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

Optimal. Leaf size=117 \[ \frac{(d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{p+1}}{2 a^2 d n (p+1) (2 p+1)}-\frac{\left (a+b x^n\right ) (d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{a d n (2 p+1)} \]

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

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Rubi [A]  time = 0.0642789, antiderivative size = 124, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {1356, 273, 264} \[ \frac{\left (\frac{b x^n}{a}+1\right )^2 (d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{2 d n \left (2 p^2+3 p+1\right )}-\frac{\left (\frac{b x^n}{a}+1\right ) (d x)^{-2 n (p+1)} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^p}{d n (2 p+1)} \]

Antiderivative was successfully verified.

[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 1356

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]

Rule 273

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] /; FreeQ[
{a, b, c, m, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

Rule 264

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]

Rubi steps

\begin{align*} \int (d x)^{-1-2 n (1+p)} \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 (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]  time = 0.0327773, size = 75, normalized size = 0.64 \[ -\frac{x (d x)^{-2 n (p+1)-1} \left (\left (a+b x^n\right )^2\right )^p \left (\frac{b x^n}{a}+1\right )^{-2 p} \, _2F_1\left (-2 p,-2 (p+1);1-2 (p+1);-\frac{b x^n}{a}\right )}{2 n (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{-1-2\,n \left ( 1+p \right ) } \left ({a}^{2}+2\,ab{x}^{n}+{b}^{2}{x}^{2\,n} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.62185, size = 401, normalized size = 3.43 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(-1-2*n*(1+p))*(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., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)