### 3.2563 $$\int (d+e x)^3 (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=327 $-\frac{2^{p-1} (2 c d-b e) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^3 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{e \left (a+b x+c x^2\right )^{p+1} \left (-2 c (2 p+3) \left (c d^2 (2 p+5)-e (a e+b d (p+2))\right )-2 c e (p+1) (p+3) x (2 c d-b e)+b e (p+2) (p+3) (2 c d-b e)\right )}{4 c^3 (p+1) (p+2) (2 p+3)}+\frac{e (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}$

[Out]

(e*(d + e*x)^2*(a + b*x + c*x^2)^(1 + p))/(2*c*(2 + p)) - (e*(b*e*(2*c*d - b*e)*(2 + p)*(3 + p) - 2*c*(3 + 2*p
)*(c*d^2*(5 + 2*p) - e*(a*e + b*d*(2 + p))) - 2*c*e*(2*c*d - b*e)*(1 + p)*(3 + p)*x)*(a + b*x + c*x^2)^(1 + p)
)/(4*c^3*(1 + p)*(2 + p)*(3 + 2*p)) - (2^(-1 + p)*(2*c*d - b*e)*(b^2*e^2*(3 + p) + 2*c^2*d^2*(3 + 2*p) - 2*c*e
*(3*a*e + b*d*(3 + 2*p)))*(-((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^3*Sqrt[b^
2 - 4*a*c]*(1 + p)*(3 + 2*p))

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Rubi [A]  time = 0.423452, antiderivative size = 327, 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, 779, 624} $-\frac{2^{p-1} (2 c d-b e) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^3 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{e \left (a+b x+c x^2\right )^{p+1} \left (-2 c (2 p+3) \left (c d^2 (2 p+5)-e (a e+b d (p+2))\right )-2 c e (p+1) (p+3) x (2 c d-b e)+b e (p+2) (p+3) (2 c d-b e)\right )}{4 c^3 (p+1) (p+2) (2 p+3)}+\frac{e (d+e x)^2 \left (a+b x+c x^2\right )^{p+1}}{2 c (p+2)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3*(a + b*x + c*x^2)^p,x]

[Out]

(e*(d + e*x)^2*(a + b*x + c*x^2)^(1 + p))/(2*c*(2 + p)) - (e*(b*e*(2*c*d - b*e)*(2 + p)*(3 + p) - 2*c*(3 + 2*p
)*(c*d^2*(5 + 2*p) - e*(a*e + b*d*(2 + p))) - 2*c*e*(2*c*d - b*e)*(1 + p)*(3 + p)*x)*(a + b*x + c*x^2)^(1 + p)
)/(4*c^3*(1 + p)*(2 + p)*(3 + 2*p)) - (2^(-1 + p)*(2*c*d - b*e)*(b^2*e^2*(3 + p) + 2*c^2*d^2*(3 + 2*p) - 2*c*e
*(3*a*e + b*d*(3 + 2*p)))*(-((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^3*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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[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)^3 \left (a+b x+c x^2\right )^p \, dx &=\frac{e (d+e x)^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}+\frac{\int (d+e x) \left (2 c d^2 (2+p)-e (2 a e+b d (1+p))+e (2 c d-b e) (3+p) x\right ) \left (a+b x+c x^2\right )^p \, dx}{2 c (2+p)}\\ &=\frac{e (d+e x)^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}-\frac{e \left (b e (2 c d-b e) (2+p) (3+p)-2 c (3+2 p) \left (c d^2 (5+2 p)-e (a e+b d (2+p))\right )-2 c e (2 c d-b e) (1+p) (3+p) x\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^3 (1+p) (2+p) (3+2 p)}+\frac{\left ((2 c d-b e) \left (b^2 e^2 (3+p)+2 c^2 d^2 (3+2 p)-2 c e (3 a e+b d (3+2 p))\right )\right ) \int \left (a+b x+c x^2\right )^p \, dx}{4 c^3 (3+2 p)}\\ &=\frac{e (d+e x)^2 \left (a+b x+c x^2\right )^{1+p}}{2 c (2+p)}-\frac{e \left (b e (2 c d-b e) (2+p) (3+p)-2 c (3+2 p) \left (c d^2 (5+2 p)-e (a e+b d (2+p))\right )-2 c e (2 c d-b e) (1+p) (3+p) x\right ) \left (a+b x+c x^2\right )^{1+p}}{4 c^3 (1+p) (2+p) (3+2 p)}-\frac{2^{-1+p} (2 c d-b e) \left (b^2 e^2 (3+p)+2 c^2 d^2 (3+2 p)-2 c e (3 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^3 \sqrt{b^2-4 a c} (1+p) (3+2 p)}\\ \end{align*}

Mathematica [C]  time = 1.52061, size = 558, normalized size = 1.71 $\frac{1}{4} (a+x (b+c x))^p \left (6 d^2 e x^2 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (2;-p,-p;3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+4 d e^2 x^3 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (3;-p,-p;4;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+e^3 x^4 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} F_1\left (4;-p,-p;5;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+\frac{d^3 2^{p+1} \left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p} \, _2F_1\left (-p,p+1;p+2;\frac{-b-2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c (p+1)}\right )$

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x)^3*(a + b*x + c*x^2)^p,x]

[Out]

((a + x*(b + c*x))^p*((6*d^2*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) + (4*d*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) + (e^3*x^4*AppellF1[4, -p, -p, 5, (-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^(1 + p)*d^3*(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)))/4

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Maple [F]  time = 1.213, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{3} \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)^3*(c*x^2+b*x+a)^p,x)

[Out]

int((e*x+d)^3*(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 )}^{3}{\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)^3*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((e*x + d)^3*(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^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\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)^3*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*(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)**3*(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 )}^{3}{\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)^3*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((e*x + d)^3*(c*x^2 + b*x + a)^p, x)