3.91 \(\int \frac{(f x)^m (a+c x^{2 n})^p}{(d+e x^n)^2} \, dx\)

Optimal. Leaf size=302 \[ \frac{x (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2 n};-p,2;\frac{m+1}{2 n}+1;-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^2 (m+1)}-\frac{2 e x^{n+1} (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+n+1}{2 n};-p,2;\frac{m+3 n+1}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3 (m+n+1)}+\frac{e^2 x^{2 n+1} (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+2 n+1}{2 n};-p,2;\frac{m+4 n+1}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (m+2 n+1)} \]

[Out]

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

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Rubi [A]  time = 0.332553, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1562, 511, 510} \[ \frac{x (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2 n};-p,2;\frac{m+1}{2 n}+1;-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^2 (m+1)}-\frac{2 e x^{n+1} (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+n+1}{2 n};-p,2;\frac{m+3 n+1}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3 (m+n+1)}+\frac{e^2 x^{2 n+1} (f x)^m \left (a+c x^{2 n}\right )^p \left (\frac{c x^{2 n}}{a}+1\right )^{-p} F_1\left (\frac{m+2 n+1}{2 n};-p,2;\frac{m+4 n+1}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (m+2 n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1562

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Dist[(f*x)^m
/x^m, Int[ExpandIntegrand[x^m*(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], x] /; FreeQ[{a, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] &&  !IntegerQ[p] && ILtQ[q, 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{(f x)^m \left (a+c x^{2 n}\right )^p}{\left (d+e x^n\right )^2} \, dx &=\left (x^{-m} (f x)^m\right ) \int \left (\frac{d^2 x^m \left (a+c x^{2 n}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^2}-\frac{2 d e x^{m+n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2}+\frac{e^2 x^{m+2 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2}\right ) \, dx\\ &=\left (d^2 x^{-m} (f x)^m\right ) \int \frac{x^m \left (a+c x^{2 n}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^2} \, dx-\left (2 d e x^{-m} (f x)^m\right ) \int \frac{x^{m+n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2} \, dx+\left (e^2 x^{-m} (f x)^m\right ) \int \frac{x^{m+2 n} \left (a+c x^{2 n}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2} \, dx\\ &=\left (d^2 x^{-m} (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^m \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (d^2-e^2 x^{2 n}\right )^2} \, dx-\left (2 d e x^{-m} (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^{m+n} \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2} \, dx+\left (e^2 x^{-m} (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p}\right ) \int \frac{x^{m+2 n} \left (1+\frac{c x^{2 n}}{a}\right )^p}{\left (-d^2+e^2 x^{2 n}\right )^2} \, dx\\ &=\frac{x (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1+m}{2 n};-p,2;1+\frac{1+m}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^2 (1+m)}-\frac{2 e x^{1+n} (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1+m+n}{2 n};-p,2;\frac{1+m+3 n}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^3 (1+m+n)}+\frac{e^2 x^{1+2 n} (f x)^m \left (a+c x^{2 n}\right )^p \left (1+\frac{c x^{2 n}}{a}\right )^{-p} F_1\left (\frac{1+m+2 n}{2 n};-p,2;\frac{1+m+4 n}{2 n};-\frac{c x^{2 n}}{a},\frac{e^2 x^{2 n}}{d^2}\right )}{d^4 (1+m+2 n)}\\ \end{align*}

Mathematica [F]  time = 0.16251, size = 0, normalized size = 0. \[ \int \frac{(f x)^m \left (a+c x^{2 n}\right )^p}{\left (d+e x^n\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.094, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{m} \left ( a+c{x}^{2\,n} \right ) ^{p}}{ \left ( d+e{x}^{n} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(a+c*x^(2*n))^p/(d+e*x^n)^2,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 (f x\right )^{m}}{{\left (e x^{n} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^(2*n) + a)^p*(f*x)^m/(e*x^n + d)^2, 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} \left (f x\right )^{m}}{e^{2} x^{2 \, n} + 2 \, d e x^{n} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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