3.334 \(\int \frac{\log (\frac{x^{-m} (-a+a c+b c x^m)}{b})}{x (a+b x^m)} \, dx\)

Optimal. Leaf size=27 \[ \frac{\text{PolyLog}\left (2,\frac{(1-c) \left (a x^{-m}+b\right )}{b}\right )}{a m} \]

[Out]

PolyLog[2, ((1 - c)*(b + a/x^m))/b]/(a*m)

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Rubi [A]  time = 0.180217, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2480, 2475, 2412, 2393, 2391} \[ \frac{\text{PolyLog}\left (2,\frac{(1-c) \left (a x^{-m}+b\right )}{b}\right )}{a m} \]

Antiderivative was successfully verified.

[In]

Int[Log[(-a + a*c + b*c*x^m)/(b*x^m)]/(x*(a + b*x^m)),x]

[Out]

PolyLog[2, ((1 - c)*(b + a/x^m))/b]/(a*m)

Rule 2480

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.)*(u_)^(r_.)*((h_.)*(x_))^(m_.), x_Symbol] :> Int[(h*x)^m*Expand
ToSum[u, x]^r*(a + b*Log[c*ExpandToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, h, m, p, q, r}, x] && BinomialQ[{u,
v}, x] &&  !BinomialMatchQ[{u, v}, x]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 2412

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

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

Mathematica [A]  time = 0.0074884, size = 29, normalized size = 1.07 \[ \frac{\text{PolyLog}\left (2,-\frac{(c-1) x^{-m} \left (a+b x^m\right )}{b}\right )}{a m} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(-a + a*c + b*c*x^m)/(b*x^m)]/(x*(a + b*x^m)),x]

[Out]

PolyLog[2, -(((-1 + c)*(a + b*x^m))/(b*x^m))]/(a*m)

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Maple [A]  time = 0.065, size = 24, normalized size = 0.9 \begin{align*}{\frac{1}{am}{\it dilog} \left ( c+{\frac{a \left ( -1+c \right ) }{b{x}^{m}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln((-a+a*c+b*c*x^m)/b/(x^m))/x/(a+b*x^m),x)

[Out]

1/m/a*dilog(c+a*(-1+c)/(x^m)/b)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\left (c m - m\right )} \int \frac{\log \left (x\right )}{b c x x^{m} + a{\left (c - 1\right )} x}\,{d x} + \frac{\log \left (b c x^{m} + a c - a\right ) \log \left (x\right ) - \log \left (b\right ) \log \left (x\right ) - \log \left (x\right ) \log \left (x^{m}\right )}{a} + \frac{\log \left (b\right ) \log \left (\frac{b x^{m} + a}{b}\right )}{a m} + \frac{\log \left (x^{m}\right ) \log \left (\frac{b x^{m}}{a} + 1\right ) +{\rm Li}_2\left (-\frac{b x^{m}}{a}\right )}{a m} - \frac{\log \left (b c x^{m} + a c - a\right ) \log \left (\frac{b c x^{m} + a{\left (c - 1\right )}}{a} + 1\right ) +{\rm Li}_2\left (-\frac{b c x^{m} + a{\left (c - 1\right )}}{a}\right )}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*m - m)*integrate(log(x)/(b*c*x*x^m + a*(c - 1)*x), x) + (log(b*c*x^m + a*c - a)*log(x) - log(b)*log(x) - lo
g(x)*log(x^m))/a + log(b)*log((b*x^m + a)/b)/(a*m) + (log(x^m)*log(b*x^m/a + 1) + dilog(-b*x^m/a))/(a*m) - (lo
g(b*c*x^m + a*c - a)*log((b*c*x^m + a*(c - 1))/a + 1) + dilog(-(b*c*x^m + a*(c - 1))/a))/(a*m)

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Fricas [A]  time = 1.57642, size = 63, normalized size = 2.33 \begin{align*} \frac{{\rm Li}_2\left (-\frac{b c x^{m} + a c - a}{b x^{m}} + 1\right )}{a m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

dilog(-(b*c*x^m + a*c - a)/(b*x^m) + 1)/(a*m)

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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(ln((-a+a*c+b*c*x**m)/b/(x**m))/x/(a+b*x**m),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((b*c*x^m + a*c - a)/(b*x^m))/((b*x^m + a)*x), x)