\(\int x^5 \log (c (a+b x^3)^p) \, dx\) [13]

Optimal result

Integrand size = 16, antiderivative size = 59 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {a p x^3}{6 b}-\frac {p x^6}{12}-\frac {a^2 p \log \left (a+b x^3\right )}{6 b^2}+\frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

1/6*a*p*x^3/b-1/12*p*x^6-1/6*a^2*p*ln(b*x^3+a)/b^2+1/6*x^6*ln(c*(b*x^3+a)^p)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2442, 45} \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {a^2 p \log \left (a+b x^3\right )}{6 b^2}+\frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right )+\frac {a p x^3}{6 b}-\frac {p x^6}{12} \]

[In]

Int[x^5*Log[c*(a + b*x^3)^p],x]

[Out]

(a*p*x^3)/(6*b) - (p*x^6)/12 - (a^2*p*Log[a + b*x^3])/(6*b^2) + (x^6*Log[c*(a + b*x^3)^p])/6

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int x \log \left (c (a+b x)^p\right ) \, dx,x,x^3\right ) \\ & = \frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{6} (b p) \text {Subst}\left (\int \frac {x^2}{a+b x} \, dx,x,x^3\right ) \\ & = \frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right )-\frac {1}{6} (b p) \text {Subst}\left (\int \left (-\frac {a}{b^2}+\frac {x}{b}+\frac {a^2}{b^2 (a+b x)}\right ) \, dx,x,x^3\right ) \\ & = \frac {a p x^3}{6 b}-\frac {p x^6}{12}-\frac {a^2 p \log \left (a+b x^3\right )}{6 b^2}+\frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {a p x^3}{6 b}-\frac {p x^6}{12}-\frac {a^2 p \log \left (a+b x^3\right )}{6 b^2}+\frac {1}{6} x^6 \log \left (c \left (a+b x^3\right )^p\right ) \]

[In]

Integrate[x^5*Log[c*(a + b*x^3)^p],x]

[Out]

(a*p*x^3)/(6*b) - (p*x^6)/12 - (a^2*p*Log[a + b*x^3])/(6*b^2) + (x^6*Log[c*(a + b*x^3)^p])/6

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97

method	result	size
parts	\(\frac {x^{6} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{6}-\frac {p b \left (-\frac {-\frac {1}{2} b \,x^{6}+x^{3} a}{3 b^{2}}+\frac {a^{2} \ln \left (b \,x^{3}+a \right )}{3 b^{3}}\right )}{2}\)	\(57\)
parallelrisch	\(-\frac {-2 x^{6} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) b^{2}+x^{6} b^{2} p -2 a b p \,x^{3}+2 \ln \left (b \,x^{3}+a \right ) a^{2} p +2 a^{2} p}{12 b^{2}}\)	\(63\)
risch	\(\text {Expression too large to display}\)	\(1190\)

[In]

int(x^5*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)

[Out]

1/6*x^6*ln(c*(b*x^3+a)^p)-1/2*p*b*(-1/3/b^2*(-1/2*b*x^6+x^3*a)+1/3*a^2/b^3*ln(b*x^3+a))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.97 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {b^{2} p x^{6} - 2 \, b^{2} x^{6} \log \left (c\right ) - 2 \, a b p x^{3} - 2 \, {\left (b^{2} p x^{6} - a^{2} p\right )} \log \left (b x^{3} + a\right )}{12 \, b^{2}} \]

[In]

integrate(x^5*log(c*(b*x^3+a)^p),x, algorithm="fricas")

[Out]

-1/12*(b^2*p*x^6 - 2*b^2*x^6*log(c) - 2*a*b*p*x^3 - 2*(b^2*p*x^6 - a^2*p)*log(b*x^3 + a))/b^2

Sympy [A] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\begin {cases} - \frac {a^{2} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{6 b^{2}} + \frac {a p x^{3}}{6 b} - \frac {p x^{6}}{12} + \frac {x^{6} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{6} & \text {for}\: b \neq 0 \\\frac {x^{6} \log {\left (a^{p} c \right )}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*ln(c*(b*x**3+a)**p),x)

[Out]

Piecewise((-a**2*log(c*(a + b*x**3)**p)/(6*b**2) + a*p*x**3/(6*b) - p*x**6/12 + x**6*log(c*(a + b*x**3)**p)/6,
Ne(b, 0)), (x**6*log(a**p*c)/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{6} \, x^{6} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{12} \, b p {\left (\frac {2 \, a^{2} \log \left (b x^{3} + a\right )}{b^{3}} + \frac {b x^{6} - 2 \, a x^{3}}{b^{2}}\right )} \]

[In]

integrate(x^5*log(c*(b*x^3+a)^p),x, algorithm="maxima")

[Out]

1/6*x^6*log((b*x^3 + a)^p*c) - 1/12*b*p*(2*a^2*log(b*x^3 + a)/b^3 + (b*x^6 - 2*a*x^3)/b^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.64 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{2} p \log \left (b x^{3} + a\right ) - {\left (b x^{3} + a\right )}^{2} p + 2 \, {\left (b x^{3} + a\right )}^{2} \log \left (c\right )}{12 \, b^{2}} + \frac {{\left (b x^{3} - {\left (b x^{3} + a\right )} \log \left (b x^{3} + a\right ) + a\right )} a p - {\left (b x^{3} + a\right )} a \log \left (c\right )}{3 \, b^{2}} \]

[In]

integrate(x^5*log(c*(b*x^3+a)^p),x, algorithm="giac")

[Out]

1/12*(2*(b*x^3 + a)^2*p*log(b*x^3 + a) - (b*x^3 + a)^2*p + 2*(b*x^3 + a)^2*log(c))/b^2 + 1/3*((b*x^3 - (b*x^3
+ a)*log(b*x^3 + a) + a)*a*p - (b*x^3 + a)*a*log(c))/b^2

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int x^5 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^6\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{6}-\frac {p\,x^6}{12}-\frac {a^2\,p\,\ln \left (b\,x^3+a\right )}{6\,b^2}+\frac {a\,p\,x^3}{6\,b} \]

[In]

int(x^5*log(c*(a + b*x^3)^p),x)

[Out]

(x^6*log(c*(a + b*x^3)^p))/6 - (p*x^6)/12 - (a^2*p*log(a + b*x^3))/(6*b^2) + (a*p*x^3)/(6*b)