\(\int \frac {\log (c (a+\frac {b}{x})^p)}{x^5} \, dx\) [35]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {p}{16 x^4}-\frac {a p}{12 b x^3}+\frac {a^2 p}{8 b^2 x^2}-\frac {a^3 p}{4 b^3 x}+\frac {a^4 p \log \left (a+\frac {b}{x}\right )}{4 b^4}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, 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} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {a^4 p \log \left (a+\frac {b}{x}\right )}{4 b^4}-\frac {a^3 p}{4 b^3 x}+\frac {a^2 p}{8 b^2 x^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}-\frac {a p}{12 b x^3}+\frac {p}{16 x^4} \]

[In]

Int[Log[c*(a + b/x)^p]/x^5,x]

[Out]

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)

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}& = -\text {Subst}\left (\int x^3 \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}+\frac {1}{4} (b p) \text {Subst}\left (\int \frac {x^4}{a+b x} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4}+\frac {1}{4} (b p) \text {Subst}\left (\int \left (-\frac {a^3}{b^4}+\frac {a^2 x}{b^3}-\frac {a x^2}{b^2}+\frac {x^3}{b}+\frac {a^4}{b^4 (a+b x)}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {p}{16 x^4}-\frac {a p}{12 b x^3}+\frac {a^2 p}{8 b^2 x^2}-\frac {a^3 p}{4 b^3 x}+\frac {a^4 p \log \left (a+\frac {b}{x}\right )}{4 b^4}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 x^4} \]

[In]

Integrate[Log[c*(a + b/x)^p]/x^5,x]

[Out]

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)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{4 x^{4}}-\frac {p b \left (-\frac {1}{4 b \,x^{4}}-\frac {a^{2}}{2 b^{3} x^{2}}+\frac {a^{4} \ln \left (x \right )}{b^{5}}+\frac {a}{3 b^{2} x^{3}}+\frac {a^{3}}{b^{4} x}-\frac {a^{4} \ln \left (a x +b \right )}{b^{5}}\right )}{4}\)	\(85\)
parallelrisch	\(-\frac {12 \ln \left (x \right ) x^{4} a^{4} p -12 \ln \left (a x +b \right ) x^{4} a^{4} p -12 x^{4} a^{4} p +12 x^{3} a^{3} b p -6 x^{2} a^{2} b^{2} p +4 x a \,b^{3} p +12 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{4}-3 b^{4} p}{48 x^{4} b^{4}}\)	\(100\)

[In]

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

[Out]

-1/4*ln(c*(a+b/x)^p)/x^4-1/4*p*b*(-1/4/b/x^4-1/2*a^2/b^3/x^2+a^4/b^5*ln(x)+1/3*a/b^2/x^3+a^3/b^4/x-a^4/b^5*ln(
a*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=-\frac {12 \, a^{3} b p x^{3} - 6 \, a^{2} b^{2} p x^{2} + 4 \, a b^{3} p x - 3 \, b^{4} p + 12 \, b^{4} \log \left (c\right ) - 12 \, {\left (a^{4} p x^{4} - b^{4} p\right )} \log \left (\frac {a x + b}{x}\right )}{48 \, b^{4} x^{4}} \]

[In]

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

[Out]

-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*log(c) - 12*(a^4*p*x^4 - b^4*p)*log((
a*x + b)/x))/(b^4*x^4)

Sympy [A] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\begin {cases} \frac {a^{4} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4 b^{4}} - \frac {a^{3} p}{4 b^{3} x} + \frac {a^{2} p}{8 b^{2} x^{2}} - \frac {a p}{12 b x^{3}} + \frac {p}{16 x^{4}} - \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4 x^{4}} & \text {for}\: b \neq 0 \\- \frac {\log {\left (a^{p} c \right )}}{4 x^{4}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {1}{48} \, b p {\left (\frac {12 \, a^{4} \log \left (a x + b\right )}{b^{5}} - \frac {12 \, a^{4} \log \left (x\right )}{b^{5}} - \frac {12 \, a^{3} x^{3} - 6 \, a^{2} b x^{2} + 4 \, a b^{2} x - 3 \, b^{3}}{b^{4} x^{4}}\right )} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (75) = 150\).

Time = 0.31 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.64 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {\frac {48 \, {\left (a x + b\right )} a^{3} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x} - \frac {48 \, {\left (a x + b\right )} a^{3} p}{b^{3} x} - \frac {72 \, {\left (a x + b\right )}^{2} a^{2} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{2}} + \frac {48 \, {\left (a x + b\right )} a^{3} \log \left (c\right )}{b^{3} x} + \frac {36 \, {\left (a x + b\right )}^{2} a^{2} p}{b^{3} x^{2}} + \frac {48 \, {\left (a x + b\right )}^{3} a p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{3}} - \frac {72 \, {\left (a x + b\right )}^{2} a^{2} \log \left (c\right )}{b^{3} x^{2}} - \frac {16 \, {\left (a x + b\right )}^{3} a p}{b^{3} x^{3}} - \frac {12 \, {\left (a x + b\right )}^{4} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{3} x^{4}} + \frac {48 \, {\left (a x + b\right )}^{3} a \log \left (c\right )}{b^{3} x^{3}} + \frac {3 \, {\left (a x + b\right )}^{4} p}{b^{3} x^{4}} - \frac {12 \, {\left (a x + b\right )}^{4} \log \left (c\right )}{b^{3} x^{4}}}{48 \, b} \]

[In]

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

[Out]

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*log(c)/(b^3*x^4))/b

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^5} \, dx=\frac {\frac {p}{4}+\frac {a^2\,p\,x^2}{2\,b^2}-\frac {a^3\,p\,x^3}{b^3}-\frac {a\,p\,x}{3\,b}}{4\,x^4}-\frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{4\,x^4}+\frac {a^4\,p\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{2\,b^4} \]

[In]

int(log(c*(a + b/x)^p)/x^5,x)

[Out]

(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) - log(c*(a + b/x)^p)/(4*x^4) + (a^4*p*at
anh((2*a*x)/b + 1))/(2*b^4)