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

Optimal. Leaf size=184 \[ -\frac{x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac{d (1-4 n) (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{8 a^3 n^2}+\frac{e (1-3 n) (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{8 a^3 n^2 (n+1)}+\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2} \]

[Out]

(x*(d + e*x^n))/(4*a*n*(a + c*x^(2*n))^2) - (x*(d*(1 - 4*n) + e*(1 - 3*n)*x^n))/(8*a^2*n^2*(a + c*x^(2*n))) +
(d*(1 - 4*n)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*n^2) + (e*(1
- 3*n)*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*n^2*(1
+ n))

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Rubi [A]  time = 0.10074, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1431, 1418, 245, 364} \[ -\frac{x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac{d (1-4 n) (1-2 n) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{8 a^3 n^2}+\frac{e (1-3 n) (1-n) x^{n+1} \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{8 a^3 n^2 (n+1)}+\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)/(a + c*x^(2*n))^3,x]

[Out]

(x*(d + e*x^n))/(4*a*n*(a + c*x^(2*n))^2) - (x*(d*(1 - 4*n) + e*(1 - 3*n)*x^n))/(8*a^2*n^2*(a + c*x^(2*n))) +
(d*(1 - 4*n)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*n^2) + (e*(1
- 3*n)*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*n^2*(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 )^3} \, dx &=\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2}-\frac{\int \frac{d (1-4 n)+e (1-3 n) x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{4 a n}\\ &=\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2}-\frac{x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac{\int \frac{d (1-4 n) (1-2 n)+e (1-3 n) (1-n) x^n}{a+c x^{2 n}} \, dx}{8 a^2 n^2}\\ &=\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2}-\frac{x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac{(d (1-4 n) (1-2 n)) \int \frac{1}{a+c x^{2 n}} \, dx}{8 a^2 n^2}+\frac{(e (1-3 n) (1-n)) \int \frac{x^n}{a+c x^{2 n}} \, dx}{8 a^2 n^2}\\ &=\frac{x \left (d+e x^n\right )}{4 a n \left (a+c x^{2 n}\right )^2}-\frac{x \left (d (1-4 n)+e (1-3 n) x^n\right )}{8 a^2 n^2 \left (a+c x^{2 n}\right )}+\frac{d (1-4 n) (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 )}{8 a^3 n^2}+\frac{e (1-3 n) (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 )}{8 a^3 n^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.047879, size = 83, normalized size = 0.45 \[ \frac{d x \, _2F_1\left (3,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a^3}+\frac{e x^{n+1} \, _2F_1\left (3,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a^3 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.079, size = 0, normalized size = 0. \begin{align*} \int{\frac{d+e{x}^{n}}{ \left ( a+c{x}^{2\,n} \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c e{\left (3 \, n - 1\right )} x x^{3 \, n} + c d{\left (4 \, n - 1\right )} x x^{2 \, n} + a e{\left (5 \, n - 1\right )} x x^{n} + a d{\left (6 \, n - 1\right )} x}{8 \,{\left (a^{2} c^{2} n^{2} x^{4 \, n} + 2 \, a^{3} c n^{2} x^{2 \, n} + a^{4} n^{2}\right )}} + \int \frac{{\left (3 \, n^{2} - 4 \, n + 1\right )} e x^{n} +{\left (8 \, n^{2} - 6 \, n + 1\right )} d}{8 \,{\left (a^{2} c n^{2} x^{2 \, n} + a^{3} n^{2}\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))^3,x, algorithm="maxima")

[Out]

1/8*(c*e*(3*n - 1)*x*x^(3*n) + c*d*(4*n - 1)*x*x^(2*n) + a*e*(5*n - 1)*x*x^n + a*d*(6*n - 1)*x)/(a^2*c^2*n^2*x
^(4*n) + 2*a^3*c*n^2*x^(2*n) + a^4*n^2) + integrate(1/8*((3*n^2 - 4*n + 1)*e*x^n + (8*n^2 - 6*n + 1)*d)/(a^2*c
*n^2*x^(2*n) + a^3*n^2), 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^{3} x^{6 \, n} + 3 \, a c^{2} x^{4 \, n} + 3 \, a^{2} c x^{2 \, n} + a^{3}}, 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))^3,x, algorithm="fricas")

[Out]

integral((e*x^n + d)/(c^3*x^(6*n) + 3*a*c^2*x^(4*n) + 3*a^2*c*x^(2*n) + a^3), 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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

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