\(\int (d x)^{-1-2 n (1+p)} (a^2+2 a b x^n+b^2 x^{2 n})^p \, dx\) [547]

Optimal result

Integrand size = 35, antiderivative size = 117 \[ \int (d x)^{-1-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=-\frac {(d x)^{-2 n (1+p)} \left (a+b x^n\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{a d n (1+2 p)}+\frac {(d x)^{-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{1+p}}{2 a^2 d n (1+p) (1+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {1370, 279, 270} \[ \int (d x)^{-1-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\frac {\left (\frac {b x^n}{a}+1\right )^2 (d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 d n \left (2 p^2+3 p+1\right )}-\frac {\left (\frac {b x^n}{a}+1\right ) (d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{d n (2 p+1)} \]

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

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*} \text {integral}& = \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 (d x)^{-1-2 n (1+p)} \left (1+\frac {b x^n}{a}\right )^{2 p} \, dx \\ & = -\frac {(d x)^{-2 n (1+p)} \left (1+\frac {b x^n}{a}\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{d n (1+2 p)}+\frac {\left ((-2 n (1+p)+n (1+2 p)) \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 (d x)^{-1-2 n (1+p)} \left (1+\frac {b x^n}{a}\right )^{1+2 p} \, dx}{n (1+2 p)} \\ & = -\frac {(d x)^{-2 n (1+p)} \left (1+\frac {b x^n}{a}\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{d n (1+2 p)}+\frac {(d x)^{-2 n (1+p)} \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 d n \left (1+3 p+2 p^2\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int (d x)^{-1-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=-\frac {x (d x)^{-1-2 n (1+p)} \left (\left (a+b x^n\right )^2\right )^p \left (1+\frac {b x^n}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-2 p,-2 (1+p),1-2 (1+p),-\frac {b x^n}{a}\right )}{2 n (1+p)} \]

[In]

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

[Out]

-1/2*(x*(d*x)^(-1 - 2*n*(1 + p))*((a + b*x^n)^2)^p*Hypergeometric2F1[-2*p, -2*(1 + p), 1 - 2*(1 + p), -((b*x^n
)/a)])/(n*(1 + p)*(1 + (b*x^n)/a)^(2*p))

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.41 \[ \int (d x)^{-1-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=-\frac {{\left (2 \, a b p x x^{n} e^{\left (-{\left (2 \, n p + 2 \, n + 1\right )} \log \left (d\right ) - {\left (2 \, n p + 2 \, n + 1\right )} \log \left (x\right )\right )} - b^{2} x x^{2 \, n} e^{\left (-{\left (2 \, n p + 2 \, n + 1\right )} \log \left (d\right ) - {\left (2 \, n p + 2 \, n + 1\right )} \log \left (x\right )\right )} + {\left (2 \, a^{2} p + a^{2}\right )} x e^{\left (-{\left (2 \, n p + 2 \, n + 1\right )} \log \left (d\right ) - {\left (2 \, n p + 2 \, n + 1\right )} \log \left (x\right )\right )}\right )} {\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{p}}{2 \, {\left (2 \, a^{2} n p^{2} + 3 \, a^{2} n p + a^{2} n\right )}} \]

[In]

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

[Out]

-1/2*(2*a*b*p*x*x^n*e^(-(2*n*p + 2*n + 1)*log(d) - (2*n*p + 2*n + 1)*log(x)) - b^2*x*x^(2*n)*e^(-(2*n*p + 2*n
+ 1)*log(d) - (2*n*p + 2*n + 1)*log(x)) + (2*a^2*p + a^2)*x*e^(-(2*n*p + 2*n + 1)*log(d) - (2*n*p + 2*n + 1)*l
og(x)))*(b^2*x^(2*n) + 2*a*b*x^n + a^2)^p/(2*a^2*n*p^2 + 3*a^2*n*p + a^2*n)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

Mupad [F(-1)]

Timed out. \[ \int (d x)^{-1-2 n (1+p)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p \, dx=\int \frac {{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^p}{{\left (d\,x\right )}^{2\,n\,\left (p+1\right )+1}} \,d x \]

[In]

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

[Out]

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