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

Optimal result

Integrand size = 18, antiderivative size = 160 \[ \int (d+e x) \left (a+b x+c x^2\right )^p \, dx=\frac {e \left (a+b x+c x^2\right )^{1+p}}{2 c (1+p)}-\frac {2^p (2 c d-b e) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {654, 638} \[ \int (d+e x) \left (a+b x+c x^2\right )^p \, dx=\frac {e \left (a+b x+c x^2\right )^{p+1}}{2 c (p+1)}-\frac {2^p (2 c d-b e) \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (a+b x+c x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c (p+1) \sqrt {b^2-4 a c}} \]

[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*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.37 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.68 \[ \int (d+e x) \left (a+b x+c x^2\right )^p \, dx=\frac {1}{2} (a+x (b+c x))^p \left (e x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {2^p d \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)}\right ) \]

[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

Maple [F]

\[\int \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )^{p}d x\]

[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)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[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)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d+e x) \left (a+b x+c x^2\right )^p \, dx=\int \left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

[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)