\(\int (A+B x) (a+b x+c x^2)^{3/2} \, dx\) [120]

Optimal result

Integrand size = 20, antiderivative size = 158 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {3 \left (b^2-4 a c\right ) (b B-2 A c) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^3}-\frac {(b B-2 A c) (b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{16 c^2}+\frac {B \left (a+b x+c x^2\right )^{5/2}}{5 c}-\frac {3 \left (b^2-4 a c\right )^2 (b B-2 A c) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2}} \] Output:

3/128*(-4*a*c+b^2)*(-2*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3-1/16*(-2 
*A*c+B*b)*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c^2+1/5*B*(c*x^2+b*x+a)^(5/2)/c-3/ 
256*(-4*a*c+b^2)^2*(-2*A*c+B*b)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a 
)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.22 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^4 B-10 b^3 c (3 A+B x)+8 b c^2 \left (25 a A+7 a B x+30 A c x^2+22 B c x^3\right )+4 b^2 c (-25 a B+c x (5 A+2 B x))+16 c^2 \left (8 a^2 B+2 c^2 x^3 (5 A+4 B x)+a c x (25 A+16 B x)\right )\right )-15 \left (b^2-4 a c\right )^2 (b B-2 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{640 c^{7/2}} \] Input:

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

Output:

(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^4*B - 10*b^3*c*(3*A + B*x) + 8*b*c^2* 
(25*a*A + 7*a*B*x + 30*A*c*x^2 + 22*B*c*x^3) + 4*b^2*c*(-25*a*B + c*x*(5*A 
 + 2*B*x)) + 16*c^2*(8*a^2*B + 2*c^2*x^3*(5*A + 4*B*x) + a*c*x*(25*A + 16* 
B*x))) - 15*(b^2 - 4*a*c)^2*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + 
Sqrt[a + x*(b + c*x)])])/(640*c^(7/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1160, 1087, 1087, 1092, 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.

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*(a + b*x + c*x^2)^(5/2))/(5*c) - ((b*B - 2*A*c)*(((b + 2*c*x)*(a + b*x 
+ c*x^2)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*x)*Sqrt[a + b*x + c*x^ 
2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c 
*x^2])])/(8*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 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, 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 = 1.18 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.48

method	result	size
risch	\(\frac {\left (128 B \,c^{4} x^{4}+160 A \,c^{4} x^{3}+176 B \,c^{3} x^{3} b +240 A b \,c^{3} x^{2}+256 B a \,c^{3} x^{2}+8 c^{2} x^{2} B \,b^{2}+400 A a \,c^{3} x +20 A \,b^{2} c^{2} x +56 B a b \,c^{2} x -10 B \,b^{3} c x +200 A a b \,c^{2}-30 A \,b^{3} c +128 B \,a^{2} c^{2}-100 B a \,b^{2} c +15 B \,b^{4}\right ) \sqrt {c \,x^{2}+b x +a}}{640 c^{3}}+\frac {3 \left (32 a^{2} A \,c^{3}-16 A a \,b^{2} c^{2}+2 A \,b^{4} c -16 B \,a^{2} b \,c^{2}+8 B a \,b^{3} c -b^{5} B \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {7}{2}}}\)	\(234\)
default	\(A \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+B \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )\)	\(236\)

Input:

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

Output:

1/640/c^3*(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+256 
*B*a*c^3*x^2+8*B*b^2*c^2*x^2+400*A*a*c^3*x+20*A*b^2*c^2*x+56*B*a*b*c^2*x-1 
0*B*b^3*c*x+200*A*a*b*c^2-30*A*b^3*c+128*B*a^2*c^2-100*B*a*b^2*c+15*B*b^4) 
*(c*x^2+b*x+a)^(1/2)+3/256*(32*A*a^2*c^3-16*A*a*b^2*c^2+2*A*b^4*c-16*B*a^2 
*b*c^2+8*B*a*b^3*c-B*b^5)/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/ 
2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.26 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (B b^{5} - 32 \, A a^{2} c^{3} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c + 8 \, {\left (16 \, B a^{2} + 25 \, A a b\right )} c^{3} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 10 \, {\left (10 \, B a b^{2} + 3 \, A b^{3}\right )} c^{2} + 8 \, {\left (B b^{2} c^{3} + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{3} c^{2} - 200 \, A a c^{4} - 2 \, {\left (14 \, B a b + 5 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2560 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 32 \, A a^{2} c^{3} + 16 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (128 \, B c^{5} x^{4} + 15 \, B b^{4} c + 8 \, {\left (16 \, B a^{2} + 25 \, A a b\right )} c^{3} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{3} - 10 \, {\left (10 \, B a b^{2} + 3 \, A b^{3}\right )} c^{2} + 8 \, {\left (B b^{2} c^{3} + 2 \, {\left (16 \, B a + 15 \, A b\right )} c^{4}\right )} x^{2} - 2 \, {\left (5 \, B b^{3} c^{2} - 200 \, A a c^{4} - 2 \, {\left (14 \, B a b + 5 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{1280 \, c^{4}}\right ] \] Input:

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

Output:

[-1/2560*(15*(B*b^5 - 32*A*a^2*c^3 + 16*(B*a^2*b + A*a*b^2)*c^2 - 2*(4*B*a 
*b^3 + A*b^4)*c)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b 
*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(128*B*c^5*x^4 + 15*B*b^4*c + 8*( 
16*B*a^2 + 25*A*a*b)*c^3 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 - 10*(10*B*a*b^2 
 + 3*A*b^3)*c^2 + 8*(B*b^2*c^3 + 2*(16*B*a + 15*A*b)*c^4)*x^2 - 2*(5*B*b^3 
*c^2 - 200*A*a*c^4 - 2*(14*B*a*b + 5*A*b^2)*c^3)*x)*sqrt(c*x^2 + b*x + a)) 
/c^4, 1/1280*(15*(B*b^5 - 32*A*a^2*c^3 + 16*(B*a^2*b + A*a*b^2)*c^2 - 2*(4 
*B*a*b^3 + A*b^4)*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b) 
*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(128*B*c^5*x^4 + 15*B*b^4*c + 8*(16 
*B*a^2 + 25*A*a*b)*c^3 + 16*(11*B*b*c^4 + 10*A*c^5)*x^3 - 10*(10*B*a*b^2 + 
 3*A*b^3)*c^2 + 8*(B*b^2*c^3 + 2*(16*B*a + 15*A*b)*c^4)*x^2 - 2*(5*B*b^3*c 
^2 - 200*A*a*c^4 - 2*(14*B*a*b + 5*A*b^2)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c 
^4]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (151) = 302\).

Time = 0.60 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.50 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((sqrt(a + b*x + c*x**2)*(B*c*x**4/5 + x**3*(A*c**2 + 11*B*b*c/10 
)/(4*c) + x**2*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/10)/ 
(8*c))/(3*c) + x*(2*A*a*c + A*b**2 + 2*B*a*b - 3*a*(A*c**2 + 11*B*b*c/10)/ 
(4*c) - 5*b*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/10)/(8* 
c))/(6*c))/(2*c) + (2*A*a*b + B*a**2 - 2*a*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 
 7*b*(A*c**2 + 11*B*b*c/10)/(8*c))/(3*c) - 3*b*(2*A*a*c + A*b**2 + 2*B*a*b 
 - 3*a*(A*c**2 + 11*B*b*c/10)/(4*c) - 5*b*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 
7*b*(A*c**2 + 11*B*b*c/10)/(8*c))/(6*c))/(4*c))/c) + (A*a**2 - a*(2*A*a*c 
+ A*b**2 + 2*B*a*b - 3*a*(A*c**2 + 11*B*b*c/10)/(4*c) - 5*b*(2*A*b*c + 6*B 
*a*c/5 + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/10)/(8*c))/(6*c))/(2*c) - b*(2*A* 
a*b + B*a**2 - 2*a*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/ 
10)/(8*c))/(3*c) - 3*b*(2*A*a*c + A*b**2 + 2*B*a*b - 3*a*(A*c**2 + 11*B*b* 
c/10)/(4*c) - 5*b*(2*A*b*c + 6*B*a*c/5 + B*b**2 - 7*b*(A*c**2 + 11*B*b*c/1 
0)/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + 
 c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2 
*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)), Ne(c, 0)), (2*(B*(a + b*x)**(7/ 
2)/(7*b) + (a + b*x)**(5/2)*(A*b - B*a)/(5*b))/b, Ne(b, 0)), (a**(3/2)*(A* 
x + B*x**2/2), True))
 

Maxima [F(-2)]

Exception generated. \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.58 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x + \frac {11 \, B b c^{4} + 10 \, A c^{5}}{c^{4}}\right )} x + \frac {B b^{2} c^{3} + 32 \, B a c^{4} + 30 \, A b c^{4}}{c^{4}}\right )} x - \frac {5 \, B b^{3} c^{2} - 28 \, B a b c^{3} - 10 \, A b^{2} c^{3} - 200 \, A a c^{4}}{c^{4}}\right )} x + \frac {15 \, B b^{4} c - 100 \, B a b^{2} c^{2} - 30 \, A b^{3} c^{2} + 128 \, B a^{2} c^{3} + 200 \, A a b c^{3}}{c^{4}}\right )} + \frac {3 \, {\left (B b^{5} - 8 \, B a b^{3} c - 2 \, A b^{4} c + 16 \, B a^{2} b c^{2} + 16 \, A a b^{2} c^{2} - 32 \, A a^{2} c^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*B*c*x + (11*B*b*c^4 + 10*A*c^5)/c^ 
4)*x + (B*b^2*c^3 + 32*B*a*c^4 + 30*A*b*c^4)/c^4)*x - (5*B*b^3*c^2 - 28*B* 
a*b*c^3 - 10*A*b^2*c^3 - 200*A*a*c^4)/c^4)*x + (15*B*b^4*c - 100*B*a*b^2*c 
^2 - 30*A*b^3*c^2 + 128*B*a^2*c^3 + 200*A*a*b*c^3)/c^4) + 3/256*(B*b^5 - 8 
*B*a*b^3*c - 2*A*b^4*c + 16*B*a^2*b*c^2 + 16*A*a*b^2*c^2 - 32*A*a^2*c^3)*l 
og(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
 

Mupad [B] (verification not implemented)

Time = 10.98 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.93 \[ \int (A+B x) \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {B\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5\,c}+\frac {A\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )\,\left (3\,a\,c-\frac {3\,b^2}{4}\right )}{4\,c}-\frac {B\,b\,\left (\frac {3\,a\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{4}+\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4}+\frac {b\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{8\,c}-\frac {3\,b^2\,\left (\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (\frac {a}{2\,\sqrt {c}}-\frac {b^2}{8\,c^{3/2}}\right )+\frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{4\,c}\right )}{16\,c}\right )}{2\,c}+\frac {A\,\left (\frac {b}{2}+c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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