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

Optimal. Leaf size=70 \[ \frac{b x^{-3 (n+1)} \left (a x^2+b x^3\right )^{n+1}}{a^2 (n+1) (n+2)}-\frac{x^{-3 n-4} \left (a x^2+b x^3\right )^{n+1}}{a (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.0543414, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2016, 2014} \[ \frac{b x^{-3 (n+1)} \left (a x^2+b x^3\right )^{n+1}}{a^2 (n+1) (n+2)}-\frac{x^{-3 n-4} \left (a x^2+b x^3\right )^{n+1}}{a (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[x^(-3 - 3*n)*(a*x^2 + b*x^3)^n,x]

[Out]

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

Mathematica [A]  time = 0.0208736, size = 44, normalized size = 0.63 \[ -\frac{x^{-3 n-4} (a n+a-b x) \left (x^2 (a+b x)\right )^{n+1}}{a^2 (n+1) (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 50, normalized size = 0.7 \begin{align*} -{\frac{{x}^{-2-3\,n} \left ( b{x}^{3}+a{x}^{2} \right ) ^{n} \left ( an-bx+a \right ) \left ( bx+a \right ) }{ \left ( 2+n \right ) \left ( 1+n \right ){a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.811785, size = 136, normalized size = 1.94 \begin{align*} -\frac{{\left (a b n x^{2} - b^{2} x^{3} +{\left (a^{2} n + a^{2}\right )} x\right )}{\left (b x^{3} + a x^{2}\right )}^{n} x^{-3 \, n - 3}}{a^{2} n^{2} + 3 \, a^{2} n + 2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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**(-3-3*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 - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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