3.75 \(\int \frac{(d+e x^n)^3}{(a+b x^n+c x^{2 n})^2} \, dx\)

Optimal. Leaf size=750 \[ \frac{x \left (\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt{b^2-4 a c}}+(1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{x \left ((1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )-\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{x \left (x^n \left (-\left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )\right )-a b e \left (a e^2+3 c d^2\right )-2 a c d \left (c d^2-3 a e^2\right )+b^2 c d^3\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac{e^2 x \left (\frac{6 c d-3 b e}{\sqrt{b^2-4 a c}}+e\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{c \left (b-\sqrt{b^2-4 a c}\right )}+\frac{e^2 x \left (e-\frac{3 (2 c d-b e)}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{c \left (\sqrt{b^2-4 a c}+b\right )} \]

[Out]

(x*(b^2*c*d^3 - 2*a*c*d*(c*d^2 - 3*a*e^2) - a*b*e*(3*c*d^2 + a*e^2) - (a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) -
b*c*d*(c*d^2 + 3*a*e^2))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e^2*(e + (6*c*d - 3*b*e)/Sqrt
[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(c*(b - Sqrt[b^
2 - 4*a*c])) + (((a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) + (b^2*c*d*(3*a*e^2
*(1 - 3*n) - c*d^2*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(
2 - 5*n) + 3*c*d^2*n))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2
- 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) + (e^2*(e - (3*(2*c*d - b*e))/Sqrt[b^2 - 4*a*c])*x*H
ypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(c*(b + Sqrt[b^2 - 4*a*c])) + (((
a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) - (b^2*c*d*(3*a*e^2*(1 - 3*n) - c*d^2
*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(2 - 5*n) + 3*c*d^2
*n))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*c*(
b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*n)

________________________________________________________________________________________

Rubi [A]  time = 2.93827, antiderivative size = 750, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1436, 1430, 1422, 245} \[ \frac{x \left (\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt{b^2-4 a c}}+(1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{x \left ((1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )-\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{x \left (x^n \left (-\left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )\right )-a b e \left (a e^2+3 c d^2\right )-2 a c d \left (c d^2-3 a e^2\right )+b^2 c d^3\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac{e^2 x \left (\frac{6 c d-3 b e}{\sqrt{b^2-4 a c}}+e\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{c \left (b-\sqrt{b^2-4 a c}\right )}+\frac{e^2 x \left (e-\frac{3 (2 c d-b e)}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{c \left (\sqrt{b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)^3/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

(x*(b^2*c*d^3 - 2*a*c*d*(c*d^2 - 3*a*e^2) - a*b*e*(3*c*d^2 + a*e^2) - (a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) -
b*c*d*(c*d^2 + 3*a*e^2))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e^2*(e + (6*c*d - 3*b*e)/Sqrt
[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(c*(b - Sqrt[b^
2 - 4*a*c])) + (((a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) + (b^2*c*d*(3*a*e^2
*(1 - 3*n) - c*d^2*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(
2 - 5*n) + 3*c*d^2*n))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2
- 4*a*c])])/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) + (e^2*(e - (3*(2*c*d - b*e))/Sqrt[b^2 - 4*a*c])*x*H
ypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(c*(b + Sqrt[b^2 - 4*a*c])) + (((
a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) - (b^2*c*d*(3*a*e^2*(1 - 3*n) - c*d^2
*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(2 - 5*n) + 3*c*d^2
*n))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*c*(
b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*n)

Rule 1436

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandInt
egrand[(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ((IntegersQ[p, q] &&  !IntegerQ[n]) || IGtQ[p, 0] ||
(IGtQ[q, 0] &&  !IntegerQ[n]))

Rule 1430

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> -Simp[(x*(d*b^2 -
a*b*e - 2*a*c*d + (b*d - 2*a*e)*c*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*n*(p + 1)*(b^2 - 4*a*c)), x] + Dist
[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[Simp[(n*p + n + 1)*d*b^2 - a*b*e - 2*a*c*d*(2*n*p + 2*n + 1) + (2*n*p + 3*
n + 1)*(d*b - 2*a*e)*c*x^n, x]*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[
n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\int \left (\frac{c^2 d^3-3 a c d e^2+a b e^3+\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3\right ) x^n}{c^2 \left (a+b x^n+c x^{2 n}\right )^2}+\frac{e^2 \left (3 c d-b e+c e x^n\right )}{c^2 \left (a+b x^n+c x^{2 n}\right )}\right ) \, dx\\ &=\frac{\int \frac{c^2 d^3-3 a c d e^2+a b e^3+\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3\right ) x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{c^2}+\frac{e^2 \int \frac{3 c d-b e+c e x^n}{a+b x^n+c x^{2 n}} \, dx}{c^2}\\ &=\frac{x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac{\left (e^2 \left (e+\frac{6 c d-3 b e}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^n} \, dx}{2 c}+\frac{\left (e^2 \left (e-\frac{3 (2 c d-b e)}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^n} \, dx}{2 c}-\frac{\int \frac{-a b c e \left (3 c d^2+a e^2 (1-4 n)\right )-2 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)-a b^3 e^3 n+b^2 c d \left (c d^2 (1-n)+3 a e^2 n\right )-c \left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n) x^n}{a+b x^n+c x^{2 n}} \, dx}{a c^2 \left (b^2-4 a c\right ) n}\\ &=\frac{x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac{e^2 \left (e+\frac{6 c d-3 b e}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{c \left (b-\sqrt{b^2-4 a c}\right )}+\frac{e^2 \left (e-\frac{3 (2 c d-b e)}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{c \left (b+\sqrt{b^2-4 a c}\right )}+\frac{\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)-\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n}+\frac{\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)+\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n}\\ &=\frac{x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac{e^2 \left (e+\frac{6 c d-3 b e}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{c \left (b-\sqrt{b^2-4 a c}\right )}+\frac{\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)+\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right ) n}+\frac{e^2 \left (e-\frac{3 (2 c d-b e)}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{c \left (b+\sqrt{b^2-4 a c}\right )}+\frac{\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)-\frac{b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt{b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b+\sqrt{b^2-4 a c}\right ) n}\\ \end{align*}

Mathematica [B]  time = 4.67146, size = 2462, normalized size = 3.28 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^n)^3/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

-((x*((b^2 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])*(b + Sqrt[b^2 - 4*a*c])*(1 + n)*(b^2*(-(c*d^3) + a*e^3*x^n) + b*(
a^2*e^3 - c^2*d^3*x^n + 3*a*c*d*e*(d - e*x^n)) - 2*a*c*(a*e^2*(3*d + e*x^n) - c*d^2*(d + 3*e*x^n))) + 2*b*c^3*
Sqrt[b^2 - 4*a*c]*d^3*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2
+ n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 +
n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 12*a*c^3*Sqrt[b^2 - 4*a*c]*d^2*e*x^n*(a + x^n*(b + c*x^n))*(-((
b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b
- Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 6*a*b
*c^2*Sqrt[b^2 - 4*a*c]*d*e^2*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(
-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1
), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 2*a*b^2*c*Sqrt[b^2 - 4*a*c]*e^3*x^n*(a + x^n*(b + c*x^n)
)*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])
+ (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) +
4*a^2*c^2*Sqrt[b^2 - 4*a*c]*e^3*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 +
n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n
^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 2*b*c^3*Sqrt[b^2 - 4*a*c]*d^3*n*x^n*(a + x^n*(b + c*
x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c
])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])
]) + 12*a*c^3*Sqrt[b^2 - 4*a*c]*d^2*e*n*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1
[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1
, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 6*a*b*c^2*Sqrt[b^2 - 4*a*c]*d*e^2*n*x^n*(a +
x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[
b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^
2 - 4*a*c])]) + 4*a^2*c^2*Sqrt[b^2 - 4*a*c]*e^3*n*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hyperge
ometric2F1[1, 1 + n^(-1), 2 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeom
etric2F1[1, 1 + n^(-1), 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 2*b^2*c^2*Sqrt[b^2 - 4*a*c]*d^3*(1
+ n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b +
Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[
b^2 - 4*a*c])]) - 4*a*c^3*Sqrt[b^2 - 4*a*c]*d^3*(1 + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hyper
geometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeomet
ric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 6*a*b*c^2*Sqrt[b^2 - 4*a*c]*d^2*e*(1 + n)
*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqr
t[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2
- 4*a*c])]) + 12*a^2*c^2*Sqrt[b^2 - 4*a*c]*d*e^2*(1 + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hype
rgeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeome
tric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 2*a^2*b*c*Sqrt[b^2 - 4*a*c]*e^3*(1 + n)*
(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt
[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 -
4*a*c])]) - 2*b^2*c^2*Sqrt[b^2 - 4*a*c]*d^3*n*(1 + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hyperg
eometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometr
ic2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 8*a*c^3*Sqrt[b^2 - 4*a*c]*d^3*n*(1 + n)*(a
+ x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt[b
^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4
*a*c])])))/(a*c*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c])*(b + Sqrt[b^2 - 4*a*c])*n*(1 + n)*(a + x^n*(b + c*x^n
))))

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d+e{x}^{n} \right ) ^{3}}{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x^n)^3/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((d+e*x^n)^3/(a+b*x^n+c*x^(2*n))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b c^{2} d^{3} + 2 \, a^{2} c e^{3} -{\left (6 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}\right )} a\right )} x x^{n} +{\left (b^{2} c d^{3} +{\left (6 \, c d e^{2} - b e^{3}\right )} a^{2} -{\left (2 \, c^{2} d^{3} + 3 \, b c d^{2} e\right )} a\right )} x}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n +{\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} +{\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}} + \int \frac{b^{2} c d^{3}{\left (n - 1\right )} -{\left (6 \, c d e^{2} - b e^{3}\right )} a^{2} -{\left (2 \, c^{2} d^{3}{\left (2 \, n - 1\right )} - 3 \, b c d^{2} e\right )} a -{\left (2 \, a^{2} c e^{3}{\left (n + 1\right )} - b c^{2} d^{3}{\left (n - 1\right )} +{\left (6 \, c^{2} d^{2} e{\left (n - 1\right )} - 3 \, b c d e^{2}{\left (n - 1\right )} - b^{2} e^{3}\right )} a\right )} x^{n}}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n +{\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} +{\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

((b*c^2*d^3 + 2*a^2*c*e^3 - (6*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*a)*x*x^n + (b^2*c*d^3 + (6*c*d*e^2 - b*e^3)*
a^2 - (2*c^2*d^3 + 3*b*c*d^2*e)*a)*x)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^
3*c*n - 4*a^2*b*c^2*n)*x^n) + integrate((b^2*c*d^3*(n - 1) - (6*c*d*e^2 - b*e^3)*a^2 - (2*c^2*d^3*(2*n - 1) -
3*b*c*d^2*e)*a - (2*a^2*c*e^3*(n + 1) - b*c^2*d^3*(n - 1) + (6*c^2*d^2*e*(n - 1) - 3*b*c*d*e^2*(n - 1) - b^2*e
^3)*a)*x^n)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a^2*b*c^2*n)*x^n
), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{e^{3} x^{3 \, n} + 3 \, d e^{2} x^{2 \, n} + 3 \, d^{2} e x^{n} + d^{3}}{c^{2} x^{4 \, n} + b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 2 \,{\left (b c x^{n} + a c\right )} x^{2 \, n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*
(b*c*x^n + a*c)*x^(2*n)), x)

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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((d+e*x**n)**3/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x^n)^3/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

integrate((e*x^n + d)^3/(c*x^(2*n) + b*x^n + a)^2, x)