3.242 \(\int \frac{x \log (c (a+\frac{b}{x})^p)}{d+e x} \, dx\)

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

[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

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Rubi [A]  time = 0.212287, antiderivative size = 151, normalized size of antiderivative = 1., 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 = {2466, 2448, 263, 31, 2462, 260, 2416, 2394, 2315, 2393, 2391} \[ \frac{d p \text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )}{e^2}-\frac{d p \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{e^2}-\frac{d \log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e^2}+\frac{x \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e}+\frac{d p \log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )}{e^2}+\frac{b p \log (a x+b)}{a e}-\frac{d p \log \left (-\frac{e x}{d}\right ) \log (d+e x)}{e^2} \]

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 2466

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]

Rule 2448

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 263

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 31

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

Rule 2462

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 260

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 2416

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 2394

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 2315

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

Rule 2393

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 2391

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]

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) \operatorname{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 = 0.0571212, size = 149, normalized size = 0.99 \[ -\frac{d p \left (-\text{PolyLog}\left (2,\frac{a (d+e x)}{a d-b e}\right )+\text{PolyLog}\left (2,\frac{d+e x}{d}\right )-\log (d+e x) \log \left (-\frac{e (a x+b)}{a d-b e}\right )+\log \left (-\frac{e x}{d}\right ) \log (d+e x)\right )}{e^2}-\frac{d \log (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e^2}+\frac{x \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{e}+\frac{b p \left (\frac{\log \left (a+\frac{b}{x}\right )}{a}+\frac{\log (x)}{a}\right )}{e} \]

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

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Maple [F]  time = 0.742, size = 0, normalized size = 0. \begin{align*} \int{\frac{x}{ex+d}\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) }\, dx \end{align*}

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)

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

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)/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \log \left (c \left (\frac{a x + b}{x}\right )^{p}\right )}{e x + d}, x\right ) \end{align*}

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)/(e*x + d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

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)/(e*x + d), x)