\(\int \frac {f+g x+h x^2+i x^3+j x^4}{(a+b x+c x^2)^{5/2}} \, dx\) [201]

Optimal result

Integrand size = 35, antiderivative size = 354 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (a b^2 c i+2 a c^2 (c g-a i)-a b^3 j-b c \left (c^2 f+a c h-3 a^2 j\right )-\left (2 c^4 f-c^3 (b g+2 a h)+b^4 j-b^2 c (b i+4 a j)+c^2 \left (b^2 h+3 a b i+2 a^2 j\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (b^4 c i+24 a^2 c^3 i+2 b^2 c^2 (2 c g-3 a i)-b^5 j-b^3 c (c h-10 a j)-4 b c^2 \left (2 c^2 f+a c h+8 a^2 j\right )-c \left (16 c^4 f-c^3 (8 b g-8 a h)-4 b^4 j+b^2 c (b i+28 a j)+2 c^2 \left (b^2 h-6 a b i-16 a^2 j\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {j \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \] Output:

2/3*(a*b^2*c*i+2*a*c^2*(-a*i+c*g)-a*b^3*j-b*c*(-3*a^2*j+a*c*h+c^2*f)-(2*c^ 
4*f-c^3*(2*a*h+b*g)+b^4*j-b^2*c*(4*a*j+b*i)+c^2*(2*a^2*j+3*a*b*i+b^2*h))*x 
)/c^3/(-4*a*c+b^2)/(c*x^2+b*x+a)^(3/2)-2/3*(b^4*c*i+24*a^2*c^3*i+2*b^2*c^2 
*(-3*a*i+2*c*g)-b^5*j-b^3*c*(-10*a*j+c*h)-4*b*c^2*(8*a^2*j+a*c*h+2*c^2*f)- 
c*(16*c^4*f-c^3*(-8*a*h+8*b*g)-4*b^4*j+b^2*c*(28*a*j+b*i)+2*c^2*(-16*a^2*j 
-6*a*b*i+b^2*h))*x)/c^3/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^(1/2)+j*arctanh(1/2*( 
2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 3.11 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.89 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (-3 b^5 j x^2-2 b^4 j x \left (3 a+2 c x^2\right )+b^3 \left (-3 a^2 j+18 a c j x^2+c^2 \left (-f-3 g x+3 h x^2+i x^3\right )\right )+2 b^2 c \left (21 a^2 j x+c^2 x \left (3 f-6 g x+h x^2\right )-a c \left (g-6 h x+3 i x^2-14 j x^3\right )\right )-8 c^2 \left (-2 c^3 f x^3+a^3 (2 i+3 j x)-a c^2 x \left (3 f+h x^2\right )+a^2 c \left (g+3 i x^2+4 j x^3\right )\right )+4 b c \left (5 a^3 j-2 c^3 x^2 (-3 f+g x)+2 a^2 c (h-3 i x)+3 a c^2 \left (f-x \left (g-h x+i x^2\right )\right )\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}+\frac {2 j \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{5/2}} \] Input:

Integrate[(f + g*x + h*x^2 + i*x^3 + j*x^4)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*(-3*b^5*j*x^2 - 2*b^4*j*x*(3*a + 2*c*x^2) + b^3*(-3*a^2*j + 18*a*c*j*x^ 
2 + c^2*(-f - 3*g*x + 3*h*x^2 + i*x^3)) + 2*b^2*c*(21*a^2*j*x + c^2*x*(3*f 
 - 6*g*x + h*x^2) - a*c*(g - 6*h*x + 3*i*x^2 - 14*j*x^3)) - 8*c^2*(-2*c^3* 
f*x^3 + a^3*(2*i + 3*j*x) - a*c^2*x*(3*f + h*x^2) + a^2*c*(g + 3*i*x^2 + 4 
*j*x^3)) + 4*b*c*(5*a^3*j - 2*c^3*x^2*(-3*f + g*x) + 2*a^2*c*(h - 3*i*x) + 
 3*a*c^2*(f - x*(g - h*x + i*x^2)))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c 
*x))^(3/2)) + (2*j*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])] 
)/c^(5/2)
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2191, 27, 2191, 27, 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.

\(\displaystyle \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2191

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {3 \left (4 a-\frac {b^2}{c}\right ) j x^2-\frac {3 \left (b^2-4 a c\right ) (c i-b j) x}{c^2}+\frac {j b^4-c (b i+a j) b^2+8 c^4 f-c^3 (4 b g-4 a h)+c^2 \left (b^2 h-4 a^2 j\right )}{c^3}}{2 \left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\int \frac {\frac {j b^4}{c^3}-\frac {(b i+a j) b^2}{c^2}-4 g b+3 \left (4 a-\frac {b^2}{c}\right ) j x^2+8 c f+4 a h+\frac {b^2 h-4 a^2 j}{c}-\frac {3 \left (b^2-4 a c\right ) (c i-b j) x}{c^2}}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 2191

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i-b^3 c (c h-10 a j)+b^2 \left (4 c^3 g-6 a c^2 i\right )+b^5 (-j)+b^4 c i\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {3 \left (b^2-4 a c\right )^2 j}{2 c^2 \sqrt {c x^2+b x+a}}dx}{b^2-4 a c}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i-b^3 c (c h-10 a j)+b^2 \left (4 c^3 g-6 a c^2 i\right )+b^5 (-j)+b^4 c i\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {3 j \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{c^2}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i-b^3 c (c h-10 a j)+b^2 \left (4 c^3 g-6 a c^2 i\right )+b^5 (-j)+b^4 c i\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {6 j \left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c^2}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )-a b^3 j+a b^2 c i+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i-b^3 c (c h-10 a j)+b^2 \left (4 c^3 g-6 a c^2 i\right )+b^5 (-j)+b^4 c i\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {3 j \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}}{3 \left (b^2-4 a c\right )}\)

Input:

Int[(f + g*x + h*x^2 + i*x^3 + j*x^4)/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*(a*b^2*c*i + 2*a*c^2*(c*g - a*i) - a*b^3*j - b*c*(c^2*f + a*c*h - 3*a^2 
*j) - (2*c^4*f - c^3*(b*g + 2*a*h) + b^4*j - b^2*c*(b*i + 4*a*j) + c^2*(b^ 
2*h + 3*a*b*i + 2*a^2*j))*x))/(3*c^3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2) 
) - ((2*(b^4*c*i + 24*a^2*c^3*i + b^2*(4*c^3*g - 6*a*c^2*i) - b^5*j - b^3* 
c*(c*h - 10*a*j) - 4*b*c^2*(2*c^2*f + a*c*h + 8*a^2*j) - c*(16*c^4*f - c^3 
*(8*b*g - 8*a*h) - 4*b^4*j + b^2*c*(b*i + 28*a*j) + 2*c^2*(b^2*h - 6*a*b*i 
 - 16*a^2*j))*x))/(c^3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) - (3*(b^2 - 4* 
a*c)*j*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2))/(3 
*(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 2191

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = 
PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P 
q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + 
c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ 
(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int 
[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* 
(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 
2 - 4*a*c, 0] && LtQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1113\) vs. \(2(338)=676\).

Time = 0.43 (sec) , antiderivative size = 1114, normalized size of antiderivative = 3.15

method	result	size
default	\(\text {Expression too large to display}\)	\(1114\)

Input:

int((j*x^4+i*x^3+h*x^2+g*x+f)/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

f*(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))+g*(-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)))+h*(-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)))+i*(-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))))+j*(-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+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 685 vs. \(2 (337) = 674\).

Time = 22.00 (sec) , antiderivative size = 1373, normalized size of antiderivative = 3.88 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((j*x^4+i*x^3+h*x^2+g*x+f)/(c*x^2+b*x+a)^(5/2),x, algorithm="fric 
as")
 

Output:

[1/6*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*j*x^4 + 2*(b^5*c - 8*a*b^3*c 
^2 + 16*a^2*b*c^3)*j*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*j*x^2 + 2*(a*b^5 
 - 8*a^2*b^3*c + 16*a^3*b*c^2)*j*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)* 
j)*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*(8*a^2*b*c^3*h - 16*a^3*c^3*i + (16*c^6*f - 8*b 
*c^5*g + 2*(b^2*c^4 + 4*a*c^5)*h + (b^3*c^3 - 12*a*b*c^4)*i - 4*(b^4*c^2 - 
 7*a*b^2*c^3 + 8*a^2*c^4)*j)*x^3 + 3*(8*b*c^5*f - 4*b^2*c^4*g + (b^3*c^3 + 
 4*a*b*c^4)*h - 2*(a*b^2*c^3 + 4*a^2*c^4)*i - (b^5*c - 6*a*b^3*c^2)*j)*x^2 
 - (b^3*c^3 - 12*a*b*c^4)*f - 2*(a*b^2*c^3 + 4*a^2*c^4)*g - (3*a^2*b^3*c - 
 20*a^3*b*c^2)*j + 3*(4*a*b^2*c^3*h - 8*a^2*b*c^3*i + 2*(b^2*c^4 + 4*a*c^5 
)*f - (b^3*c^3 + 4*a*b*c^4)*g - 2*(a*b^4*c - 7*a^2*b^2*c^2 + 4*a^3*c^3)*j) 
*x)*sqrt(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^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^ 
4)*j*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*j*x^3 + (b^6 - 6*a*b^4*c 
 + 32*a^3*c^3)*j*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*j*x + (a^2*b 
^4 - 8*a^3*b^2*c + 16*a^4*c^2)*j)*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*(8*a^2*b*c^3*h - 16*a^ 
3*c^3*i + (16*c^6*f - 8*b*c^5*g + 2*(b^2*c^4 + 4*a*c^5)*h + (b^3*c^3 - ...
 

Sympy [F]

\[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {f + g x + h x^{2} + i x^{3} + j x^{4}}{\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((j*x**4+i*x**3+h*x**2+g*x+f)/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Integral((f + g*x + h*x**2 + i*x**3 + j*x**4)/(a + b*x + c*x**2)**(5/2), x 
)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((j*x^4+i*x^3+h*x^2+g*x+f)/(c*x^2+b*x+a)^(5/2),x, algorithm="maxi 
ma")
 

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 = 0.17 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.31 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (\frac {{\left (16 \, c^{5} f - 8 \, b c^{4} g + 2 \, b^{2} c^{3} h + 8 \, a c^{4} h + b^{3} c^{2} i - 12 \, a b c^{3} i - 4 \, b^{4} c j + 28 \, a b^{2} c^{2} j - 32 \, a^{2} c^{3} j\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (8 \, b c^{4} f - 4 \, b^{2} c^{3} g + b^{3} c^{2} h + 4 \, a b c^{3} h - 2 \, a b^{2} c^{2} i - 8 \, a^{2} c^{3} i - b^{5} j + 6 \, a b^{3} c j\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {3 \, {\left (2 \, b^{2} c^{3} f + 8 \, a c^{4} f - b^{3} c^{2} g - 4 \, a b c^{3} g + 4 \, a b^{2} c^{2} h - 8 \, a^{2} b c^{2} i - 2 \, a b^{4} j + 14 \, a^{2} b^{2} c j - 8 \, a^{3} c^{2} j\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {b^{3} c^{2} f - 12 \, a b c^{3} f + 2 \, a b^{2} c^{2} g + 8 \, a^{2} c^{3} g - 8 \, a^{2} b c^{2} h + 16 \, a^{3} c^{2} i + 3 \, a^{2} b^{3} j - 20 \, a^{3} b c j}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} - \frac {j \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} \] Input:

integrate((j*x^4+i*x^3+h*x^2+g*x+f)/(c*x^2+b*x+a)^(5/2),x, algorithm="giac 
")
 

Output:

2/3*((((16*c^5*f - 8*b*c^4*g + 2*b^2*c^3*h + 8*a*c^4*h + b^3*c^2*i - 12*a* 
b*c^3*i - 4*b^4*c*j + 28*a*b^2*c^2*j - 32*a^2*c^3*j)*x/(b^4*c^2 - 8*a*b^2* 
c^3 + 16*a^2*c^4) + 3*(8*b*c^4*f - 4*b^2*c^3*g + b^3*c^2*h + 4*a*b*c^3*h - 
 2*a*b^2*c^2*i - 8*a^2*c^3*i - b^5*j + 6*a*b^3*c*j)/(b^4*c^2 - 8*a*b^2*c^3 
 + 16*a^2*c^4))*x + 3*(2*b^2*c^3*f + 8*a*c^4*f - b^3*c^2*g - 4*a*b*c^3*g + 
 4*a*b^2*c^2*h - 8*a^2*b*c^2*i - 2*a*b^4*j + 14*a^2*b^2*c*j - 8*a^3*c^2*j) 
/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - (b^3*c^2*f - 12*a*b*c^3*f + 2*a 
*b^2*c^2*g + 8*a^2*c^3*g - 8*a^2*b*c^2*h + 16*a^3*c^2*i + 3*a^2*b^3*j - 20 
*a^3*b*c*j)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2) 
- j*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {j\,x^4+i\,x^3+h\,x^2+g\,x+f}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((f + g*x + h*x^2 + i*x^3 + j*x^4)/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((f + g*x + h*x^2 + i*x^3 + j*x^4)/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 2208, normalized size of antiderivative = 6.24 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int((j*x^4+i*x^3+h*x^2+g*x+f)/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(40*sqrt(a + b*x + c*x**2)*a**3*b*c**2*j - 32*sqrt(a + b*x + c*x**2)*a**3* 
c**3*i - 48*sqrt(a + b*x + c*x**2)*a**3*c**3*j*x - 6*sqrt(a + b*x + c*x**2 
)*a**2*b**3*c*j + 84*sqrt(a + b*x + c*x**2)*a**2*b**2*c**2*j*x + 16*sqrt(a 
 + b*x + c*x**2)*a**2*b*c**3*h - 48*sqrt(a + b*x + c*x**2)*a**2*b*c**3*i*x 
 - 16*sqrt(a + b*x + c*x**2)*a**2*c**4*g - 48*sqrt(a + b*x + c*x**2)*a**2* 
c**4*i*x**2 - 64*sqrt(a + b*x + c*x**2)*a**2*c**4*j*x**3 - 12*sqrt(a + b*x 
 + c*x**2)*a*b**4*c*j*x + 36*sqrt(a + b*x + c*x**2)*a*b**3*c**2*j*x**2 - 4 
*sqrt(a + b*x + c*x**2)*a*b**2*c**3*g + 24*sqrt(a + b*x + c*x**2)*a*b**2*c 
**3*h*x - 12*sqrt(a + b*x + c*x**2)*a*b**2*c**3*i*x**2 + 56*sqrt(a + b*x + 
 c*x**2)*a*b**2*c**3*j*x**3 + 24*sqrt(a + b*x + c*x**2)*a*b*c**4*f - 24*sq 
rt(a + b*x + c*x**2)*a*b*c**4*g*x + 24*sqrt(a + b*x + c*x**2)*a*b*c**4*h*x 
**2 - 24*sqrt(a + b*x + c*x**2)*a*b*c**4*i*x**3 + 48*sqrt(a + b*x + c*x**2 
)*a*c**5*f*x + 16*sqrt(a + b*x + c*x**2)*a*c**5*h*x**3 - 6*sqrt(a + b*x + 
c*x**2)*b**5*c*j*x**2 - 8*sqrt(a + b*x + c*x**2)*b**4*c**2*j*x**3 - 2*sqrt 
(a + b*x + c*x**2)*b**3*c**3*f - 6*sqrt(a + b*x + c*x**2)*b**3*c**3*g*x + 
6*sqrt(a + b*x + c*x**2)*b**3*c**3*h*x**2 + 2*sqrt(a + b*x + c*x**2)*b**3* 
c**3*i*x**3 + 12*sqrt(a + b*x + c*x**2)*b**2*c**4*f*x - 24*sqrt(a + b*x + 
c*x**2)*b**2*c**4*g*x**2 + 4*sqrt(a + b*x + c*x**2)*b**2*c**4*h*x**3 + 48* 
sqrt(a + b*x + c*x**2)*b*c**5*f*x**2 - 16*sqrt(a + b*x + c*x**2)*b*c**5*g* 
x**3 + 32*sqrt(a + b*x + c*x**2)*c**6*f*x**3 + 48*sqrt(c)*log((2*sqrt(c...