3.2850 \(\int (c+d x)^2 (a+b (c+d x)^2)^p \, dx\)

Optimal. Leaf size=55 \[ \frac{(c+d x)^3 \left (a+b (c+d x)^2\right )^{p+1} \, _2F_1\left (1,p+\frac{5}{2};\frac{5}{2};-\frac{b (c+d x)^2}{a}\right )}{3 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0470593, antiderivative size = 68, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {372, 365, 364} \[ \frac{(c+d x)^3 \left (a+b (c+d x)^2\right )^p \left (\frac{b (c+d x)^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b (c+d x)^2}{a}\right )}{3 d} \]

Antiderivative was successfully verified.

[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 372

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]

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 (c+d x)^2 \left (a+b (c+d x)^2\right )^p \, dx &=\frac{\operatorname{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 ) \operatorname{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]  time = 0.0167402, size = 68, normalized size = 1.24 \[ \frac{(c+d x)^3 \left (a+b (c+d x)^2\right )^p \left (\frac{b (c+d x)^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b (c+d x)^2}{a}\right )}{3 d} \]

Antiderivative was successfully verified.

[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]  time = 0.121, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{2} \left ( a+b \left ( dx+c \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)