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

Optimal. Leaf size=543 \[ -\frac{x \left ((1-n) \left (a b e^2-4 a c d e+b c d^2\right )-\frac{b^2 \left (a e^2 (1-3 n)-c d^2 (1-n)\right )+4 a b c d e n+4 a c (1-2 n) \left (c d^2-a e^2\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}-\frac{x \left (\frac{b^2 \left (a e^2 (1-3 n)-c d^2 (1-n)\right )+4 a b c d e n+4 a c (1-2 n) \left (c d^2-a e^2\right )}{\sqrt{b^2-4 a c}}+(1-n) \left (a b e^2-4 a c d e+b c d^2\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{x \left (x^n \left (a b e^2-4 a c d e+b c d^2\right )-2 a b d e-2 a \left (c d^2-a e^2\right )+b^2 d^2\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac{2 e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2} \]

[Out]

(x*(b^2*d^2 - 2*a*b*d*e - 2*a*(c*d^2 - a*e^2) + (b*c*d^2 - 4*a*c*d*e + a*b*e^2)*x^n))/(a*(b^2 - 4*a*c)*n*(a +
b*x^n + c*x^(2*n))) - (2*e^2*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(
b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (((b*c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - n) - (b^2*(a*e^2*(1 - 3*n) - c*d^2
*(1 - n)) + 4*a*c*(c*d^2 - a*e^2)*(1 - 2*n) + 4*a*b*c*d*e*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*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) - (2*e^2*x*Hyper
geometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) -
(((b*c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - n) + (b^2*(a*e^2*(1 - 3*n) - c*d^2*(1 - n)) + 4*a*c*(c*d^2 - a*e^2)*(1
- 2*n) + 4*a*b*c*d*e*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*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*n)

________________________________________________________________________________________

Rubi [A]  time = 1.81832, antiderivative size = 543, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1436, 1430, 1422, 245, 1347} \[ -\frac{x \left ((1-n) \left (a b e^2-4 a c d e+b c d^2\right )-\frac{b^2 \left (a e^2 (1-3 n)-c d^2 (1-n)\right )+4 a b c d e n+4 a c (1-2 n) \left (c d^2-a e^2\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}-\frac{x \left (\frac{b^2 \left (a e^2 (1-3 n)-c d^2 (1-n)\right )+4 a b c d e n+4 a c (1-2 n) \left (c d^2-a e^2\right )}{\sqrt{b^2-4 a c}}+(1-n) \left (a b e^2-4 a c d e+b c d^2\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{x \left (x^n \left (a b e^2-4 a c d e+b c d^2\right )-2 a b d e-2 a \left (c d^2-a e^2\right )+b^2 d^2\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac{2 e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{2 e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b^2*d^2 - 2*a*b*d*e - 2*a*(c*d^2 - a*e^2) + (b*c*d^2 - 4*a*c*d*e + a*b*e^2)*x^n))/(a*(b^2 - 4*a*c)*n*(a +
b*x^n + c*x^(2*n))) - (2*e^2*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(
b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (((b*c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - n) - (b^2*(a*e^2*(1 - 3*n) - c*d^2
*(1 - n)) + 4*a*c*(c*d^2 - a*e^2)*(1 - 2*n) + 4*a*b*c*d*e*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*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) - (2*e^2*x*Hyper
geometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) -
(((b*c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - n) + (b^2*(a*e^2*(1 - 3*n) - c*d^2*(1 - n)) + 4*a*c*(c*d^2 - a*e^2)*(1
- 2*n) + 4*a*b*c*d*e*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*(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])

Rule 1347

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

Rubi steps

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

Mathematica [B]  time = 2.78323, size = 1835, normalized size = 3.38 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^n)^2/(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*d^2 + 2*a^2*e^2 + b*c*d^2*
x^n + a*b*e*(-2*d + e*x^n) - 2*a*c*d*(d + 2*e*x^n))) + 2*b*c^2*Sqrt[b^2 - 4*a*c]*d^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])]) -
8*a*c^2*Sqrt[b^2 - 4*a*c]*d*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])]) + 2*a*b*c*Sqrt[b^2 - 4*a*c]*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*b*c^2*Sqrt[b^2 - 4*a*c]*d^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])]) + 8*a*c^2*Sqrt[b^2 - 4*a*c]*d*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])]
) - 2*a*b*c*Sqrt[b^2 - 4*a*c]*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])]) + 2*b^2*c*Sqrt[b^2 - 4*a*c]*d^2*(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^2*Sqrt[b^2 - 4*a*c]*d^2*(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*b*c*Sqrt[b^2 - 4*a*c]*d*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 + 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^2*
c*Sqrt[b^2 - 4*a*c]*e^2*(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*Sqrt[b^2 - 4*a*c]*d^2*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 - S
qrt[b^2 - 4*a*c])*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 8*a*c^2*Sqrt
[b^2 - 4*a*c]*d^2*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*(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 = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d+e{x}^{n} \right ) ^{2}}{ \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)^2/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((d+e*x^n)^2/(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 d^{2} -{\left (4 \, c d e - b e^{2}\right )} a\right )} x x^{n} +{\left (b^{2} d^{2} + 2 \, a^{2} e^{2} - 2 \,{\left (c d^{2} + b d e\right )} a\right )} x}{a^{2} b^{2} n - 4 \, a^{3} c n +{\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} +{\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}} - \int -\frac{b^{2} d^{2}{\left (n - 1\right )} - 2 \, a^{2} e^{2} - 2 \,{\left (c d^{2}{\left (2 \, n - 1\right )} - b d e\right )} a +{\left (b c d^{2}{\left (n - 1\right )} -{\left (4 \, c d e{\left (n - 1\right )} - b e^{2}{\left (n - 1\right )}\right )} a\right )} x^{n}}{a^{2} b^{2} n - 4 \, a^{3} c n +{\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} +{\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral((e^2*x^(2*n) + 2*d*e*x^n + d^2)/(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)**2/(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 )}^{2}}{{\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)^2/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

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