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

Optimal. Leaf size=152 \[ -\frac{c e x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )}+\frac{c d x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )}+\frac{e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2+c d^2\right )} \]

[Out]

(c*d*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*(c*d^2 + a*e^2)) + (e^2*x*Hypergeom
etric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d*(c*d^2 + a*e^2)) - (c*e*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)*(1 + n))

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Rubi [A]  time = 0.110592, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {1425, 245, 1418, 364} \[ -\frac{c e x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )}+\frac{c d x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )}+\frac{e^2 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )}{d \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*d*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(a*(c*d^2 + a*e^2)) + (e^2*x*Hypergeom
etric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d*(c*d^2 + a*e^2)) - (c*e*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)*(1 + n))

Rule 1425

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q/(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] && 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 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 )} \, dx &=\int \left (\frac{e^2}{\left (c d^2+a e^2\right ) \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 )}\right ) \, dx\\ &=-\frac{c \int \frac{-d+e x^n}{a+c x^{2 n}} \, dx}{c d^2+a e^2}+\frac{e^2 \int \frac{1}{d+e x^n} \, dx}{c d^2+a e^2}\\ &=\frac{e^2 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 )}+\frac{(c d) \int \frac{1}{a+c x^{2 n}} \, dx}{c d^2+a e^2}-\frac{(c e) \int \frac{x^n}{a+c x^{2 n}} \, dx}{c d^2+a e^2}\\ &=\frac{c d 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 )}+\frac{e^2 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 )}-\frac{c e 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 ) (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.143869, size = 131, normalized size = 0.86 \[ \frac{x \left (c d^2 (n+1) \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )+e \left (a e (n+1) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{e x^n}{d}\right )-c d x^n \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )\right )\right )}{a d (n+1) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*d^2*(1 + n)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)] + e*(a*e*(1 + n)*Hypergeomet
ric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)] - c*d*x^n*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c
*x^(2*n))/a)])))/(a*d*(c*d^2 + a*e^2)*(1 + n))

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Maple [F]  time = 0.079, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d+e{x}^{n} \right ) \left ( a+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)/(a+c*x^(2*n)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2 \, n} + a\right )}{\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)),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}{\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)),x, algorithm="giac")

[Out]

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