3.1097 \(\int x (A+B x) (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=211 \[ \frac{2^p \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \left (2 a B c+A b c (2 p+3)+b^2 (-B) (p+2)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{\left (a+b x+c x^2\right )^{p+1} (-A c (2 p+3)+b B (p+2)-2 B c (p+1) x)}{2 c^2 (p+1) (2 p+3)} \]

[Out]

-((b*B*(2 + p) - A*c*(3 + 2*p) - 2*B*c*(1 + p)*x)*(a + b*x + c*x^2)^(1 + p))/(2*c^2*(1 + p)*(3 + 2*p)) + (2^p*
(2*a*B*c - b^2*B*(2 + p) + A*b*c*(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^2*Sqrt[b^2 - 4*a*c]*(1 + p)*(3 + 2*p))

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Rubi [A]  time = 0.0893853, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {779, 624} \[ \frac{2^p \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \left (2 a B c+A b c (2 p+3)+b^2 (-B) (p+2)\right ) \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt{b^2-4 a c}}-\frac{\left (a+b x+c x^2\right )^{p+1} (-A c (2 p+3)+b B (p+2)-2 B c (p+1) x)}{2 c^2 (p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + b*x + c*x^2)^p,x]

[Out]

-((b*B*(2 + p) - A*c*(3 + 2*p) - 2*B*c*(1 + p)*x)*(a + b*x + c*x^2)^(1 + p))/(2*c^2*(1 + p)*(3 + 2*p)) + (2^p*
(2*a*B*c - b^2*B*(2 + p) + A*b*c*(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^2*Sqrt[b^2 - 4*a*c]*(1 + p)*(3 + 2*p))

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 x (A+B x) \left (a+b x+c x^2\right )^p \, dx &=-\frac{(b B (2+p)-A c (3+2 p)-2 B c (1+p) x) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}-\frac{\left (2 a B c-b^2 B (2+p)+A b c (3+2 p)\right ) \int \left (a+b x+c x^2\right )^p \, dx}{2 c^2 (3+2 p)}\\ &=-\frac{(b B (2+p)-A c (3+2 p)-2 B c (1+p) x) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac{2^p \left (2 a B c-b^2 B (2+p)+A b c (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 = 0.296914, size = 210, normalized size = 1. \[ \frac{1}{6} x^2 \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} (a+x (b+c x))^p \left (3 A F_1\left (2;-p,-p;3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+2 B x F_1\left (3;-p,-p;4;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(A + B*x)*(a + b*x + c*x^2)^p,x]

[Out]

(x^2*(a + x*(b + c*x))^p*(3*A*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])] + 2*B*x*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])])
)/(6*((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)

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Maple [F]  time = 0.086, size = 0, normalized size = 0. \begin{align*} \int x \left ( Bx+A \right ) \left ( c{x}^{2}+bx+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)*(c*x^2+b*x+a)^p,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((B*x + A)*(c*x^2 + b*x + a)^p*x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B x^{2} + A x\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(x*(B*x+A)*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((B*x + A)*(c*x^2 + b*x + a)^p*x, x)