3.2161 \(\int (a+b x) (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=32 \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0081101, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {629} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0120167, size = 23, normalized size = 0.72 \[ \frac{\left ((a+b x)^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^2)^(1 + p)/(2*b*(1 + p))

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Maple [A]  time = 0.001, size = 36, normalized size = 1.1 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{2\,b \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x+a)^2/b/(1+p)*(b^2*x^2+2*a*b*x+a^2)^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((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^p,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54481, size = 93, normalized size = 2.91 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.478004, size = 119, normalized size = 3.72 \begin{align*} \begin{cases} \frac{x}{a} & \text{for}\: b = 0 \wedge p = -1 \\a x \left (a^{2}\right )^{p} & \text{for}\: b = 0 \\\frac{\log{\left (\frac{a}{b} + x \right )}}{b} & \text{for}\: p = -1 \\\frac{a^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{2 b p + 2 b} + \frac{2 a b x \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{2 b p + 2 b} + \frac{b^{2} x^{2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{2 b p + 2 b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/a, Eq(b, 0) & Eq(p, -1)), (a*x*(a**2)**p, Eq(b, 0)), (log(a/b + x)/b, Eq(p, -1)), (a**2*(a**2 + 2
*a*b*x + b**2*x**2)**p/(2*b*p + 2*b) + 2*a*b*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(2*b*p + 2*b) + b**2*x**2*(a**2
+ 2*a*b*x + b**2*x**2)**p/(2*b*p + 2*b), True))

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Giac [B]  time = 1.1604, size = 108, normalized size = 3.38 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{2} x^{2} + 2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b x +{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2}}{2 \,{\left (b p + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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