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

Optimal. Leaf size=43 \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{m+3}}{m+3}+\frac{b^2 x^{m+5}}{m+5} \]

[Out]

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

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Rubi [A]  time = 0.0151615, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {270} \[ \frac{a^2 x^{m+1}}{m+1}+\frac{2 a b x^{m+3}}{m+3}+\frac{b^2 x^{m+5}}{m+5} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(a + b*x^2)^2,x]

[Out]

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

Mathematica [A]  time = 0.0194217, size = 40, normalized size = 0.93 \[ x^{m+1} \left (\frac{a^2}{m+1}+\frac{2 a b x^2}{m+3}+\frac{b^2 x^4}{m+5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(a + b*x^2)^2,x]

[Out]

x^(1 + m)*(a^2/(1 + m) + (2*a*b*x^2)/(3 + m) + (b^2*x^4)/(5 + m))

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Maple [B]  time = 0.003, size = 93, normalized size = 2.2 \begin{align*}{\frac{{x}^{1+m} \left ({b}^{2}{m}^{2}{x}^{4}+4\,{b}^{2}m{x}^{4}+2\,ab{m}^{2}{x}^{2}+3\,{b}^{2}{x}^{4}+12\,abm{x}^{2}+{a}^{2}{m}^{2}+10\,ab{x}^{2}+8\,{a}^{2}m+15\,{a}^{2} \right ) }{ \left ( 5+m \right ) \left ( 3+m \right ) \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1+m)*(b^2*m^2*x^4+4*b^2*m*x^4+2*a*b*m^2*x^2+3*b^2*x^4+12*a*b*m*x^2+a^2*m^2+10*a*b*x^2+8*a^2*m+15*a^2)/(5+m)
/(3+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^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57655, size = 181, normalized size = 4.21 \begin{align*} \frac{{\left ({\left (b^{2} m^{2} + 4 \, b^{2} m + 3 \, b^{2}\right )} x^{5} + 2 \,{\left (a b m^{2} + 6 \, a b m + 5 \, a b\right )} x^{3} +{\left (a^{2} m^{2} + 8 \, a^{2} m + 15 \, a^{2}\right )} x\right )} x^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*m^2 + 4*b^2*m + 3*b^2)*x^5 + 2*(a*b*m^2 + 6*a*b*m + 5*a*b)*x^3 + (a^2*m^2 + 8*a^2*m + 15*a^2)*x)*x^m/(m^
3 + 9*m^2 + 23*m + 15)

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Sympy [A]  time = 0.841628, size = 306, normalized size = 7.12 \begin{align*} \begin{cases} - \frac{a^{2}}{4 x^{4}} - \frac{a b}{x^{2}} + b^{2} \log{\left (x \right )} & \text{for}\: m = -5 \\- \frac{a^{2}}{2 x^{2}} + 2 a b \log{\left (x \right )} + \frac{b^{2} x^{2}}{2} & \text{for}\: m = -3 \\a^{2} \log{\left (x \right )} + a b x^{2} + \frac{b^{2} x^{4}}{4} & \text{for}\: m = -1 \\\frac{a^{2} m^{2} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{8 a^{2} m x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{15 a^{2} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{2 a b m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{12 a b m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{10 a b x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{b^{2} m^{2} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{4 b^{2} m x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{3 b^{2} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2/(4*x**4) - a*b/x**2 + b**2*log(x), Eq(m, -5)), (-a**2/(2*x**2) + 2*a*b*log(x) + b**2*x**2/2,
Eq(m, -3)), (a**2*log(x) + a*b*x**2 + b**2*x**4/4, Eq(m, -1)), (a**2*m**2*x*x**m/(m**3 + 9*m**2 + 23*m + 15) +
8*a**2*m*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 15*a**2*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 2*a*b*m**2*x**3*x*
*m/(m**3 + 9*m**2 + 23*m + 15) + 12*a*b*m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 10*a*b*x**3*x**m/(m**3 + 9*m
**2 + 23*m + 15) + b**2*m**2*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15) + 4*b**2*m*x**5*x**m/(m**3 + 9*m**2 + 23*m
+ 15) + 3*b**2*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15), True))

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Giac [B]  time = 2.97042, size = 158, normalized size = 3.67 \begin{align*} \frac{b^{2} m^{2} x^{5} x^{m} + 4 \, b^{2} m x^{5} x^{m} + 2 \, a b m^{2} x^{3} x^{m} + 3 \, b^{2} x^{5} x^{m} + 12 \, a b m x^{3} x^{m} + a^{2} m^{2} x x^{m} + 10 \, a b x^{3} x^{m} + 8 \, a^{2} m x x^{m} + 15 \, a^{2} x x^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*m^2*x^5*x^m + 4*b^2*m*x^5*x^m + 2*a*b*m^2*x^3*x^m + 3*b^2*x^5*x^m + 12*a*b*m*x^3*x^m + a^2*m^2*x*x^m + 10
*a*b*x^3*x^m + 8*a^2*m*x*x^m + 15*a^2*x*x^m)/(m^3 + 9*m^2 + 23*m + 15)