3.2541 \(\int x^{-1+2 n} (a+b x^n)^3 \, dx\)

Optimal. Leaf size=40 \[ \frac {\left (a+b x^n\right )^5}{5 b^2 n}-\frac {a \left (a+b x^n\right )^4}{4 b^2 n} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 40, 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 = {266, 43} \[ \frac {\left (a+b x^n\right )^5}{5 b^2 n}-\frac {a \left (a+b x^n\right )^4}{4 b^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*(a + b*x^n)^3,x]

[Out]

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

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]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 27, normalized size = 0.68 \[ -\frac {\left (a-4 b x^n\right ) \left (a+b x^n\right )^4}{20 b^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/20*((a - 4*b*x^n)*(a + b*x^n)^4)/(b^2*n)

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fricas [A]  time = 0.51, size = 48, normalized size = 1.20 \[ \frac {4 \, b^{3} x^{5 \, n} + 15 \, a b^{2} x^{4 \, n} + 20 \, a^{2} b x^{3 \, n} + 10 \, a^{3} x^{2 \, n}}{20 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3,x, algorithm="fricas")

[Out]

1/20*(4*b^3*x^(5*n) + 15*a*b^2*x^(4*n) + 20*a^2*b*x^(3*n) + 10*a^3*x^(2*n))/n

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{n} + a\right )}^{3} x^{2 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 63, normalized size = 1.58 \[ \frac {a^{3} {\mathrm e}^{2 n \ln \relax (x )}}{2 n}+\frac {a^{2} b \,{\mathrm e}^{3 n \ln \relax (x )}}{n}+\frac {3 a \,b^{2} {\mathrm e}^{4 n \ln \relax (x )}}{4 n}+\frac {b^{3} {\mathrm e}^{5 n \ln \relax (x )}}{5 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*b/n*exp(n*ln(x))^3+1/2*a^3/n*exp(n*ln(x))^2+1/5*b^3/n*exp(n*ln(x))^5+3/4*a*b^2/n*exp(n*ln(x))^4

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maxima [A]  time = 0.46, size = 54, normalized size = 1.35 \[ \frac {b^{3} x^{5 \, n}}{5 \, n} + \frac {3 \, a b^{2} x^{4 \, n}}{4 \, n} + \frac {a^{2} b x^{3 \, n}}{n} + \frac {a^{3} x^{2 \, n}}{2 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)*(a+b*x^n)^3,x, algorithm="maxima")

[Out]

1/5*b^3*x^(5*n)/n + 3/4*a*b^2*x^(4*n)/n + a^2*b*x^(3*n)/n + 1/2*a^3*x^(2*n)/n

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mupad [B]  time = 1.33, size = 54, normalized size = 1.35 \[ \frac {a^3\,x^{2\,n}}{2\,n}+\frac {b^3\,x^{5\,n}}{5\,n}+\frac {a^2\,b\,x^{3\,n}}{n}+\frac {3\,a\,b^2\,x^{4\,n}}{4\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)*(a + b*x^n)^3,x)

[Out]

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

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sympy [A]  time = 26.60, size = 58, normalized size = 1.45 \[ \begin {cases} \frac {a^{3} x^{2 n}}{2 n} + \frac {a^{2} b x^{3 n}}{n} + \frac {3 a b^{2} x^{4 n}}{4 n} + \frac {b^{3} x^{5 n}}{5 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{3} \log {\relax (x )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)*(a+b*x**n)**3,x)

[Out]

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

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