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

Optimal. Leaf size=357 \[ -\frac{3 e x^{n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{n+1}{2 n};-p,3;\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (n+1)}+\frac{3 e^2 x^{2 n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2} \left (2+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (4+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^5 (2 n+1)}-\frac{e^3 x^{3 n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2} \left (3+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (5+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^6 (3 n+1)}+\frac{x \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2 n};-p,3;\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3} \]

[Out]

(3*e^2*x^(1 + 2*n)*(a + c*x^(2*n))^p*AppellF1[(2 + n^(-1))/2, -p, 3, (4 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^
(2*n))/d^2])/(d^5*(1 + 2*n)*(1 + (c*x^(2*n))/a)^p) - (e^3*x^(1 + 3*n)*(a + c*x^(2*n))^p*AppellF1[(3 + n^(-1))/
2, -p, 3, (5 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^6*(1 + 3*n)*(1 + (c*x^(2*n))/a)^p) + (x*(a
+ c*x^(2*n))^p*AppellF1[1/(2*n), -p, 3, (2 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^3*(1 + (c*x^(
2*n))/a)^p) - (3*e*x^(1 + n)*(a + c*x^(2*n))^p*AppellF1[(1 + n)/(2*n), -p, 3, (3 + n^(-1))/2, -((c*x^(2*n))/a)
, (e^2*x^(2*n))/d^2])/(d^4*(1 + n)*(1 + (c*x^(2*n))/a)^p)

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Rubi [A]  time = 0.338036, antiderivative size = 357, 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 = {1438, 430, 429, 511, 510} \[ -\frac{3 e x^{n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{n+1}{2 n};-p,3;\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (n+1)}+\frac{3 e^2 x^{2 n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2} \left (2+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (4+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^5 (2 n+1)}-\frac{e^3 x^{3 n+1} \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2} \left (3+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (5+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^6 (3 n+1)}+\frac{x \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{1}{2 n};-p,3;\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*e^2*x^(1 + 2*n)*(a + c*x^(2*n))^p*AppellF1[(2 + n^(-1))/2, -p, 3, (4 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^
(2*n))/d^2])/(d^5*(1 + 2*n)*(1 + (c*x^(2*n))/a)^p) - (e^3*x^(1 + 3*n)*(a + c*x^(2*n))^p*AppellF1[(3 + n^(-1))/
2, -p, 3, (5 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^6*(1 + 3*n)*(1 + (c*x^(2*n))/a)^p) + (x*(a
+ c*x^(2*n))^p*AppellF1[1/(2*n), -p, 3, (2 + n^(-1))/2, -((c*x^(2*n))/a), (e^2*x^(2*n))/d^2])/(d^3*(1 + (c*x^(
2*n))/a)^p) - (3*e*x^(1 + n)*(a + c*x^(2*n))^p*AppellF1[(1 + n)/(2*n), -p, 3, (3 + n^(-1))/2, -((c*x^(2*n))/a)
, (e^2*x^(2*n))/d^2])/(d^4*(1 + n)*(1 + (c*x^(2*n))/a)^p)

Rule 1438

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - (e*x^n)/(d^2 - e^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+c x^{2 n}\right )^p}{\left (d+e x^n\right )^3} \, dx &=\int \left (\frac{d^3 \left (a+c x^{2 n}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^3}+\frac{3 d^2 e x^n \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3}-\frac{3 d e^2 x^{2 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3}+\frac{e^3 x^{3 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3}\right ) \, dx\\ &=d^3 \int \frac{\left (a+c x^{2 n}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^3} \, dx+\left (3 d^2 e\right ) \int \frac{x^n \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx-\left (3 d e^2\right ) \int \frac{x^{2 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx+e^3 \int \frac{x^{3 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx\\ &=\left (d^3 \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^3} \, dx+\left (3 d^2 e \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^n \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx-\left (3 d e^2 \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^{2 n} \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx+\left (e^3 \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^{3 n} \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^3} \, dx\\ &=\frac{3 e^2 x^{1+2 n} \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1}{2} \left (2+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (4+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^5 (1+2 n)}-\frac{e^3 x^{1+3 n} \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1}{2} \left (3+\frac{1}{n}\right );-p,3;\frac{1}{2} \left (5+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^6 (1+3 n)}+\frac{x \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1}{2 n};-p,3;\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3}-\frac{3 e x^{1+n} \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1+n}{2 n};-p,3;\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (1+n)}\\ \end{align*}

Mathematica [F]  time = 0.294008, size = 0, normalized size = 0. \[ \int \frac{\left (a+c x^{2 n}\right )^p}{\left (d+e x^n\right )^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.086, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+c{x}^{2\,n} \right ) ^{p}}{ \left ( d+e{x}^{n} \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+c*x^(2*n))^p/(d+e*x^n)^3,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+c*x^(2*n))^p/(d+e*x^n)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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