3.432 \(\int (g x)^m (d+e x) (a+c x^2)^p \, dx\)

Optimal. Leaf size=135 \[ \frac{d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{g (m+1)}+\frac{e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{g^2 (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.0626675, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {808, 365, 364} \[ \frac{d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{g (m+1)}+\frac{e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{g^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(g*x)^(1 + m)*(a + c*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)])/(g*(1 + m)*(1 + (c*x
^2)/a)^p) + (e*(g*x)^(2 + m)*(a + c*x^2)^p*Hypergeometric2F1[(2 + m)/2, -p, (4 + m)/2, -((c*x^2)/a)])/(g^2*(2
+ m)*(1 + (c*x^2)/a)^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 (a+c x^2\right )^p \, dx &=d \int (g x)^m \left (a+c x^2\right )^p \, dx+\frac{e \int (g x)^{1+m} \left (a+c x^2\right )^p \, dx}{g}\\ &=\left (d \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int (g x)^m \left (1+\frac{c x^2}{a}\right )^p \, dx+\frac{\left (e \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int (g x)^{1+m} \left (1+\frac{c x^2}{a}\right )^p \, dx}{g}\\ &=\frac{d (g x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},-p;\frac{3+m}{2};-\frac{c x^2}{a}\right )}{g (1+m)}+\frac{e (g x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{2+m}{2},-p;\frac{4+m}{2};-\frac{c x^2}{a}\right )}{g^2 (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.03681, size = 106, normalized size = 0.79 \[ \frac{x (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (d (m+2) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )+e (m+1) x \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(g*x)^m*(a + c*x^2)^p*(d*(2 + m)*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((c*x^2)/a)] + e*(1 + m)*x*Hy
pergeometric2F1[(2 + m)/2, -p, (4 + m)/2, -((c*x^2)/a)]))/((1 + m)*(2 + m)*(1 + (c*x^2)/a)^p)

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Maple [F]  time = 0.381, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(e*x+d)*(c*x^2+a)^p,x)

[Out]

int((g*x)^m*(e*x+d)*(c*x^2+a)^p,x)

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

[Out]

integrate((e*x + d)*(c*x^2 + a)^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 (c x^{2} + a\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)*(c*x^2+a)^p,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

[Out]

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