\(\int \frac {a+b \log (c x^n)}{x (d+e x)^2} \, dx\) [43]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)}-\frac {\log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {b n \log (d+e x)}{d^2}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^2} \]

[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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2389, 2379, 2438, 2351, 31} \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2}-\frac {e x \left (a+b \log \left (c x^n\right )\right )}{d^2 (d+e x)}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d^2}+\frac {b n \log (d+e x)}{d^2} \]

[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))) - (Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/d^2 + (b*n*Log[d + e*x])/d
^2 + (b*n*PolyLog[2, -(d/(e*x))])/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 2351

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 2379

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

Rule 2389

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 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]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\frac {\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{d+e x}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{b n}-2 b n (\log (x)-\log (d+e x))-2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )-2 b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 d^2} \]

[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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.44 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.86

method	result	size
risch	\(-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d^{2}}+\frac {b \ln \left (x^{n}\right )}{d \left (e x +d \right )}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d^{2}}-\frac {b n \ln \left (x \right )^{2}}{2 d^{2}}+\frac {b n \ln \left (e x +d \right )}{d^{2}}-\frac {b n \ln \left (x \right )}{d^{2}}+\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d^{2}}+\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d^{2}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (e x +d \right )}{d^{2}}+\frac {1}{d \left (e x +d \right )}+\frac {\ln \left (x \right )}{d^{2}}\right )\)	\(229\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x \left (d + e x\right )^{2}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x} \,d x } \]

[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)

Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )}^{2} x} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (d+e\,x\right )}^2} \,d x \]

[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)