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

Optimal. Leaf size=60 \[ \frac{a^2 (a+b x)^{p+1}}{b^3 (p+1)}-\frac{2 a (a+b x)^{p+2}}{b^3 (p+2)}+\frac{(a+b x)^{p+3}}{b^3 (p+3)} \]

[Out]

(a^2*(a + b*x)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x)^(2 + p))/(b^3*(2 + p)) + (a + b*x)^(3 + p)/(b^3*(3 + p)
)

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Rubi [A]  time = 0.0213359, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {43} \[ \frac{a^2 (a+b x)^{p+1}}{b^3 (p+1)}-\frac{2 a (a+b x)^{p+2}}{b^3 (p+2)}+\frac{(a+b x)^{p+3}}{b^3 (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^p,x]

[Out]

(a^2*(a + b*x)^(1 + p))/(b^3*(1 + p)) - (2*a*(a + b*x)^(2 + p))/(b^3*(2 + p)) + (a + b*x)^(3 + p)/(b^3*(3 + p)
)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0267356, size = 57, normalized size = 0.95 \[ \frac{(a+b x)^{p+1} \left (2 a^2-2 a b (p+1) x+b^2 \left (p^2+3 p+2\right ) x^2\right )}{b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x)^p,x]

[Out]

((a + b*x)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x + b^2*(2 + 3*p + p^2)*x^2))/(b^3*(1 + p)*(2 + p)*(3 + p))

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Maple [A]  time = 0.005, size = 73, normalized size = 1.2 \begin{align*}{\frac{ \left ( bx+a \right ) ^{1+p} \left ({b}^{2}{p}^{2}{x}^{2}+3\,{b}^{2}p{x}^{2}-2\,abpx+2\,{x}^{2}{b}^{2}-2\,axb+2\,{a}^{2} \right ) }{{b}^{3} \left ({p}^{3}+6\,{p}^{2}+11\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)^(1+p)*(b^2*p^2*x^2+3*b^2*p*x^2-2*a*b*p*x+2*b^2*x^2-2*a*b*x+2*a^2)/b^3/(p^3+6*p^2+11*p+6)

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Maxima [A]  time = 0.954102, size = 92, normalized size = 1.53 \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{3} +{\left (p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + 2 \, a^{3}\right )}{\left (b x + a\right )}^{p}}{{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((p^2 + 3*p + 2)*b^3*x^3 + (p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + 2*a^3)*(b*x + a)^p/((p^3 + 6*p^2 + 11*p + 6)*b^
3)

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Fricas [A]  time = 2.10934, size = 188, normalized size = 3.13 \begin{align*} -\frac{{\left (2 \, a^{2} b p x -{\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} -{\left (a b^{2} p^{2} + a b^{2} p\right )} x^{2}\right )}{\left (b x + a\right )}^{p}}{b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b*p*x - (b^3*p^2 + 3*b^3*p + 2*b^3)*x^3 - 2*a^3 - (a*b^2*p^2 + a*b^2*p)*x^2)*(b*x + a)^p/(b^3*p^3 + 6*
b^3*p^2 + 11*b^3*p + 6*b^3)

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Sympy [A]  time = 1.18536, size = 597, normalized size = 9.95 \begin{align*} \begin{cases} \frac{a^{p} x^{3}}{3} & \text{for}\: b = 0 \\\frac{2 a^{2} \log{\left (\frac{a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac{a^{2}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac{4 a b x \log{\left (\frac{a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac{2 b^{2} x^{2} \log{\left (\frac{a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} - \frac{2 b^{2} x^{2}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} & \text{for}\: p = -3 \\- \frac{2 a^{2} \log{\left (\frac{a}{b} + x \right )}}{a b^{3} + b^{4} x} - \frac{2 a^{2}}{a b^{3} + b^{4} x} - \frac{2 a b x \log{\left (\frac{a}{b} + x \right )}}{a b^{3} + b^{4} x} + \frac{b^{2} x^{2}}{a b^{3} + b^{4} x} & \text{for}\: p = -2 \\\frac{a^{2} \log{\left (\frac{a}{b} + x \right )}}{b^{3}} - \frac{a x}{b^{2}} + \frac{x^{2}}{2 b} & \text{for}\: p = -1 \\\frac{2 a^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} - \frac{2 a^{2} b p x \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac{a b^{2} p^{2} x^{2} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac{a b^{2} p x^{2} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac{b^{3} p^{2} x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac{3 b^{3} p x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac{2 b^{3} x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**p*x**3/3, Eq(b, 0)), (2*a**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + a**2/(2*a**
2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*b**2*x*
*2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) - 2*b**2*x**2/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2
), Eq(p, -3)), (-2*a**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a**2/(a*b**3 + b**4*x) - 2*a*b*x*log(a/b + x)/(a*b*
*3 + b**4*x) + b**2*x**2/(a*b**3 + b**4*x), Eq(p, -2)), (a**2*log(a/b + x)/b**3 - a*x/b**2 + x**2/(2*b), Eq(p,
-1)), (2*a**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) - 2*a**2*b*p*x*(a + b*x)**p/(b**3*p
**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + a*b**2*p**2*x**2*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p +
6*b**3) + a*b**2*p*x**2*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + b**3*p**2*x**3*(a + b*x
)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + 3*b**3*p*x**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11
*b**3*p + 6*b**3) + 2*b**3*x**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3), True))

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Giac [B]  time = 1.0708, size = 189, normalized size = 3.15 \begin{align*} \frac{{\left (b x + a\right )}^{p} b^{3} p^{2} x^{3} +{\left (b x + a\right )}^{p} a b^{2} p^{2} x^{2} + 3 \,{\left (b x + a\right )}^{p} b^{3} p x^{3} +{\left (b x + a\right )}^{p} a b^{2} p x^{2} + 2 \,{\left (b x + a\right )}^{p} b^{3} x^{3} - 2 \,{\left (b x + a\right )}^{p} a^{2} b p x + 2 \,{\left (b x + a\right )}^{p} a^{3}}{b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)^p*b^3*p^2*x^3 + (b*x + a)^p*a*b^2*p^2*x^2 + 3*(b*x + a)^p*b^3*p*x^3 + (b*x + a)^p*a*b^2*p*x^2 + 2*(
b*x + a)^p*b^3*x^3 - 2*(b*x + a)^p*a^2*b*p*x + 2*(b*x + a)^p*a^3)/(b^3*p^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)