3.314 \(\int (g x)^m (d+e x)^n (d^2-e^2 x^2)^p \, dx\)

Optimal. Leaf size=96 \[ \frac{(g x)^{m+1} (d+e x)^n \left (1-\frac{e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (\frac{e x}{d}+1\right )^{-n-p} F_1\left (m+1;-p,-n-p;m+2;\frac{e x}{d},-\frac{e x}{d}\right )}{g (m+1)} \]

[Out]

((g*x)^(1 + m)*(d + e*x)^n*(1 + (e*x)/d)^(-n - p)*(d^2 - e^2*x^2)^p*AppellF1[1 + m, -p, -n - p, 2 + m, (e*x)/d
, -((e*x)/d)])/(g*(1 + m)*(1 - (e*x)/d)^p)

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Rubi [A]  time = 0.0829346, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {892, 135, 133} \[ \frac{(g x)^{m+1} (d+e x)^n \left (1-\frac{e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p \left (\frac{e x}{d}+1\right )^{-n-p} F_1\left (m+1;-p,-n-p;m+2;\frac{e x}{d},-\frac{e x}{d}\right )}{g (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(g*x)^m*(d + e*x)^n*(d^2 - e^2*x^2)^p,x]

[Out]

((g*x)^(1 + m)*(d + e*x)^n*(1 + (e*x)/d)^(-n - p)*(d^2 - e^2*x^2)^p*AppellF1[1 + m, -p, -n - p, 2 + m, (e*x)/d
, -((e*x)/d)])/(g*(1 + m)*(1 - (e*x)/d)^p)

Rule 892

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + c*x^
2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (
c*x)/e)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !Int
egerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (g x)^m (d+e x)^n \left (d^2-e^2 x^2\right )^p \, dx &=\left ((d-e x)^{-p} (d+e x)^{-p} \left (d^2-e^2 x^2\right )^p\right ) \int (g x)^m (d-e x)^p (d+e x)^{n+p} \, dx\\ &=\left ((d+e x)^{-p} \left (1-\frac{e x}{d}\right )^{-p} \left (d^2-e^2 x^2\right )^p\right ) \int (g x)^m (d+e x)^{n+p} \left (1-\frac{e x}{d}\right )^p \, dx\\ &=\left ((d+e x)^n \left (1-\frac{e x}{d}\right )^{-p} \left (1+\frac{e x}{d}\right )^{-n-p} \left (d^2-e^2 x^2\right )^p\right ) \int (g x)^m \left (1-\frac{e x}{d}\right )^p \left (1+\frac{e x}{d}\right )^{n+p} \, dx\\ &=\frac{(g x)^{1+m} (d+e x)^n \left (1-\frac{e x}{d}\right )^{-p} \left (1+\frac{e x}{d}\right )^{-n-p} \left (d^2-e^2 x^2\right )^p F_1\left (1+m;-p,-n-p;2+m;\frac{e x}{d},-\frac{e x}{d}\right )}{g (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.110527, size = 90, normalized size = 0.94 \[ \frac{x (g x)^m (d-e x)^p \left (\frac{d-e x}{d}\right )^{-p} (d+e x)^{n+p} \left (\frac{d+e x}{d}\right )^{-n-p} F_1\left (m+1;-p,-n-p;m+2;\frac{e x}{d},-\frac{e x}{d}\right )}{m+1} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*x)^m*(d + e*x)^n*(d^2 - e^2*x^2)^p,x]

[Out]

(x*(g*x)^m*(d - e*x)^p*(d + e*x)^(n + p)*((d + e*x)/d)^(-n - p)*AppellF1[1 + m, -p, -n - p, 2 + m, (e*x)/d, -(
(e*x)/d)])/((1 + m)*((d - e*x)/d)^p)

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Maple [F]  time = 0.729, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) ^{n} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(e*x+d)^n*(-e^2*x^2+d^2)^p,x)

[Out]

int((g*x)^m*(e*x+d)^n*(-e^2*x^2+d^2)^p,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(e*x+d)^n*(-e^2*x^2+d^2)^p,x, algorithm="maxima")

[Out]

integrate((-e^2*x^2 + d^2)^p*(e*x + d)^n*(g*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(e*x+d)^n*(-e^2*x^2+d^2)^p,x, algorithm="fricas")

[Out]

integral((-e^2*x^2 + d^2)^p*(e*x + d)^n*(g*x)^m, 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((g*x)**m*(e*x+d)**n*(-e**2*x**2+d**2)**p,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^m*(e*x+d)^n*(-e^2*x^2+d^2)^p,x, algorithm="giac")

[Out]

integrate((-e^2*x^2 + d^2)^p*(e*x + d)^n*(g*x)^m, x)