\(\int \frac {x^4 (A+B x)}{(a+b x+c x^2)^{5/2}} \, dx\) [162]

Optimal result

Integrand size = 23, antiderivative size = 285 \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 x^3 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 x \left (a \left (5 b^3 B-2 A b^2 c-28 a b B c+24 a A c^2\right )+\left (5 b^4 B-2 A b^3 c-32 a b^2 B c+16 a A b c^2+32 a^2 B c^2\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {\left (15 b^4 B-6 A b^3 c-100 a b^2 B c+40 a A b c^2+128 a^2 B c^2\right ) \sqrt {a+b x+c x^2}}{3 c^3 \left (b^2-4 a c\right )^2}-\frac {(5 b B-2 A c) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{7/2}} \] Output:

-2/3*x^3*(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*x*(a*(24*A*a*c^2-2*A*b^2*c-28*B*a*b*c+5*B*b^3)+(16*A*a*b*c 
^2-2*A*b^3*c+32*B*a^2*c^2-32*B*a*b^2*c+5*B*b^4)*x)/c^2/(-4*a*c+b^2)^2/(c*x 
^2+b*x+a)^(1/2)+1/3*(40*A*a*b*c^2-6*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)/c^3/(-4*a*c+b^2)^2-1/2*(-2*A*c+5*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 = 2.41 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.01 \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {128 a^4 B c^2+2 a b^2 x \left (15 b^3 B+2 b c^2 x (9 A-37 B x)+4 c^3 x^2 (7 A-3 B x)-3 b^2 c (2 A+15 B x)\right )+a^2 \left (15 b^4 B+256 b B c^3 x^3+16 c^4 x^3 (-4 A+3 B x)+12 b^2 c^2 x (7 A+4 B x)-6 b^3 c (A+35 B x)\right )-4 a^3 c \left (25 b^2 B+12 c^2 x (A-4 B x)-2 b c (5 A+39 B x)\right )+b^4 x^2 \left (15 b^2 B+c^2 x (-8 A+3 B x)+b (-6 A c+20 B c x)\right )}{3 c^3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}+\frac {(5 b B-2 A c) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{2 c^{7/2}} \] Input:

Integrate[(x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(128*a^4*B*c^2 + 2*a*b^2*x*(15*b^3*B + 2*b*c^2*x*(9*A - 37*B*x) + 4*c^3*x^ 
2*(7*A - 3*B*x) - 3*b^2*c*(2*A + 15*B*x)) + a^2*(15*b^4*B + 256*b*B*c^3*x^ 
3 + 16*c^4*x^3*(-4*A + 3*B*x) + 12*b^2*c^2*x*(7*A + 4*B*x) - 6*b^3*c*(A + 
35*B*x)) - 4*a^3*c*(25*b^2*B + 12*c^2*x*(A - 4*B*x) - 2*b*c*(5*A + 39*B*x) 
) + b^4*x^2*(15*b^2*B + c^2*x*(-8*A + 3*B*x) + b*(-6*A*c + 20*B*c*x)))/(3* 
c^3*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) + ((5*b*B - 2*A*c)*Log[b + 2* 
c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(2*c^(7/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {1233, 27, 1233, 27, 1160, 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.

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

\(\Big \downarrow \) 1233

\(\displaystyle \frac {2 \int \frac {x^2 \left (6 a (b B-2 A c)+\left (5 B b^2-2 A c b-16 a B c\right ) x\right )}{2 \left (c x^2+b x+a\right )^{3/2}}dx}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {x^2 \left (6 a (b B-2 A c)+\left (5 B b^2-2 A c b-16 a B c\right ) x\right )}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1233

\(\displaystyle \frac {\frac {2 \int \frac {2 a \left (5 B b^3-2 A c b^2-28 a B c b+24 a A c^2\right )+\left (15 B b^4-6 A c b^3-100 a B c b^2+40 a A c^2 b+128 a^2 B c^2\right ) x}{2 \sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {2 a \left (5 B b^3-2 A c b^2-28 a B c b+24 a A c^2\right )+\left (15 B b^4-6 A c b^3-100 a B c b^2+40 a A c^2 b+128 a^2 B c^2\right ) x}{\sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+40 a A b c^2-100 a b^2 B c-6 A b^3 c+15 b^4 B\right )}{c}-\frac {3 \left (b^2-4 a c\right )^2 (5 b B-2 A c) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{2 c}}{c \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+40 a A b c^2-100 a b^2 B c-6 A b^3 c+15 b^4 B\right )}{c}-\frac {3 \left (b^2-4 a c\right )^2 (5 b B-2 A c) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}}{c \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (128 a^2 B c^2+40 a A b c^2-100 a b^2 B c-6 A b^3 c+15 b^4 B\right )}{c}-\frac {3 \left (b^2-4 a c\right )^2 (5 b B-2 A c) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}}}{c \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (32 a^2 B c^2+16 a A b c^2-32 a b^2 B c-2 A b^3 c+5 b^4 B\right )+a \left (24 a A c^2-28 a b B c-2 A b^2 c+5 b^3 B\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}}{3 c \left (b^2-4 a c\right )}-\frac {2 x^3 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{3 c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\)

Input:

Int[(x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(-2*x^3*(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*x*(a*(5*b^3*B - 2*A*b^2*c - 28*a*b*B*c + 
 24*a*A*c^2) + (5*b^4*B - 2*A*b^3*c - 32*a*b^2*B*c + 16*a*A*b*c^2 + 32*a^2 
*B*c^2)*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (((15*b^4*B - 6*A*b^ 
3*c - 100*a*b^2*B*c + 40*a*A*b*c^2 + 128*a^2*B*c^2)*Sqrt[a + b*x + c*x^2]) 
/c - (3*(b^2 - 4*a*c)^2*(5*b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr 
t[a + b*x + c*x^2])])/(2*c^(3/2)))/(c*(b^2 - 4*a*c)))/(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 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]
 

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. \(1245\) vs. \(2(263)=526\).

Time = 1.34 (sec) , antiderivative size = 1246, normalized size of antiderivative = 4.37

method	result	size
default	\(\text {Expression too large to display}\)	\(1246\)
risch	\(\text {Expression too large to display}\)	\(5483\)

Input:

int(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

A*(-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) 
)))+B*(x^4/c/(c*x^2+b*x+a)^(3/2)-5/2*b/c*(-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))))-4*a/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)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (263) = 526\).

Time = 0.65 (sec) , antiderivative size = 1601, normalized size of antiderivative = 5.62 \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(5*B*a^2*b^5 - 32*A*a^4*c^3 + (5*B*b^5*c^2 - 32*A*a^2*c^5 + 16*( 
5*B*a^2*b + A*a*b^2)*c^4 - 2*(20*B*a*b^3 + A*b^4)*c^3)*x^4 + 2*(5*B*b^6*c 
- 32*A*a^2*b*c^4 + 16*(5*B*a^2*b^2 + A*a*b^3)*c^3 - 2*(20*B*a*b^4 + A*b^5) 
*c^2)*x^3 + 16*(5*B*a^4*b + A*a^3*b^2)*c^2 + (5*B*b^7 + 12*A*a*b^4*c^2 + 1 
60*B*a^3*b*c^3 - 64*A*a^3*c^4 - 2*(15*B*a*b^5 + A*b^6)*c)*x^2 - 2*(20*B*a^ 
3*b^3 + A*a^2*b^4)*c + 2*(5*B*a*b^6 - 32*A*a^3*b*c^3 + 16*(5*B*a^3*b^2 + A 
*a^2*b^3)*c^2 - 2*(20*B*a^2*b^4 + A*a*b^5)*c)*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*( 
15*B*a^2*b^4*c + 3*(B*b^4*c^3 - 8*B*a*b^2*c^4 + 16*B*a^2*c^5)*x^4 + 8*(16* 
B*a^4 + 5*A*a^3*b)*c^3 + 4*(5*B*b^5*c^2 - 16*A*a^2*c^5 + 2*(32*B*a^2*b + 7 
*A*a*b^2)*c^4 - (37*B*a*b^3 + 2*A*b^4)*c^3)*x^3 - 2*(50*B*a^3*b^2 + 3*A*a^ 
2*b^3)*c^2 + 3*(5*B*b^6*c + 64*B*a^3*c^4 + 4*(4*B*a^2*b^2 + 3*A*a*b^3)*c^3 
 - 2*(15*B*a*b^4 + A*b^5)*c^2)*x^2 + 6*(5*B*a*b^5*c - 8*A*a^3*c^4 + 2*(26* 
B*a^3*b + 7*A*a^2*b^2)*c^3 - (35*B*a^2*b^3 + 2*A*a*b^4)*c^2)*x)*sqrt(c*x^2 
 + b*x + a))/(a^2*b^4*c^4 - 8*a^3*b^2*c^5 + 16*a^4*c^6 + (b^4*c^6 - 8*a*b^ 
2*c^7 + 16*a^2*c^8)*x^4 + 2*(b^5*c^5 - 8*a*b^3*c^6 + 16*a^2*b*c^7)*x^3 + ( 
b^6*c^4 - 6*a*b^4*c^5 + 32*a^3*c^7)*x^2 + 2*(a*b^5*c^4 - 8*a^2*b^3*c^5 + 1 
6*a^3*b*c^6)*x), 1/6*(3*(5*B*a^2*b^5 - 32*A*a^4*c^3 + (5*B*b^5*c^2 - 32*A* 
a^2*c^5 + 16*(5*B*a^2*b + A*a*b^2)*c^4 - 2*(20*B*a*b^3 + A*b^4)*c^3)*x^4 + 
 2*(5*B*b^6*c - 32*A*a^2*b*c^4 + 16*(5*B*a^2*b^2 + A*a*b^3)*c^3 - 2*(20...
 

Sympy [F]

\[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4*(B*x+A)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Integral(x**4*(A + B*x)/(a + b*x + c*x**2)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4*(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 = 0.25 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.60 \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {{\left ({\left ({\left (\frac {3 \, {\left (B b^{4} c^{2} - 8 \, B a b^{2} c^{3} + 16 \, B a^{2} c^{4}\right )} x}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}} + \frac {4 \, {\left (5 \, B b^{5} c - 37 \, B a b^{3} c^{2} - 2 \, A b^{4} c^{2} + 64 \, B a^{2} b c^{3} + 14 \, A a b^{2} c^{3} - 16 \, A a^{2} c^{4}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {3 \, {\left (5 \, B b^{6} - 30 \, B a b^{4} c - 2 \, A b^{5} c + 16 \, B a^{2} b^{2} c^{2} + 12 \, A a b^{3} c^{2} + 64 \, B a^{3} c^{3}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {6 \, {\left (5 \, B a b^{5} - 35 \, B a^{2} b^{3} c - 2 \, A a b^{4} c + 52 \, B a^{3} b c^{2} + 14 \, A a^{2} b^{2} c^{2} - 8 \, A a^{3} c^{3}\right )}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}\right )} x + \frac {15 \, B a^{2} b^{4} - 100 \, B a^{3} b^{2} c - 6 \, A a^{2} b^{3} c + 128 \, B a^{4} c^{2} + 40 \, A a^{3} b c^{2}}{b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, B b - 2 \, A c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {7}{2}}} \] Input:

integrate(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

1/3*((((3*(B*b^4*c^2 - 8*B*a*b^2*c^3 + 16*B*a^2*c^4)*x/(b^4*c^3 - 8*a*b^2* 
c^4 + 16*a^2*c^5) + 4*(5*B*b^5*c - 37*B*a*b^3*c^2 - 2*A*b^4*c^2 + 64*B*a^2 
*b*c^3 + 14*A*a*b^2*c^3 - 16*A*a^2*c^4)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^ 
5))*x + 3*(5*B*b^6 - 30*B*a*b^4*c - 2*A*b^5*c + 16*B*a^2*b^2*c^2 + 12*A*a* 
b^3*c^2 + 64*B*a^3*c^3)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5))*x + 6*(5*B*a 
*b^5 - 35*B*a^2*b^3*c - 2*A*a*b^4*c + 52*B*a^3*b*c^2 + 14*A*a^2*b^2*c^2 - 
8*A*a^3*c^3)/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5))*x + (15*B*a^2*b^4 - 100 
*B*a^3*b^2*c - 6*A*a^2*b^3*c + 128*B*a^4*c^2 + 40*A*a^3*b*c^2)/(b^4*c^3 - 
8*a*b^2*c^4 + 16*a^2*c^5))/(c*x^2 + b*x + a)^(3/2) + 1/2*(5*B*b - 2*A*c)*l 
og(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((x^4*(A + B*x))/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^4 (A+B x)}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (B x +A \right )}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}d x \] Input:

int(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

int(x^4*(B*x+A)/(c*x^2+b*x+a)^(5/2),x)