\(\int \frac {\log (c (a+\frac {b}{x})^p)}{x^4} \, dx\) [34]

Optimal result

Integrand size = 16, antiderivative size = 73 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=\frac {p}{9 x^3}-\frac {a p}{6 b x^2}+\frac {a^2 p}{3 b^2 x}-\frac {a^3 p \log \left (a+\frac {b}{x}\right )}{3 b^3}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 x^3} \]

[Out]

1/9*p/x^3-1/6*a*p/b/x^2+1/3*a^2*p/b^2/x-1/3*a^3*p*ln(a+b/x)/b^3-1/3*ln(c*(a+b/x)^p)/x^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, 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} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=-\frac {a^3 p \log \left (a+\frac {b}{x}\right )}{3 b^3}+\frac {a^2 p}{3 b^2 x}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 x^3}-\frac {a p}{6 b x^2}+\frac {p}{9 x^3} \]

[In]

Int[Log[c*(a + b/x)^p]/x^4,x]

[Out]

p/(9*x^3) - (a*p)/(6*b*x^2) + (a^2*p)/(3*b^2*x) - (a^3*p*Log[a + b/x])/(3*b^3) - Log[c*(a + b/x)^p]/(3*x^3)

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^2 \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 )}{3 x^3}+\frac {1}{3} (b p) \text {Subst}\left (\int \frac {x^3}{a+b x} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 x^3}+\frac {1}{3} (b p) \text {Subst}\left (\int \left (\frac {a^2}{b^3}-\frac {a x}{b^2}+\frac {x^2}{b}-\frac {a^3}{b^3 (a+b x)}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = \frac {p}{9 x^3}-\frac {a p}{6 b x^2}+\frac {a^2 p}{3 b^2 x}-\frac {a^3 p \log \left (a+\frac {b}{x}\right )}{3 b^3}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=\frac {p}{9 x^3}-\frac {a p}{6 b x^2}+\frac {a^2 p}{3 b^2 x}-\frac {a^3 p \log \left (a+\frac {b}{x}\right )}{3 b^3}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 x^3} \]

[In]

Integrate[Log[c*(a + b/x)^p]/x^4,x]

[Out]

p/(9*x^3) - (a*p)/(6*b*x^2) + (a^2*p)/(3*b^2*x) - (a^3*p*Log[a + b/x])/(3*b^3) - Log[c*(a + b/x)^p]/(3*x^3)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03

method	result	size
parts	\(-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{3 x^{3}}-\frac {p b \left (-\frac {1}{3 b \,x^{3}}-\frac {a^{2}}{b^{3} x}+\frac {a}{2 b^{2} x^{2}}-\frac {a^{3} \ln \left (x \right )}{b^{4}}+\frac {a^{3} \ln \left (a x +b \right )}{b^{4}}\right )}{3}\)	\(75\)
parallelrisch	\(-\frac {6 x^{3} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} p +6 x^{3} a^{3} p^{2}-6 x^{2} a^{2} b \,p^{2}+3 x a \,b^{2} p^{2}+6 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) b^{3} p -2 b^{3} p^{2}}{18 x^{3} b^{3} p}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=\frac {6 \, a^{2} b p x^{2} - 3 \, a b^{2} p x + 2 \, b^{3} p - 6 \, b^{3} \log \left (c\right ) - 6 \, {\left (a^{3} p x^{3} + b^{3} p\right )} \log \left (\frac {a x + b}{x}\right )}{18 \, b^{3} x^{3}} \]

[In]

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

[Out]

1/18*(6*a^2*b*p*x^2 - 3*a*b^2*p*x + 2*b^3*p - 6*b^3*log(c) - 6*(a^3*p*x^3 + b^3*p)*log((a*x + b)/x))/(b^3*x^3)

Sympy [A] (verification not implemented)

Time = 1.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=\begin {cases} - \frac {a^{3} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{3 b^{3}} + \frac {a^{2} p}{3 b^{2} x} - \frac {a p}{6 b x^{2}} + \frac {p}{9 x^{3}} - \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{3 x^{3}} & \text {for}\: b \neq 0 \\- \frac {\log {\left (a^{p} c \right )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a**3*log(c*(a + b/x)**p)/(3*b**3) + a**2*p/(3*b**2*x) - a*p/(6*b*x**2) + p/(9*x**3) - log(c*(a + b
/x)**p)/(3*x**3), Ne(b, 0)), (-log(a**p*c)/(3*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=-\frac {1}{18} \, b p {\left (\frac {6 \, a^{3} \log \left (a x + b\right )}{b^{4}} - \frac {6 \, a^{3} \log \left (x\right )}{b^{4}} - \frac {6 \, a^{2} x^{2} - 3 \, a b x + 2 \, b^{2}}{b^{3} x^{3}}\right )} - \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{3 \, x^{3}} \]

[In]

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

[Out]

-1/18*b*p*(6*a^3*log(a*x + b)/b^4 - 6*a^3*log(x)/b^4 - (6*a^2*x^2 - 3*a*b*x + 2*b^2)/(b^3*x^3)) - 1/3*log((a +
b/x)^p*c)/x^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (63) = 126\).

Time = 0.32 (sec) , antiderivative size = 234, normalized size of antiderivative = 3.21 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=-\frac {\frac {18 \, {\left (a x + b\right )} a^{2} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{2} x} - \frac {18 \, {\left (a x + b\right )} a^{2} p}{b^{2} x} - \frac {18 \, {\left (a x + b\right )}^{2} a p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{2} x^{2}} + \frac {18 \, {\left (a x + b\right )} a^{2} \log \left (c\right )}{b^{2} x} + \frac {9 \, {\left (a x + b\right )}^{2} a p}{b^{2} x^{2}} + \frac {6 \, {\left (a x + b\right )}^{3} p \log \left (-b {\left (\frac {a}{b} - \frac {a x + b}{b x}\right )} + a\right )}{b^{2} x^{3}} - \frac {18 \, {\left (a x + b\right )}^{2} a \log \left (c\right )}{b^{2} x^{2}} - \frac {2 \, {\left (a x + b\right )}^{3} p}{b^{2} x^{3}} + \frac {6 \, {\left (a x + b\right )}^{3} \log \left (c\right )}{b^{2} x^{3}}}{18 \, b} \]

[In]

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

[Out]

-1/18*(18*(a*x + b)*a^2*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^2*x) - 18*(a*x + b)*a^2*p/(b^2*x) - 18*(a*x +
b)^2*a*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^2*x^2) + 18*(a*x + b)*a^2*log(c)/(b^2*x) + 9*(a*x + b)^2*a*p/
(b^2*x^2) + 6*(a*x + b)^3*p*log(-b*(a/b - (a*x + b)/(b*x)) + a)/(b^2*x^3) - 18*(a*x + b)^2*a*log(c)/(b^2*x^2)
- 2*(a*x + b)^3*p/(b^2*x^3) + 6*(a*x + b)^3*log(c)/(b^2*x^3))/b

Mupad [B] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^4} \, dx=\frac {\frac {p}{3}+\frac {a^2\,p\,x^2}{b^2}-\frac {a\,p\,x}{2\,b}}{3\,x^3}-\frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{3\,x^3}-\frac {2\,a^3\,p\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{3\,b^3} \]

[In]

int(log(c*(a + b/x)^p)/x^4,x)

[Out]

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