### 3.2564 $$\int (d+e x)^2 (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=248 $-\frac{2^p \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}+\frac{e (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1}}{2 c^2 (p+1) (2 p+3)}+\frac{e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)}$

[Out]

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

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Rubi [A]  time = 0.209372, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.15, Rules used = {742, 640, 624} $-\frac{2^p \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}+\frac{e (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1}}{2 c^2 (p+1) (2 p+3)}+\frac{e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 624

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, -Simp[((a + b*x + c*
x^2)^(p + 1)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps

\begin{align*} \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx &=\frac{e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}+\frac{\int \left (c d^2 (3+2 p)-e (a e+b d (1+p))+e (2 c d-b e) (2+p) x\right ) \left (a+b x+c x^2\right )^p \, dx}{c (3+2 p)}\\ &=\frac{e (2 c d-b e) (2+p) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac{e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}+\frac{\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \int \left (a+b x+c x^2\right )^p \, dx}{2 c^2 (3+2 p)}\\ &=\frac{e (2 c d-b e) (2+p) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac{e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}-\frac{2^p \left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (-\frac{b-\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac{b+\sqrt{b^2-4 a c}+2 c x}{2 \sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} (1+p) (3+2 p)}\\ \end{align*}

Mathematica [C]  time = 0.811754, size = 414, normalized size = 1.67 $\frac{1}{6} (a+x (b+c x))^p \left (6 d e x^2 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (2;-p,-p;3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+2 e^2 x^3 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (3;-p,-p;4;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+\frac{3 d^2 2^p \left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p} \, _2F_1\left (-p,p+1;p+2;\frac{-b-2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1)}\right )$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + x*(b + c*x))^p*((6*d*e*x^2*AppellF1[2, -p, -p, 3, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b
^2 - 4*a*c])])/(((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(
b + Sqrt[b^2 - 4*a*c]))^p) + (2*e^2*x^3*AppellF1[3, -p, -p, 4, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b +
Sqrt[b^2 - 4*a*c])])/(((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2
*c*x)/(b + Sqrt[b^2 - 4*a*c]))^p) + (3*2^p*d^2*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*Hypergeometric2F1[-p, 1 + p, 2
+ p, (-b + Sqrt[b^2 - 4*a*c] - 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(c*(1 + p)*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt
[b^2 - 4*a*c])^p)))/6

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Maple [F]  time = 1.245, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^2*(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 (e x + d\right )}^{2}{\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((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)^2*(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 (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}{\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((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((e^2*x^2 + 2*d*e*x + d^2)*(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((e*x+d)**2*(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 (e x + d\right )}^{2}{\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((e*x+d)^2*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

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