3.3.46 \(\int \frac {a+b \log (c (d+e x)^n)}{x (f+g x)} \, dx\) [246]

Optimal. Leaf size=107 \[ \frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{f}-\frac {b n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f}+\frac {b n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {36, 29, 31, 2463, 2441, 2352, 2440, 2438} \begin {gather*} -\frac {b n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f}+\frac {b n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f}-\frac {\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}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 85, normalized size = 0.79 \begin {gather*} \frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (\log \left (-\frac {e x}{d}\right )-\log \left (\frac {e (f+g x)}{e f-d g}\right )\right )-b n \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )+b n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.37, size = 455, normalized size = 4.25

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{f}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f}-\frac {b n \dilog \left (\frac {e x +d}{d}\right )}{f}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f}+\frac {b n \dilog \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f}+\frac {b n \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{f}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g x +f \right )}{2 f}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{2 f}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (x \right )}{2 f}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g x +f \right )}{2 f}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g x +f \right )}{2 f}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (x \right )}{2 f}-\frac {b \ln \left (c \right ) \ln \left (g x +f \right )}{f}+\frac {b \ln \left (c \right ) \ln \left (x \right )}{f}-\frac {a \ln \left (g x +f \right )}{f}+\frac {a \ln \left (x \right )}{f}\)	\(455\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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