\(\int \frac {\log (c (a+\frac {b}{x})^p)}{x^2 (d+e x)} \, dx\) [245]

Optimal result

Integrand size = 23, antiderivative size = 198 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {p}{d x}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^2}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,1+\frac {b}{a x}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {2516, 2504, 2436, 2332, 2441, 2352, 2512, 266, 2463, 2440, 2438} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {e \log \left (-\frac {b}{a x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2}+\frac {e \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d^2}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e p \operatorname {PolyLog}\left (2,\frac {b}{a x}+1\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2}-\frac {e p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}+\frac {p}{d x} \]

[In]

Int[Log[c*(a + b/x)^p]/(x^2*(d + e*x)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {p}{d x}-\frac {\left (a+\frac {b}{x}\right ) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{b d}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log \left (-\frac {b}{a x}\right )}{d^2}+\frac {e \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{d^2}+\frac {e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {a+\frac {b}{x}}{a}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{d^2} \]

[In]

Integrate[Log[c*(a + b/x)^p]/(x^2*(d + e*x)),x]

[Out]

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

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.33

method	result	size
parts	\(\frac {e \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{d x}-\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e \ln \left (x \right )}{d^{2}}+p b \left (\frac {e \left (\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b}-\frac {a \left (\frac {\operatorname {dilog}\left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{a}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{a}\right )}{b}\right )}{d^{2}}+\frac {1}{d b x}+\frac {a \ln \left (x \right )}{d \,b^{2}}-\frac {a \ln \left (a x +b \right )}{d \,b^{2}}-\frac {e \ln \left (x \right )^{2}}{2 d^{2} b}+\frac {e \operatorname {dilog}\left (\frac {a x +b}{b}\right )}{d^{2} b}+\frac {e \ln \left (x \right ) \ln \left (\frac {a x +b}{b}\right )}{d^{2} b}\right )\)	\(264\)

[In]

int(ln(c*(a+b/x)^p)/x^2/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}} \,d x } \]

[In]

integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="fricas")

[Out]

integral(log(c*((a*x + b)/x)^p)/(e*x^3 + d*x^2), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\text {Timed out} \]

[In]

integrate(ln(c*(a+b/x)**p)/x**2/(e*x+d),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.16 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {1}{2} \, b p {\left (\frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} e}{b d^{2}} - \frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} e}{b d^{2}} - \frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )\right )} e}{b d^{2}} - \frac {2 \, a \log \left (a x + b\right )}{b^{2} d} + \frac {2 \, a \log \left (x\right )}{b^{2} d} + \frac {2 \, e \log \left (e x + d\right ) \log \left (x\right ) - e \log \left (x\right )^{2}}{b d^{2}} + \frac {2}{b d x}\right )} + {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

[In]

integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}} \,d x } \]

[In]

integrate(log(c*(a+b/x)^p)/x^2/(e*x+d),x, algorithm="giac")

[Out]

integrate(log((a + b/x)^p*c)/((e*x + d)*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{x^2 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{x^2\,\left (d+e\,x\right )} \,d x \]

[In]

int(log(c*(a + b/x)^p)/(x^2*(d + e*x)),x)

[Out]

int(log(c*(a + b/x)^p)/(x^2*(d + e*x)), x)