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

Optimal. Leaf size=353 \[ -\frac{2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+2 b^2 c^2 (3 a i+2 c g)+b^3 c (10 a j+c h)+b^4 c i+b^5 j\right )}{3 c^3 \left (4 a c+b^2\right )^2 \sqrt{a+b x-c x^2}}+\frac{2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^2 c i+a b^3 j+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}-\frac{j \tan ^{-1}\left (\frac{b-2 c x}{2 \sqrt{c} \sqrt{a+b x-c x^2}}\right )}{c^{5/2}} \]

[Out]

(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))/(3*c^3*(b^2 + 4*a*c)^2*Sqrt[a + b*x - c*x^2]) - (j*ArcTan[(b - 2*c*x)/(2*Sqrt[c]*Sqrt[a + b
*x - c*x^2])])/c^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.3857, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1660, 12, 621, 204} \[ -\frac{2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )-b^2 c (28 a j+b i)+8 c^3 (b g-a h)-4 b^4 j+16 c^4 f\right )+4 b c^2 \left (8 a^2 j-a c h+2 c^2 f\right )+24 a^2 c^3 i+2 b^2 c^2 (3 a i+2 c g)+b^3 c (10 a j+c h)+b^4 c i+b^5 j\right )}{3 c^3 \left (4 a c+b^2\right )^2 \sqrt{a+b x-c x^2}}+\frac{2 \left (x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )+b^2 c (4 a j+b i)+c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j-a c h+c^2 f\right )+a b^2 c i+a b^3 j+2 a c^2 (a i+c g)\right )}{3 c^3 \left (4 a c+b^2\right ) \left (a+b x-c x^2\right )^{3/2}}-\frac{j \tan ^{-1}\left (\frac{b-2 c x}{2 \sqrt{c} \sqrt{a+b x-c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))/(3*c^3*(b^2 + 4*a*c)^2*Sqrt[a + b*x - c*x^2]) - (j*ArcTan[(b - 2*c*x)/(2*Sqrt[c]*Sqrt[a + b
*x - c*x^2])])/c^(5/2)

Rule 1660

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[Pq, 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] + Dist[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
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{f+g x+h x^2+366 x^3+j x^4}{\left (a+b x-c x^2\right )^{5/2}} \, dx &=-\frac{2 \left (c^3 \left (b f-\frac{3 a^2 (244 c+b j)}{c^2}-\frac{a \left (366 b^2 c+2 c^3 g+b c^2 h+b^3 j\right )}{c^3}\right )-\left (366 b^3 c+b c^2 (1098 a+c g)+b^4 j+b^2 c (c h+4 a j)+2 c^2 \left (c^2 f+a c h+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 \int \frac{-\frac{366 b^3 c+4 b c^3 g+b^4 j+b^2 c (c h+a j)+4 c^2 \left (2 c^2 f-a c h-a^2 j\right )}{2 c^3}+\frac{3 \left (b^2+4 a c\right ) (366 c+b j) x}{2 c^2}+\frac{3 \left (b^2+4 a c\right ) j x^2}{2 c}}{\left (a+b x-c x^2\right )^{3/2}} \, dx}{3 \left (b^2+4 a c\right )}\\ &=-\frac{2 \left (c^3 \left (b f-\frac{3 a^2 (244 c+b j)}{c^2}-\frac{a \left (366 b^2 c+2 c^3 g+b c^2 h+b^3 j\right )}{c^3}\right )-\left (366 b^3 c+b c^2 (1098 a+c g)+b^4 j+b^2 c (c h+4 a j)+2 c^2 \left (c^2 f+a c h+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 (366 b^4 c+8784 a^2 c^3+4 b^2 c^2 (549 a+c g)+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 )+2 c \left (183 b^3 c+4 b c^2 (549 a-c g)+2 b^4 j-b^2 c (c h-14 a j)-4 c^2 \left (2 c^2 f-a c h-4 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{4 \int \frac{3 \left (b^2+4 a c\right )^2 j}{4 c^2 \sqrt{a+b x-c x^2}} \, dx}{3 \left (b^2+4 a c\right )^2}\\ &=-\frac{2 \left (c^3 \left (b f-\frac{3 a^2 (244 c+b j)}{c^2}-\frac{a \left (366 b^2 c+2 c^3 g+b c^2 h+b^3 j\right )}{c^3}\right )-\left (366 b^3 c+b c^2 (1098 a+c g)+b^4 j+b^2 c (c h+4 a j)+2 c^2 \left (c^2 f+a c h+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 (366 b^4 c+8784 a^2 c^3+4 b^2 c^2 (549 a+c g)+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 )+2 c \left (183 b^3 c+4 b c^2 (549 a-c g)+2 b^4 j-b^2 c (c h-14 a j)-4 c^2 \left (2 c^2 f-a c h-4 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 \int \frac{1}{\sqrt{a+b x-c x^2}} \, dx}{c^2}\\ &=-\frac{2 \left (c^3 \left (b f-\frac{3 a^2 (244 c+b j)}{c^2}-\frac{a \left (366 b^2 c+2 c^3 g+b c^2 h+b^3 j\right )}{c^3}\right )-\left (366 b^3 c+b c^2 (1098 a+c g)+b^4 j+b^2 c (c h+4 a j)+2 c^2 \left (c^2 f+a c h+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 (366 b^4 c+8784 a^2 c^3+4 b^2 c^2 (549 a+c g)+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 )+2 c \left (183 b^3 c+4 b c^2 (549 a-c g)+2 b^4 j-b^2 c (c h-14 a j)-4 c^2 \left (2 c^2 f-a c h-4 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{(2 j) \operatorname{Subst}\left (\int \frac{1}{-4 c-x^2} \, dx,x,\frac{b-2 c x}{\sqrt{a+b x-c x^2}}\right )}{c^2}\\ &=-\frac{2 \left (c^3 \left (b f-\frac{3 a^2 (244 c+b j)}{c^2}-\frac{a \left (366 b^2 c+2 c^3 g+b c^2 h+b^3 j\right )}{c^3}\right )-\left (366 b^3 c+b c^2 (1098 a+c g)+b^4 j+b^2 c (c h+4 a j)+2 c^2 \left (c^2 f+a c h+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 (366 b^4 c+8784 a^2 c^3+4 b^2 c^2 (549 a+c g)+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 )+2 c \left (183 b^3 c+4 b c^2 (549 a-c g)+2 b^4 j-b^2 c (c h-14 a j)-4 c^2 \left (2 c^2 f-a c h-4 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 \tan ^{-1}\left (\frac{b-2 c x}{2 \sqrt{c} \sqrt{a+b x-c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 1.17812, size = 319, normalized size = 0.9 \[ -\frac{2 \left (b^3 \left (3 a^2 j+18 a c j x^2+c^2 \left (f+3 g x+x^2 (-(3 h+i x))\right )\right )+2 b^2 c \left (21 a^2 j x+a c \left (g+x \left (-6 h+3 i x-14 j x^2\right )\right )+c^2 x (3 f+x (h x-6 g))\right )+4 b c \left (-2 a^2 c (h-3 i x)+5 a^3 j+3 a c^2 \left (f-x \left (g-h x+i x^2\right )\right )+2 c^3 x^2 (g x-3 f)\right )+8 c^2 \left (-a^2 c \left (g+x^2 (3 i+4 j x)\right )+a^3 (2 i+3 j x)-a c^2 x \left (3 f+h x^2\right )+2 c^3 f x^3\right )+b^4 \left (6 a j x-4 c j x^3\right )+3 b^5 j x^2\right )}{3 c^2 \left (4 a c+b^2\right )^2 (a+x (b-c x))^{3/2}}+\frac{i j \log \left (2 \sqrt{a+x (b-c x)}+\frac{i (b-2 c x)}{\sqrt{c}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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)) + (I*j*Log[(I*(b - 2*c*x))/Sqrt[c] + 2*Sqrt[a + x*(b - c*x)]])/c^(5/2)

________________________________________________________________________________________

Maple [B]  time = 0.059, size = 1453, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*i*b^4/c^3/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)+1/3*i*b^4/c^2/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)+i*b/c*a/(-
4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x-4*j*b^2/c*a/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x+1/6*j*b^5/c^3/(-4*a*c-b^2)
^2/(-c*x^2+b*x+a)^(1/2)-1/3*j*b/c^3*a/(-c*x^2+b*x+a)^(3/2)-1/2*j/c^3*b^3/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(1/2)-2/3
*g*b/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x+1/3*g*b^2/c/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)+2/3*h*a/(-4*a*c-b^2)/(-
c*x^2+b*x+a)^(3/2)*x+8/3*h*a/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*b-4/3*f/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x*c
+32/3*f*c^2/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x-1/4*i*b/c^2*x/(-c*x^2+b*x+a)^(3/2)+1/2*j*b/c^2*x^2/(-c*x^2+b
*x+a)^(3/2)-1/8*j*b^2/c^3*x/(-c*x^2+b*x+a)^(3/2)-1/48*j*b^5/c^4/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)+1/2*j*b^2/c^
2*a/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x+1/2*h*x/c/(-c*x^2+b*x+a)^(3/2)+1/12*h*b/c^2/(-c*x^2+b*x+a)^(3/2)+1/3*j
*x^3/c/(-c*x^2+b*x+a)^(3/2)-1/48*j*b^3/c^4/(-c*x^2+b*x+a)^(3/2)-2/3*h*b^3/c/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2
)+1/12*h*b^3/c^2/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)-j/c^2*x/(-c*x^2+b*x+a)^(1/2)-1/2*j/c^3*b/(-c*x^2+b*x+a)^(1/
2)-2/3*i*a/c^2/(-c*x^2+b*x+a)^(3/2)+i*x^2/c/(-c*x^2+b*x+a)^(3/2)-1/24*i*b^2/c^3/(-c*x^2+b*x+a)^(3/2)+2/3*f/(-4
*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*b-8/3*g*b^2/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)+4/3*h*b^2/(-4*a*c-b^2)^2/(-c*x^
2+b*x+a)^(1/2)*x+j/c^(5/2)*arctan(c^(1/2)*(x-1/2*b/c)/(-c*x^2+b*x+a)^(1/2))+1/3*g/c/(-c*x^2+b*x+a)^(3/2)-16/3*
f*c/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*b+1/24*j*b^4/c^3/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x-1/3*j*b^4/c^2/(-4
*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x-1/4*j*b^3/c^3*a/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)+2*j*b^3/c^2*a/(-4*a*c-b^2
)^2/(-c*x^2+b*x+a)^(1/2)+j/c^2*b^2/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(1/2)*x+1/12*i*b^3/c^2/(-4*a*c-b^2)/(-c*x^2+b*x
+a)^(3/2)*x-2/3*i*b^3/c/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x-1/2*i*b^2/c^2*a/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2
)-8*i*b*a/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x+4*i*b^2/c*a/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)-1/6*h*b^2/c/(-
4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*x-1/3*h*a/c/(-4*a*c-b^2)/(-c*x^2+b*x+a)^(3/2)*b-16/3*h*a*c/(-4*a*c-b^2)^2/(-c*
x^2+b*x+a)^(1/2)*x+16/3*g*b*c/(-4*a*c-b^2)^2/(-c*x^2+b*x+a)^(1/2)*x

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.1967, size = 659, normalized size = 1.87 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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 + 1
4*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))*sqrt(-c) + b))/(sqrt(-c)*c^2)