3.2528 \(\int x^{-1+4 n} (a+b x^n)^2 \, dx\)

Optimal. Leaf size=45 \[ \frac{a^2 x^{4 n}}{4 n}+\frac{2 a b x^{5 n}}{5 n}+\frac{b^2 x^{6 n}}{6 n} \]

[Out]

(a^2*x^(4*n))/(4*n) + (2*a*b*x^(5*n))/(5*n) + (b^2*x^(6*n))/(6*n)

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Rubi [A]  time = 0.019128, 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} \[ \frac{a^2 x^{4 n}}{4 n}+\frac{2 a b x^{5 n}}{5 n}+\frac{b^2 x^{6 n}}{6 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 4*n)*(a + b*x^n)^2,x]

[Out]

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

Mathematica [A]  time = 0.0171584, size = 35, normalized size = 0.78 \[ \frac{x^{4 n} \left (15 a^2+24 a b x^n+10 b^2 x^{2 n}\right )}{60 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 4*n)*(a + b*x^n)^2,x]

[Out]

(x^(4*n)*(15*a^2 + 24*a*b*x^n + 10*b^2*x^(2*n)))/(60*n)

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Maple [A]  time = 0.016, size = 40, normalized size = 0.9 \begin{align*}{\frac{{b}^{2} \left ({x}^{n} \right ) ^{6}}{6\,n}}+{\frac{2\,ab \left ({x}^{n} \right ) ^{5}}{5\,n}}+{\frac{{a}^{2} \left ({x}^{n} \right ) ^{4}}{4\,n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+4*n)*(a+b*x^n)^2,x)

[Out]

1/6*b^2/n*(x^n)^6+2/5*a*b/n*(x^n)^5+1/4*a^2/n*(x^n)^4

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.05528, size = 80, normalized size = 1.78 \begin{align*} \frac{10 \, b^{2} x^{6 \, n} + 24 \, a b x^{5 \, n} + 15 \, a^{2} x^{4 \, n}}{60 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+4*n)*(a+b*x^n)^2,x, algorithm="fricas")

[Out]

1/60*(10*b^2*x^(6*n) + 24*a*b*x^(5*n) + 15*a^2*x^(4*n))/n

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Sympy [A]  time = 11.5647, size = 44, normalized size = 0.98 \begin{align*} \begin{cases} \frac{a^{2} x^{4 n}}{4 n} + \frac{2 a b x^{5 n}}{5 n} + \frac{b^{2} x^{6 n}}{6 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+4*n)*(a+b*x**n)**2,x)

[Out]

Piecewise((a**2*x**(4*n)/(4*n) + 2*a*b*x**(5*n)/(5*n) + b**2*x**(6*n)/(6*n), Ne(n, 0)), ((a + b)**2*log(x), Tr
ue))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{n} + a\right )}^{2} x^{4 \, n - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+4*n)*(a+b*x^n)^2,x, algorithm="giac")

[Out]

integrate((b*x^n + a)^2*x^(4*n - 1), x)