3.3.80 \(\int x^{-2-3 n} (a x^2+b x^3)^n \, dx\) [280]

Optimal. Leaf size=32 \[ -\frac {x^{-3 (1+n)} \left (a x^2+b x^3\right )^{1+n}}{a (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2039} \begin {gather*} -\frac {x^{-3 (n+1)} \left (a x^2+b x^3\right )^{n+1}}{a (n+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-2 - 3*n)*(a*x^2 + b*x^3)^n,x]

[Out]

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

Rule 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 30, normalized size = 0.94 \begin {gather*} -\frac {x^{-3 (1+n)} \left (x^2 (a+b x)\right )^{1+n}}{a (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.40, size = 36, normalized size = 1.12

method	result	size
gosper	\(-\frac {x^{-1-3 n} \left (b x +a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{n}}{a \left (1+n \right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.59, size = 38, normalized size = 1.19 \begin {gather*} -\frac {{\left (b x^{2} + a x\right )} {\left (b x^{3} + a x^{2}\right )}^{n} x^{-3 \, n - 2}}{a n + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{- 3 n - 2} \left (x^{2} \left (a + b x\right )\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.28, size = 54, normalized size = 1.69 \begin {gather*} -{\left (b\,x^3+a\,x^2\right )}^n\,\left (\frac {x}{x^{3\,n+2}\,\left (n+1\right )}+\frac {b\,x^2}{a\,x^{3\,n+2}\,\left (n+1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^n/x^(3*n + 2),x)

[Out]

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

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