\(\int x^{-1+2 n} (a+b x^n)^3 \, dx\) [2541]

Optimal result

Integrand size = 17, antiderivative size = 40 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=-\frac {a \left (a+b x^n\right )^4}{4 b^2 n}+\frac {\left (a+b x^n\right )^5}{5 b^2 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 45} \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\frac {\left (a+b x^n\right )^5}{5 b^2 n}-\frac {a \left (a+b x^n\right )^4}{4 b^2 n} \]

[In]

Int[x^(-1 + 2*n)*(a + b*x^n)^3,x]

[Out]

-1/4*(a*(a + b*x^n)^4)/(b^2*n) + (a + b*x^n)^5/(5*b^2*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*} \text {integral}& = \frac {\text {Subst}\left (\int x (a+b x)^3 \, dx,x,x^n\right )}{n} \\ & = \frac {\text {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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\frac {x^{2 n} \left (10 a^3+20 a^2 b x^n+15 a b^2 x^{2 n}+4 b^3 x^{3 n}\right )}{20 n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38

method	result	size
risch	\(\frac {a^{2} b \,x^{3 n}}{n}+\frac {a^{3} x^{2 n}}{2 n}+\frac {b^{3} x^{5 n}}{5 n}+\frac {3 a \,b^{2} x^{4 n}}{4 n}\)	\(55\)
norman	\(\frac {a^{2} b \,{\mathrm e}^{3 n \ln \left (x \right )}}{n}+\frac {a^{3} {\mathrm e}^{2 n \ln \left (x \right )}}{2 n}+\frac {b^{3} {\mathrm e}^{5 n \ln \left (x \right )}}{5 n}+\frac {3 a \,b^{2} {\mathrm e}^{4 n \ln \left (x \right )}}{4 n}\)	\(63\)
parallelrisch	\(\frac {4 x \,x^{3 n} x^{-1+2 n} b^{3}+15 x \,x^{2 n} x^{-1+2 n} a \,b^{2}+20 x \,x^{n} x^{-1+2 n} a^{2} b +10 x \,x^{-1+2 n} a^{3}}{20 n}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\frac {4 \, b^{3} x^{5 \, n} + 15 \, a b^{2} x^{4 \, n} + 20 \, a^{2} b x^{3 \, n} + 10 \, a^{3} x^{2 \, n}}{20 \, n} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (31) = 62\).

Time = 0.38 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.12 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\begin {cases} \frac {a^{3} x x^{2 n - 1}}{2 n} + \frac {a^{2} b x x^{n} x^{2 n - 1}}{n} + \frac {3 a b^{2} x x^{2 n} x^{2 n - 1}}{4 n} + \frac {b^{3} x x^{3 n} x^{2 n - 1}}{5 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{3} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**3*x*x**(2*n - 1)/(2*n) + a**2*b*x*x**n*x**(2*n - 1)/n + 3*a*b**2*x*x**(2*n)*x**(2*n - 1)/(4*n) +
b**3*x*x**(3*n)*x**(2*n - 1)/(5*n), Ne(n, 0)), ((a + b)**3*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.35 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\frac {b^{3} x^{5 \, n}}{5 \, n} + \frac {3 \, a b^{2} x^{4 \, n}}{4 \, n} + \frac {a^{2} b x^{3 \, n}}{n} + \frac {a^{3} x^{2 \, n}}{2 \, n} \]

[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

Giac [F]

\[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\int { {\left (b x^{n} + a\right )}^{3} x^{2 \, n - 1} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.35 \[ \int x^{-1+2 n} \left (a+b x^n\right )^3 \, dx=\frac {a^3\,x^{2\,n}}{2\,n}+\frac {b^3\,x^{5\,n}}{5\,n}+\frac {a^2\,b\,x^{3\,n}}{n}+\frac {3\,a\,b^2\,x^{4\,n}}{4\,n} \]

[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)