3.4.14 \(\int (g x)^m (d+e x)^n (d^2-e^2 x^2)^p \, dx\) [314]

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

[Out]

(g*x)^(1+m)*(e*x+d)^n*(1+e*x/d)^(-n-p)*(-e^2*x^2+d^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 = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 = {906, 140, 138} \begin {gather*} \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)} \end {gather*}

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 138

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

Rule 140

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 906

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/e)*x)^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]

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*}

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Mathematica [A]
time = 0.15, size = 90, normalized size = 0.94 \begin {gather*} \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 (1+m;-p,-n-p;2+m;\frac {e x}{d},-\frac {e x}{d}\right )}{1+m} \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \left (g x \right )^{m} \left (e x +d \right )^{n} \left (-e^{2} x^{2}+d^{2}\right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (g x\right )^{m} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{p} \left (d + e x\right )^{n}\, dx \end {gather*}

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]

Integral((g*x)**m*(-(-d + e*x)*(d + e*x))**p*(d + e*x)**n, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d^2-e^2\,x^2\right )}^p\,{\left (g\,x\right )}^m\,{\left (d+e\,x\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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