3.3.42 \(\int \frac {x \log (c (a+\frac {b}{x})^p)}{d+e x} \, dx\) [242]

Optimal. Leaf size=151 \[ \frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}+\frac {b p \log (b+a x)}{a e}-\frac {d \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^2}-\frac {d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^2}+\frac {d p \text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )}{e^2}-\frac {d p \text {Li}_2\left (1+\frac {e x}{d}\right )}{e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2516, 2498, 269, 31, 2512, 266, 2463, 2441, 2352, 2440, 2438} \begin {gather*} \frac {d p \text {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^2}-\frac {d p \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e^2}-\frac {d \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^2}+\frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}+\frac {d p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{e^2}+\frac {b p \log (a x+b)}{a e}-\frac {d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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 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 2498

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 149, normalized size = 0.99 \begin {gather*} \frac {x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}+\frac {b p \left (\frac {\log \left (a+\frac {b}{x}\right )}{a}+\frac {\log (x)}{a}\right )}{e}-\frac {d \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^2}-\frac {d p \left (\log \left (-\frac {e x}{d}\right ) \log (d+e x)-\log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)+\text {Li}_2\left (\frac {d+e x}{d}\right )-\text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \frac {x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e x +d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________