3.78 \(\int \frac{1}{(d+e x^n) (a+b x^n+c x^{2 n})^2} \, dx\)

Optimal. Leaf size=726 \[ -\frac{c x \left (\frac{2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)+b^3 (-e) (1-n)}{\sqrt{b^2-4 a c}}+(1-n) \left (2 a c e+b^2 (-e)+b c d\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right ) \left (a e^2-b d e+c d^2\right )}+\frac{x \left (c x^n \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d+b^2 c d-b^3 e\right )}{a n \left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac{c e^2 x \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{\left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac{c e^2 x \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{\left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac{c x \left (b^2 (1-n) \left (e \sqrt{b^2-4 a c}+c d\right )+b c \left (2 a e (2-3 n)-d (1-n) \sqrt{b^2-4 a c}\right )-2 a c \left (e (1-n) \sqrt{b^2-4 a c}+2 c d (1-2 n)\right )+b^3 (-e) (1-n)\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^4 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^2} \]

[Out]

(x*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e + c*(b*c*d - b^2*e + 2*a*c*e)*x^n))/(a*(b^2 - 4*a*c)*(c*d^2 - b*d*
e + a*e^2)*n*(a + b*x^n + c*x^(2*n))) - (c*e^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*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])*(c*d^2 - b*d*e + a*e
^2)^2) - (c*((2*a*b*c*e*(2 - 3*n) - 4*a*c^2*d*(1 - 2*n) + b^2*c*d*(1 - n) - b^3*e*(1 - n))/Sqrt[b^2 - 4*a*c] +
(b*c*d - b^2*e + 2*a*c*e)*(1 - n))*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])*(c*d^2 - b*d*e + a*e^2)*n) - (c*e^2*(2*c*d - (b - Sqrt[b^2 - 4*
a*c])*e)*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*Sqr
t[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^2) + (c*(b*c*(2*a*e*(2 - 3*n) - Sqrt[b^2 - 4*a*c]*d*(1 - n)) - 2*a*c*(
2*c*d*(1 - 2*n) + Sqrt[b^2 - 4*a*c]*e*(1 - n)) - b^3*e*(1 - n) + b^2*(c*d + Sqrt[b^2 - 4*a*c]*e)*(1 - n))*x*Hy
pergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*
Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)*n) + (e^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])
/(d*(c*d^2 - b*d*e + a*e^2)^2)

________________________________________________________________________________________

Rubi [A]  time = 1.92679, antiderivative size = 726, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1436, 245, 1430, 1422} \[ -\frac{c x \left (\frac{2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)+b^3 (-e) (1-n)}{\sqrt{b^2-4 a c}}+(1-n) \left (2 a c e+b^2 (-e)+b c d\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right ) \left (a e^2-b d e+c d^2\right )}+\frac{x \left (c x^n \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d+b^2 c d-b^3 e\right )}{a n \left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac{c e^2 x \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{\left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac{c e^2 x \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{\left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac{c x \left (b^2 (1-n) \left (e \sqrt{b^2-4 a c}+c d\right )+b c \left (2 a e (2-3 n)-d (1-n) \sqrt{b^2-4 a c}\right )-2 a c \left (e (1-n) \sqrt{b^2-4 a c}+2 c d (1-2 n)\right )+b^3 (-e) (1-n)\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac{e^4 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]

[Out]

(x*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e + c*(b*c*d - b^2*e + 2*a*c*e)*x^n))/(a*(b^2 - 4*a*c)*(c*d^2 - b*d*
e + a*e^2)*n*(a + b*x^n + c*x^(2*n))) - (c*e^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*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])*(c*d^2 - b*d*e + a*e
^2)^2) - (c*((2*a*b*c*e*(2 - 3*n) - 4*a*c^2*d*(1 - 2*n) + b^2*c*d*(1 - n) - b^3*e*(1 - n))/Sqrt[b^2 - 4*a*c] +
(b*c*d - b^2*e + 2*a*c*e)*(1 - n))*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])*(c*d^2 - b*d*e + a*e^2)*n) - (c*e^2*(2*c*d - (b - Sqrt[b^2 - 4*
a*c])*e)*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*Sqr
t[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^2) + (c*(b*c*(2*a*e*(2 - 3*n) - Sqrt[b^2 - 4*a*c]*d*(1 - n)) - 2*a*c*(
2*c*d*(1 - 2*n) + Sqrt[b^2 - 4*a*c]*e*(1 - n)) - b^3*e*(1 - n) + b^2*(c*d + Sqrt[b^2 - 4*a*c]*e)*(1 - n))*x*Hy
pergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*
Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)*n) + (e^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])
/(d*(c*d^2 - b*d*e + a*e^2)^2)

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 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 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])

Rubi steps

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

Mathematica [B]  time = 7.27299, size = 8045, normalized size = 11.08 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]

[Out]

Result too large to show

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d+e{x}^{n} \right ) \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(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

e^4*integrate(1/(c^2*d^5 - 2*b*c*d^4*e + b^2*d^3*e^2 + a^2*d*e^4 + 2*(c*d^3*e^2 - b*d^2*e^3)*a + (c^2*d^4*e -
2*b*c*d^3*e^2 + b^2*d^2*e^3 + a^2*e^5 + 2*(c*d^2*e^3 - b*d*e^4)*a)*x^n), x) - ((b*c^2*d - b^2*c*e + 2*a*c^2*e)
*x*x^n + (b^2*c*d - b^3*e - (2*c^2*d - 3*b*c*e)*a)*x)/(4*a^4*c*e^2*n + (4*c^2*d^2*n - 4*b*c*d*e*n - b^2*e^2*n)
*a^3 - (b^2*c*d^2*n - b^3*d*e*n)*a^2 + (4*a^3*c^2*e^2*n + (4*c^3*d^2*n - 4*b*c^2*d*e*n - b^2*c*e^2*n)*a^2 - (b
^2*c^2*d^2*n - b^3*c*d*e*n)*a)*x^(2*n) + (4*a^3*b*c*e^2*n + (4*b*c^2*d^2*n - 4*b^2*c*d*e*n - b^3*e^2*n)*a^2 -
(b^3*c*d^2*n - b^4*d*e*n)*a)*x^n) - integrate((b^2*c^2*d^3*(n - 1) - 2*b^3*c*d^2*e*(n - 1) + b^4*d*e^2*(n - 1)
+ (b*c*e^3*(8*n - 3) - 2*c^2*d*e^2*(4*n - 1))*a^2 + (b*c^2*d^2*e*(8*n - 5) - 2*c^3*d^3*(2*n - 1) - b^3*e^3*(2
*n - 1) - 2*b^2*c*d*e^2*(n - 1))*a + (2*a^2*c^2*e^3*(3*n - 1) + b*c^3*d^3*(n - 1) - 2*b^2*c^2*d^2*e*(n - 1) +
b^3*c*d*e^2*(n - 1) - (b^2*c*e^3*(2*n - 1) - 2*c^3*d^2*e*(n - 1) + b*c^2*d*e^2*(n - 1))*a)*x^n)/(4*a^5*c*e^4*n
+ (8*c^2*d^2*e^2*n - 8*b*c*d*e^3*n - b^2*e^4*n)*a^4 + 2*(2*c^3*d^4*n - 4*b*c^2*d^3*e*n + b^2*c*d^2*e^2*n + b^
3*d*e^3*n)*a^3 - (b^2*c^2*d^4*n - 2*b^3*c*d^3*e*n + b^4*d^2*e^2*n)*a^2 + (4*a^4*c^2*e^4*n + (8*c^3*d^2*e^2*n -
8*b*c^2*d*e^3*n - b^2*c*e^4*n)*a^3 + 2*(2*c^4*d^4*n - 4*b*c^3*d^3*e*n + b^2*c^2*d^2*e^2*n + b^3*c*d*e^3*n)*a^
2 - (b^2*c^3*d^4*n - 2*b^3*c^2*d^3*e*n + b^4*c*d^2*e^2*n)*a)*x^(2*n) + (4*a^4*b*c*e^4*n + (8*b*c^2*d^2*e^2*n -
8*b^2*c*d*e^3*n - b^3*e^4*n)*a^3 + 2*(2*b*c^3*d^4*n - 4*b^2*c^2*d^3*e*n + b^3*c*d^2*e^2*n + b^4*d*e^3*n)*a^2
- (b^3*c^2*d^4*n - 2*b^4*c*d^3*e*n + b^5*d^2*e^2*n)*a)*x^n), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

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