\(\int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx\) [144]

Optimal result

Integrand size = 23, antiderivative size = 206 \[ \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=-\frac {(7 b B-8 A c) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (105 b^3 B-120 A b^2 c-220 a b B c+128 a A c^2-2 c \left (35 b^2 B-40 A b c-36 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}+\frac {\left (35 b^4 B-40 A b^3 c-120 a b^2 B c+96 a A b c^2+48 a^2 B c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}} \] Output:

-1/24*(-8*A*c+7*B*b)*x^2*(c*x^2+b*x+a)^(1/2)/c^2+1/4*B*x^3*(c*x^2+b*x+a)^( 
1/2)/c-1/192*(105*B*b^3-120*A*b^2*c-220*B*a*b*c+128*A*a*c^2-2*c*(-40*A*b*c 
-36*B*a*c+35*B*b^2)*x)*(c*x^2+b*x+a)^(1/2)/c^4+1/128*(96*A*a*b*c^2-40*A*b^ 
3*c+48*B*a^2*c^2-120*B*a*b^2*c+35*B*b^4)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c* 
x^2+b*x+a)^(1/2))/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.84 \[ \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-105 b^3 B+10 b^2 c (12 A+7 B x)+8 c^2 \left (-16 a A-9 a B x+8 A c x^2+6 B c x^3\right )+4 b c (55 a B-2 c x (10 A+7 B x))\right )-3 \left (35 b^4 B-40 A b^3 c-120 a b^2 B c+96 a A b c^2+48 a^2 B c^2\right ) \log \left (c^4 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{384 c^{9/2}} \] Input:

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

Output:

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-105*b^3*B + 10*b^2*c*(12*A + 7*B*x) + 8 
*c^2*(-16*a*A - 9*a*B*x + 8*A*c*x^2 + 6*B*c*x^3) + 4*b*c*(55*a*B - 2*c*x*( 
10*A + 7*B*x))) - 3*(35*b^4*B - 40*A*b^3*c - 120*a*b^2*B*c + 96*a*A*b*c^2 
+ 48*a^2*B*c^2)*Log[c^4*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(3 
84*c^(9/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1236, 27, 1236, 27, 1225, 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.

\(\displaystyle \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {\int -\frac {x^2 (6 a B+(7 b B-8 A c) x)}{2 \sqrt {c x^2+b x+a}}dx}{4 c}+\frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\int \frac {x^2 (6 a B+(7 b B-8 A c) x)}{\sqrt {c x^2+b x+a}}dx}{8 c}\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\frac {\int -\frac {x \left (4 a (7 b B-8 A c)+\left (35 B b^2-40 A c b-36 a B c\right ) x\right )}{2 \sqrt {c x^2+b x+a}}dx}{3 c}+\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{3 c}}{8 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{3 c}-\frac {\int \frac {x \left (4 a (7 b B-8 A c)+\left (35 B b^2-40 A c b-36 a B c\right ) x\right )}{\sqrt {c x^2+b x+a}}dx}{6 c}}{8 c}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{3 c}-\frac {\frac {3 \left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c^2}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-36 a B c-40 A b c+35 b^2 B\right )+128 a A c^2-220 a b B c-120 A b^2 c+105 b^3 B\right )}{4 c^2}}{6 c}}{8 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{3 c}-\frac {\frac {3 \left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c^2}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-36 a B c-40 A b c+35 b^2 B\right )+128 a A c^2-220 a b B c-120 A b^2 c+105 b^3 B\right )}{4 c^2}}{6 c}}{8 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {B x^3 \sqrt {a+b x+c x^2}}{4 c}-\frac {\frac {x^2 \sqrt {a+b x+c x^2} (7 b B-8 A c)}{3 c}-\frac {\frac {3 \left (48 a^2 B c^2+96 a A b c^2-120 a b^2 B c-40 A b^3 c+35 b^4 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-36 a B c-40 A b c+35 b^2 B\right )+128 a A c^2-220 a b B c-120 A b^2 c+105 b^3 B\right )}{4 c^2}}{6 c}}{8 c}\)

Input:

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

Output:

(B*x^3*Sqrt[a + b*x + c*x^2])/(4*c) - (((7*b*B - 8*A*c)*x^2*Sqrt[a + b*x + 
 c*x^2])/(3*c) - (-1/4*((105*b^3*B - 120*A*b^2*c - 220*a*b*B*c + 128*a*A*c 
^2 - 2*c*(35*b^2*B - 40*A*b*c - 36*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/c^2 + 
(3*(35*b^4*B - 40*A*b^3*c - 120*a*b^2*B*c + 96*a*A*b*c^2 + 48*a^2*B*c^2)*A 
rcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2)))/(6*c)) 
/(8*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 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]
 

rule 1236

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[1/(c*(m + 2*p + 2))   Int[(d + e*x)^(m - 1 
)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m 
*(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\left (-48 B \,c^{3} x^{3}-64 A \,c^{3} x^{2}+56 B b \,c^{2} x^{2}+80 A b \,c^{2} x +72 B a \,c^{2} x -70 B \,b^{2} c x +128 A a \,c^{2}-120 A \,b^{2} c -220 B a b c +105 B \,b^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{4}}+\frac {\left (96 A a b \,c^{2}-40 A \,b^{3} c +48 B \,a^{2} c^{2}-120 B a \,b^{2} c +35 B \,b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {9}{2}}}\)	\(167\)
default	\(A \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+B \left (\frac {x^{3} \sqrt {c \,x^{2}+b x +a}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )}{8 c}-\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )\)	\(513\)

Input:

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/192*(-48*B*c^3*x^3-64*A*c^3*x^2+56*B*b*c^2*x^2+80*A*b*c^2*x+72*B*a*c^2* 
x-70*B*b^2*c*x+128*A*a*c^2-120*A*b^2*c-220*B*a*b*c+105*B*b^3)/c^4*(c*x^2+b 
*x+a)^(1/2)+1/128*(96*A*a*b*c^2-40*A*b^3*c+48*B*a^2*c^2-120*B*a*b^2*c+35*B 
*b^4)/c^(9/2)*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 = 395, normalized size of antiderivative = 1.92 \[ \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=\left [\frac {3 \, {\left (35 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c - 128 \, A a c^{3} + 20 \, {\left (11 \, B a b + 6 \, A b^{2}\right )} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{2} c^{2} - 4 \, {\left (9 \, B a + 10 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{5}}, -\frac {3 \, {\left (35 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (48 \, B c^{4} x^{3} - 105 \, B b^{3} c - 128 \, A a c^{3} + 20 \, {\left (11 \, B a b + 6 \, A b^{2}\right )} c^{2} - 8 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{2} c^{2} - 4 \, {\left (9 \, B a + 10 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{5}}\right ] \] Input:

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[1/768*(3*(35*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2 + A*b^3)*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*(48*B*c^4*x^3 - 105*B*b^3*c - 128*A*a*c^3 + 20*(1 
1*B*a*b + 6*A*b^2)*c^2 - 8*(7*B*b*c^3 - 8*A*c^4)*x^2 + 2*(35*B*b^2*c^2 - 4 
*(9*B*a + 10*A*b)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/384*(3*(35*B*b^4 
+ 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2 + A*b^3)*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*( 
48*B*c^4*x^3 - 105*B*b^3*c - 128*A*a*c^3 + 20*(11*B*a*b + 6*A*b^2)*c^2 - 8 
*(7*B*b*c^3 - 8*A*c^4)*x^2 + 2*(35*B*b^2*c^2 - 4*(9*B*a + 10*A*b)*c^3)*x)* 
sqrt(c*x^2 + b*x + a))/c^5]
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.05 \[ \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (- \frac {a \left (- \frac {3 B a}{4 c} - \frac {5 b \left (A - \frac {7 B b}{8 c}\right )}{6 c}\right )}{2 c} - \frac {b \left (- \frac {2 a \left (A - \frac {7 B b}{8 c}\right )}{3 c} - \frac {3 b \left (- \frac {3 B a}{4 c} - \frac {5 b \left (A - \frac {7 B b}{8 c}\right )}{6 c}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x + c x^{2}} \left (\frac {B x^{3}}{4 c} + \frac {x^{2} \left (A - \frac {7 B b}{8 c}\right )}{3 c} + \frac {x \left (- \frac {3 B a}{4 c} - \frac {5 b \left (A - \frac {7 B b}{8 c}\right )}{6 c}\right )}{2 c} + \frac {- \frac {2 a \left (A - \frac {7 B b}{8 c}\right )}{3 c} - \frac {3 b \left (- \frac {3 B a}{4 c} - \frac {5 b \left (A - \frac {7 B b}{8 c}\right )}{6 c}\right )}{4 c}}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {B \left (a + b x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (A b - 4 B a\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 3 A a b + 6 B a^{2}\right )}{5 b} + \frac {\left (a + b x\right )^{\frac {3}{2}} \cdot \left (3 A a^{2} b - 4 B a^{3}\right )}{3 b} + \frac {\sqrt {a + b x} \left (- A a^{3} b + B a^{4}\right )}{b}\right )}{b^{4}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate(x**3*(B*x+A)/(c*x**2+b*x+a)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),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 = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 (A+B x)}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (\frac {6 \, B x}{c} - \frac {7 \, B b c^{2} - 8 \, A c^{3}}{c^{4}}\right )} x + \frac {35 \, B b^{2} c - 36 \, B a c^{2} - 40 \, A b c^{2}}{c^{4}}\right )} x - \frac {105 \, B b^{3} - 220 \, B a b c - 120 \, A b^{2} c + 128 \, A a c^{2}}{c^{4}}\right )} - \frac {{\left (35 \, B b^{4} - 120 \, B a b^{2} c - 40 \, A b^{3} c + 48 \, B a^{2} c^{2} + 96 \, A a b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {9}{2}}} \] Input:

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*x/c - (7*B*b*c^2 - 8*A*c^3)/c^4)*x 
+ (35*B*b^2*c - 36*B*a*c^2 - 40*A*b*c^2)/c^4)*x - (105*B*b^3 - 220*B*a*b*c 
 - 120*A*b^2*c + 128*A*a*c^2)/c^4) - 1/128*(35*B*b^4 - 120*B*a*b^2*c - 40* 
A*b^3*c + 48*B*a^2*c^2 + 96*A*a*b*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + 
 b*x + a))*sqrt(c) + b))/c^(9/2)
 

Mupad [F(-1)]

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

int((x^3*(A + B*x))/(a + b*x + c*x^2)^(1/2),x)
 

Output:

int((x^3*(A + B*x))/(a + b*x + c*x^2)^(1/2), x)
 

Reduce [F]

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

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)
 

Output:

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)