3.1098 \(\int (A+B x) (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=158 \[ \frac {2^p (b B-2 A c) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1) \sqrt {b^2-4 a c}}+\frac {B \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)} \]

[Out]

1/2*B*(c*x^2+b*x+a)^(1+p)/c/(1+p)+2^p*(-2*A*c+B*b)*(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 = 0.05, antiderivative size = 158, 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 = {640, 624} \[ \frac {2^p (b B-2 A c) \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1) \sqrt {b^2-4 a c}}+\frac {B \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(a + b*x + c*x^2)^p,x]

[Out]

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

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]

Rubi steps

\begin {align*} \int (A+B x) \left (a+b x+c x^2\right )^p \, dx &=\frac {B \left (a+b x+c x^2\right )^{1+p}}{2 c (1+p)}+\frac {(-b B+2 A c) \int \left (a+b x+c x^2\right )^p \, dx}{2 c}\\ &=\frac {B \left (a+b x+c x^2\right )^{1+p}}{2 c (1+p)}+\frac {2^p (b B-2 A c) \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]  time = 0.40, size = 268, normalized size = 1.70 \[ \frac {1}{2} (a+x (b+c x))^p \left (\frac {A 2^p \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \, _2F_1\left (-p,p+1;p+2;\frac {-b-2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1)}+B x^2 \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )^{-p} F_1\left (2;-p,-p;3;-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + x*(b + c*x))^p*((B*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*A*(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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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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maple [F]  time = 1.18, size = 0, normalized size = 0.00 \[ \int \left (B x +A \right ) \left (c \,x^{2}+b x +a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)*(a + b*x + c*x^2)^p,x)

[Out]

int((A + B*x)*(a + b*x + c*x^2)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B x\right ) \left (a + b x + c x^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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