\(\int (d+e x)^2 (a+b x+c x^2)^p \, dx\) [2564]

Optimal result

Integrand size = 20, antiderivative size = 248 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\frac {e (2 c d-b e) (2+p) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}-\frac {2^p \left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \]

[Out]

1/2*e*(-b*e+2*c*d)*(2+p)*(c*x^2+b*x+a)^(p+1)/c^2/(p+1)/(3+2*p)+e*(e*x+d)*(c*x^2+b*x+a)^(p+1)/c/(3+2*p)-2^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)))*(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^2/(p+1
)/(3+2*p)/(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {756, 654, 638} \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=-\frac {2^p \left (a+b x+c x^2\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p-1} \left (-2 c e (a e+b d (2 p+3))+b^2 e^2 (p+2)+2 c^2 d^2 (2 p+3)\right ) \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {b+2 c x+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt {b^2-4 a c}}+\frac {e (p+2) (2 c d-b e) \left (a+b x+c x^2\right )^{p+1}}{2 c^2 (p+1) (2 p+3)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{p+1}}{c (2 p+3)} \]

[In]

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

[Out]

(e*(2*c*d - b*e)*(2 + p)*(a + b*x + c*x^2)^(1 + p))/(2*c^2*(1 + p)*(3 + 2*p)) + (e*(d + e*x)*(a + b*x + c*x^2)
^(1 + p))/(c*(3 + 2*p)) - (2^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 - S
qrt[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 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]

Rule 756

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)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}+\frac {\int \left (c d^2 (3+2 p)-e (a e+b d (1+p))+e (2 c d-b e) (2+p) x\right ) \left (a+b x+c x^2\right )^p \, dx}{c (3+2 p)} \\ & = \frac {e (2 c d-b e) (2+p) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (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 \left (a+b x+c x^2\right )^p \, dx}{2 c^2 (3+2 p)} \\ & = \frac {e (2 c d-b e) (2+p) \left (a+b x+c x^2\right )^{1+p}}{2 c^2 (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+b x+c x^2\right )^{1+p}}{c (3+2 p)}-\frac {2^p \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 (-\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] (verified)

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

Time = 0.78 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.67 \[ \int (d+e x)^2 \left (a+b x+c x^2\right )^p \, dx=\frac {1}{6} (a+x (b+c x))^p \left (6 d e x^2 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (2,-p,-p,3,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+2 e^2 x^3 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\frac {3\ 2^p d^2 \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {-b+\sqrt {b^2-4 a c}-2 c x}{2 \sqrt {b^2-4 a c}}\right )}{c (1+p)}\right ) \]

[In]

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

[Out]

((a + x*(b + c*x))^p*((6*d*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*e^2*x^3*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])])/(((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) + (3*2^p*d^2*(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)))/6

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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