Integrand size = 36, antiderivative size = 353 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=\frac {2 \left (a b^2 c i+2 a c^2 (c g+a i)+a b^3 j-b c \left (c^2 f-a c h-3 a^2 j\right )+\left (2 c^4 f+c^3 (b g+2 a h)+b^4 j+b^2 c (b i+4 a j)+c^2 \left (b^2 h+3 a b i+2 a^2 j\right )\right ) x\right )}{3 c^3 \left (b^2+4 a c\right ) \left (a+b x-c x^2\right )^{3/2}}-\frac {2 \left (b^4 c i+24 a^2 c^3 i+2 b^2 c^2 (2 c g+3 a i)+b^5 j+b^3 c (c h+10 a j)+4 b c^2 \left (2 c^2 f-a c h+8 a^2 j\right )-c \left (16 c^4 f+8 c^3 (b g-a h)-4 b^4 j-b^2 c (b i+28 a j)+2 c^2 \left (b^2 h-6 a b i-16 a^2 j\right )\right ) x\right )}{3 c^3 \left (b^2+4 a c\right )^2 \sqrt {a+b x-c x^2}}-\frac {j \arctan \left (\frac {b-2 c x}{2 \sqrt {c} \sqrt {a+b x-c x^2}}\right )}{c^{5/2}} \]
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)-j*arctan(1/2*(-2*c*x+b)/c^(1/2)/(-c* x^2+b*x+a)^(1/2))/c^(5/2)-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+8*c ^3*(-a*h+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)
Time = 1.97 (sec) , antiderivative size = 316, normalized size of antiderivative = 0.90 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=-\frac {2 \left (3 b^5 j x^2+b^4 \left (6 a j x-4 c j x^3\right )+b^3 \left (3 a^2 j+18 a c j x^2+c^2 \left (f+3 g x-x^2 (3 h+i x)\right )\right )+8 c^2 \left (2 c^3 f x^3+a^3 (2 i+3 j x)-a c^2 x \left (3 f+h x^2\right )-a^2 c \left (g+x^2 (3 i+4 j x)\right )\right )+4 b c \left (5 a^3 j+2 c^3 x^2 (-3 f+g x)-2 a^2 c (h-3 i x)+3 a c^2 \left (f-x \left (g-h x+i x^2\right )\right )\right )+2 b^2 c \left (21 a^2 j x+c^2 x (3 f+x (-6 g+h x))+a c \left (g+x \left (-6 h+3 i x-14 j x^2\right )\right )\right )\right )}{3 c^2 \left (b^2+4 a c\right )^2 (a+x (b-c x))^{3/2}}+\frac {2 j \arctan \left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b-c x)}}\right )}{c^{5/2}} \]
(-2*(3*b^5*j*x^2 + b^4*(6*a*j*x - 4*c*j*x^3) + b^3*(3*a^2*j + 18*a*c*j*x^2 + c^2*(f + 3*g*x - x^2*(3*h + i*x))) + 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 + x^2*(3*i + 4*j*x))) + 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))) + 2*b^2*c*(21*a^2*j*x + c^2*x*(3*f + x*(-6*g + h*x)) + a*c* (g + x*(-6*h + 3*i*x - 14*j*x^2)))))/(3*c^2*(b^2 + 4*a*c)^2*(a + x*(b - c* x))^(3/2)) + (2*j*ArcTan[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b - c*x)])])/ c^(5/2)
Time = 0.71 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2191, 27, 2191, 27, 1092, 217}
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 |
| \(\displaystyle \frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}-\frac {2 \int -\frac {-\frac {3 \left (b^2+4 a c\right ) j x^2}{c}-\frac {3 \left (b^2+4 a c\right ) (c i+b j) x}{c^2}+\frac {j b^4+c (b i+a j) b^2+8 c^4 f+4 c^3 (b g-a h)+c^2 \left (b^2 h-4 a^2 j\right )}{c^3}}{2 \left (-c x^2+b x+a\right )^{3/2}}dx}{3 \left (4 a c+b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\frac {j b^4}{c^3}+\frac {(b i+a j) b^2}{c^2}+4 g b-\frac {3 \left (b^2+4 a c\right ) j x^2}{c}+8 c f-4 a h+\frac {b^2 h-4 a^2 j}{c}-\frac {3 \left (b^2+4 a c\right ) (c i+b j) x}{c^2}}{\left (-c x^2+b x+a\right )^{3/2}}dx}{3 \left (4 a c+b^2\right )}+\frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2191 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {3 \left (b^2+4 a c\right )^2 j}{2 c^2 \sqrt {-c x^2+b x+a}}dx}{4 a c+b^2}-\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+b^3 c (10 a j+c h)+2 b^2 c^2 (3 a i+2 c g)+b^5 j+b^4 c i\right )}{c^3 \left (4 a c+b^2\right ) \sqrt {a+b x-c x^2}}}{3 \left (4 a c+b^2\right )}+\frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 j \left (4 a c+b^2\right ) \int \frac {1}{\sqrt {-c x^2+b x+a}}dx}{c^2}-\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+b^3 c (10 a j+c h)+2 b^2 c^2 (3 a i+2 c g)+b^5 j+b^4 c i\right )}{c^3 \left (4 a c+b^2\right ) \sqrt {a+b x-c x^2}}}{3 \left (4 a c+b^2\right )}+\frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\frac {6 j \left (4 a c+b^2\right ) \int \frac {1}{-\frac {(b-2 c x)^2}{-c x^2+b x+a}-4 c}d\frac {b-2 c x}{\sqrt {-c x^2+b x+a}}}{c^2}-\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+b^3 c (10 a j+c h)+2 b^2 c^2 (3 a i+2 c g)+b^5 j+b^4 c i\right )}{c^3 \left (4 a c+b^2\right ) \sqrt {a+b x-c x^2}}}{3 \left (4 a c+b^2\right )}+\frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {-\frac {2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+b^3 c (10 a j+c h)+2 b^2 c^2 (3 a i+2 c g)+b^5 j+b^4 c i\right )}{c^3 \left (4 a c+b^2\right ) \sqrt {a+b x-c x^2}}-\frac {3 j \left (4 a c+b^2\right ) \arctan \left (\frac {b-2 c x}{2 \sqrt {c} \sqrt {a+b x-c x^2}}\right )}{c^{5/2}}}{3 \left (4 a c+b^2\right )}+\frac {2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^3 j+a b^2 c i+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}\) |
(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 + 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*(2*c^2*f - a*c*h + 8*a^2*j) - c*(16*c^4*f + 8* c^3*(b*g - 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*ArcTan[(b - 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x - c*x^2])])/c^(5/2))/(3*( b^2 + 4*a*c))
3.4.66.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1146\) vs. \(2(337)=674\).
Time = 2.46 (sec) , antiderivative size = 1147, normalized size of antiderivative = 3.25
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))+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+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)*arctan(c^(1/2)*(x-1/2*b/c)/(-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))))+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/...
Leaf count of result is larger than twice the leaf count of optimal. 689 vs. \(2 (339) = 678\).
Time = 23.13 (sec) , antiderivative size = 1385, normalized size of antiderivative = 3.92 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
[-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...
Timed out. \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=\int { \frac {j x^{4} + i x^{3} + h x^{2} + g x + f}{{\left (-c x^{2} + b x + a\right )}^{\frac {5}{2}}} \,d x } \]
-1/3*i*(32*a*b*x/(sqrt(-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2) - 16*a*b^2/(sqrt (-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2*c) + b^3*x/((-c*x^2 + b*x + a)^(3/2)*(b ^2 + 4*a*c)*c^2) + 2*(b^2 - 4*a*c)*b*x/(sqrt(-c*x^2 + b*x + a)*(b^2 + 4*a* c)^2*c) + 6*a*b*x/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)*c) - 3*x^2/((-c* x^2 + b*x + a)^(3/2)*c) - (b^2 - 4*a*c)*b^2/(sqrt(-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2*c^2) - a*b^2/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)*c^2) + 2*a/ ((-c*x^2 + b*x + a)^(3/2)*c^2)) + 1/3*g*(16*b*c*x/(sqrt(-c*x^2 + b*x + a)* (b^2 + 4*a*c)^2) - 8*b^2/(sqrt(-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2) + 2*b*x/ ((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)) - b^2/((-c*x^2 + b*x + a)^(3/2)*( b^2 + 4*a*c)*c) + 1/((-c*x^2 + b*x + a)^(3/2)*c)) + 2/3*f*(16*c^2*x/(sqrt( -c*x^2 + b*x + a)*(b^2 + 4*a*c)^2) - 8*b*c/(sqrt(-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2) + 2*c*x/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)) - b/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c))) + 2/3*h*(2*(b^2 - 4*a*c)*x/(sqrt(-c*x^2 + b *x + a)*(b^2 + 4*a*c)^2) + 2*a*x/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)) + b^2*x/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)*c) - (b^2 - 4*a*c)*b/(sqrt (-c*x^2 + b*x + a)*(b^2 + 4*a*c)^2*c) + a*b/((-c*x^2 + b*x + a)^(3/2)*(b^2 + 4*a*c)*c)) + j*integrate(x^4/((c^2*x^4 - 2*b*c*x^3 + 2*a*b*x + (b^2 - 2 *a*c)*x^2 + a^2)*sqrt(-c*x^2 + b*x + a)), x)
Time = 0.30 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.38 \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=-\frac {2 \, \sqrt {-c x^{2} + b x + a} {\left ({\left ({\left (\frac {{\left (16 \, c^{5} f + 8 \, b c^{4} g + 2 \, b^{2} c^{3} h - 8 \, a c^{4} h - b^{3} c^{2} i - 12 \, a b c^{3} i - 4 \, b^{4} c j - 28 \, a b^{2} c^{2} j - 32 \, a^{2} c^{3} j\right )} x}{b^{4} c^{2} + 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} - \frac {3 \, {\left (8 \, b c^{4} f + 4 \, b^{2} c^{3} g + b^{3} c^{2} h - 4 \, a b c^{3} h - 2 \, a b^{2} c^{2} i + 8 \, a^{2} c^{3} i - b^{5} j - 6 \, a b^{3} c j\right )}}{b^{4} c^{2} + 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {3 \, {\left (2 \, b^{2} c^{3} f - 8 \, a c^{4} f + b^{3} c^{2} g - 4 \, a b c^{3} g - 4 \, a b^{2} c^{2} h + 8 \, a^{2} b c^{2} i + 2 \, a b^{4} j + 14 \, a^{2} b^{2} c j + 8 \, a^{3} c^{2} j\right )}}{b^{4} c^{2} + 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {b^{3} c^{2} f + 12 \, a b c^{3} f + 2 \, a b^{2} c^{2} g - 8 \, a^{2} c^{3} g - 8 \, a^{2} b c^{2} h + 16 \, a^{3} c^{2} i + 3 \, a^{2} b^{3} j + 20 \, a^{3} b c j}{b^{4} c^{2} + 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} - b x - a\right )}^{2}} - \frac {j \log \left ({\left | 2 \, {\left (\sqrt {-c} x - \sqrt {-c x^{2} + b x + a}\right )} \sqrt {-c} + b \right |}\right )}{\sqrt {-c} c^{2}} \]
-2/3*sqrt(-c*x^2 + b*x + a)*((((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)^2 - j*log(abs(2*(sqrt(-c)*x - sqrt(-c*x^2 + b*x + a))*sq rt(-c) + b))/(sqrt(-c)*c^2)
Timed out. \[ \int \frac {f+g x+h x^2+i x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx=\int \frac {j\,x^4+i\,x^3+h\,x^2+g\,x+f}{{\left (-c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]