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

Optimal. Leaf size=80 \[ -\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)}+\frac {b n \text {Li}_2\left (-\frac {d}{e x}\right )}{d^2}+\frac {b n \log (d+e x)}{d^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 102, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2347, 2344, 2301, 2317, 2391, 2314, 31} \[ -\frac {b n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2}-\frac {\log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^2 n}+\frac {b n \log (d+e x)}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

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

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Mathematica [A]  time = 0.08, size = 96, normalized size = 1.20 \[ \frac {-2 \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b n \text {Li}_2\left (-\frac {e x}{d}\right )-2 b n (\log (x)-\log (d+e x))}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.19, size = 521, normalized size = 6.51 \[ \frac {b n \ln \left (-\frac {e x}{d}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 \left (e x +d \right ) d}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2 d^{2}}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +d \right )}{2 d^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +d \right )}{2 d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2 d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 \left (e x +d \right ) d}-\frac {b n \ln \relax (x )^{2}}{2 d^{2}}-\frac {b n \ln \relax (x )}{d^{2}}+\frac {b \ln \left (x^{n}\right )}{\left (e x +d \right ) d}+\frac {b \ln \relax (x ) \ln \left (x^{n}\right )}{d^{2}}-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{2}}+\frac {b \ln \relax (c ) \ln \relax (x )}{d^{2}}-\frac {b \ln \relax (c ) \ln \left (e x +d \right )}{d^{2}}+\frac {b n \dilog \left (-\frac {e x}{d}\right )}{d^{2}}+\frac {b \ln \relax (c )}{\left (e x +d \right ) d}+\frac {a \ln \relax (x )}{d^{2}}-\frac {a \ln \left (e x +d \right )}{d^{2}}+\frac {a}{\left (e x +d \right ) d}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +d \right )}{2 d^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) d}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2 d^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 \left (e x +d \right ) d}+\frac {b n \ln \left (e x +d \right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*log(c*x**n))/(x*(d + e*x)**2), x)

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