) x^{3} c^{\frac {9}{2}}}{5}+\frac {7 B \sqrt {c}\, b^{4}}{8}\right ) b \right )}{64 c^{\frac {9}{2}}}\)	\(133\)
risch	\(\frac {\left (1280 B \,c^{5} x^{5}+1536 A \,c^{5} x^{4}+1664 B b \,c^{4} x^{4}+2112 A b \,c^{4} x^{3}+48 B \,b^{2} c^{3} x^{3}+96 A \,b^{2} c^{3} x^{2}-56 B \,b^{3} c^{2} x^{2}-120 A \,b^{3} c^{2} x +70 B \,b^{4} c x +180 A \,b^{4} c -105 b^{5} B \right ) x \left (c x +b \right )}{7680 c^{4} \sqrt {x \left (c x +b \right )}}-\frac {b^{5} \left (12 A c -7 B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {9}{2}}}\)	\(169\)
default	\(A \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )+B \left (\frac {x \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}\right )\)	\(248\)

Input:

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

Output:

-1/64/c^(9/2)*((3/2*A*b^5*c-7/8*B*b^6)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2) 
)+(x*(c*x+b))^(1/2)*(-64/5*x^4*(5/6*B*x+A)*c^(11/2)+(-3/2*(7/18*B*x+A)*b^3 
*c^(3/2)+b^2*x*(7/15*B*x+A)*c^(5/2)-4/5*(1/2*B*x+A)*x^2*b*c^(7/2)-88/5*(26 
/33*B*x+A)*x^3*c^(9/2)+7/8*B*c^(1/2)*b^4)*b))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.55 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{15360 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (1280 \, B c^{6} x^{5} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{4} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{3} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{2} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{7680 \, c^{5}}\right ] \] Input:

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

Output:

[-1/15360*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 
+ b*x)*sqrt(c)) - 2*(1280*B*c^6*x^5 - 105*B*b^5*c + 180*A*b^4*c^2 + 128*(1 
3*B*b*c^5 + 12*A*c^6)*x^4 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^3 - 8*(7*B*b^3*c 
^3 - 12*A*b^2*c^4)*x^2 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c*x^2 + b 
*x))/c^5, -1/7680*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^2 + 
b*x)*sqrt(-c)/(c*x + b)) - (1280*B*c^6*x^5 - 105*B*b^5*c + 180*A*b^4*c^2 + 
 128*(13*B*b*c^5 + 12*A*c^6)*x^4 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^3 - 8*(7* 
B*b^3*c^3 - 12*A*b^2*c^4)*x^2 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x)*sqrt(c* 
x^2 + b*x))/c^5]
 

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.70 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} - \frac {5 b^{3} \left (A b^{2} - \frac {7 b \left (2 A b c + B b^{2} - \frac {9 b \left (A c^{2} + \frac {13 B b c}{12}\right )}{10 c}\right )}{8 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 c^{3}} + \sqrt {b x + c x^{2}} \left (\frac {B c x^{5}}{6} + \frac {5 b^{2} \left (A b^{2} - \frac {7 b \left (2 A b c + B b^{2} - \frac {9 b \left (A c^{2} + \frac {13 B b c}{12}\right )}{10 c}\right )}{8 c}\right )}{8 c^{3}} - \frac {5 b x \left (A b^{2} - \frac {7 b \left (2 A b c + B b^{2} - \frac {9 b \left (A c^{2} + \frac {13 B b c}{12}\right )}{10 c}\right )}{8 c}\right )}{12 c^{2}} + \frac {x^{4} \left (A c^{2} + \frac {13 B b c}{12}\right )}{5 c} + \frac {x^{3} \cdot \left (2 A b c + B b^{2} - \frac {9 b \left (A c^{2} + \frac {13 B b c}{12}\right )}{10 c}\right )}{4 c} + \frac {x^{2} \left (A b^{2} - \frac {7 b \left (2 A b c + B b^{2} - \frac {9 b \left (A c^{2} + \frac {13 B b c}{12}\right )}{10 c}\right )}{8 c}\right )}{3 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {7}{2}}}{7} + \frac {B \left (b x\right )^{\frac {9}{2}}}{9 b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-5*b**3*(A*b**2 - 7*b*(2*A*b*c + B*b**2 - 9*b*(A*c**2 + 13*B*b* 
c/12)/(10*c))/(8*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c 
*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2* 
c) + x)**2), True))/(16*c**3) + sqrt(b*x + c*x**2)*(B*c*x**5/6 + 5*b**2*(A 
*b**2 - 7*b*(2*A*b*c + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(8*c))/ 
(8*c**3) - 5*b*x*(A*b**2 - 7*b*(2*A*b*c + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/ 
12)/(10*c))/(8*c))/(12*c**2) + x**4*(A*c**2 + 13*B*b*c/12)/(5*c) + x**3*(2 
*A*b*c + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(4*c) + x**2*(A*b**2 
- 7*b*(2*A*b*c + B*b**2 - 9*b*(A*c**2 + 13*B*b*c/12)/(10*c))/(8*c))/(3*c)) 
, Ne(c, 0)), (2*(A*(b*x)**(7/2)/7 + B*(b*x)**(9/2)/(9*b))/b**2, Ne(b, 0)), 
 (0, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.23 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=-\frac {7 \, \sqrt {c x^{2} + b x} B b^{4} x}{256 \, c^{3}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2} x}{96 \, c^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} A b^{3} x}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B x}{6 \, c} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b x}{8 \, c} + \frac {7 \, B b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {9}{2}}} - \frac {3 \, A b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} - \frac {7 \, \sqrt {c x^{2} + b x} B b^{5}}{512 \, c^{4}} + \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{3}}{192 \, c^{3}} + \frac {3 \, \sqrt {c x^{2} + b x} A b^{4}}{128 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} B b}{60 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b^{2}}{16 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{5 \, c} \] Input:

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

Output:

-7/256*sqrt(c*x^2 + b*x)*B*b^4*x/c^3 + 7/96*(c*x^2 + b*x)^(3/2)*B*b^2*x/c^ 
2 + 3/64*sqrt(c*x^2 + b*x)*A*b^3*x/c^2 + 1/6*(c*x^2 + b*x)^(5/2)*B*x/c - 1 
/8*(c*x^2 + b*x)^(3/2)*A*b*x/c + 7/1024*B*b^6*log(2*c*x + b + 2*sqrt(c*x^2 
 + b*x)*sqrt(c))/c^(9/2) - 3/256*A*b^5*log(2*c*x + b + 2*sqrt(c*x^2 + b*x) 
*sqrt(c))/c^(7/2) - 7/512*sqrt(c*x^2 + b*x)*B*b^5/c^4 + 7/192*(c*x^2 + b*x 
)^(3/2)*B*b^3/c^3 + 3/128*sqrt(c*x^2 + b*x)*A*b^4/c^3 - 7/60*(c*x^2 + b*x) 
^(5/2)*B*b/c^2 - 1/16*(c*x^2 + b*x)^(3/2)*A*b^2/c^2 + 1/5*(c*x^2 + b*x)^(5 
/2)*A/c
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.85 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{7680} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x + \frac {13 \, B b c^{5} + 12 \, A c^{6}}{c^{5}}\right )} x + \frac {3 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )}}{c^{5}}\right )} x - \frac {7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}}{c^{5}}\right )} x + \frac {5 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )}}{c^{5}}\right )} x - \frac {15 \, {\left (7 \, B b^{5} c - 12 \, A b^{4} c^{2}\right )}}{c^{5}}\right )} - \frac {{\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{1024 \, c^{\frac {9}{2}}} \] Input:

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

Output:

1/7680*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*B*c*x + (13*B*b*c^5 + 12*A*c^6)/c 
^5)*x + 3*(B*b^2*c^4 + 44*A*b*c^5)/c^5)*x - (7*B*b^3*c^3 - 12*A*b^2*c^4)/c 
^5)*x + 5*(7*B*b^4*c^2 - 12*A*b^3*c^3)/c^5)*x - 15*(7*B*b^5*c - 12*A*b^4*c 
^2)/c^5) - 1/1024*(7*B*b^6 - 12*A*b^5*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 
 + b*x))*sqrt(c) + b))/c^(9/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.11 \[ \int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {180 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{4} c^{2}-120 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{3} c^{3} x +96 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{2} c^{4} x^{2}+2112 \sqrt {x}\, \sqrt {c x +b}\, a b \,c^{5} x^{3}+1536 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{6} x^{4}-105 \sqrt {x}\, \sqrt {c x +b}\, b^{6} c +70 \sqrt {x}\, \sqrt {c x +b}\, b^{5} c^{2} x -56 \sqrt {x}\, \sqrt {c x +b}\, b^{4} c^{3} x^{2}+48 \sqrt {x}\, \sqrt {c x +b}\, b^{3} c^{4} x^{3}+1664 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c^{5} x^{4}+1280 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{6} x^{5}-180 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,b^{5} c +105 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{7}}{7680 c^{5}} \] Input:

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

Output:

(180*sqrt(x)*sqrt(b + c*x)*a*b**4*c**2 - 120*sqrt(x)*sqrt(b + c*x)*a*b**3* 
c**3*x + 96*sqrt(x)*sqrt(b + c*x)*a*b**2*c**4*x**2 + 2112*sqrt(x)*sqrt(b + 
 c*x)*a*b*c**5*x**3 + 1536*sqrt(x)*sqrt(b + c*x)*a*c**6*x**4 - 105*sqrt(x) 
*sqrt(b + c*x)*b**6*c + 70*sqrt(x)*sqrt(b + c*x)*b**5*c**2*x - 56*sqrt(x)* 
sqrt(b + c*x)*b**4*c**3*x**2 + 48*sqrt(x)*sqrt(b + c*x)*b**3*c**4*x**3 + 1 
664*sqrt(x)*sqrt(b + c*x)*b**2*c**5*x**4 + 1280*sqrt(x)*sqrt(b + c*x)*b*c* 
*6*x**5 - 180*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*a*b** 
5*c + 105*sqrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*b**7)/(76 
80*c**5)