### 3.132 $$\int x^{-1+n} (b+2 c x^n) (a+b x^n+c x^{2 n})^p \, dx$$

Optimal. Leaf size=27 $\frac{\left (a+b x^n+c x^{2 n}\right )^{p+1}}{n (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0284039, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.067, Rules used = {1468, 629} $\frac{\left (a+b x^n+c x^{2 n}\right )^{p+1}}{n (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1468

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x]
&& EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0338035, size = 26, normalized size = 0.96 $\frac{\left (a+x^n \left (b+c x^n\right )\right )^{p+1}}{n (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 40, normalized size = 1.5 \begin{align*}{\frac{ \left ( a+b{x}^{n}+c \left ({x}^{n} \right ) ^{2} \right ) \left ( a+b{x}^{n}+c \left ({x}^{n} \right ) ^{2} \right ) ^{p}}{n \left ( 1+p \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.28956, size = 53, normalized size = 1.96 \begin{align*} \frac{{\left (c x^{2 \, n} + b x^{n} + a\right )}{\left (c x^{2 \, n} + b x^{n} + a\right )}^{p}}{n{\left (p + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.18037, size = 82, normalized size = 3.04 \begin{align*} \frac{{\left (c x^{2 \, n} + b x^{n} + a\right )}{\left (c x^{2 \, n} + b x^{n} + a\right )}^{p}}{n p + n} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.15221, size = 36, normalized size = 1.33 \begin{align*} \frac{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{p + 1}}{n{\left (p + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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