Integrand size = 16, antiderivative size = 87 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {p}{16 x^4}-\frac {a p}{12 b x^3}+\frac {a^2 p}{8 b^2 x^2}-\frac {a^3 p}{4 b^3 x}+\frac {a^4 p \log \left (a+\frac {b}{x}\right )}{4 b^4}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4} \] Output:
1/16*p/x^4-1/12*a*p/b/x^3+1/8*a^2*p/b^2/x^2-1/4*a^3*p/b^3/x+1/4*a^4*p*ln(a +b/x)/b^4-1/4*ln(c*(a+b/x)^p)/x^4
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {p}{16 x^4}-\frac {a p}{12 b x^3}+\frac {a^2 p}{8 b^2 x^2}-\frac {a^3 p}{4 b^3 x}+\frac {a^4 p \log \left (a+\frac {b}{x}\right )}{4 b^4}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4} \] Input:
Integrate[Log[c*(a + b/x)^p]/x^5,x]
Output:
p/(16*x^4) - (a*p)/(12*b*x^3) + (a^2*p)/(8*b^2*x^2) - (a^3*p)/(4*b^3*x) + (a^4*p*Log[a + b/x])/(4*b^4) - Log[c*(a + b/x)^p]/(4*x^4)
Time = 0.45 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2904, 2842, 49, 2009}
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.
\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -\int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3}d\frac {1}{x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{4} b p \int \frac {1}{\left (a+\frac {b}{x}\right ) x^4}d\frac {1}{x}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{4} b p \int \left (\frac {a^4}{b^4 \left (a+\frac {b}{x}\right )}-\frac {a^3}{b^4}+\frac {a^2}{b^3 x}-\frac {a}{b^2 x^2}+\frac {1}{b x^3}\right )d\frac {1}{x}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} b p \left (\frac {a^4 \log \left (a+\frac {b}{x}\right )}{b^5}-\frac {a^3}{b^4 x}+\frac {a^2}{2 b^3 x^2}-\frac {a}{3 b^2 x^3}+\frac {1}{4 b x^4}\right )-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}\) |
Input:
Int[Log[c*(a + b/x)^p]/x^5,x]
Output:
(b*p*(1/(4*b*x^4) - a/(3*b^2*x^3) + a^2/(2*b^3*x^2) - a^3/(b^4*x) + (a^4*L og[a + b/x])/b^5))/4 - Log[c*(a + b/x)^p]/(4*x^4)
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98
method | result | size |
parts | \(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{4 x^{4}}-\frac {p b \left (-\frac {a^{4} \ln \left (a x +b \right )}{b^{5}}-\frac {1}{4 b \,x^{4}}+\frac {a^{4} \ln \left (x \right )}{b^{5}}-\frac {a^{2}}{2 b^{3} x^{2}}+\frac {a^{3}}{b^{4} x}+\frac {a}{3 b^{2} x^{3}}\right )}{4}\) | \(85\) |
parallelrisch | \(-\frac {12 \ln \left (x \right ) x^{4} a^{4} p -12 \ln \left (a x +b \right ) x^{4} a^{4} p -12 x^{4} a^{4} p +12 x^{3} a^{3} b p -6 x^{2} a^{2} b^{2} p +4 x a \,b^{3} p +12 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{4}-3 b^{4} p}{48 x^{4} b^{4}}\) | \(100\) |
Input:
int(ln(c*(a+b/x)^p)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/4*ln(c*(a+b/x)^p)/x^4-1/4*p*b*(-1/b^5*a^4*ln(a*x+b)-1/4/b/x^4+1/b^5*a^4 *ln(x)-1/2/b^3*a^2/x^2+1/b^4*a^3/x+1/3/b^2*a/x^3)
Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=-\frac {12 \, a^{3} b p x^{3} - 6 \, a^{2} b^{2} p x^{2} + 4 \, a b^{3} p x - 3 \, b^{4} p + 12 \, b^{4} \log \left (c\right ) - 12 \, {\left (a^{4} p x^{4} - b^{4} p\right )} \log \left (\frac {a x + b}{x}\right )}{48 \, b^{4} x^{4}} \] Input:
integrate(log(c*(a+b/x)^p)/x^5,x, algorithm="fricas")
Output:
-1/48*(12*a^3*b*p*x^3 - 6*a^2*b^2*p*x^2 + 4*a*b^3*p*x - 3*b^4*p + 12*b^4*l og(c) - 12*(a^4*p*x^4 - b^4*p)*log((a*x + b)/x))/(b^4*x^4)
Time = 1.62 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\begin {cases} \frac {a^{4} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4 b^{4}} - \frac {a^{3} p}{4 b^{3} x} + \frac {a^{2} p}{8 b^{2} x^{2}} - \frac {a p}{12 b x^{3}} + \frac {p}{16 x^{4}} - \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4 x^{4}} & \text {for}\: b \neq 0 \\- \frac {\log {\left (a^{p} c \right )}}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:
integrate(ln(c*(a+b/x)**p)/x**5,x)
Output:
Piecewise((a**4*log(c*(a + b/x)**p)/(4*b**4) - a**3*p/(4*b**3*x) + a**2*p/ (8*b**2*x**2) - a*p/(12*b*x**3) + p/(16*x**4) - log(c*(a + b/x)**p)/(4*x** 4), Ne(b, 0)), (-log(a**p*c)/(4*x**4), True))
Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {1}{48} \, b p {\left (\frac {12 \, a^{4} \log \left (a x + b\right )}{b^{5}} - \frac {12 \, a^{4} \log \left (x\right )}{b^{5}} - \frac {12 \, a^{3} x^{3} - 6 \, a^{2} b x^{2} + 4 \, a b^{2} x - 3 \, b^{3}}{b^{4} x^{4}}\right )} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{4 \, x^{4}} \] Input:
integrate(log(c*(a+b/x)^p)/x^5,x, algorithm="maxima")
Output:
1/48*b*p*(12*a^4*log(a*x + b)/b^5 - 12*a^4*log(x)/b^5 - (12*a^3*x^3 - 6*a^ 2*b*x^2 + 4*a*b^2*x - 3*b^3)/(b^4*x^4)) - 1/4*log((a + b/x)^p*c)/x^4
Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (75) = 150\).
Time = 0.13 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.64 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {\frac {48 \, {\left (a x + b\right )} a^{3} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x} - \frac {48 \, {\left (a x + b\right )} a^{3} p}{b^{3} x} - \frac {72 \, {\left (a x + b\right )}^{2} a^{2} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{2}} + \frac {48 \, {\left (a x + b\right )} a^{3} \log \left (c\right )}{b^{3} x} + \frac {36 \, {\left (a x + b\right )}^{2} a^{2} p}{b^{3} x^{2}} + \frac {48 \, {\left (a x + b\right )}^{3} a p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{3}} - \frac {72 \, {\left (a x + b\right )}^{2} a^{2} \log \left (c\right )}{b^{3} x^{2}} - \frac {16 \, {\left (a x + b\right )}^{3} a p}{b^{3} x^{3}} - \frac {12 \, {\left (a x + b\right )}^{4} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{4}} + \frac {48 \, {\left (a x + b\right )}^{3} a \log \left (c\right )}{b^{3} x^{3}} + \frac {3 \, {\left (a x + b\right )}^{4} p}{b^{3} x^{4}} - \frac {12 \, {\left (a x + b\right )}^{4} \log \left (c\right )}{b^{3} x^{4}}}{48 \, b} \] Input:
integrate(log(c*(a+b/x)^p)/x^5,x, algorithm="giac")
Output:
1/48*(48*(a*x + b)*a^3*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^3*x) - 48* (a*x + b)*a^3*p/(b^3*x) - 72*(a*x + b)^2*a^2*p*log(-b*(a/b - (a*x + b)/(b* x)) + a)/(b^3*x^2) + 48*(a*x + b)*a^3*log(c)/(b^3*x) + 36*(a*x + b)^2*a^2* p/(b^3*x^2) + 48*(a*x + b)^3*a*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^3* x^3) - 72*(a*x + b)^2*a^2*log(c)/(b^3*x^2) - 16*(a*x + b)^3*a*p/(b^3*x^3) - 12*(a*x + b)^4*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^3*x^4) + 48*(a*x + b)^3*a*log(c)/(b^3*x^3) + 3*(a*x + b)^4*p/(b^3*x^4) - 12*(a*x + b)^4*lo g(c)/(b^3*x^4))/b
Time = 26.00 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {\frac {p}{4}+\frac {a^2\,p\,x^2}{2\,b^2}-\frac {a^3\,p\,x^3}{b^3}-\frac {a\,p\,x}{3\,b}}{4\,x^4}-\frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{4\,x^4}+\frac {a^4\,p\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{2\,b^4} \] Input:
int(log(c*(a + b/x)^p)/x^5,x)
Output:
(p/4 + (a^2*p*x^2)/(2*b^2) - (a^3*p*x^3)/b^3 - (a*p*x)/(3*b))/(4*x^4) - lo g(c*(a + b/x)^p)/(4*x^4) + (a^4*p*atanh((2*a*x)/b + 1))/(2*b^4)
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {12 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) a^{4} x^{4}-12 \,\mathrm {log}\left (\frac {\left (a x +b \right )^{p} c}{x^{p}}\right ) b^{4}-12 a^{3} b p \,x^{3}+6 a^{2} b^{2} p \,x^{2}-4 a \,b^{3} p x +3 b^{4} p}{48 b^{4} x^{4}} \] Input:
int(log(c*(a+b/x)^p)/x^5,x)
Output:
(12*log(((a*x + b)**p*c)/x**p)*a**4*x**4 - 12*log(((a*x + b)**p*c)/x**p)*b **4 - 12*a**3*b*p*x**3 + 6*a**2*b**2*p*x**2 - 4*a*b**3*p*x + 3*b**4*p)/(48 *b**4*x**4)