3.505 \(\int x (d+e x) (a+c x^2)^p \, dx\)

Optimal. Leaf size=75 \[ \frac{d \left (a+c x^2\right )^{p+1}}{2 c (p+1)}+\frac{1}{3} e x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{c x^2}{a}\right ) \]

[Out]

(d*(a + c*x^2)^(1 + p))/(2*c*(1 + p)) + (e*x^3*(a + c*x^2)^p*Hypergeometric2F1[3/2, -p, 5/2, -((c*x^2)/a)])/(3
*(1 + (c*x^2)/a)^p)

________________________________________________________________________________________

Rubi [A]  time = 0.0283265, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {764, 261, 365, 364} \[ \frac{d \left (a+c x^2\right )^{p+1}}{2 c (p+1)}+\frac{1}{3} e x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{c x^2}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*(d + e*x)*(a + c*x^2)^p,x]

[Out]

(d*(a + c*x^2)^(1 + p))/(2*c*(1 + p)) + (e*x^3*(a + c*x^2)^p*Hypergeometric2F1[3/2, -p, 5/2, -((c*x^2)/a)])/(3
*(1 + (c*x^2)/a)^p)

Rule 764

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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 x (d+e x) \left (a+c x^2\right )^p \, dx &=d \int x \left (a+c x^2\right )^p \, dx+e \int x^2 \left (a+c x^2\right )^p \, dx\\ &=\frac{d \left (a+c x^2\right )^{1+p}}{2 c (1+p)}+\left (e \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac{c x^2}{a}\right )^p \, dx\\ &=\frac{d \left (a+c x^2\right )^{1+p}}{2 c (1+p)}+\frac{1}{3} e x^3 \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{c x^2}{a}\right )\\ \end{align*}

Mathematica [A]  time = 0.0327172, size = 71, normalized size = 0.95 \[ \frac{1}{6} \left (a+c x^2\right )^p \left (\frac{3 d \left (a+c x^2\right )}{c (p+1)}+2 e x^3 \left (\frac{c x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{c x^2}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + c*x^2)^p*((3*d*(a + c*x^2))/(c*(1 + p)) + (2*e*x^3*Hypergeometric2F1[3/2, -p, 5/2, -((c*x^2)/a)])/(1 + (
c*x^2)/a)^p))/6

________________________________________________________________________________________

Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int x \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(x*(e*x+d)*(c*x^2+a)^p,x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 9.82812, size = 65, normalized size = 0.87 \begin{align*} \frac{a^{p} e x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{c x^{2} e^{i \pi }}{a}} \right )}}{3} + d \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\begin{cases} \frac{\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + c x^{2} \right )} & \text{otherwise} \end{cases}}{2 c} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**p*e*x**3*hyper((3/2, -p), (5/2,), c*x**2*exp_polar(I*pi)/a)/3 + d*Piecewise((a**p*x**2/2, Eq(c, 0)), (Piece
wise(((a + c*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + c*x**2), True))/(2*c), True))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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