3.1435 \(\int (b d+2 c d x)^m (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=107 \[ -\frac{2 \left (\frac{1}{4} \left (4 a-\frac{b^2}{c}\right )+\frac{(b+2 c x)^2}{4 c}\right )^{p+1} (b d+2 c d x)^{m+1} \, _2F_1\left (1,\frac{1}{2} (m+2 p+3);\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{d (m+1) \left (b^2-4 a c\right )} \]

[Out]

(-2*(b*d + 2*c*d*x)^(1 + m)*((4*a - b^2/c)/4 + (b + 2*c*x)^2/(4*c))^(1 + p)*Hypergeometric2F1[1, (3 + m + 2*p)
/2, (3 + m)/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/((b^2 - 4*a*c)*d*(1 + m))

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Rubi [A]  time = 0.074085, antiderivative size = 102, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {694, 365, 364} \[ \frac{\left (a+b x+c x^2\right )^p \left (1-\frac{(b+2 c x)^2}{b^2-4 a c}\right )^{-p} (d (b+2 c x))^{m+1} \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d*(b + 2*c*x))^(1 + m)*(a + b*x + c*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (b + 2*c*x)^2/(b^2 -
4*a*c)])/(2*c*d*(1 + m)*(1 - (b + 2*c*x)^2/(b^2 - 4*a*c))^p)

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 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 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 (b d+2 c d x)^m \left (a+b x+c x^2\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int x^m \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )^p \, dx,x,b d+2 c d x\right )}{2 c d}\\ &=\frac{\left (2^{-1+2 p} \left (a+b x+c x^2\right )^p \left (4+\frac{(b d+2 c d x)^2}{\left (a-\frac{b^2}{4 c}\right ) c d^2}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^m \left (1+\frac{x^2}{4 \left (a-\frac{b^2}{4 c}\right ) c d^2}\right )^p \, dx,x,b d+2 c d x\right )}{c d}\\ &=\frac{2^{-1+2 p} (d (b+2 c x))^{1+m} \left (a+b x+c x^2\right )^p \left (4-\frac{4 (b+2 c x)^2}{b^2-4 a c}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},-p;\frac{3+m}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{c d (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0606184, size = 107, normalized size = 1. \[ \frac{2^{-2 p-1} (b+2 c x) (a+x (b+c x))^p \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{-p} (d (b+2 c x))^m \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 - 2*p)*(b + 2*c*x)*(d*(b + 2*c*x))^m*(a + x*(b + c*x))^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, (b
+ 2*c*x)^2/(b^2 - 4*a*c)])/(c*(1 + m)*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^p)

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Maple [F]  time = 1.272, size = 0, normalized size = 0. \begin{align*} \int \left ( 2\,cdx+bd \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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