3.247 \(\int \frac{a+b \log (c (d+e x)^n)}{x^2 (f+g x)} \, dx\)

Optimal. Leaf size=162 \[ \frac{b g n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{f^2}-\frac{b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{g \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac{a+b \log \left (c (d+e x)^n\right )}{f x}+\frac{b e n \log (x)}{d f}-\frac{b e n \log (d+e x)}{d f} \]

[Out]

(b*e*n*Log[x])/(d*f) - (b*e*n*Log[d + e*x])/(d*f) - (a + b*Log[c*(d + e*x)^n])/(f*x) - (g*Log[-((e*x)/d)]*(a +
b*Log[c*(d + e*x)^n]))/f^2 + (g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^2 + (b*g*n*PolyL
og[2, -((g*(d + e*x))/(e*f - d*g))])/f^2 - (b*g*n*PolyLog[2, 1 + (e*x)/d])/f^2

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Rubi [A]  time = 0.191103, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {44, 2416, 2395, 36, 29, 31, 2394, 2315, 2393, 2391} \[ \frac{b g n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{f^2}-\frac{b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )}{f^2}-\frac{g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac{g \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac{a+b \log \left (c (d+e x)^n\right )}{f x}+\frac{b e n \log (x)}{d f}-\frac{b e n \log (d+e x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*n*Log[x])/(d*f) - (b*e*n*Log[d + e*x])/(d*f) - (a + b*Log[c*(d + e*x)^n])/(f*x) - (g*Log[-((e*x)/d)]*(a +
b*Log[c*(d + e*x)^n]))/f^2 + (g*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/f^2 + (b*g*n*PolyL
og[2, -((g*(d + e*x))/(e*f - d*g))])/f^2 - (b*g*n*PolyLog[2, 1 + (e*x)/d])/f^2

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 2395

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

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

Mathematica [A]  time = 0.0835287, size = 141, normalized size = 0.87 \[ \frac{b g n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )-b g n \text{PolyLog}\left (2,\frac{e x}{d}+1\right )+g \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right )}{x}-g \log \left (-\frac{e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b e f n (\log (x)-\log (d+e x))}{d}}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.547, size = 669, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/x^2/(g*x+f),x)

[Out]

b*e*n*ln(x)/d/f-b*e*n*ln(e*x+d)/d/f-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(x)-1/2*I*b*Pi*csgn(I*c
)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f^2*g*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d
)^n)/f^2*g*ln(x)-1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f/x+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f/x-b*n/f^2*g
*dilog(((g*x+f)*e+d*g-f*e)/(d*g-e*f))+b*n/f^2*g*dilog((e*x+d)/d)+b*ln(c)/f^2*g*ln(g*x+f)-b*ln(c)/f^2*g*ln(x)-b
*ln((e*x+d)^n)/f/x-b*n/f^2*g*ln(g*x+f)*ln(((g*x+f)*e+d*g-f*e)/(d*g-e*f))+b*n/f^2*g*ln(x)*ln((e*x+d)/d)-a/f/x+a
/f^2*g*ln(g*x+f)-a/f^2*g*ln(x)-b*ln(c)/f/x+1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/f/x+1/2*
I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(g*x+f)+1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f^2*g*ln(x)+1/2*I*b*Pi
*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f^2*g*ln(g*x+f)-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f^
2*g*ln(x)-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/f/x-1/2*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/f^2*g*ln(g*x
+f)+b*ln((e*x+d)^n)/f^2*g*ln(g*x+f)-b*ln((e*x+d)^n)/f^2*g*ln(x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{g \log \left (g x + f\right )}{f^{2}} - \frac{g \log \left (x\right )}{f^{2}} - \frac{1}{f x}\right )} + b \int \frac{\log \left ({\left (e x + d\right )}^{n}\right ) + \log \left (c\right )}{g x^{3} + f x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(g*log(g*x + f)/f^2 - g*log(x)/f^2 - 1/(f*x)) + b*integrate((log((e*x + d)^n) + log(c))/(g*x^3 + f*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g x^{3} + f x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((e*x + d)^n*c) + a)/(g*x^3 + f*x^2), 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((a+b*ln(c*(e*x+d)**n))/x**2/(g*x+f),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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