\(\int x (d+e x) (a+b x^2)^p \, dx\) [385]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\frac {d \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac {1}{3} e x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {778, 267, 372, 371} \[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\frac {d \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac {1}{3} e x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right ) \]

[In]

Int[x*(d + e*x)*(a + b*x^2)^p,x]

[Out]

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

Rule 267

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 778

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\frac {1}{6} \left (a+b x^2\right )^p \left (\frac {3 d \left (a+b x^2\right )}{b (1+p)}+2 e x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int x \left (e x +d \right ) \left (b \,x^{2}+a \right )^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x \,d x } \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.69 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87 \[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + d \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

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

Maxima [F]

\[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x (d+e x) \left (a+b x^2\right )^p \, dx=\int x\,{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]

[In]

int(x*(a + b*x^2)^p*(d + e*x),x)

[Out]

int(x*(a + b*x^2)^p*(d + e*x), x)