3.26.30 \(\int x^{-1+2 n} (a+b x^n)^2 \, dx\) [2530]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 = {272, 45} \begin {gather*} \frac {a^2 x^{2 n}}{2 n}+\frac {2 a b x^{3 n}}{3 n}+\frac {b^2 x^{4 n}}{4 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*(a + b*x^n)^2,x]

[Out]

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

Rule 45

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])

Rule 272

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]]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.78 \begin {gather*} \frac {x^{2 n} \left (6 a^2+8 a b x^n+3 b^2 x^{2 n}\right )}{12 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(2*n)*(6*a^2 + 8*a*b*x^n + 3*b^2*x^(2*n)))/(12*n)

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Maple [A]
time = 0.21, size = 40, normalized size = 0.89

method	result	size
risch	\(\frac {a^{2} x^{2 n}}{2 n}+\frac {2 a b \,x^{3 n}}{3 n}+\frac {b^{2} x^{4 n}}{4 n}\)	\(40\)
norman	\(\frac {a^{2} {\mathrm e}^{2 n \ln \left (x \right )}}{2 n}+\frac {b^{2} {\mathrm e}^{4 n \ln \left (x \right )}}{4 n}+\frac {2 a b \,{\mathrm e}^{3 n \ln \left (x \right )}}{3 n}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+2*n)*(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 39, normalized size = 0.87 \begin {gather*} \frac {b^{2} x^{4 \, n}}{4 \, n} + \frac {2 \, a b x^{3 \, n}}{3 \, n} + \frac {a^{2} x^{2 \, n}}{2 \, n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 35, normalized size = 0.78 \begin {gather*} \frac {3 \, b^{2} x^{4 \, n} + 8 \, a b x^{3 \, n} + 6 \, a^{2} x^{2 \, n}}{12 \, n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^2*x^(4*n) + 8*a*b*x^(3*n) + 6*a^2*x^(2*n))/n

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Sympy [A]
time = 1.12, size = 44, normalized size = 0.98 \begin {gather*} \begin {cases} \frac {a^{2} x^{2 n}}{2 n} + \frac {2 a b x^{3 n}}{3 n} + \frac {b^{2} x^{4 n}}{4 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{2} \log {\left (x \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**(2*n)/(2*n) + 2*a*b*x**(3*n)/(3*n) + b**2*x**(4*n)/(4*n), Ne(n, 0)), ((a + b)**2*log(x), Tr
ue))

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

[Out]

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

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Mupad [B]
time = 1.29, size = 33, normalized size = 0.73 \begin {gather*} \frac {x^{2\,n}\,\left (6\,a^2+3\,b^2\,x^{2\,n}+8\,a\,b\,x^n\right )}{12\,n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)*(a + b*x^n)^2,x)

[Out]

(x^(2*n)*(6*a^2 + 3*b^2*x^(2*n) + 8*a*b*x^n))/(12*n)

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