\(\int \frac {x^3 (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\) [163]

Optimal result

Integrand size = 23, antiderivative size = 189 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (a \left (3 b^3 B-20 a b B c+16 a A c^2\right )+\left (3 b^4 B-22 a b^2 B c+8 a A b c^2+24 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {B \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \] Output:

-2/3*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)^(3/2)-2/3*(a*(16*A*a*c^2-20*B*a*b*c+3*B*b^3)+(8*A*a*b*c^2+24*B*a^2*c 
^2-22*B*a*b^2*c+3*B*b^4)*x)/c^2/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)+B*arcta 
nh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 1.76 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.08 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \left (4 a^3 c (-5 b B+4 A c+6 B c x)+b^3 x^2 \left (3 b^2 B+4 b B c x-A c^2 x\right )+2 a b x \left (3 b^3 B-9 b^2 B c x+6 A c^3 x^2+b c^2 x (3 A-14 B x)\right )+a^2 \left (3 b^3 B-42 b^2 B c x+24 A b c^2 x+8 c^3 x^2 (3 A+4 B x)\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}-\frac {B \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{c^{5/2}} \] Input:

Integrate[(x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*(4*a^3*c*(-5*b*B + 4*A*c + 6*B*c*x) + b^3*x^2*(3*b^2*B + 4*b*B*c*x - A 
*c^2*x) + 2*a*b*x*(3*b^3*B - 9*b^2*B*c*x + 6*A*c^3*x^2 + b*c^2*x*(3*A - 14 
*B*x)) + a^2*(3*b^3*B - 42*b^2*B*c*x + 24*A*b*c^2*x + 8*c^3*x^2*(3*A + 4*B 
*x))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) - (B*Log[c^2*(b + 2 
*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(5/2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1233, 27, 1224, 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.

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

\(\Big \downarrow \) 1233

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1224

\(\displaystyle \frac {\frac {3 B \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{c}-\frac {2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {6 B \left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}-\frac {2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 B \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 \left (x \left (24 a^2 B c^2+8 a A b c^2-22 a b^2 B c+3 b^4 B\right )+a \left (16 a A c^2-20 a b B c+3 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[(x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*x^2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(3*c*(b^2 - 4*a*c 
)*(a + b*x + c*x^2)^(3/2)) + ((-2*(a*(3*b^3*B - 20*a*b*B*c + 16*a*A*c^2) + 
 (3*b^4*B - 22*a*b^2*B*c + 8*a*A*b*c^2 + 24*a^2*B*c^2)*x))/(c*(b^2 - 4*a*c 
)*Sqrt[a + b*x + c*x^2]) + (3*B*(b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[ 
c]*Sqrt[a + b*x + c*x^2])])/c^(3/2))/(3*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 1224

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - ( 
b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x))*((a + b*x + c*x^2)^(p 
+ 1)/(c*(p + 1)*(b^2 - 4*a*c))), 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))/(c*(p + 1)*(b^2 - 4*a*c))   Int[(a + 
b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, - 
1] &&  !(IntegerQ[p] && NeQ[a, 0] && NiceSqrtQ[b^2 - 4*a*c])
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(764\) vs. \(2(173)=346\).

Time = 1.22 (sec) , antiderivative size = 765, normalized size of antiderivative = 4.05

method	result	size
default	\(A \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{2 c}+\frac {2 a \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{c}\right )+B \left (-\frac {x^{3}}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{2 c}+\frac {2 a \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{c}\right )}{2 c}+\frac {-\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}}}}{c}\right )\)	\(765\)

Input:

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (173) = 346\).

Time = 0.61 (sec) , antiderivative size = 1061, normalized size of antiderivative = 5.61 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B*a^4*c^2 + (B*b^4*c^2 - 8*B*a*b^2 
*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*c - 8*B*a*b^3*c^2 + 16*B*a^2*b*c^3)*x^ 
3 + (B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2*(B*a*b^5 - 8*B*a^2*b^3*c 
+ 16*B*a^3*b*c^2)*x)*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*(3*B*a^2*b^3*c - 20*B*a^3*b*c 
^2 + 16*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b)*c^4 - (28*B*a*b^2 
 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3*c^2 + 2*A*a*b^2*c^3 + 8*A*a^2* 
c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^2*b^2*c^2 + 4*(B*a^3 + A*a^2*b)*c^3)*x)*sq 
rt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 
- 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6) 
*x^3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3 
*c^4 + 16*a^3*b*c^5)*x), -1/3*(3*(B*a^2*b^4 - 8*B*a^3*b^2*c + 16*B*a^4*c^2 
 + (B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x^4 + 2*(B*b^5*c - 8*B*a*b^3 
*c^2 + 16*B*a^2*b*c^3)*x^3 + (B*b^6 - 6*B*a*b^4*c + 32*B*a^3*c^3)*x^2 + 2* 
(B*a*b^5 - 8*B*a^2*b^3*c + 16*B*a^3*b*c^2)*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*(3*B*a^2*b 
^3*c - 20*B*a^3*b*c^2 + 16*A*a^3*c^3 + (4*B*b^4*c^2 + 4*(8*B*a^2 + 3*A*a*b 
)*c^4 - (28*B*a*b^2 + A*b^3)*c^3)*x^3 + 3*(B*b^5*c - 6*B*a*b^3*c^2 + 2*A*a 
*b^2*c^3 + 8*A*a^2*c^4)*x^2 + 6*(B*a*b^4*c - 7*B*a^2*b^2*c^2 + 4*(B*a^3 + 
A*a^2*b)*c^3)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + ...
 

Sympy [F]

\[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**3*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Integral(x**3*(A + B*x)/(a + b*x + c*x**2)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),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 = 0.27 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.65 \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left ({\left ({\left (\frac {{\left (4 \, B b^{4} c - 28 \, B a b^{2} c^{2} - A b^{3} c^{2} + 32 \, B a^{2} c^{3} + 12 \, A a b c^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (B b^{5} - 6 \, B a b^{3} c + 2 \, A a b^{2} c^{2} + 8 \, A a^{2} c^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {6 \, {\left (B a b^{4} - 7 \, B a^{2} b^{2} c + 4 \, B a^{3} c^{2} + 4 \, A a^{2} b c^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {3 \, B a^{2} b^{3} - 20 \, B a^{3} b c + 16 \, A a^{3} c^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} - \frac {B \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} \] Input:

integrate(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

-2/3*((((4*B*b^4*c - 28*B*a*b^2*c^2 - A*b^3*c^2 + 32*B*a^2*c^3 + 12*A*a*b* 
c^3)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(B*b^5 - 6*B*a*b^3*c + 2*A 
*a*b^2*c^2 + 8*A*a^2*c^3)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(B*a 
*b^4 - 7*B*a^2*b^2*c + 4*B*a^3*c^2 + 4*A*a^2*b*c^2)/(b^4*c^2 - 8*a*b^2*c^3 
 + 16*a^2*c^4))*x + (3*B*a^2*b^3 - 20*B*a^3*b*c + 16*A*a^3*c^2)/(b^4*c^2 - 
 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2) - B*log(abs(2*(sqrt(c) 
*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((x^3*(A + B*x))/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^3 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{3} \left (B x +A \right )}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}d x \] Input:

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

int(x^3*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)