### 3.2578 $$\int (d+e x)^{-4-2 p} (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=442 $\frac{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}\right )^{-p} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt{b^2-4 a c}\right )}\right )}{2 (2 p+1) (2 p+3) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac{e (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac{e (p+2) (2 c d-b e) (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) (2 p+3) \left (a e^2-b d e+c d^2\right )^2}$

[Out]

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

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Rubi [A]  time = 0.352557, antiderivative size = 442, normalized size of antiderivative = 1., 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 = {744, 806, 726} $\frac{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}\right )^{-p} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt{b^2-4 a c}\right )}\right )}{2 (2 p+1) (2 p+3) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac{e (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac{e (p+2) (2 c d-b e) (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) (2 p+3) \left (a e^2-b d e+c d^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^(-4 - 2*p)*(a + b*x + c*x^2)^p,x]

[Out]

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

Rule 744

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))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 726

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b - Rt[b^2 - 4*a*
c, 2] + 2*c*x)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Hypergeometric2F1[m + 1, -p, m + 2, (-4*c*Rt[b^2 - 4*a*c,
2]*(d + e*x))/((2*c*d - b*e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x))])/((m + 1)*(2*c*d - b*e
+ e*Rt[b^2 - 4*a*c, 2])*(((2*c*d - b*e + e*Rt[b^2 - 4*a*c, 2])*(b + Rt[b^2 - 4*a*c, 2] + 2*c*x))/((2*c*d - b*
e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x)))^p), 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] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0
]

Rubi steps

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

Mathematica [A]  time = 1.48216, size = 399, normalized size = 0.9 $-\frac{(d+e x)^{-2 p-3} (a+x (b+c x))^p \left (\frac{(d+e x)^2 \left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right )}\right )^{-p-1} \, _2F_1\left (-2 p-1,-p;-2 p;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (\left (b+\sqrt{b^2-4 a c}\right ) e-2 c d\right ) \left (-b-2 c x+\sqrt{b^2-4 a c}\right )}\right )}{(2 p+1) \left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right ) \left (e (a e-b d)+c d^2\right )}+\frac{e (p+2) (d+e x) (a+x (b+c x)) (2 c d-b e)}{(p+1) \left (e (a e-b d)+c d^2\right )}+2 e (a+x (b+c x))\right )}{2 (2 p+3) \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^(-4 - 2*p)*(a + b*x + c*x^2)^p,x]

[Out]

-((d + e*x)^(-3 - 2*p)*(a + x*(b + c*x))^p*(2*e*(a + x*(b + c*x)) + (e*(2*c*d - b*e)*(2 + p)*(d + e*x)*(a + x*
(b + c*x)))/((c*d^2 + e*(-(b*d) + a*e))*(1 + 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 + Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*
c*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))^(-1 - p)*(d + e*x)^2*Hypergeome
tric2F1[-1 - 2*p, -p, -2*p, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqr
t[b^2 - 4*a*c] - 2*c*x))])/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(c*d^2 + e*(-(b*d) + a*e))*(1 + 2*p))))/(2*(c
*d^2 + e*(-(b*d) + a*e))*(3 + 2*p))

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Maple [F]  time = 1.299, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{-4-2\,p} \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)^(-4-2*p)*(c*x^2+b*x+a)^p,x)

[Out]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(-4-2*p)*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 4}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(-4-2*p)*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 4), 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)**(-4-2*p)*(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 (c x^{2} + b x + a\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 4}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(-4-2*p)*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 4), x)