\(\int x^{-1+4 n} (a+b x^n)^2 \, dx\) [2528]

Optimal result

Integrand size = 17, antiderivative size = 45 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {a^2 x^{4 n}}{4 n}+\frac {2 a b x^{5 n}}{5 n}+\frac {b^2 x^{6 n}}{6 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, 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} \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {a^2 x^{4 n}}{4 n}+\frac {2 a b x^{5 n}}{5 n}+\frac {b^2 x^{6 n}}{6 n} \]

[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 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^3 (a+b x)^2 \, dx,x,x^n\right )}{n} \\ & = \frac {\text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {x^{4 n} \left (15 a^2+24 a b x^n+10 b^2 x^{2 n}\right )}{60 n} \]

[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)

Maple [A] (verified)

Time = 3.74 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {a^{2} x^{4 n}}{4 n}+\frac {2 a b \,x^{5 n}}{5 n}+\frac {b^{2} x^{6 n}}{6 n}\)	\(40\)
parallelrisch	\(\frac {10 x \,x^{2 n} x^{-1+4 n} b^{2}+24 x \,x^{n} x^{-1+4 n} a b +15 x \,x^{-1+4 n} a^{2}}{60 n}\)	\(53\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {10 \, b^{2} x^{6 \, n} + 24 \, a b x^{5 \, n} + 15 \, a^{2} x^{4 \, n}}{60 \, n} \]

[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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.40 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\begin {cases} \frac {a^{2} x x^{4 n - 1}}{4 n} + \frac {2 a b x x^{n} x^{4 n - 1}}{5 n} + \frac {b^{2} x x^{2 n} x^{4 n - 1}}{6 n} & \text {for}\: n \neq 0 \\\left (a + b\right )^{2} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x*x**(4*n - 1)/(4*n) + 2*a*b*x*x**n*x**(4*n - 1)/(5*n) + b**2*x*x**(2*n)*x**(4*n - 1)/(6*n), N
e(n, 0)), ((a + b)**2*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {b^{2} x^{6 \, n}}{6 \, n} + \frac {2 \, a b x^{5 \, n}}{5 \, n} + \frac {a^{2} x^{4 \, n}}{4 \, n} \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\int { {\left (b x^{n} + a\right )}^{2} x^{4 \, n - 1} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 5.97 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int x^{-1+4 n} \left (a+b x^n\right )^2 \, dx=\frac {x^{4\,n}\,\left (15\,a^2+10\,b^2\,x^{2\,n}+24\,a\,b\,x^n\right )}{60\,n} \]

[In]

int(x^(4*n - 1)*(a + b*x^n)^2,x)

[Out]

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