3.2537 \(\int x^{-1-5 n} (a+b x^n)^2 \, dx\)

Optimal. Leaf size=45 \[ -\frac{a^2 x^{-5 n}}{5 n}-\frac{a b x^{-4 n}}{2 n}-\frac{b^2 x^{-3 n}}{3 n} \]

[Out]

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

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Rubi [A]  time = 0.0180822, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {266, 43} \[ -\frac{a^2 x^{-5 n}}{5 n}-\frac{a b x^{-4 n}}{2 n}-\frac{b^2 x^{-3 n}}{3 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 5*n)*(a + b*x^n)^2,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^{-1-5 n} \left (a+b x^n\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^6} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2}{x^6}+\frac{2 a b}{x^5}+\frac{b^2}{x^4}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac{a^2 x^{-5 n}}{5 n}-\frac{a b x^{-4 n}}{2 n}-\frac{b^2 x^{-3 n}}{3 n}\\ \end{align*}

Mathematica [A]  time = 0.0137462, size = 35, normalized size = 0.78 \[ -\frac{x^{-5 n} \left (6 a^2+15 a b x^n+10 b^2 x^{2 n}\right )}{30 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - 5*n)*(a + b*x^n)^2,x]

[Out]

-(6*a^2 + 15*a*b*x^n + 10*b^2*x^(2*n))/(30*n*x^(5*n))

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Maple [A]  time = 0.015, size = 45, normalized size = 1. \begin{align*}{\frac{1}{ \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{5}} \left ( -{\frac{{a}^{2}}{5\,n}}-{\frac{{b}^{2} \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}}{3\,n}}-{\frac{a{{\rm e}^{n\ln \left ( x \right ) }}b}{2\,n}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-5*n)*(a+b*x^n)^2,x)

[Out]

(-1/5*a^2/n-1/3*b^2/n*exp(n*ln(x))^2-1/2*a*b/n*exp(n*ln(x)))/exp(n*ln(x))^5

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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^(-1-5*n)*(a+b*x^n)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35583, size = 77, normalized size = 1.71 \begin{align*} -\frac{10 \, b^{2} x^{2 \, n} + 15 \, a b x^{n} + 6 \, a^{2}}{30 \, n x^{5 \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-5*n)*(a+b*x^n)^2,x, algorithm="fricas")

[Out]

-1/30*(10*b^2*x^(2*n) + 15*a*b*x^n + 6*a^2)/(n*x^(5*n))

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Sympy [A]  time = 11.5997, size = 44, normalized size = 0.98 \begin{align*} \begin{cases} - \frac{a^{2} x^{- 5 n}}{5 n} - \frac{a b x^{- 4 n}}{2 n} - \frac{b^{2} x^{- 3 n}}{3 n} & \text{for}\: n \neq 0 \\\left (a + b\right )^{2} \log{\left (x \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-5*n)*(a+b*x**n)**2,x)

[Out]

Piecewise((-a**2*x**(-5*n)/(5*n) - a*b*x**(-4*n)/(2*n) - b**2*x**(-3*n)/(3*n), Ne(n, 0)), ((a + b)**2*log(x),
True))

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Giac [A]  time = 1.23868, size = 47, normalized size = 1.04 \begin{align*} -\frac{10 \, b^{2} x^{2 \, n} + 15 \, a b x^{n} + 6 \, a^{2}}{30 \, n x^{5 \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-5*n)*(a+b*x^n)^2,x, algorithm="giac")

[Out]

-1/30*(10*b^2*x^(2*n) + 15*a*b*x^n + 6*a^2)/(n*x^(5*n))