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

Optimal. Leaf size=153 \[ \frac{e (g x)^{m+2} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2)}+\frac{d (g x)^{m+1} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1)} \]

[Out]

(d*(g*x)^(1 + m)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*(1 -
(e^2*x^2)/d^2)^p) + (e*(g*x)^(2 + m)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, (e^2*x^2)/
d^2])/(g^2*(2 + m)*(1 - (e^2*x^2)/d^2)^p)

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Rubi [A]  time = 0.0676501, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {808, 365, 364} \[ \frac{e (g x)^{m+2} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (m+2)}+\frac{d (g x)^{m+1} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )}{g (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(g*x)^(1 + m)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (e^2*x^2)/d^2])/(g*(1 + m)*(1 -
(e^2*x^2)/d^2)^p) + (e*(g*x)^(2 + m)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, (e^2*x^2)/
d^2])/(g^2*(2 + m)*(1 - (e^2*x^2)/d^2)^p)

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || 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 (g x)^m (d+e x) \left (d^2-e^2 x^2\right )^p \, dx &=d \int (g x)^m \left (d^2-e^2 x^2\right )^p \, dx+\frac{e \int (g x)^{1+m} \left (d^2-e^2 x^2\right )^p \, dx}{g}\\ &=\left (d \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int (g x)^m \left (1-\frac{e^2 x^2}{d^2}\right )^p \, dx+\frac{\left (e \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p}\right ) \int (g x)^{1+m} \left (1-\frac{e^2 x^2}{d^2}\right )^p \, dx}{g}\\ &=\frac{d (g x)^{1+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},-p;\frac{3+m}{2};\frac{e^2 x^2}{d^2}\right )}{g (1+m)}+\frac{e (g x)^{2+m} \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac{2+m}{2},-p;\frac{4+m}{2};\frac{e^2 x^2}{d^2}\right )}{g^2 (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0363992, size = 116, normalized size = 0.76 \[ \frac{x (g x)^m \left (d^2-e^2 x^2\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^{-p} \left (d (m+2) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};\frac{e^2 x^2}{d^2}\right )+e (m+1) x \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};\frac{e^2 x^2}{d^2}\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(d^2 - e^2*x^2)^p*(d*(2 + m)*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (e^2*x^2)/d^2] + e*(1 + m)
*x*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, (e^2*x^2)/d^2]))/((1 + m)*(2 + m)*(1 - (e^2*x^2)/d^2)^p)

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Maple [F]  time = 0.465, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) \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)*(-e^2*x^2+d^2)^p,x)

[Out]

int((g*x)^m*(e*x+d)*(-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 x + d\right )}{\left (-e^{2} x^{2} + d^{2}\right )}^{p} \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)*(-e^2*x^2+d^2)^p,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [C]  time = 19.3651, size = 122, normalized size = 0.8 \begin{align*} \frac{d d^{2 p} g^{m} x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{d^{2 p} e g^{m} x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - p, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{\frac{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 \Gamma \left (\frac{m}{2} + 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*d**(2*p)*g**m*x*x**m*gamma(m/2 + 1/2)*hyper((-p, m/2 + 1/2), (m/2 + 3/2,), e**2*x**2*exp_polar(2*I*pi)/d**2)
/(2*gamma(m/2 + 3/2)) + d**(2*p)*e*g**m*x**2*x**m*gamma(m/2 + 1)*hyper((-p, m/2 + 1), (m/2 + 2,), e**2*x**2*ex
p_polar(2*I*pi)/d**2)/(2*gamma(m/2 + 2))

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

[Out]

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