### 3.129 $$\int (b+2 c x) (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=20 $\frac{\left (a+b x+c x^2\right )^{p+1}}{p+1}$

[Out]

(a + b*x + c*x^2)^(1 + p)/(1 + p)

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Rubi [A]  time = 0.0054882, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.053, Rules used = {629} $\frac{\left (a+b x+c x^2\right )^{p+1}}{p+1}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a + b*x + c*x^2)^(1 + p)/(1 + p)

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 (b+2 c x) \left (a+b x+c x^2\right )^p \, dx &=\frac{\left (a+b x+c x^2\right )^{1+p}}{1+p}\\ \end{align*}

Mathematica [A]  time = 0.0078453, size = 19, normalized size = 0.95 $\frac{(a+x (b+c x))^{p+1}}{p+1}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(a + x*(b + c*x))^(1 + p)/(1 + p)

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Maple [A]  time = 0.004, size = 21, normalized size = 1.1 \begin{align*}{\frac{ \left ( c{x}^{2}+bx+a \right ) ^{1+p}}{1+p}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2+b*x+a)^(1+p)/(1+p)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.07846, size = 63, normalized size = 3.15 \begin{align*} \frac{{\left (c x^{2} + b x + a\right )}{\left (c x^{2} + b x + a\right )}^{p}}{p + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*x^2 + b*x + a)*(c*x^2 + b*x + a)^p/(p + 1)

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Sympy [B]  time = 51.315, size = 104, normalized size = 5.2 \begin{align*} \begin{cases} \frac{a \left (a + b x + c x^{2}\right )^{p}}{p + 1} + \frac{b x \left (a + b x + c x^{2}\right )^{p}}{p + 1} + \frac{c x^{2} \left (a + b x + c x^{2}\right )^{p}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (\frac{b}{2 c} + x - \frac{\sqrt{- 4 a c + b^{2}}}{2 c} \right )} + \log{\left (\frac{b}{2 c} + x + \frac{\sqrt{- 4 a c + b^{2}}}{2 c} \right )} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*(a + b*x + c*x**2)**p/(p + 1) + b*x*(a + b*x + c*x**2)**p/(p + 1) + c*x**2*(a + b*x + c*x**2)**p/
(p + 1), Ne(p, -1)), (log(b/(2*c) + x - sqrt(-4*a*c + b**2)/(2*c)) + log(b/(2*c) + x + sqrt(-4*a*c + b**2)/(2*
c)), True))

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Giac [B]  time = 1.11774, size = 72, normalized size = 3.6 \begin{align*} \frac{{\left (c x^{2} + b x + a\right )}^{p} c x^{2} +{\left (c x^{2} + b x + a\right )}^{p} b x +{\left (c x^{2} + b x + a\right )}^{p} a}{p + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c*x^2 + b*x + a)^p*c*x^2 + (c*x^2 + b*x + a)^p*b*x + (c*x^2 + b*x + a)^p*a)/(p + 1)