\(\int \frac {\log (c (a+\frac {b}{x})^p)}{x^3 (d+e x)} \, dx\) [246]

Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {2516, 2504, 2442, 45, 2436, 2332, 2441, 2352, 2512, 266, 2463, 2440, 2438} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}-\frac {e^2 \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}-\frac {e^2 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^3}+\frac {e^2 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^3}-\frac {a p}{2 b d x}-\frac {e^2 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}-\frac {e p}{d^2 x}+\frac {p}{4 d x^2} \]

[In]

Int[Log[c*(a + b/x)^p]/(x^3*(d + e*x)),x]

[Out]

p/(4*d*x^2) - (a*p)/(2*b*d*x) - (e*p)/(d^2*x) + (a^2*p*Log[a + b/x])/(2*b^2*d) + (e*(a + b/x)*Log[c*(a + b/x)^
p])/(b*d^2) - Log[c*(a + b/x)^p]/(2*d*x^2) - (e^2*Log[c*(a + b/x)^p]*Log[-(b/(a*x))])/d^3 - (e^2*Log[c*(a + b/
x)^p]*Log[d + e*x])/d^3 - (e^2*p*Log[-((e*x)/d)]*Log[d + e*x])/d^3 + (e^2*p*Log[-((e*(b + a*x))/(a*d - b*e))]*
Log[d + e*x])/d^3 - (e^2*p*PolyLog[2, 1 + b/(a*x)])/d^3 + (e^2*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)])/d^3 -
(e^2*p*PolyLog[2, 1 + (e*x)/d])/d^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 266

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 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 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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])

Rule 2512

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[f +
g*x]*((a + b*Log[c*(d + e*x^n)^p])/g), x] - Dist[b*e*n*(p/g), Int[x^(n - 1)*(Log[f + g*x]/(d + e*x^n)), x], x]
/; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]

Rule 2516

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_))^(r_.), x_S
ymbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e,
f, g, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d x^3}-\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2 x^2}+\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 x}-\frac {e^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^3 (d+e x)}\right ) \, dx \\ & = \frac {\int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3} \, dx}{d}-\frac {e \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2} \, dx}{d^2}+\frac {e^2 \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x} \, dx}{d^3}-\frac {e^3 \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx}{d^3} \\ & = -\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {\text {Subst}\left (\int x \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x}\right )}{d}+\frac {e \text {Subst}\left (\int \log \left (c (a+b x)^p\right ) \, dx,x,\frac {1}{x}\right )}{d^2}-\frac {e^2 \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,\frac {1}{x}\right )}{d^3}-\frac {\left (b e^2 p\right ) \int \frac {\log (d+e x)}{\left (a+\frac {b}{x}\right ) x^2} \, dx}{d^3} \\ & = -\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}+\frac {e \text {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,a+\frac {b}{x}\right )}{b d^2}+\frac {(b p) \text {Subst}\left (\int \frac {x^2}{a+b x} \, dx,x,\frac {1}{x}\right )}{2 d}-\frac {\left (b e^2 p\right ) \int \left (\frac {\log (d+e x)}{b x}-\frac {a \log (d+e x)}{b (b+a x)}\right ) \, dx}{d^3}+\frac {\left (b e^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,\frac {1}{x}\right )}{d^3} \\ & = -\frac {e p}{d^2 x}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {(b p) \text {Subst}\left (\int \left (-\frac {a}{b^2}+\frac {x}{b}+\frac {a^2}{b^2 (a+b x)}\right ) \, dx,x,\frac {1}{x}\right )}{2 d}-\frac {\left (e^2 p\right ) \int \frac {\log (d+e x)}{x} \, dx}{d^3}+\frac {\left (a e^2 p\right ) \int \frac {\log (d+e x)}{b+a x} \, dx}{d^3} \\ & = \frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {\left (e^3 p\right ) \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx}{d^3}-\frac {\left (e^3 p\right ) \int \frac {\log \left (\frac {e (b+a x)}{-a d+b e}\right )}{d+e x} \, dx}{d^3} \\ & = \frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {e x}{d}\right )}{d^3}-\frac {\left (e^2 p\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {a x}{-a d+b e}\right )}{x} \, dx,x,d+e x\right )}{d^3} \\ & = \frac {p}{4 d x^2}-\frac {a p}{2 b d x}-\frac {e p}{d^2 x}+\frac {a^2 p \log \left (a+\frac {b}{x}\right )}{2 b^2 d}+\frac {e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d^2}-\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 d x^2}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^3}-\frac {e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^3}-\frac {e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^3}+\frac {e^2 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {b}{a x}\right )}{d^3}+\frac {e^2 p \text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )}{d^3}-\frac {e^2 p \text {Li}_2\left (1+\frac {e x}{d}\right )}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.92 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=-\frac {-\frac {d^2 p}{x^2}+\frac {2 a d^2 p}{b x}+\frac {4 d e p}{x}-\frac {2 a^2 d^2 p \log \left (a+\frac {b}{x}\right )}{b^2}-\frac {4 d e \left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b}+\frac {2 d^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2}+4 e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )+4 e^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)+4 e^2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)-4 e^2 p \log \left (\frac {e (b+a x)}{-a d+b e}\right ) \log (d+e x)+4 e^2 p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )-4 e^2 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )+4 e^2 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{4 d^3} \]

[In]

Integrate[Log[c*(a + b/x)^p]/(x^3*(d + e*x)),x]

[Out]

-1/4*(-((d^2*p)/x^2) + (2*a*d^2*p)/(b*x) + (4*d*e*p)/x - (2*a^2*d^2*p*Log[a + b/x])/b^2 - (4*d*e*(a + b/x)*Log
[c*(a + b/x)^p])/b + (2*d^2*Log[c*(a + b/x)^p])/x^2 + 4*e^2*Log[c*(a + b/x)^p]*Log[-(b/(a*x))] + 4*e^2*Log[c*(
a + b/x)^p]*Log[d + e*x] + 4*e^2*p*Log[-((e*x)/d)]*Log[d + e*x] - 4*e^2*p*Log[(e*(b + a*x))/(-(a*d) + b*e)]*Lo
g[d + e*x] + 4*e^2*p*PolyLog[2, 1 + b/(a*x)] - 4*e^2*p*PolyLog[2, (a*(d + e*x))/(a*d - b*e)] + 4*e^2*p*PolyLog
[2, 1 + (e*x)/d])/d^3

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.20

method	result	size
parts	\(-\frac {e^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e}{d^{2} x}+\frac {p b \left (-\frac {-\frac {d}{2 b \,x^{2}}-\frac {-a d -2 b e}{b^{2} x}+\frac {\left (a d +2 b e \right ) a \ln \left (x \right )}{b^{3}}-\frac {\left (a d +2 b e \right ) a \ln \left (a x +b \right )}{b^{3}}}{d^{2}}+\frac {e^{2} \ln \left (x \right )^{2}}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (\frac {a x +b}{b}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (x \right ) \ln \left (\frac {a x +b}{b}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{3} b}+\frac {2 e^{2} \operatorname {dilog}\left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{d^{3} b}+\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{d^{3} b}\right )}{2}\)	\(345\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]

[In]

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

[Out]

integral(log(c*((a*x + b)/x)^p)/(e*x^4 + d*x^3), x)

Sympy [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{x^{3} \left (d + e x\right )}\, dx \]

[In]

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

[Out]

Integral(log(c*(a + b/x)**p)/(x**3*(d + e*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\frac {1}{4} \, {\left (4 \, e {\left (\frac {a \log \left (a x + b\right )}{b^{2} d^{2}} - \frac {a \log \left (x\right )}{b^{2} d^{2}} - \frac {1}{b d^{2} x}\right )} - \frac {4 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} e^{2}}{b d^{3}} + \frac {4 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )\right )} e^{2}}{b d^{3}} + \frac {2 \, a^{2} \log \left (a x + b\right )}{b^{3} d} - \frac {2 \, a^{2} \log \left (x\right )}{b^{3} d} - \frac {2 \, {\left (2 \, e^{2} \log \left (e x + d\right ) \log \left (x\right ) - e^{2} \log \left (x\right )^{2}\right )}}{b d^{3}} - \frac {2 \, a x - b}{b^{2} d x^{2}}\right )} b p - \frac {1}{2} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

[In]

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

[Out]

1/4*(4*e*(a*log(a*x + b)/(b^2*d^2) - a*log(x)/(b^2*d^2) - 1/(b*d^2*x)) - 4*(log(a*x/b + 1)*log(x) + dilog(-a*x
/b))*e^2/(b*d^3) + 4*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*e^2/(b*d^3) + 4*(log(e*x + d)*log(-(a*e*x + a*d)/
(a*d - b*e) + 1) + dilog((a*e*x + a*d)/(a*d - b*e)))*e^2/(b*d^3) + 2*a^2*log(a*x + b)/(b^3*d) - 2*a^2*log(x)/(
b^3*d) - 2*(2*e^2*log(e*x + d)*log(x) - e^2*log(x)^2)/(b*d^3) - (2*a*x - b)/(b^2*d*x^2))*b*p - 1/2*(2*e^2*log(
e*x + d)/d^3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2))*log((a + b/x)^p*c)

Giac [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{3}} \,d x } \]

[In]

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

[Out]

integrate(log((a + b/x)^p*c)/((e*x + d)*x^3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^3 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{x^3\,\left (d+e\,x\right )} \,d x \]

[In]

int(log(c*(a + b/x)^p)/(x^3*(d + e*x)),x)

[Out]

int(log(c*(a + b/x)^p)/(x^3*(d + e*x)), x)