### 3.1440 $$\int (b d+2 c d x) (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=21 $\frac{d \left (a+b x+c x^2\right )^{p+1}}{p+1}$

[Out]

(d*(a + b*x + c*x^2)^(1 + p))/(1 + p)

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Rubi [A]  time = 0.0055643, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.045, Rules used = {629} $\frac{d \left (a+b x+c x^2\right )^{p+1}}{p+1}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0066071, size = 20, normalized size = 0.95 $\frac{d (a+x (b+c x))^{p+1}}{p+1}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(d*(a + x*(b + c*x))^(1 + p))/(1 + p)

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Maple [A]  time = 0.042, size = 22, normalized size = 1.1 \begin{align*}{\frac{d \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*d*x+b*d)*(c*x^2+b*x+a)^p,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.43547, size = 72, normalized size = 3.43 \begin{align*} \frac{{\left (c d x^{2} + b d x + a d\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*d*x+b*d)*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 90.2458, size = 112, normalized size = 5.33 \begin{align*} \begin{cases} \frac{a d \left (a + b x + c x^{2}\right )^{p}}{p + 1} + \frac{b d x \left (a + b x + c x^{2}\right )^{p}}{p + 1} + \frac{c d x^{2} \left (a + b x + c x^{2}\right )^{p}}{p + 1} & \text{for}\: p \neq -1 \\d \log{\left (\frac{b}{2 c} + x - \frac{\sqrt{- 4 a c + b^{2}}}{2 c} \right )} + d \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*d*x+b*d)*(c*x**2+b*x+a)**p,x)

[Out]

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

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Giac [B]  time = 1.23159, size = 76, normalized size = 3.62 \begin{align*} \frac{{\left (c x^{2} + b x + a\right )}^{p} c d x^{2} +{\left (c x^{2} + b x + a\right )}^{p} b d x +{\left (c x^{2} + b x + a\right )}^{p} a d}{p + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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