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

Optimal. Leaf size=179 \[ -\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f (f+g x)}-\frac {b n \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{f^2}+\frac {b n \text {Li}_2\left (\frac {e x}{d}+1\right )}{f^2}-\frac {b e n \log (d+e x)}{f (e f-d g)}+\frac {b e n \log (f+g x)}{f (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {44, 2416, 2394, 2315, 2395, 36, 31, 2393, 2391} \[ -\frac {b n \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{f^2}+\frac {b n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}-\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{f (f+g x)}-\frac {b e n \log (d+e x)}{f (e f-d g)}+\frac {b e n \log (f+g x)}{f (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 152, normalized size = 0.85 \[ \frac {-\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x}+\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-b n \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )-\frac {b e f n (\log (d+e x)-\log (f+g x))}{e f-d g}+b n \text {Li}_2\left (\frac {e x}{d}+1\right )}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g^{2} x^{3} + 2 \, f g x^{2} + f^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.26, size = 694, normalized size = 3.88 \[ -\frac {b n \dilog \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {b n \dilog \left (\frac {d g -e f +\left (g x +f \right ) e}{d g -e f}\right )}{f^{2}}-\frac {b e n \ln \left (g x +f \right )}{\left (d g -e f \right ) f}+\frac {b e n \ln \left (e x +d \right )}{\left (d g -e f \right ) f}+\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{\left (g x +f \right ) f}+\frac {b \ln \relax (x ) \ln \left (\left (e x +d \right )^{n}\right )}{f^{2}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{f^{2}}-\frac {a \ln \left (g x +f \right )}{f^{2}}+\frac {a}{\left (g x +f \right ) f}+\frac {a \ln \relax (x )}{f^{2}}-\frac {b \ln \relax (c ) \ln \left (g x +f \right )}{f^{2}}+\frac {b \ln \relax (c )}{\left (g x +f \right ) f}+\frac {b \ln \relax (c ) \ln \relax (x )}{f^{2}}-\frac {b n \ln \relax (x ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {i \pi b \,\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 )}{2 \left (g x +f \right ) f}-\frac {i \pi b \,\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 \relax (x )}{2 f^{2}}+\frac {i \pi b \,\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^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 \left (g x +f \right ) f}+\frac {i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 \left (g x +f \right ) f}+\frac {b n \ln \left (\frac {d g -e f +\left (g x +f \right ) e}{d g -e f}\right ) \ln \left (g x +f \right )}{f^{2}}+\frac {i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g x +f \right )}{2 f^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \relax (x )}{2 f^{2}}-\frac {i \pi b \,\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^{2}}+\frac {i \pi b \,\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 \relax (x )}{2 f^{2}}-\frac {i \pi b \,\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^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 \left (g x +f \right ) f}-\frac {i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \relax (x )}{2 f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {1}{f g x + f^{2}} - \frac {\log \left (g x + f\right )}{f^{2}} + \frac {\log \relax (x)}{f^{2}}\right )} + b \int \frac {\log \left ({\left (e x + d\right )}^{n}\right ) + \log \relax (c)}{g^{2} x^{3} + 2 \, f g x^{2} + f^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x\,{\left (f+g\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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