3.2.40 \(\int x^{-1+n} (b+2 c x^n) (b x^n+c x^{2 n})^p \, dx\) [140]

Optimal. Leaf size=26 \[ \frac {\left (b x^n+c x^{2 n}\right )^{1+p}}{n (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2059, 643} \begin {gather*} \frac {\left (b x^n+c x^{2 n}\right )^{p+1}}{n (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 643

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 2059

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 24, normalized size = 0.92 \begin {gather*} \frac {\left (x^n \left (b+c x^n\right )\right )^{1+p}}{n (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^n*(b + c*x^n))^(1 + p)/(n*(1 + p))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.22, size = 155, normalized size = 5.96

method	result	size
risch	\(\frac {x^{n} \left (b +c \,x^{n}\right ) {\mathrm e}^{\frac {p \left (-i \pi \mathrm {csgn}\left (i x^{n} \left (b +c \,x^{n}\right )\right )^{3}+i \pi \mathrm {csgn}\left (i x^{n} \left (b +c \,x^{n}\right )\right )^{2} \mathrm {csgn}\left (i x^{n}\right )+i \pi \mathrm {csgn}\left (i x^{n} \left (b +c \,x^{n}\right )\right )^{2} \mathrm {csgn}\left (i \left (b +c \,x^{n}\right )\right )-i \pi \,\mathrm {csgn}\left (i x^{n} \left (b +c \,x^{n}\right )\right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i \left (b +c \,x^{n}\right )\right )+2 \ln \left (x^{n}\right )+2 \ln \left (b +c \,x^{n}\right )\right )}{2}}}{n \left (1+p \right )}\)	\(155\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 40, normalized size = 1.54 \begin {gather*} \frac {{\left (c x^{2 \, n} + b x^{n}\right )} e^{\left (p \log \left (c x^{n} + b\right ) + p \log \left (x^{n}\right )\right )}}{n {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 36, normalized size = 1.38 \begin {gather*} \frac {{\left (c x^{2 \, n} + b x^{n}\right )} {\left (c x^{2 \, n} + b x^{n}\right )}^{p}}{n p + n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]
time = 3.69, size = 26, normalized size = 1.00 \begin {gather*} \frac {{\left (c x^{2 \, n} + b x^{n}\right )}^{p + 1}}{n {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.13, size = 34, normalized size = 1.31 \begin {gather*} \frac {x^n\,\left (b+c\,x^n\right )\,{\left (b\,x^n+c\,x^{2\,n}\right )}^p}{n\,\left (p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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