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

Optimal. Leaf size=104 \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^2}-\frac{f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac{a x}{g}+\frac{b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac{b n x}{g} \]

[Out]

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

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Rubi [A]  time = 0.130702, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {43, 2416, 2389, 2295, 2394, 2393, 2391} \[ -\frac{b f n \text{PolyLog}\left (2,-\frac{g (d+e x)}{e f-d g}\right )}{g^2}-\frac{f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac{a x}{g}+\frac{b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac{b n x}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 2389

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, 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 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 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx &=\int \left (\frac{a+b \log \left (c (d+e x)^n\right )}{g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (f+g x)}\right ) \, dx\\ &=\frac{\int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}-\frac{f \int \frac{a+b \log \left (c (d+e x)^n\right )}{f+g x} \, dx}{g}\\ &=\frac{a x}{g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g^2}+\frac{b \int \log \left (c (d+e x)^n\right ) \, dx}{g}+\frac{(b e f n) \int \frac{\log \left (\frac{e (f+g x)}{e f-d g}\right )}{d+e x} \, dx}{g^2}\\ &=\frac{a x}{g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g^2}+\frac{b \operatorname{Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g}+\frac{(b f n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{g^2}\\ &=\frac{a x}{g}-\frac{b n x}{g}+\frac{b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac{f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac{e (f+g x)}{e f-d g}\right )}{g^2}-\frac{b f n \text{Li}_2\left (-\frac{g (d+e x)}{e f-d g}\right )}{g^2}\\ \end{align*}

Mathematica [A]  time = 0.0719131, size = 95, normalized size = 0.91 \[ \frac{-b f n \text{PolyLog}\left (2,\frac{g (d+e x)}{d g-e f}\right )-f \log \left (\frac{e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+a g x+\frac{b g (d+e x) \log \left (c (d+e x)^n\right )}{e}-b g n x}{g^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.588, size = 463, normalized size = 4.5 \begin{align*}{\frac{b\ln \left ( \left ( ex+d \right ) ^{n} \right ) x}{g}}-{\frac{b\ln \left ( \left ( ex+d \right ) ^{n} \right ) f\ln \left ( gx+f \right ) }{{g}^{2}}}-{\frac{bnx}{g}}-{\frac{bnf}{{g}^{2}}}+{\frac{bdn\ln \left ( \left ( gx+f \right ) e+dg-fe \right ) }{eg}}+{\frac{bnf}{{g}^{2}}{\it dilog} \left ({\frac{ \left ( gx+f \right ) e+dg-fe}{dg-fe}} \right ) }+{\frac{bnf\ln \left ( gx+f \right ) }{{g}^{2}}\ln \left ({\frac{ \left ( gx+f \right ) e+dg-fe}{dg-fe}} \right ) }+{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}x}{g}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}f\ln \left ( gx+f \right ) }{{g}^{2}}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ){\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) x}{g}}+{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{3}f\ln \left ( gx+f \right ) }{{g}^{2}}}+{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( ic \right ){\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ){\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) f\ln \left ( gx+f \right ) }{{g}^{2}}}-{\frac{{\frac{i}{2}}b\pi \, \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{3}x}{g}}+{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( ic \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}x}{g}}-{\frac{{\frac{i}{2}}b\pi \,{\it csgn} \left ( i \left ( ex+d \right ) ^{n} \right ) \left ({\it csgn} \left ( ic \left ( ex+d \right ) ^{n} \right ) \right ) ^{2}f\ln \left ( gx+f \right ) }{{g}^{2}}}+{\frac{\ln \left ( c \right ) bx}{g}}-{\frac{b\ln \left ( c \right ) f\ln \left ( gx+f \right ) }{{g}^{2}}}+{\frac{ax}{g}}-{\frac{af\ln \left ( gx+f \right ) }{{g}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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