3.11.80 \(\int x^{-1+3 (1+p)} (b+c x^3)^p (b+2 c x^3) \, dx\) [1080]

Optimal. Leaf size=27 \[ \frac {x^{3 (1+p)} \left (b+c x^3\right )^{1+p}}{3 (1+p)} \]

[Out]

1/3*x^(3+3*p)*(c*x^3+b)^(1+p)/(1+p)

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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {460} \begin {gather*} \frac {x^{3 (p+1)} \left (b+c x^3\right )^{p+1}}{3 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*(1 + p))*(b + c*x^3)^p*(b + 2*c*x^3),x]

[Out]

(x^(3*(1 + p))*(b + c*x^3)^(1 + p))/(3*(1 + p))

Rule 460

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

Rubi steps

\begin {align*} \int x^{-1+3 (1+p)} \left (b+c x^3\right )^p \left (b+2 c x^3\right ) \, dx &=\frac {x^{3 (1+p)} \left (b+c x^3\right )^{1+p}}{3 (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 26, normalized size = 0.96 \begin {gather*} \frac {x^{3+3 p} \left (b+c x^3\right )^{1+p}}{3+3 p} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 3*(1 + p))*(b + c*x^3)^p*(b + 2*c*x^3),x]

[Out]

(x^(3 + 3*p)*(b + c*x^3)^(1 + p))/(3 + 3*p)

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Maple [A]
time = 0.32, size = 26, normalized size = 0.96

method	result	size
gosper	\(\frac {x^{3+3 p} \left (c \,x^{3}+b \right )^{1+p}}{3+3 p}\)	\(26\)
risch	\(\frac {x \left (c \,x^{3}+b \right ) x^{2+3 p} \left (c \,x^{3}+b \right )^{p}}{3+3 p}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2+3*p)*(c*x^3+b)^p*(2*c*x^3+b),x,method=_RETURNVERBOSE)

[Out]

1/3*x^(3+3*p)*(c*x^3+b)^(1+p)/(1+p)

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Maxima [A]
time = 0.34, size = 35, normalized size = 1.30 \begin {gather*} \frac {{\left (c x^{6} + b x^{3}\right )} e^{\left (p \log \left (c x^{3} + b\right ) + 3 \, p \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(2+3*p)*(c*x^3+b)^p*(2*c*x^3+b),x, algorithm="maxima")

[Out]

1/3*(c*x^6 + b*x^3)*e^(p*log(c*x^3 + b) + 3*p*log(x))/(p + 1)

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Fricas [A]
time = 1.31, size = 32, normalized size = 1.19 \begin {gather*} \frac {{\left (c x^{4} + b x\right )} {\left (c x^{3} + b\right )}^{p} x^{3 \, p + 2}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(2+3*p)*(c*x^3+b)^p*(2*c*x^3+b),x, algorithm="fricas")

[Out]

1/3*(c*x^4 + b*x)*(c*x^3 + b)^p*x^(3*p + 2)/(p + 1)

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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**(2+3*p)*(c*x**3+b)**p*(2*c*x**3+b),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
time = 0.68, size = 56, normalized size = 2.07 \begin {gather*} \frac {{\left (c x^{3} + b\right )}^{p} c x^{4} e^{\left (3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )} + {\left (c x^{3} + b\right )}^{p} b x e^{\left (3 \, p \log \left (x\right ) + 2 \, \log \left (x\right )\right )}}{3 \, {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(2+3*p)*(c*x^3+b)^p*(2*c*x^3+b),x, algorithm="giac")

[Out]

1/3*((c*x^3 + b)^p*c*x^4*e^(3*p*log(x) + 2*log(x)) + (c*x^3 + b)^p*b*x*e^(3*p*log(x) + 2*log(x)))/(p + 1)

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Mupad [B]
time = 4.90, size = 47, normalized size = 1.74 \begin {gather*} {\left (c\,x^3+b\right )}^p\,\left (\frac {c\,x^{3\,p+2}\,x^4}{3\,p+3}+\frac {b\,x\,x^{3\,p+2}}{3\,p+3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*p + 2)*(b + c*x^3)^p*(b + 2*c*x^3),x)

[Out]

(b + c*x^3)^p*((c*x^(3*p + 2)*x^4)/(3*p + 3) + (b*x*x^(3*p + 2))/(3*p + 3))

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