\(\int \frac {x^3 (A+B x)}{(a+b x+c x^2)^{3/2}} \, dx\) [154]

Optimal result

Integrand size = 23, antiderivative size = 197 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {\left (15 b^3 B-12 A b^2 c-52 a b B c+32 a A c^2-2 c \left (5 b^2 B-4 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 \left (5 b^2 B-4 A b c-4 a B c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}} \] Output:

-2*x^2*(a*(-2*A*c+B*b)+(-A*b*c-2*B*a*c+B*b^2)*x)/c/(-4*a*c+b^2)/(c*x^2+b*x 
+a)^(1/2)-1/4*(15*B*b^3-12*A*b^2*c-52*B*a*b*c+32*A*a*c^2-2*c*(-4*A*b*c-12* 
B*a*c+5*B*b^2)*x)*(c*x^2+b*x+a)^(1/2)/c^3/(-4*a*c+b^2)+3/8*(-4*A*b*c-4*B*a 
*c+5*B*b^2)*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 = 1.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {c} \left (4 a^2 c (-13 b B+8 A c+6 B c x)+a \left (15 b^3 B-20 b c^2 x (-2 A+B x)+8 c^3 x^2 (2 A+B x)-2 b^2 c (6 A+31 B x)\right )+b^2 x \left (15 b^2 B-2 c^2 x (2 A+B x)+b (-12 A c+5 B c x)\right )\right )+3 \left (b^2-4 a c\right ) \left (5 b^2 B-4 A b c-4 a B c\right ) \sqrt {a+x (b+c x)} \log \left (c^3 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{8 c^{7/2} \left (-b^2+4 a c\right ) \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(2*Sqrt[c]*(4*a^2*c*(-13*b*B + 8*A*c + 6*B*c*x) + a*(15*b^3*B - 20*b*c^2*x 
*(-2*A + B*x) + 8*c^3*x^2*(2*A + B*x) - 2*b^2*c*(6*A + 31*B*x)) + b^2*x*(1 
5*b^2*B - 2*c^2*x*(2*A + B*x) + b*(-12*A*c + 5*B*c*x))) + 3*(b^2 - 4*a*c)* 
(5*b^2*B - 4*A*b*c - 4*a*B*c)*Sqrt[a + x*(b + c*x)]*Log[c^3*(b + 2*c*x - 2 
*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b 
+ c*x)])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1233, 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.

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

\(\Big \downarrow \) 1233

\(\displaystyle \frac {2 \int \frac {x \left (4 a (b B-2 A c)+\left (5 B b^2-4 A c b-12 a B c\right ) x\right )}{2 \sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {x \left (4 a (b B-2 A c)+\left (5 B b^2-4 A c b-12 a B c\right ) x\right )}{\sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\frac {3 \left (b^2-4 a c\right ) \left (-4 a B c-4 A b c+5 b^2 B\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c^2}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-12 a B c-4 A b c+5 b^2 B\right )+32 a A c^2-52 a b B c-12 A b^2 c+15 b^3 B\right )}{4 c^2}}{c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {3 \left (b^2-4 a c\right ) \left (-4 a B c-4 A b c+5 b^2 B\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c^2}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-12 a B c-4 A b c+5 b^2 B\right )+32 a A c^2-52 a b B c-12 A b^2 c+15 b^3 B\right )}{4 c^2}}{c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \left (b^2-4 a c\right ) \left (-4 a B c-4 A b c+5 b^2 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2}}-\frac {\sqrt {a+b x+c x^2} \left (-2 c x \left (-12 a B c-4 A b c+5 b^2 B\right )+32 a A c^2-52 a b B c-12 A b^2 c+15 b^3 B\right )}{4 c^2}}{c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\)

Input:

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

Output:

(-2*x^2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(c*(b^2 - 4*a*c)* 
Sqrt[a + b*x + c*x^2]) + (-1/4*((15*b^3*B - 12*A*b^2*c - 52*a*b*B*c + 32*a 
*A*c^2 - 2*c*(5*b^2*B - 4*A*b*c - 12*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/c^2 
+ (3*(b^2 - 4*a*c)*(5*b^2*B - 4*A*b*c - 4*a*B*c)*ArcTanh[(b + 2*c*x)/(2*Sq 
rt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2)))/(c*(b^2 - 4*a*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 1233

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.82

method	result	size
risch	\(\frac {\left (2 B c x +4 A c -7 B b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{3}}-\frac {3 c \left (4 A b c +4 a B c -5 B \,b^{2}\right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )+\left (8 A a \,c^{2}+4 A \,b^{2} c -4 B a b c -7 B \,b^{3}\right ) \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )-\frac {14 B a \,b^{2} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {8 B \,a^{2} c \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {8 A a b c \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}}{8 c^{3}}\)	\(359\)
default	\(A \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}-\frac {2 a \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{c}\right )+B \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x +a}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}-\frac {2 a \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{c}\right )}{4 c}-\frac {3 a \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\)	\(535\)

Input:

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

Output:

1/4*(2*B*c*x+4*A*c-7*B*b)/c^3*(c*x^2+b*x+a)^(1/2)-1/8/c^3*(3*c*(4*A*b*c+4* 
B*a*c-5*B*b^2)*(-x/c/(c*x^2+b*x+a)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^(1/2) 
-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c 
^(1/2)+(c*x^2+b*x+a)^(1/2)))+(8*A*a*c^2+4*A*b^2*c-4*B*a*b*c-7*B*b^3)*(-1/c 
/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2))-14*B*a 
*b^2*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+8*B*a^2*c*(2*c*x+b)/(4*a*c- 
b^2)/(c*x^2+b*x+a)^(1/2)+8*A*a*b*c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/ 
2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (181) = 362\).

Time = 0.25 (sec) , antiderivative size = 793, normalized size of antiderivative = 4.03 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/16*(3*(5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 + (5*B*b^4*c + 16*(B*a^2 + 
 A*a*b)*c^3 - 4*(6*B*a*b^2 + A*b^3)*c^2)*x^2 - 4*(6*B*a^2*b^2 + A*a*b^3)*c 
 + (5*B*b^5 + 16*(B*a^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*x)*sqr 
t(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*(15*B*a*b^3*c + 32*A*a^2*c^3 - 2*(B*b^2*c^3 - 4*B*a*c 
^4)*x^3 - 4*(13*B*a^2*b + 3*A*a*b^2)*c^2 + (5*B*b^3*c^2 + 16*A*a*c^4 - 4*( 
5*B*a*b + A*b^2)*c^3)*x^2 + (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(3 
1*B*a*b^2 + 6*A*b^3)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 
 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*(3*(5*B*a*b^4 
+ 16*(B*a^3 + A*a^2*b)*c^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(6*B* 
a*b^2 + A*b^3)*c^2)*x^2 - 4*(6*B*a^2*b^2 + A*a*b^3)*c + (5*B*b^5 + 16*(B*a 
^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*x)*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*(15*B* 
a*b^3*c + 32*A*a^2*c^3 - 2*(B*b^2*c^3 - 4*B*a*c^4)*x^3 - 4*(13*B*a^2*b + 3 
*A*a*b^2)*c^2 + (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*b^2)*c^3)*x^2 + 
 (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^3)*c^2)*x 
)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 
+ (b^3*c^4 - 4*a*b*c^5)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^3*(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 = 0.23 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} - \frac {5 \, B b^{3} c - 20 \, B a b c^{2} - 4 \, A b^{2} c^{2} + 16 \, A a c^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {15 \, B b^{4} - 62 \, B a b^{2} c - 12 \, A b^{3} c + 24 \, B a^{2} c^{2} + 40 \, A a b c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {15 \, B a b^{3} - 52 \, B a^{2} b c - 12 \, A a b^{2} c + 32 \, A a^{2} c^{2}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, B a c - 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/4*(((2*(B*b^2*c^2 - 4*B*a*c^3)*x/(b^2*c^3 - 4*a*c^4) - (5*B*b^3*c - 20*B 
*a*b*c^2 - 4*A*b^2*c^2 + 16*A*a*c^3)/(b^2*c^3 - 4*a*c^4))*x - (15*B*b^4 - 
62*B*a*b^2*c - 12*A*b^3*c + 24*B*a^2*c^2 + 40*A*a*b*c^2)/(b^2*c^3 - 4*a*c^ 
4))*x - (15*B*a*b^3 - 52*B*a^2*b*c - 12*A*a*b^2*c + 32*A*a^2*c^2)/(b^2*c^3 
 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 3/8*(5*B*b^2 - 4*B*a*c - 4*A*b*c)*log 
(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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