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

Optimal. Leaf size=29 \[ \frac {\left (-a+b x^n+c x^{2 n}\right )^{p+1}}{n (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, 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 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]

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]

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*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.97 \[ \frac {\left (x^n \left (b+c x^n\right )-a\right )^{p+1}}{n (p+1)} \]

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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fricas [A]  time = 0.93, size = 42, normalized size = 1.45 \[ \frac {{\left (c x^{2 \, n} + b x^{n} - a\right )} {\left (c x^{2 \, n} + b x^{n} - a\right )}^{p}}{n p + n} \]

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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giac [A]  time = 0.82, size = 29, normalized size = 1.00 \[ \frac {{\left (c x^{2 \, n} + b x^{n} - a\right )}^{p + 1}}{n {\left (p + 1\right )}} \]

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))

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maple [A]  time = 0.06, size = 45, normalized size = 1.55 \[ -\frac {\left (-b \,x^{n}-c \,x^{2 n}+a \right ) \left (b \,x^{n}+c \,x^{2 n}-a \right )^{p}}{\left (p +1\right ) n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 43, normalized size = 1.48 \[ \frac {{\left (c x^{2 \, n} + b x^{n} - a\right )} {\left (c x^{2 \, n} + b x^{n} - a\right )}^{p}}{n {\left (p + 1\right )}} \]

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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mupad [B]  time = 2.54, size = 59, normalized size = 2.03 \[ \left (\frac {b\,x^n}{n\,\left (p+1\right )}-\frac {a}{n\,\left (p+1\right )}+\frac {c\,x^{2\,n}}{n\,\left (p+1\right )}\right )\,{\left (b\,x^n-a+c\,x^{2\,n}\right )}^p \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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