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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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])

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 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]

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 = 0.03, 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

________________________________________________________________________________________

fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{2} + d x\right )} {\left (c x^{2} + a\right )}^{p}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )} {\left (c x^{2} + a\right )}^{p} x\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 0.86, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right ) x \left (c \,x^{2}+a \right )^{p}\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ e \int {\left (c x^{2} + a\right )}^{p} x^{2}\,{d x} + \frac {{\left (c x^{2} + a\right )}^{p + 1} d}{2 \, c {\left (p + 1\right )}} \]

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]

e*integrate((c*x^2 + a)^p*x^2, x) + 1/2*(c*x^2 + a)^(p + 1)*d/(c*(p + 1))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (c\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + c*x^2)^p*(d + e*x),x)

[Out]

int(x*(a + c*x^2)^p*(d + e*x), x)

________________________________________________________________________________________

sympy [A]  time = 8.82, size = 65, normalized size = 0.87 \[ \frac {a^{p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{3} + d \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\begin {cases} \frac {\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + c x^{2} \right )} & \text {otherwise} \end {cases}}{2 c} & \text {otherwise} \end {cases}\right ) \]

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))

________________________________________________________________________________________