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

Optimal. Leaf size=134 \[ -\frac{d (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n}-\frac{e (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n (n+1)}+\frac{x \left (d+e x^n\right )}{2 a n \left (a+c x^{2 n}\right )} \]

[Out]

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

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Rubi [A]  time = 0.0543761, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1431, 1418, 245, 364} \[ -\frac{d (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n}-\frac{e (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{2 a^2 n (n+1)}+\frac{x \left (d+e x^n\right )}{2 a n \left (a+c x^{2 n}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)/(a + c*x^(2*n))^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0423645, size = 83, normalized size = 0.62 \[ \frac{d x \, _2F_1\left (2,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a^2}+\frac{e x^{n+1} \, _2F_1\left (2,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{d+e{x}^{n}}{ \left ( a+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)/(a+c*x^(2*n))^2,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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