3.26.65 \(\int (d+e x) (a+b x+c x^2)^p \, dx\) [2565]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 160, normalized size of antiderivative = 1.00, 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 = {654, 638} \begin {gather*} \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}} \end {gather*}

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 638

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)/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/
(2*q)], x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rule 654

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]

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.38, size = 268, normalized size = 1.68 \begin {gather*} \frac {1}{2} (a+x (b+c x))^p \left (e x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} F_1\left (2;-p,-p;3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {2^p d \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-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 (1+p)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 = 0.36, size = 0, normalized size = 0.00 \[\int \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((x*e + d)*(c*x^2 + b*x + a)^p, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x*e + d)*(c*x^2 + b*x + a)^p, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right ) \left (a + b x + c x^{2}\right )^{p}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x*e + d)*(c*x^2 + b*x + a)^p, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)*(a + b*x + c*x^2)^p,x)

[Out]

int((d + e*x)*(a + b*x + c*x^2)^p, x)

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