3.6.48 \(\int x^{-1+2 n} (a^2+2 a b x^n+b^2 x^{2 n})^p \, dx\) [548]

Optimal. Leaf size=103 \[ -\frac {a^2 \left (1+\frac {b x^n}{a}\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{b^2 n (1+2 p)}+\frac {a^2 \left (1+\frac {b x^n}{a}\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 b^2 n (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1370, 272, 45} \begin {gather*} \frac {a^2 \left (\frac {b x^n}{a}+1\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 b^2 n (p+1)}-\frac {a^2 \left (\frac {b x^n}{a}+1\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{b^2 n (2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

Rule 1370

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a
+ b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2*FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/b))^(2*p), x],
x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 54, normalized size = 0.52 \begin {gather*} \frac {\left (a+b x^n\right ) \left (\left (a+b x^n\right )^2\right )^p \left (-a+b (1+2 p) x^n\right )}{2 b^2 n (1+p) (1+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.08, size = 148, normalized size = 1.44

method	result	size
risch	\(-\frac {\left (-2 b^{2} p \,x^{2 n}-2 a p \,x^{n} b -b^{2} x^{2 n}+a^{2}\right ) {\mathrm e}^{\frac {p \left (-i \pi \mathrm {csgn}\left (i \left (a +b \,x^{n}\right )^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i \left (a +b \,x^{n}\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (a +b \,x^{n}\right )\right )-i \pi \,\mathrm {csgn}\left (i \left (a +b \,x^{n}\right )^{2}\right ) \mathrm {csgn}\left (i \left (a +b \,x^{n}\right )\right )^{2}+4 \ln \left (a +b \,x^{n}\right )\right )}{2}}}{2 \left (1+2 p \right ) \left (1+p \right ) n \,b^{2}}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-2*b^2*p*(x^n)^2-2*a*p*x^n*b-b^2*(x^n)^2+a^2)/(1+2*p)/(1+p)/n/b^2*exp(1/2*p*(-I*Pi*csgn(I*(a+b*x^n)^2)^3
+2*I*Pi*csgn(I*(a+b*x^n)^2)^2*csgn(I*(a+b*x^n))-I*Pi*csgn(I*(a+b*x^n)^2)*csgn(I*(a+b*x^n))^2+4*ln(a+b*x^n)))

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Maxima [A]
time = 0.30, size = 59, normalized size = 0.57 \begin {gather*} \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2 \, n} + 2 \, a b p x^{n} - a^{2}\right )} {\left (b x^{n} + a\right )}^{2 \, p}}{2 \, {\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 78, normalized size = 0.76 \begin {gather*} \frac {{\left (2 \, a b p x^{n} - a^{2} + {\left (2 \, b^{2} p + b^{2}\right )} x^{2 \, n}\right )} {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{p}}{2 \, {\left (2 \, b^{2} n p^{2} + 3 \, b^{2} n p + b^{2} n\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{2\,n-1}\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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