3.277 \(\int (d+e x+f x^2)^p (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2) \, dx\)

Optimal. Leaf size=57 \[ 2 c f x \left (d+e x+f x^2\right )^{p+1}-\frac{(c e (p+2)-b f (2 p+3)) \left (d+e x+f x^2\right )^{p+1}}{p+1} \]

[Out]

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

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Rubi [A]  time = 0.121235, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 69, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.029, Rules used = {1661, 629} \[ 2 c f x \left (d+e x+f x^2\right )^{p+1}-\frac{(c e (p+2)-b f (2 p+3)) \left (d+e x+f x^2\right )^{p+1}}{p+1} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2)^p*(-2*c*e^2 + 2*c*d*f + 3*b*e*f - c*e^2*p + 2*b*e*f*p + 2*b*f^2*(3 + 2*p)*x + 2*c*f^2*(3
+ 2*p)*x^2),x]

[Out]

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

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx &=2 c f x \left (d+e x+f x^2\right )^{1+p}+\frac{\int \left (-e f (3+2 p) (c e (2+p)-b f (3+2 p))-2 f^2 (3+2 p) (c e (2+p)-b f (3+2 p)) x\right ) \left (d+e x+f x^2\right )^p \, dx}{f (3+2 p)}\\ &=-\frac{(c e (2+p)-b f (3+2 p)) \left (d+e x+f x^2\right )^{1+p}}{1+p}+2 c f x \left (d+e x+f x^2\right )^{1+p}\\ \end{align*}

Mathematica [A]  time = 0.300149, size = 43, normalized size = 0.75 \[ \frac{(d+x (e+f x))^{p+1} (b f (2 p+3)-c e (p+2)+2 c f (p+1) x)}{p+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2)^p*(-2*c*e^2 + 2*c*d*f + 3*b*e*f - c*e^2*p + 2*b*e*f*p + 2*b*f^2*(3 + 2*p)*x + 2*c*
f^2*(3 + 2*p)*x^2),x]

[Out]

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

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Maple [A]  time = 0.049, size = 51, normalized size = 0.9 \begin{align*}{\frac{ \left ( f{x}^{2}+ex+d \right ) ^{1+p} \left ( 2\,cfxp+2\,bfp-cep+2\,cfx+3\,bf-2\,ce \right ) }{1+p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b*f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x)

[Out]

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

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Maxima [A]  time = 1.14622, size = 132, normalized size = 2.32 \begin{align*} \frac{{\left (2 \, c f^{2}{\left (p + 1\right )} x^{3} + b d f{\left (2 \, p + 3\right )} - c d e{\left (p + 2\right )} +{\left (b f^{2}{\left (2 \, p + 3\right )} + c e f p\right )} x^{2} +{\left (b e f{\left (2 \, p + 3\right )} -{\left (e^{2}{\left (p + 2\right )} - 2 \, d f{\left (p + 1\right )}\right )} c\right )} x\right )}{\left (f x^{2} + e x + d\right )}^{p}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b*f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x
, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.37041, size = 269, normalized size = 4.72 \begin{align*} \frac{{\left (2 \,{\left (c f^{2} p + c f^{2}\right )} x^{3} - 2 \, c d e + 3 \, b d f +{\left (3 \, b f^{2} +{\left (c e f + 2 \, b f^{2}\right )} p\right )} x^{2} -{\left (c d e - 2 \, b d f\right )} p -{\left (2 \, c e^{2} -{\left (2 \, c d + 3 \, b e\right )} f +{\left (c e^{2} - 2 \,{\left (c d + b e\right )} f\right )} p\right )} x\right )}{\left (f x^{2} + e x + d\right )}^{p}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b*f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x
, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 113.673, size = 483, normalized size = 8.47 \begin{align*} \begin{cases} \frac{2 b d f p \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{3 b d f \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 b e f p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{3 b e f x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 b f^{2} p x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{3 b f^{2} x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac{c d e p \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac{2 c d e \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 c d f p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 c d f x \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac{c e^{2} p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac{2 c e^{2} x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{c e f p x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 c f^{2} p x^{3} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac{2 c f^{2} x^{3} \left (d + e x + f x^{2}\right )^{p}}{p + 1} & \text{for}\: p \neq -1 \\b f \log{\left (\frac{e}{2 f} + x - \frac{\sqrt{- 4 d f + e^{2}}}{2 f} \right )} + b f \log{\left (\frac{e}{2 f} + x + \frac{\sqrt{- 4 d f + e^{2}}}{2 f} \right )} - c e \log{\left (\frac{e}{2 f} + x - \frac{\sqrt{- 4 d f + e^{2}}}{2 f} \right )} - c e \log{\left (\frac{e}{2 f} + x + \frac{\sqrt{- 4 d f + e^{2}}}{2 f} \right )} + 2 c f x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e*x+d)**p*(-2*c*e**2+2*c*d*f+3*b*e*f-c*e**2*p+2*b*e*f*p+2*b*f**2*(3+2*p)*x+2*c*f**2*(3+2*p)*
x**2),x)

[Out]

Piecewise((2*b*d*f*p*(d + e*x + f*x**2)**p/(p + 1) + 3*b*d*f*(d + e*x + f*x**2)**p/(p + 1) + 2*b*e*f*p*x*(d +
e*x + f*x**2)**p/(p + 1) + 3*b*e*f*x*(d + e*x + f*x**2)**p/(p + 1) + 2*b*f**2*p*x**2*(d + e*x + f*x**2)**p/(p
+ 1) + 3*b*f**2*x**2*(d + e*x + f*x**2)**p/(p + 1) - c*d*e*p*(d + e*x + f*x**2)**p/(p + 1) - 2*c*d*e*(d + e*x
+ f*x**2)**p/(p + 1) + 2*c*d*f*p*x*(d + e*x + f*x**2)**p/(p + 1) + 2*c*d*f*x*(d + e*x + f*x**2)**p/(p + 1) - c
*e**2*p*x*(d + e*x + f*x**2)**p/(p + 1) - 2*c*e**2*x*(d + e*x + f*x**2)**p/(p + 1) + c*e*f*p*x**2*(d + e*x + f
*x**2)**p/(p + 1) + 2*c*f**2*p*x**3*(d + e*x + f*x**2)**p/(p + 1) + 2*c*f**2*x**3*(d + e*x + f*x**2)**p/(p + 1
), Ne(p, -1)), (b*f*log(e/(2*f) + x - sqrt(-4*d*f + e**2)/(2*f)) + b*f*log(e/(2*f) + x + sqrt(-4*d*f + e**2)/(
2*f)) - c*e*log(e/(2*f) + x - sqrt(-4*d*f + e**2)/(2*f)) - c*e*log(e/(2*f) + x + sqrt(-4*d*f + e**2)/(2*f)) +
2*c*f*x, True))

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Giac [B]  time = 1.22389, size = 424, normalized size = 7.44 \begin{align*} \frac{2 \,{\left (f x^{2} + x e + d\right )}^{p} c f^{2} p x^{3} + 2 \,{\left (f x^{2} + x e + d\right )}^{p} b f^{2} p x^{2} + 2 \,{\left (f x^{2} + x e + d\right )}^{p} c f^{2} x^{3} +{\left (f x^{2} + x e + d\right )}^{p} c f p x^{2} e + 2 \,{\left (f x^{2} + x e + d\right )}^{p} c d f p x + 3 \,{\left (f x^{2} + x e + d\right )}^{p} b f^{2} x^{2} + 2 \,{\left (f x^{2} + x e + d\right )}^{p} b f p x e + 2 \,{\left (f x^{2} + x e + d\right )}^{p} b d f p + 2 \,{\left (f x^{2} + x e + d\right )}^{p} c d f x -{\left (f x^{2} + x e + d\right )}^{p} c p x e^{2} -{\left (f x^{2} + x e + d\right )}^{p} c d p e + 3 \,{\left (f x^{2} + x e + d\right )}^{p} b f x e + 3 \,{\left (f x^{2} + x e + d\right )}^{p} b d f - 2 \,{\left (f x^{2} + x e + d\right )}^{p} c x e^{2} - 2 \,{\left (f x^{2} + x e + d\right )}^{p} c d e}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b*f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x
, algorithm="giac")

[Out]

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