\(\int (c+d x)^2 (a+b (c+d x)^2)^p \, dx\) [2850]

Optimal result

Integrand size = 21, antiderivative size = 55 \[ \int (c+d x)^2 \left (a+b (c+d x)^2\right )^p \, dx=\frac {(c+d x)^3 \left (a+b (c+d x)^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2}+p,\frac {5}{2},-\frac {b (c+d x)^2}{a}\right )}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {379, 372, 371} \[ \int (c+d x)^2 \left (a+b (c+d x)^2\right )^p \, dx=\frac {(c+d x)^3 \left (a+b (c+d x)^2\right )^p \left (\frac {b (c+d x)^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b (c+d x)^2}{a}\right )}{3 d} \]

[In]

Int[(c + d*x)^2*(a + b*(c + d*x)^2)^p,x]

[Out]

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

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 379

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int (c+d x)^2 \left (a+b (c+d x)^2\right )^p \, dx=\frac {(c+d x)^3 \left (a+b (c+d x)^2\right )^p \left (1+\frac {b (c+d x)^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b (c+d x)^2}{a}\right )}{3 d} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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