3.2.7 \(\int (b+2 c x) (a+b x+c x^2)^p \, dx\)

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

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {\left (a+b x+c x^2\right )^{p+1}}{p+1} \end {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.95 \begin {gather*} \frac {(a+x (b+c x))^{p+1}}{p+1} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int (b+2 c x) \left (a+b x+c x^2\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 28, normalized size = 1.40 \begin {gather*} \frac {{\left (c x^{2} + b x + a\right )} {\left (c x^{2} + b x + a\right )}^{p}}{p + 1} \end {gather*}

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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giac [A]  time = 0.41, size = 20, normalized size = 1.00 \begin {gather*} \frac {{\left (c x^{2} + b x + a\right )}^{p + 1}}{p + 1} \end {gather*}

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 + 1)/(p + 1)

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maple [A]  time = 0.00, size = 21, normalized size = 1.05 \begin {gather*} \frac {\left (c \,x^{2}+b x +a \right )^{p +1}}{p +1} \end {gather*}

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)^(p+1)/(p+1)

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maxima [A]  time = 0.42, size = 20, normalized size = 1.00 \begin {gather*} \frac {{\left (c x^{2} + b x + a\right )}^{p + 1}}{p + 1} \end {gather*}

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]

(c*x^2 + b*x + a)^(p + 1)/(p + 1)

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mupad [B]  time = 2.04, size = 39, normalized size = 1.95 \begin {gather*} \left (\frac {a}{p+1}+\frac {b\,x}{p+1}+\frac {c\,x^2}{p+1}\right )\,{\left (c\,x^2+b\,x+a\right )}^p \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 57.11, size = 104, normalized size = 5.20 \begin {gather*} \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 {gather*}

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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