\(\int (A+B x) (b x+c x^2)^{3/2} \, dx\) [116]

Optimal result

Integrand size = 19, antiderivative size = 186 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {3 b^3 (b B-2 A c) \sqrt {b x+c x^2}}{128 c^3}-\frac {b^2 (b B-2 A c) x \sqrt {b x+c x^2}}{64 c^2}-\frac {3 b (b B-2 A c) x^2 \sqrt {b x+c x^2}}{16 c}-\frac {1}{8} (b B-2 A c) x^3 \sqrt {b x+c x^2}+\frac {B \left (b x+c x^2\right )^{5/2}}{5 c}-\frac {3 b^4 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{128 c^{7/2}} \] Output:

3/128*b^3*(-2*A*c+B*b)*(c*x^2+b*x)^(1/2)/c^3-1/64*b^2*(-2*A*c+B*b)*x*(c*x^ 
2+b*x)^(1/2)/c^2-3/16*b*(-2*A*c+B*b)*x^2*(c*x^2+b*x)^(1/2)/c-1/8*(-2*A*c+B 
*b)*x^3*(c*x^2+b*x)^(1/2)+1/5*B*(c*x^2+b*x)^(5/2)/c-3/128*b^4*(-2*A*c+B*b) 
*arctanh(c^(1/2)*x/(c*x^2+b*x)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.90 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {(x (b+c x))^{3/2} \left (\frac {\sqrt {c} \sqrt {x} \left (15 b^4 B-10 b^3 c (3 A+B x)+4 b^2 c^2 x (5 A+2 B x)+32 c^4 x^3 (5 A+4 B x)+16 b c^3 x^2 (15 A+11 B x)\right )}{b+c x}+\frac {30 b^4 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{(b+c x)^{3/2}}\right )}{640 c^{7/2} x^{3/2}} \] Input:

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

Output:

((x*(b + c*x))^(3/2)*((Sqrt[c]*Sqrt[x]*(15*b^4*B - 10*b^3*c*(3*A + B*x) + 
4*b^2*c^2*x*(5*A + 2*B*x) + 32*c^4*x^3*(5*A + 4*B*x) + 16*b*c^3*x^2*(15*A 
+ 11*B*x)))/(b + c*x) + (30*b^4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*Sqrt[x])/(S 
qrt[b] - Sqrt[b + c*x])])/(b + c*x)^(3/2)))/(640*c^(7/2)*x^(3/2))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1160, 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.

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*(b*x + c*x^2)^(5/2))/(5*c) - ((b*B - 2*A*c)*(((b + 2*c*x)*(b*x + c*x^2) 
^(3/2))/(8*c) - (3*b^2*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTa 
nh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(3/2))))/(16*c)))/(2*c)
 

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 1160

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.62

method	result	size
pseudoelliptic	\(\frac {\left (\frac {3}{2} A \,b^{4} c -\frac {3}{4} b^{5} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\left (-\frac {3 \left (\frac {B x}{3}+A \right ) b^{3} c^{\frac {3}{2}}}{2}+b^{2} x \left (\frac {2 B x}{5}+A \right ) c^{\frac {5}{2}}+12 \left (\frac {11 B x}{15}+A \right ) x^{2} b \,c^{\frac {7}{2}}+8 x^{3} \left (\frac {4 B x}{5}+A \right ) c^{\frac {9}{2}}+\frac {3 B \sqrt {c}\, b^{4}}{4}\right ) \sqrt {x \left (c x +b \right )}}{32 c^{\frac {7}{2}}}\)	\(116\)
risch	\(-\frac {\left (-128 B \,c^{4} x^{4}-160 A \,c^{4} x^{3}-176 B b \,c^{3} x^{3}-240 A b \,c^{3} x^{2}-8 B \,b^{2} c^{2} x^{2}-20 A \,b^{2} c^{2} x +10 B \,b^{3} c x +30 A \,b^{3} c -15 B \,b^{4}\right ) x \left (c x +b \right )}{640 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {3 b^{4} \left (2 A c -B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{256 c^{\frac {7}{2}}}\)	\(145\)
default	\(A \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+B \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )\)	\(201\)

Input:

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

Output:

1/32/c^(7/2)*((3/2*A*b^4*c-3/4*b^5*B)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2)) 
+(-3/2*(1/3*B*x+A)*b^3*c^(3/2)+b^2*x*(2/5*B*x+A)*c^(5/2)+12*(11/15*B*x+A)* 
x^2*b*c^(7/2)+8*x^3*(4/5*B*x+A)*c^(9/2)+3/4*B*c^(1/2)*b^4)*(x*(c*x+b))^(1/ 
2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.60 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{1280 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{2} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{640 \, c^{4}}\right ] \] Input:

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

Output:

[-1/1280*(15*(B*b^5 - 2*A*b^4*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b* 
x)*sqrt(c)) - 2*(128*B*c^5*x^4 + 15*B*b^4*c - 30*A*b^3*c^2 + 16*(11*B*b*c^ 
4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c^4)*x^2 - 10*(B*b^3*c^2 - 2*A*b 
^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/640*(15*(B*b^5 - 2*A*b^4*c)*sqrt(-c)* 
arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x + b)) + (128*B*c^5*x^4 + 15*B*b^4*c 
 - 30*A*b^3*c^2 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 + 8*(B*b^2*c^3 + 30*A*b*c 
^4)*x^2 - 10*(B*b^3*c^2 - 2*A*b^2*c^3)*x)*sqrt(c*x^2 + b*x))/c^4]
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.75 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 b^{2} \left (A b^{2} - \frac {5 b \left (2 A b c + B b^{2} - \frac {7 b \left (A c^{2} + \frac {11 B b c}{10}\right )}{8 c}\right )}{6 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 c^{2}} + \sqrt {b x + c x^{2}} \left (\frac {B c x^{4}}{5} - \frac {3 b \left (A b^{2} - \frac {5 b \left (2 A b c + B b^{2} - \frac {7 b \left (A c^{2} + \frac {11 B b c}{10}\right )}{8 c}\right )}{6 c}\right )}{4 c^{2}} + \frac {x^{3} \left (A c^{2} + \frac {11 B b c}{10}\right )}{4 c} + \frac {x^{2} \cdot \left (2 A b c + B b^{2} - \frac {7 b \left (A c^{2} + \frac {11 B b c}{10}\right )}{8 c}\right )}{3 c} + \frac {x \left (A b^{2} - \frac {5 b \left (2 A b c + B b^{2} - \frac {7 b \left (A c^{2} + \frac {11 B b c}{10}\right )}{8 c}\right )}{6 c}\right )}{2 c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {A \left (b x\right )^{\frac {5}{2}}}{5} + \frac {B \left (b x\right )^{\frac {7}{2}}}{7 b}\right )}{b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((3*b**2*(A*b**2 - 5*b*(2*A*b*c + B*b**2 - 7*b*(A*c**2 + 11*B*b*c 
/10)/(8*c))/(6*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))/(8*c**2) + sqrt(b*x + c*x**2)*(B*c*x**4/5 - 3*b*(A*b**2 
- 5*b*(2*A*b*c + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/10)/(8*c))/(6*c))/(4*c**2 
) + x**3*(A*c**2 + 11*B*b*c/10)/(4*c) + x**2*(2*A*b*c + B*b**2 - 7*b*(A*c* 
*2 + 11*B*b*c/10)/(8*c))/(3*c) + x*(A*b**2 - 5*b*(2*A*b*c + B*b**2 - 7*b*( 
A*c**2 + 11*B*b*c/10)/(8*c))/(6*c))/(2*c)), Ne(c, 0)), (2*(A*(b*x)**(5/2)/ 
5 + B*(b*x)**(7/2)/(7*b))/b, Ne(b, 0)), (0, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.27 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A x + \frac {3 \, \sqrt {c x^{2} + b x} B b^{3} x}{64 \, c^{2}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b x}{8 \, c} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{2} x}{32 \, c} - \frac {3 \, B b^{5} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{256 \, c^{\frac {7}{2}}} + \frac {3 \, A b^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {5}{2}}} + \frac {3 \, \sqrt {c x^{2} + b x} B b^{4}}{128 \, c^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b^{2}}{16 \, c^{2}} - \frac {3 \, \sqrt {c x^{2} + b x} A b^{3}}{64 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{5 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{8 \, c} \] Input:

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

Output:

1/4*(c*x^2 + b*x)^(3/2)*A*x + 3/64*sqrt(c*x^2 + b*x)*B*b^3*x/c^2 - 1/8*(c* 
x^2 + b*x)^(3/2)*B*b*x/c - 3/32*sqrt(c*x^2 + b*x)*A*b^2*x/c - 3/256*B*b^5* 
log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 3/128*A*b^4*log(2*c 
*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 3/128*sqrt(c*x^2 + b*x)*B* 
b^4/c^3 - 1/16*(c*x^2 + b*x)^(3/2)*B*b^2/c^2 - 3/64*sqrt(c*x^2 + b*x)*A*b^ 
3/c^2 + 1/5*(c*x^2 + b*x)^(5/2)*B/c + 1/8*(c*x^2 + b*x)^(3/2)*A*b/c
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.86 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {1}{640} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x + \frac {11 \, B b c^{4} + 10 \, A c^{5}}{c^{4}}\right )} x + \frac {B b^{2} c^{3} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac {5 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )}}{c^{4}}\right )} x + \frac {15 \, {\left (B b^{4} c - 2 \, A b^{3} c^{2}\right )}}{c^{4}}\right )} + \frac {3 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/640*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^4)*x 
 + (B*b^2*c^3 + 30*A*b*c^4)/c^4)*x - 5*(B*b^3*c^2 - 2*A*b^2*c^3)/c^4)*x + 
15*(B*b^4*c - 2*A*b^3*c^2)/c^4) + 3/256*(B*b^5 - 2*A*b^4*c)*log(abs(2*(sqr 
t(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
 

Mupad [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.12 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {B\,{\left (c\,x^2+b\,x\right )}^{5/2}}{5\,c}+\frac {A\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {B\,b\,\left (\frac {x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\frac {\sqrt {c\,x^2+b\,x}\,\left (b+2\,c\,x\right )}{4\,c}-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{2\,c}-\frac {3\,A\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c} \] Input:

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

Output:

(B*(b*x + c*x^2)^(5/2))/(5*c) + (A*(b*x + c*x^2)^(3/2)*(b/2 + c*x))/(4*c) 
- (B*b*((x*(b*x + c*x^2)^(3/2))/4 + (b*(b*x + c*x^2)^(3/2))/(8*c) - (3*b^2 
*(((b*x + c*x^2)^(1/2)*(b + 2*c*x))/(4*c) - (b^2*log((b/2 + c*x)/c^(1/2) + 
 (b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)))/(2*c) - (3*A*b^2*((b*x + c*x 
^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*log((b/2 + c*x)/c^(1/2) + (b*x + c*x^2)^( 
1/2)))/(8*c^(3/2))))/(16*c)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.15 \[ \int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx=\frac {-30 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{3} c^{2}+20 \sqrt {x}\, \sqrt {c x +b}\, a \,b^{2} c^{3} x +240 \sqrt {x}\, \sqrt {c x +b}\, a b \,c^{4} x^{2}+160 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{5} x^{3}+15 \sqrt {x}\, \sqrt {c x +b}\, b^{5} c -10 \sqrt {x}\, \sqrt {c x +b}\, b^{4} c^{2} x +8 \sqrt {x}\, \sqrt {c x +b}\, b^{3} c^{3} x^{2}+176 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c^{4} x^{3}+128 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{5} x^{4}+30 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,b^{4} c -15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{6}}{640 c^{4}} \] Input:

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

Output:

( - 30*sqrt(x)*sqrt(b + c*x)*a*b**3*c**2 + 20*sqrt(x)*sqrt(b + c*x)*a*b**2 
*c**3*x + 240*sqrt(x)*sqrt(b + c*x)*a*b*c**4*x**2 + 160*sqrt(x)*sqrt(b + c 
*x)*a*c**5*x**3 + 15*sqrt(x)*sqrt(b + c*x)*b**5*c - 10*sqrt(x)*sqrt(b + c* 
x)*b**4*c**2*x + 8*sqrt(x)*sqrt(b + c*x)*b**3*c**3*x**2 + 176*sqrt(x)*sqrt 
(b + c*x)*b**2*c**4*x**3 + 128*sqrt(x)*sqrt(b + c*x)*b*c**5*x**4 + 30*sqrt 
(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*a*b**4*c - 15*sqrt(c)*l 
og((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*b**6)/(640*c**4)