3.1.16 \(\int x^2 \log (c (a+b x^3)^p) \, dx\) [16]

Optimal. Leaf size=35 \[ -\frac {p x^3}{3}+\frac {\left (a+b x^3\right ) \log \left (c \left (a+b x^3\right )^p\right )}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2436, 2332} \begin {gather*} \frac {\left (a+b x^3\right ) \log \left (c \left (a+b x^3\right )^p\right )}{3 b}-\frac {p x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b*x^3)^p],x]

[Out]

-1/3*(p*x^3) + ((a + b*x^3)*Log[c*(a + b*x^3)^p])/(3*b)

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 34, normalized size = 0.97 \begin {gather*} \frac {1}{3} \left (-p x^3+\frac {\left (a+b x^3\right ) \log \left (c \left (a+b x^3\right )^p\right )}{b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b*x^3)^p],x]

[Out]

(-(p*x^3) + ((a + b*x^3)*Log[c*(a + b*x^3)^p])/b)/3

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Maple [A]
time = 0.47, size = 37, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {\ln \left (c \left (x^{3} b +a \right )^{p}\right ) \left (x^{3} b +a \right )-\left (x^{3} b +a \right ) p}{3 b}\)	\(37\)
default	\(\frac {\ln \left (c \left (x^{3} b +a \right )^{p}\right ) \left (x^{3} b +a \right )-\left (x^{3} b +a \right ) p}{3 b}\)	\(37\)
risch	\(\frac {x^{3} \ln \left (\left (x^{3} b +a \right )^{p}\right )}{3}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{6}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{6}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}}{6}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}}{6}+\frac {\ln \left (c \right ) x^{3}}{3}-\frac {p \,x^{3}}{3}+\frac {a p \ln \left (x^{3} b +a \right )}{3 b}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 44, normalized size = 1.26 \begin {gather*} \frac {1}{3} \, x^{3} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{3} \, {\left (\frac {x^{3}}{b} - \frac {a \log \left (b x^{3} + a\right )}{b^{2}}\right )} b p \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((b*x^3 + a)^p*c) - 1/3*(x^3/b - a*log(b*x^3 + a)/b^2)*b*p

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Fricas [A]
time = 0.38, size = 40, normalized size = 1.14 \begin {gather*} -\frac {b p x^{3} - b x^{3} \log \left (c\right ) - {\left (b p x^{3} + a p\right )} \log \left (b x^{3} + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.83, size = 51, normalized size = 1.46 \begin {gather*} \begin {cases} \frac {a \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3 b} - \frac {p x^{3}}{3} + \frac {x^{3} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3} & \text {for}\: b \neq 0 \\\frac {x^{3} \log {\left (a^{p} c \right )}}{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*ln(c*(b*x**3+a)**p),x)

[Out]

Piecewise((a*log(c*(a + b*x**3)**p)/(3*b) - p*x**3/3 + x**3*log(c*(a + b*x**3)**p)/3, Ne(b, 0)), (x**3*log(a**
p*c)/3, True))

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Giac [A]
time = 2.81, size = 43, normalized size = 1.23 \begin {gather*} -\frac {{\left (b x^{3} - {\left (b x^{3} + a\right )} \log \left (b x^{3} + a\right ) + a\right )} p - {\left (b x^{3} + a\right )} \log \left (c\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.23, size = 39, normalized size = 1.11 \begin {gather*} \frac {x^3\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{3}-\frac {p\,x^3}{3}+\frac {a\,p\,\ln \left (b\,x^3+a\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(c*(a + b*x^3)^p),x)

[Out]

(x^3*log(c*(a + b*x^3)^p))/3 - (p*x^3)/3 + (a*p*log(a + b*x^3))/(3*b)

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