Integrand size = 16, antiderivative size = 45 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^3\right )}{3 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3} \] Output:
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
Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b p \log (x)}{a}-\frac {b p \log \left (a+b x^3\right )}{3 a}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/x^4,x]
Output:
(b*p*Log[x])/a - (b*p*Log[a + b*x^3])/(3*a) - Log[c*(a + b*x^3)^p]/(3*x^3)
Time = 0.34 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2904, 2842, 47, 14, 16}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{3} \int \frac {\log \left (c \left (b x^3+a\right )^p\right )}{x^6}dx^3\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{3} \left (b p \int \frac {1}{x^3 \left (b x^3+a\right )}dx^3-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^3}\right )\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {1}{3} \left (b p \left (\frac {\int \frac {1}{x^3}dx^3}{a}-\frac {b \int \frac {1}{b x^3+a}dx^3}{a}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^3}\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle \frac {1}{3} \left (b p \left (\frac {\log \left (x^3\right )}{a}-\frac {b \int \frac {1}{b x^3+a}dx^3}{a}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^3}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \left (b p \left (\frac {\log \left (x^3\right )}{a}-\frac {\log \left (a+b x^3\right )}{a}\right )-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^3}\right )\) |
Input:
Int[Log[c*(a + b*x^3)^p]/x^4,x]
Output:
(b*p*(Log[x^3]/a - Log[a + b*x^3]/a) - Log[c*(a + b*x^3)^p]/x^3)/3
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[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])
Time = 0.63 (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 | \(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{3 x^{3}}-\frac {-6 b p \ln \left (x \right ) x^{3}+2 b p \ln \left (b \,x^{3}+a \right ) x^{3}+i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}-i \pi a \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}+i \pi a {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right ) a}{6 a \,x^{3}}\) | \(173\) |
Input:
int(ln(c*(b*x^3+a)^p)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*ln(c*(b*x^3+a)^p)/x^3+p*b*(ln(x)/a-1/3/a*ln(b*x^3+a))
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {3 \, b p x^{3} \log \left (x\right ) - {\left (b p x^{3} + a p\right )} \log \left (b x^{3} + a\right ) - a \log \left (c\right )}{3 \, a x^{3}} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="fricas")
Output:
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)
Time = 1.90 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.44 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\begin {cases} - \frac {\log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3 x^{3}} + \frac {b p \log {\left (x \right )}}{a} - \frac {b \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3 a} & \text {for}\: a \neq 0 \\- \frac {p}{3 x^{3}} - \frac {\log {\left (c \left (b x^{3}\right )^{p} \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(ln(c*(b*x**3+a)**p)/x**4,x)
Output:
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))
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=-\frac {1}{3} \, b p {\left (\frac {\log \left (b x^{3} + a\right )}{a} - \frac {\log \left (x^{3}\right )}{a}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{3 \, x^{3}} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="maxima")
Output:
-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
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=-\frac {\frac {b^{2} p \log \left (b x^{3} + a\right )}{a} - \frac {b^{2} p \log \left (b x^{3}\right )}{a} + \frac {b p \log \left (b x^{3} + a\right )}{x^{3}} + \frac {b \log \left (c\right )}{x^{3}}}{3 \, b} \] Input:
integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="giac")
Output:
-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
Time = 26.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {b\,p\,\ln \left (x\right )}{a}-\frac {b\,p\,\ln \left (b\,x^3+a\right )}{3\,a}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{3\,x^3} \] Input:
int(log(c*(a + b*x^3)^p)/x^4,x)
Output:
(b*p*log(x))/a - (b*p*log(a + b*x^3))/(3*a) - log(c*(a + b*x^3)^p)/(3*x^3)
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x^4} \, dx=\frac {-\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) a -\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) b \,x^{3}+3 \,\mathrm {log}\left (x \right ) b p \,x^{3}}{3 a \,x^{3}} \] Input:
int(log(c*(b*x^3+a)^p)/x^4,x)
Output:
( - log((a + b*x**3)**p*c)*a - log((a + b*x**3)**p*c)*b*x**3 + 3*log(x)*b* p*x**3)/(3*a*x**3)