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

Optimal. Leaf size=354 \[ -\frac{2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i+2 b^2 c^2 (2 c g-3 a i)-b^3 c (c h-10 a j)+b^4 c i+b^5 (-j)\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )+a b^2 c i-a b^3 j+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{j \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}} \]

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

________________________________________________________________________________________

Rubi [A]  time = 0.377244, antiderivative size = 354, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {1660, 12, 621, 206} \[ -\frac{2 \left (-c x \left (2 c^2 \left (-16 a^2 j-6 a b i+b^2 h\right )+b^2 c (28 a j+b i)-c^3 (8 b g-8 a h)-4 b^4 j+16 c^4 f\right )-4 b c^2 \left (8 a^2 j+a c h+2 c^2 f\right )+24 a^2 c^3 i+2 b^2 c^2 (2 c g-3 a i)-b^3 c (c h-10 a j)+b^4 c i+b^5 (-j)\right )}{3 c^3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}+\frac{2 \left (-x \left (c^2 \left (2 a^2 j+3 a b i+b^2 h\right )-b^2 c (4 a j+b i)-c^3 (2 a h+b g)+b^4 j+2 c^4 f\right )-b c \left (-3 a^2 j+a c h+c^2 f\right )+a b^2 c i-a b^3 j+2 a c^2 (c g-a i)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac{j \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}} \]

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 - 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))/(3*c^3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2]) + (j*ArcTanh[(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 206

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

Rubi steps

\begin{align*} \int \frac{f+g x+h x^2+365 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{a^2 (730 c-3 b j)}{c^2}-\frac{a \left (365 b^2 c+2 c^3 g-b c^2 h-b^3 j\right )}{c^3}\right )-\left (365 b^3 c-b c^2 (1095 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{365 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 ) (365 c-b j) x}{2 c^2}+\frac{3}{2} \left (4 a-\frac{b^2}{c}\right ) j x^2}{\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{a^2 (730 c-3 b j)}{c^2}-\frac{a \left (365 b^2 c+2 c^3 g-b c^2 h-b^3 j\right )}{c^3}\right )-\left (365 b^3 c-b c^2 (1095 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 (365 b^4 c+8760 a^2 c^3-b^2 \left (2190 a c^2-4 c^3 g\right )-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 (365 b^3 c-4 b c^2 (1095 a+2 c g)-4 b^4 j+2 b^2 c (c h+14 a j)+8 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{a^2 (730 c-3 b j)}{c^2}-\frac{a \left (365 b^2 c+2 c^3 g-b c^2 h-b^3 j\right )}{c^3}\right )-\left (365 b^3 c-b c^2 (1095 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 (365 b^4 c+8760 a^2 c^3-b^2 \left (2190 a c^2-4 c^3 g\right )-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 (365 b^3 c-4 b c^2 (1095 a+2 c g)-4 b^4 j+2 b^2 c (c h+14 a j)+8 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{a^2 (730 c-3 b j)}{c^2}-\frac{a \left (365 b^2 c+2 c^3 g-b c^2 h-b^3 j\right )}{c^3}\right )-\left (365 b^3 c-b c^2 (1095 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 (365 b^4 c+8760 a^2 c^3-b^2 \left (2190 a c^2-4 c^3 g\right )-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 (365 b^3 c-4 b c^2 (1095 a+2 c g)-4 b^4 j+2 b^2 c (c h+14 a j)+8 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{a^2 (730 c-3 b j)}{c^2}-\frac{a \left (365 b^2 c+2 c^3 g-b c^2 h-b^3 j\right )}{c^3}\right )-\left (365 b^3 c-b c^2 (1095 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 (365 b^4 c+8760 a^2 c^3-b^2 \left (2190 a c^2-4 c^3 g\right )-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 (365 b^3 c-4 b c^2 (1095 a+2 c g)-4 b^4 j+2 b^2 c (c h+14 a j)+8 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 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.23512, size = 316, normalized size = 0.89 \[ \frac{-\frac{2 \left (b c \left (-3 a^2 j+a c (h+3 i x)+c^2 (f-g x)\right )+2 c^2 \left (a^2 (i+j x)-a c (g+h x)+c^2 f x\right )+b^2 c (c h x-a (i+4 j x))+b^3 (a j-c i x)+b^4 j x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))^{3/2}}+\frac{2 \left (4 b c^2 \left (8 a^2 j+a c (h-3 i x)+2 c^2 (f-g x)\right )+8 c^3 \left (a^2 (-(3 i+4 j x))+a c h x+2 c^2 f x\right )+2 b^2 c^2 (3 a i+14 a j x-2 c g+c h x)+b^3 c (c (h+i x)-10 a j)-b^4 c (i+4 j x)+b^5 j\right )}{\left (b^2-4 a c\right )^2 \sqrt{a+x (b+c x)}}+3 \sqrt{c} j \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{3 c^3} \]

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*(b^4*j*x + b^3*(a*j - c*i*x) + b*c*(-3*a^2*j + c^2*(f - g*x) + a*c*(h + 3*i*x)) + 2*c^2*(c^2*f*x - a*c*(g
+ h*x) + a^2*(i + j*x)) + b^2*c*(c*h*x - a*(i + 4*j*x))))/((b^2 - 4*a*c)*(a + x*(b + c*x))^(3/2)) + (2*(b^5*j
- b^4*c*(i + 4*j*x) + 2*b^2*c^2*(-2*c*g + 3*a*i + c*h*x + 14*a*j*x) + 4*b*c^2*(8*a^2*j + 2*c^2*(f - g*x) + a*
c*(h - 3*i*x)) + b^3*c*(-10*a*j + c*(h + i*x)) + 8*c^3*(2*c^2*f*x + a*c*h*x - a^2*(3*i + 4*j*x))))/((b^2 - 4*a
*c)^2*Sqrt[a + x*(b + c*x)]) + 3*Sqrt[c]*j*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(3*c^3)

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Maple [B]  time = 0.056, size = 1406, normalized size = 4. \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]

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

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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

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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

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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

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Giac [A]  time = 1.16615, size = 628, normalized size = 1.77 \begin{align*} \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}}} \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*((((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)