Integrand size = 21, antiderivative size = 44 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=-\frac {\log \left (1+\frac {d}{e x}\right ) \left (a+b \log \left (c x^n\right )\right )}{d}+\frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d} \] Output:
-ln(1+d/e/x)*(a+b*ln(c*x^n))/d+b*n*polylog(2,-d/e/x)/d
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.43 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-2 b n \log \left (1+\frac {e x}{d}\right )\right )}{2 b d n}-\frac {b n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d} \] Input:
Integrate[(a + b*Log[c*x^n])/(x*(d + e*x)),x]
Output:
((a + b*Log[c*x^n])*(a + b*Log[c*x^n] - 2*b*n*Log[1 + (e*x)/d]))/(2*b*d*n) - (b*n*PolyLog[2, -((e*x)/d)])/d
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2779, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 2779 |
\(\displaystyle \frac {b n \int \frac {\log \left (\frac {d}{e x}+1\right )}{x}dx}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {b n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{d}-\frac {\log \left (\frac {d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\) |
Input:
Int[(a + b*Log[c*x^n])/(x*(d + e*x)),x]
Output:
-((Log[1 + d/(e*x)]*(a + b*Log[c*x^n]))/d) + (b*n*PolyLog[2, -(d/(e*x))])/ d
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] + Simp[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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.41 (sec) , antiderivative size = 181, normalized size of antiderivative = 4.11
method | result | size |
risch | \(-\frac {b \ln \left (x^{n}\right ) \ln \left (e x +d \right )}{d}+\frac {b \ln \left (x^{n}\right ) \ln \left (x \right )}{d}-\frac {b n \ln \left (x \right )^{2}}{2 d}+\frac {b n \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d}+\frac {b n \operatorname {dilog}\left (-\frac {e x}{d}\right )}{d}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {\ln \left (e x +d \right )}{d}+\frac {\ln \left (x \right )}{d}\right )\) | \(181\) |
Input:
int((a+b*ln(c*x^n))/x/(e*x+d),x,method=_RETURNVERBOSE)
Output:
-b*ln(x^n)/d*ln(e*x+d)+b*ln(x^n)/d*ln(x)-1/2*b*n/d*ln(x)^2+b*n/d*ln(e*x+d) *ln(-e*x/d)+b*n/d*dilog(-e*x/d)+(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/ 2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c*x^n)^3+1/ 2*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)*(-1/d*ln(e*x+d)+1/d*ln(x))
\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x/(e*x+d),x, algorithm="fricas")
Output:
integral((b*log(c*x^n) + a)/(e*x^2 + d*x), x)
Result contains complex when optimal does not.
Time = 6.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 3.98 \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=- \frac {2 a e \left (\begin {cases} \frac {1}{2 e} + \frac {x}{d} & \text {for}\: e = 0 \\- \frac {\log {\left (- 2 e x \right )}}{2 e} & \text {otherwise} \end {cases}\right )}{d} - \frac {2 a e \left (\begin {cases} \frac {1}{2 e} + \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (2 d + 2 e x \right )}}{2 e} & \text {otherwise} \end {cases}\right )}{d} + b n \left (\begin {cases} - \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\begin {cases} \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + \operatorname {Li}_{2}\left (\frac {d e^{i \pi }}{e x}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) - b \left (\begin {cases} \frac {1}{e x} & \text {for}\: d = 0 \\\frac {\log {\left (\frac {d}{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \] Input:
integrate((a+b*ln(c*x**n))/x/(e*x+d),x)
Output:
-2*a*e*Piecewise((1/(2*e) + x/d, Eq(e, 0)), (-log(-2*e*x)/(2*e), True))/d - 2*a*e*Piecewise((1/(2*e) + x/d, Eq(e, 0)), (log(2*d + 2*e*x)/(2*e), True ))/d + b*n*Piecewise((-1/(e*x), Eq(d, 0)), (Piecewise((polylog(2, d*exp_po lar(I*pi)/(e*x)), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) + polylog (2, d*exp_polar(I*pi)/(e*x)), Abs(x) < 1), (-log(e)*log(1/x) + polylog(2, d*exp_polar(I*pi)/(e*x)), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(e) + polylog(2 , d*exp_polar(I*pi)/(e*x)), True))/d, True)) - b*Piecewise((1/(e*x), Eq(d, 0)), (log(d/x + e)/d, True))*log(c*x**n)
\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x/(e*x+d),x, algorithm="maxima")
Output:
-a*(log(e*x + d)/d - log(x)/d) + b*integrate((log(c) + log(x^n))/(e*x^2 + d*x), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate((a+b*log(c*x^n))/x/(e*x+d),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)/((e*x + d)*x), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,\left (d+e\,x\right )} \,d x \] Input:
int((a + b*log(c*x^n))/(x*(d + e*x)),x)
Output:
int((a + b*log(c*x^n))/(x*(d + e*x)), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x (d+e x)} \, dx=\frac {\left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{2}+d x}d x \right ) b d -\mathrm {log}\left (e x +d \right ) a +\mathrm {log}\left (x \right ) a}{d} \] Input:
int((a+b*log(c*x^n))/x/(e*x+d),x)
Output:
(int(log(x**n*c)/(d*x + e*x**2),x)*b*d - log(d + e*x)*a + log(x)*a)/d