### 3.2565 $$\int (d+e x) (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=160 $\frac{e \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)}-\frac{2^p (2 c d-b e) \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1) \sqrt{b^2-4 a c}}$

[Out]

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

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Rubi [A]  time = 0.0451696, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {640, 624} $\frac{e \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)}-\frac{2^p (2 c d-b e) \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1) \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)*(a + b*x + c*x^2)^p,x]

[Out]

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

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

Mathematica [C]  time = 0.384858, size = 268, normalized size = 1.68 $\frac{1}{2} (a+x (b+c x))^p \left (e x^2 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (2;-p,-p;3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+\frac{d 2^p \left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p} \, _2F_1\left (-p,p+1;p+2;\frac{-b-2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1)}\right )$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + x*(b + c*x))^p*((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^p*d*(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)))/2

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Maple [F]  time = 1.055, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) \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)*(c*x^2+b*x+a)^p,x)

[Out]

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

[Out]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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