\(\int \frac {\log (c (a+b x^3)^p)}{x^4} \, dx\) [22]

Optimal result

Integrand size = 16, antiderivative size = 45 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^3\right )}{3 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2504, 2442, 36, 29, 31} \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}-\frac {b p \log \left (a+b x^3\right )}{3 a}+\frac {b p \log (x)}{a} \]

[In]

Int[Log[c*(a + b*x^3)^p]/x^4,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 \frac {\log \left (c (a+b x)^p\right )}{x^2} \, dx,x,x^3\right ) \\ & = -\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}+\frac {1}{3} (b p) \text {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,x^3\right ) \\ & = -\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}+\frac {(b p) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )}{3 a}-\frac {\left (b^2 p\right ) \text {Subst}\left (\int \frac {1}{a+b x} \, dx,x,x^3\right )}{3 a} \\ & = \frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^3\right )}{3 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^3\right )}{3 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3} \]

[In]

Integrate[Log[c*(a + b*x^3)^p]/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93

method	result	size
parts	\(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{3 x^{3}}+p b \left (\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b \,x^{3}+a \right )}{3 a}\right )\)	\(42\)
parallelrisch	\(\frac {3 p^{2} b \ln \left (x \right ) x^{3}-x^{3} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) b p -\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) a p}{3 x^{3} a p}\)	\(59\)
risch	\(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{3 x^{3}}-\frac {i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-6 b p \ln \left (x \right ) x^{3}+2 b p \ln \left (b \,x^{3}+a \right ) x^{3}+2 \ln \left (c \right ) a}{6 x^{3} a}\)	\(173\)

[In]

int(ln(c*(b*x^3+a)^p)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {3 \, b p x^{3} \log \left (x\right ) - {\left (b p x^{3} + a p\right )} \log \left (b x^{3} + a\right ) - a \log \left (c\right )}{3 \, a x^{3}} \]

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.44 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3 x^{3}} + \frac {b p \log {\left (x \right )}}{a} - \frac {b \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3 a} & \text {for}\: a \neq 0 \\- \frac {p}{3 x^{3}} - \frac {\log {\left (c \left (b x^{3}\right )^{p} \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(ln(c*(b*x**3+a)**p)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=-\frac {1}{3} \, b p {\left (\frac {\log \left (b x^{3} + a\right )}{a} - \frac {\log \left (x^{3}\right )}{a}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{3 \, x^{3}} \]

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=-\frac {\frac {b^{2} p \log \left (b x^{3} + a\right )}{a} - \frac {b^{2} p \log \left (b x^{3}\right )}{a} + \frac {b p \log \left (b x^{3} + a\right )}{x^{3}} + \frac {b \log \left (c\right )}{x^{3}}}{3 \, b} \]

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.34 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b\,p\,\ln \left (x\right )}{a}-\frac {b\,p\,\ln \left (b\,x^3+a\right )}{3\,a}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{3\,x^3} \]

[In]

int(log(c*(a + b*x^3)^p)/x^4,x)

[Out]

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