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

Optimal. Leaf size=333 \[ \frac{c e (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n (n+1) \left (a e^2+c d^2\right )}-\frac{c d (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n \left (a e^2+c d^2\right )}-\frac{c e^3 x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )^2}+\frac{c d e^2 x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )^2}+\frac{e^4 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2+c d^2\right )^2}+\frac{c x \left (d-e x^n\right )}{2 a n \left (a e^2+c d^2\right ) \left (a+c x^{2 n}\right )} \]

[Out]

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

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Rubi [A]  time = 0.224394, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {1437, 245, 1431, 1418, 364} \[ \frac{c e (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n (n+1) \left (a e^2+c d^2\right )}-\frac{c d (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n \left (a e^2+c d^2\right )}-\frac{c e^3 x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )^2}+\frac{c d e^2 x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )^2}+\frac{e^4 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2+c d^2\right )^2}+\frac{c x \left (d-e x^n\right )}{2 a n \left (a e^2+c d^2\right ) \left (a+c x^{2 n}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1437

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)
^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + 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 1431

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

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Dist[d, Int[1/(a + c*x^(2*n)), x], x] + D
ist[e, Int[x^n/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &
& (PosQ[a*c] ||  !IntegerQ[n])

Rule 364

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

Rubi steps

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

Mathematica [A]  time = 0.290809, size = 227, normalized size = 0.68 \[ \frac{x \left (a^2 e^4 (n+1) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )+a c d^2 e^2 (n+1) \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )+c d \left (\left (a e^2+c d^2\right ) \left (d (n+1) \, _2F_1\left (2,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )-e x^n \, _2F_1\left (2,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )\right )-a e^3 x^n \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )\right )\right )}{a^2 d (n+1) \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*c*d^2*e^2*(1 + n)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)] + a^2*e^4*(1 + n)*Hype
rgeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)] + c*d*(-(a*e^3*x^n*Hypergeometric2F1[1, (1 + n)/(2*n), (3 +
n^(-1))/2, -((c*x^(2*n))/a)]) + (c*d^2 + a*e^2)*(d*(1 + n)*Hypergeometric2F1[2, 1/(2*n), (2 + n^(-1))/2, -((c
*x^(2*n))/a)] - e*x^n*Hypergeometric2F1[2, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)]))))/(a^2*d*(c*d^2
+ a*e^2)^2*(1 + n))

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Maple [F]  time = 0.139, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d+e{x}^{n} \right ) \left ( a+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+c*x^(2*n))^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{4} \int \frac{1}{c^{2} d^{5} + 2 \, a c d^{3} e^{2} + a^{2} d e^{4} +{\left (c^{2} d^{4} e + 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x^{n}}\,{d x} - \frac{c e x x^{n} - c d x}{2 \,{\left (a^{2} c d^{2} n + a^{3} e^{2} n +{\left (a c^{2} d^{2} n + a^{2} c e^{2} n\right )} x^{2 \, n}\right )}} - \int -\frac{a c d e^{2}{\left (4 \, n - 1\right )} + c^{2} d^{3}{\left (2 \, n - 1\right )} -{\left (a c e^{3}{\left (3 \, n - 1\right )} + c^{2} d^{2} e{\left (n - 1\right )}\right )} x^{n}}{2 \,{\left (a^{2} c^{2} d^{4} n + 2 \, a^{3} c d^{2} e^{2} n + a^{4} e^{4} n +{\left (a c^{3} d^{4} n + 2 \, a^{2} c^{2} d^{2} e^{2} n + a^{3} c e^{4} n\right )} x^{2 \, n}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/(a^2*e*x^n + a^2*d + (c^2*e*x^n + c^2*d)*x^(4*n) + 2*(a*c*e*x^n + a*c*d)*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(1/(d+e*x**n)/(a+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} + 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+c*x^(2*n))^2,x, algorithm="giac")

[Out]

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