3.2741 \(\int x^m (a+b x^{2+2 m})^2 \, dx\)

Optimal. Leaf size=52 \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{3 (m+1)}}{3 (m+1)}+\frac{b^2 x^{5 (m+1)}}{5 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.021432, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {270} \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{3 (m+1)}}{3 (m+1)}+\frac{b^2 x^{5 (m+1)}}{5 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^(2 + 2*m))^2,x]

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x^m \left (a+b x^{2+2 m}\right )^2 \, dx &=\int \left (a^2 x^m+2 a b x^{2+3 m}+b^2 x^{4+5 m}\right ) \, dx\\ &=\frac{a^2 x^{1+m}}{1+m}+\frac{2 a b x^{3 (1+m)}}{3 (1+m)}+\frac{b^2 x^{5 (1+m)}}{5 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0238763, size = 42, normalized size = 0.81 \[ \frac{15 a^2 x^{m+1}+10 a b x^{3 m+3}+3 b^2 x^{5 m+5}}{15 m+15} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*x^(1 + m) + 10*a*b*x^(3 + 3*m) + 3*b^2*x^(5 + 5*m))/(15 + 15*m)

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Maple [A]  time = 0.014, size = 50, normalized size = 1. \begin{align*}{\frac{{b}^{2}{x}^{5} \left ({x}^{m} \right ) ^{5}}{5+5\,m}}+{\frac{2\,{x}^{3}ab \left ({x}^{m} \right ) ^{3}}{3+3\,m}}+{\frac{{a}^{2}x{x}^{m}}{1+m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5/(1+m)*(x^m)^5+2/3*x^3*a*b/(1+m)*(x^m)^3+a^2/(1+m)*x*x^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*(a+b*x^(2+2*m))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32061, size = 95, normalized size = 1.83 \begin{align*} \frac{3 \, b^{2} x^{5} x^{5 \, m} + 10 \, a b x^{3} x^{3 \, m} + 15 \, a^{2} x x^{m}}{15 \,{\left (m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*x^5*x^(5*m) + 10*a*b*x^3*x^(3*m) + 15*a^2*x*x^m)/(m + 1)

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Sympy [A]  time = 140.369, size = 61, normalized size = 1.17 \begin{align*} \begin{cases} \frac{15 a^{2} x x^{m}}{15 m + 15} + \frac{10 a b x^{3} x^{3 m}}{15 m + 15} + \frac{3 b^{2} x^{5} x^{5 m}}{15 m + 15} & \text{for}\: m \neq -1 \\\left (a + b\right )^{2} \log{\left (x \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((15*a**2*x*x**m/(15*m + 15) + 10*a*b*x**3*x**(3*m)/(15*m + 15) + 3*b**2*x**5*x**(5*m)/(15*m + 15), N
e(m, -1)), ((a + b)**2*log(x), True))

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Giac [A]  time = 1.12214, size = 57, normalized size = 1.1 \begin{align*} \frac{3 \, b^{2} x^{5} x^{5 \, m} + 10 \, a b x^{3} x^{3 \, m} + 15 \, a^{2} x x^{m}}{15 \,{\left (m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*x^5*x^(5*m) + 10*a*b*x^3*x^(3*m) + 15*a^2*x*x^m)/(m + 1)