3.282 \(\int x^{-4-3 n} (a x^2+b x^3)^n \, dx\)

Optimal. Leaf size=116 \[ -\frac{2 b^2 x^{-3 (n+1)} \left (a x^2+b x^3\right )^{n+1}}{a^3 (n+1) (n+2) (n+3)}+\frac{2 b x^{-3 n-4} \left (a x^2+b x^3\right )^{n+1}}{a^2 (n+2) (n+3)}-\frac{x^{-3 n-5} \left (a x^2+b x^3\right )^{n+1}}{a (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.0914168, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2016, 2014} \[ -\frac{2 b^2 x^{-3 (n+1)} \left (a x^2+b x^3\right )^{n+1}}{a^3 (n+1) (n+2) (n+3)}+\frac{2 b x^{-3 n-4} \left (a x^2+b x^3\right )^{n+1}}{a^2 (n+2) (n+3)}-\frac{x^{-3 n-5} \left (a x^2+b x^3\right )^{n+1}}{a (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-4 - 3*n)*(a*x^2 + b*x^3)^n,x]

[Out]

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

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

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

Mathematica [A]  time = 0.0286695, size = 72, normalized size = 0.62 \[ -\frac{x^{-3 (n+1)} (a+b x) \left (x^2 (a+b x)\right )^n \left (a^2 \left (n^2+3 n+2\right )-2 a b (n+1) x+2 b^2 x^2\right )}{a^3 (n+1) (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-4 - 3*n)*(a*x^2 + b*x^3)^n,x]

[Out]

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

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Maple [A]  time = 0.005, size = 84, normalized size = 0.7 \begin{align*} -{\frac{ \left ( bx+a \right ){x}^{-3-3\,n} \left ({a}^{2}{n}^{2}-2\,abnx+2\,{b}^{2}{x}^{2}+3\,{a}^{2}n-2\,abx+2\,{a}^{2} \right ) \left ( b{x}^{3}+a{x}^{2} \right ) ^{n}}{ \left ( 3+n \right ) \left ( 2+n \right ) \left ( 1+n \right ){a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-4-3*n)*(b*x^3+a*x^2)^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a x^{2}\right )}^{n} x^{-3 \, n - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-4-3*n)*(b*x^3+a*x^2)^n,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^n*x^(-3*n - 4), x)

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Fricas [A]  time = 0.91543, size = 217, normalized size = 1.87 \begin{align*} \frac{{\left (2 \, a b^{2} n x^{3} - 2 \, b^{3} x^{4} -{\left (a^{2} b n^{2} + a^{2} b n\right )} x^{2} -{\left (a^{3} n^{2} + 3 \, a^{3} n + 2 \, a^{3}\right )} x\right )}{\left (b x^{3} + a x^{2}\right )}^{n} x^{-3 \, n - 4}}{a^{3} n^{3} + 6 \, a^{3} n^{2} + 11 \, a^{3} n + 6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-4-3*n)*(b*x^3+a*x^2)^n,x, algorithm="fricas")

[Out]

(2*a*b^2*n*x^3 - 2*b^3*x^4 - (a^2*b*n^2 + a^2*b*n)*x^2 - (a^3*n^2 + 3*a^3*n + 2*a^3)*x)*(b*x^3 + a*x^2)^n*x^(-
3*n - 4)/(a^3*n^3 + 6*a^3*n^2 + 11*a^3*n + 6*a^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-4-3*n)*(b*x**3+a*x**2)**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a x^{2}\right )}^{n} x^{-3 \, n - 4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-4-3*n)*(b*x^3+a*x^2)^n,x, algorithm="giac")

[Out]

integrate((b*x^3 + a*x^2)^n*x^(-3*n - 4), x)