3.435 \(\int x^m (a+b x)^2 \, dx\)

Optimal. Leaf size=43 \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{m+2}}{m+2}+\frac{b^2 x^{m+3}}{m+3} \]

[Out]

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

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Rubi [A]  time = 0.0142616, antiderivative size = 43, 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} \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{m+2}}{m+2}+\frac{b^2 x^{m+3}}{m+3} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0315192, size = 38, normalized size = 0.88 \[ x^{m+1} \left (\frac{a^2}{m+1}+\frac{2 a b x}{m+2}+\frac{b^2 x^2}{m+3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x)^2,x]

[Out]

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

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Maple [A]  time = 0.004, size = 87, normalized size = 2. \begin{align*}{\frac{{x}^{1+m} \left ({b}^{2}{m}^{2}{x}^{2}+2\,ab{m}^{2}x+3\,{b}^{2}m{x}^{2}+{a}^{2}{m}^{2}+8\,abmx+2\,{b}^{2}{x}^{2}+5\,{a}^{2}m+6\,abx+6\,{a}^{2} \right ) }{ \left ( 3+m \right ) \left ( 2+m \right ) \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1+m)*(b^2*m^2*x^2+2*a*b*m^2*x+3*b^2*m*x^2+a^2*m^2+8*a*b*m*x+2*b^2*x^2+5*a^2*m+6*a*b*x+6*a^2)/(3+m)/(2+m)/(1
+m)

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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(x^m*(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73606, size = 178, normalized size = 4.14 \begin{align*} \frac{{\left ({\left (b^{2} m^{2} + 3 \, b^{2} m + 2 \, b^{2}\right )} x^{3} + 2 \,{\left (a b m^{2} + 4 \, a b m + 3 \, a b\right )} x^{2} +{\left (a^{2} m^{2} + 5 \, a^{2} m + 6 \, a^{2}\right )} x\right )} x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*m^2 + 3*b^2*m + 2*b^2)*x^3 + 2*(a*b*m^2 + 4*a*b*m + 3*a*b)*x^2 + (a^2*m^2 + 5*a^2*m + 6*a^2)*x)*x^m/(m^3
+ 6*m^2 + 11*m + 6)

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Sympy [A]  time = 0.719145, size = 299, normalized size = 6.95 \begin{align*} \begin{cases} - \frac{a^{2}}{2 x^{2}} - \frac{2 a b}{x} + b^{2} \log{\left (x \right )} & \text{for}\: m = -3 \\- \frac{a^{2}}{x} + 2 a b \log{\left (x \right )} + b^{2} x & \text{for}\: m = -2 \\a^{2} \log{\left (x \right )} + 2 a b x + \frac{b^{2} x^{2}}{2} & \text{for}\: m = -1 \\\frac{a^{2} m^{2} x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{5 a^{2} m x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{6 a^{2} x x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{2 a b m^{2} x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{8 a b m x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{6 a b x^{2} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{b^{2} m^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{3 b^{2} m x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} + \frac{2 b^{2} x^{3} x^{m}}{m^{3} + 6 m^{2} + 11 m + 6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2/(2*x**2) - 2*a*b/x + b**2*log(x), Eq(m, -3)), (-a**2/x + 2*a*b*log(x) + b**2*x, Eq(m, -2)), (
a**2*log(x) + 2*a*b*x + b**2*x**2/2, Eq(m, -1)), (a**2*m**2*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 5*a**2*m*x*x**
m/(m**3 + 6*m**2 + 11*m + 6) + 6*a**2*x*x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*a*b*m**2*x**2*x**m/(m**3 + 6*m**2
+ 11*m + 6) + 8*a*b*m*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + 6*a*b*x**2*x**m/(m**3 + 6*m**2 + 11*m + 6) + b**2
*m**2*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 3*b**2*m*x**3*x**m/(m**3 + 6*m**2 + 11*m + 6) + 2*b**2*x**3*x**m/
(m**3 + 6*m**2 + 11*m + 6), True))

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Giac [B]  time = 1.23421, size = 158, normalized size = 3.67 \begin{align*} \frac{b^{2} m^{2} x^{3} x^{m} + 2 \, a b m^{2} x^{2} x^{m} + 3 \, b^{2} m x^{3} x^{m} + a^{2} m^{2} x x^{m} + 8 \, a b m x^{2} x^{m} + 2 \, b^{2} x^{3} x^{m} + 5 \, a^{2} m x x^{m} + 6 \, a b x^{2} x^{m} + 6 \, a^{2} x x^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*m^2*x^3*x^m + 2*a*b*m^2*x^2*x^m + 3*b^2*m*x^3*x^m + a^2*m^2*x*x^m + 8*a*b*m*x^2*x^m + 2*b^2*x^3*x^m + 5*a
^2*m*x*x^m + 6*a*b*x^2*x^m + 6*a^2*x*x^m)/(m^3 + 6*m^2 + 11*m + 6)