3.74 \(\int \frac{1}{(d+e x^n)^3 (a+b x^n+c x^{2 n})} \, dx\)

Optimal. Leaf size=552 \[ \frac{e^2 x \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^3}-\frac{c x \left (-3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{\left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}-\frac{c x \left (-3 c^2 d e \left (-d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{\left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac{e^2 x (2 c d-b e) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^2 \left (a e^2-b d e+c d^2\right )^2}+\frac{e^2 x \, _2F_1\left (3,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^3 \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

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Rubi [A]  time = 1.02335, antiderivative size = 552, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1424, 245, 1422} \[ \frac{e^2 x \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^3}-\frac{c x \left (-3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{\left (-b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}-\frac{c x \left (-3 c^2 d e \left (-d \sqrt{b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{\left (b \sqrt{b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac{e^2 x (2 c d-b e) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^2 \left (a e^2-b d e+c d^2\right )^2}+\frac{e^2 x \, _2F_1\left (3,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^3 \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

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

Rule 1424

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)^q/(a + b*x^n + c*x^(2*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] && IntegerQ[q]

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

Mathematica [A]  time = 1.87317, size = 509, normalized size = 0.92 \[ \frac{x \left (\frac{e^2 \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d}+\frac{c \left (3 c^2 d e \left (d \sqrt{b^2-4 a c}+2 a e+b d\right )-c e^2 \left (3 b \left (d \sqrt{b^2-4 a c}+a e\right )+a e \sqrt{b^2-4 a c}+3 b^2 d\right )+b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )-2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};\frac{2 c x^n}{\sqrt{b^2-4 a c}-b}\right )}{-b \sqrt{b^2-4 a c}-4 a c+b^2}-\frac{c \left (3 c^2 d e \left (d \sqrt{b^2-4 a c}-2 a e-b d\right )+c e^2 \left (-3 b d \sqrt{b^2-4 a c}-a e \sqrt{b^2-4 a c}+3 a b e+3 b^2 d\right )+b^2 e^3 \left (\sqrt{b^2-4 a c}-b\right )+2 c^3 d^3\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{b \sqrt{b^2-4 a c}-4 a c+b^2}+\frac{e^2 (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^2}+\frac{e^2 \left (e (a e-b d)+c d^2\right )^2 \, _2F_1\left (3,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d^3}\right )}{\left (e (a e-b d)+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

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

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Maple [F]  time = 0.163, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d+e{x}^{n} \right ) ^{3} \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

((12*n^2 - 7*n + 1)*c^2*d^4*e^2 - 2*(8*n^2 - 6*n + 1)*b*c*d^3*e^3 + (6*n^2 - 5*n + 1)*b^2*d^2*e^4 + (2*n^2 - 3
*n + 1)*a^2*e^6 + 2*((3*n^2 - 5*n + 1)*c*d^2*e^4 - (3*n^2 - 4*n + 1)*b*d*e^5)*a)*integrate(1/2/(c^3*d^9*n^2 -
3*b*c^2*d^8*e*n^2 + 3*b^2*c*d^7*e^2*n^2 - b^3*d^6*e^3*n^2 + a^3*d^3*e^6*n^2 + 3*(c*d^5*e^4*n^2 - b*d^4*e^5*n^2
)*a^2 + 3*(c^2*d^7*e^2*n^2 - 2*b*c*d^6*e^3*n^2 + b^2*d^5*e^4*n^2)*a + (c^3*d^8*e*n^2 - 3*b*c^2*d^7*e^2*n^2 + 3
*b^2*c*d^6*e^3*n^2 - b^3*d^5*e^4*n^2 + a^3*d^2*e^7*n^2 + 3*(c*d^4*e^5*n^2 - b*d^3*e^6*n^2)*a^2 + 3*(c^2*d^6*e^
3*n^2 - 2*b*c*d^5*e^4*n^2 + b^2*d^4*e^5*n^2)*a)*x^n), x) + 1/2*((c*d^2*e^3*(6*n - 1) - b*d*e^4*(4*n - 1) + a*e
^5*(2*n - 1))*x*x^n + (c*d^3*e^2*(7*n - 1) - b*d^2*e^3*(5*n - 1) + a*d*e^4*(3*n - 1))*x)/(c^2*d^8*n^2 - 2*b*c*
d^7*e*n^2 + b^2*d^6*e^2*n^2 + a^2*d^4*e^4*n^2 + 2*(c*d^6*e^2*n^2 - b*d^5*e^3*n^2)*a + (c^2*d^6*e^2*n^2 - 2*b*c
*d^5*e^3*n^2 + b^2*d^4*e^4*n^2 + a^2*d^2*e^6*n^2 + 2*(c*d^4*e^4*n^2 - b*d^3*e^5*n^2)*a)*x^(2*n) + 2*(c^2*d^7*e
*n^2 - 2*b*c*d^6*e^2*n^2 + b^2*d^5*e^3*n^2 + a^2*d^3*e^5*n^2 + 2*(c*d^5*e^3*n^2 - b*d^4*e^4*n^2)*a)*x^n) + int
egrate((c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3 - (3*c^2*d*e^2 - 2*b*c*e^3)*a - (3*c^3*d^2*e - 3*b*c
^2*d*e^2 + b^2*c*e^3 - a*c^2*e^3)*x^n)/(a^4*e^6 + 3*(c*d^2*e^4 - b*d*e^5)*a^3 + 3*(c^2*d^4*e^2 - 2*b*c*d^3*e^3
+ b^2*d^2*e^4)*a^2 + (c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*a + (c^4*d^6 - 3*b*c^3*d^5*e +
3*b^2*c^2*d^4*e^2 - b^3*c*d^3*e^3 + a^3*c*e^6 + 3*(c^2*d^2*e^4 - b*c*d*e^5)*a^2 + 3*(c^3*d^4*e^2 - 2*b*c^2*d^
3*e^3 + b^2*c*d^2*e^4)*a)*x^(2*n) + (b*c^3*d^6 - 3*b^2*c^2*d^5*e + 3*b^3*c*d^4*e^2 - b^4*d^3*e^3 + a^3*b*e^6 +
3*(b*c*d^2*e^4 - b^2*d*e^5)*a^2 + 3*(b*c^2*d^4*e^2 - 2*b^2*c*d^3*e^3 + b^3*d^2*e^4)*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 e^{3} x^{4 \, n} + a d^{3} +{\left (3 \, b d e^{2} + a e^{3}\right )} x^{3 \, n} +{\left (c e^{3} x^{3 \, n} + 3 \, c d e^{2} x^{2 \, n} + 3 \, c d^{2} e x^{n} + c d^{3}\right )} x^{2 \, n} + 3 \,{\left (b d^{2} e + a d e^{2}\right )} x^{2 \, n} +{\left (b d^{3} + 3 \, a d^{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)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

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