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

Optimal. Leaf size=48 \[ \frac{x^{-3 n-1} \left (a x^2+b x^3\right )^{n+1} \, _2F_1\left (1,2;2-n;-\frac{b x}{a}\right )}{a (1-n)} \]

[Out]

(x^(-1 - 3*n)*(a*x^2 + b*x^3)^(1 + n)*Hypergeometric2F1[1, 2, 2 - n, -((b*x)/a)])/(a*(1 - n))

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Rubi [A]  time = 0.0371099, antiderivative size = 61, normalized size of antiderivative = 1.27, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2032, 66, 64} \[ \frac{x^{1-3 n} \left (\frac{b x}{a}+1\right )^{-n} \left (a x^2+b x^3\right )^n \, _2F_1\left (1-n,-n;2-n;-\frac{b x}{a}\right )}{1-n} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^n/x^(3*n),x]

[Out]

(x^(1 - 3*n)*(a*x^2 + b*x^3)^n*Hypergeometric2F1[1 - n, -n, 2 - n, -((b*x)/a)])/((1 - n)*(1 + (b*x)/a)^n)

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c^IntPart[n]*(c + d*x)^FracPart[n])/(1 + (d
*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0]))
||  !RationalQ[n])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int x^{-3 n} \left (a x^2+b x^3\right )^n \, dx &=\left (x^{-2 n} (a+b x)^{-n} \left (a x^2+b x^3\right )^n\right ) \int x^{-n} (a+b x)^n \, dx\\ &=\left (x^{-2 n} \left (1+\frac{b x}{a}\right )^{-n} \left (a x^2+b x^3\right )^n\right ) \int x^{-n} \left (1+\frac{b x}{a}\right )^n \, dx\\ &=\frac{x^{1-3 n} \left (1+\frac{b x}{a}\right )^{-n} \left (a x^2+b x^3\right )^n \, _2F_1\left (1-n,-n;2-n;-\frac{b x}{a}\right )}{1-n}\\ \end{align*}

Mathematica [A]  time = 0.012427, size = 59, normalized size = 1.23 \[ \frac{x^{1-3 n} \left (x^2 (a+b x)\right )^n \left (\frac{b x}{a}+1\right )^{-n} \, _2F_1\left (1-n,-n;2-n;-\frac{b x}{a}\right )}{1-n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 - 3*n)*(x^2*(a + b*x))^n*Hypergeometric2F1[1 - n, -n, 2 - n, -((b*x)/a)])/((1 - n)*(1 + (b*x)/a)^n)

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Maple [F]  time = 0.252, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{3}+a{x}^{2} \right ) ^{n}}{{x}^{3\,n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{3} + a x^{2}\right )}^{n}}{x^{3 \, n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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