Integrand size = 19, antiderivative size = 69 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {a x}{e}-\frac {b n x}{e}+\frac {b x \log \left (c x^n\right )}{e}-\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{e^2}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \] Output:
a*x/e-b*n*x/e+b*x*ln(c*x^n)/e-d*(a+b*ln(c*x^n))*ln(1+e*x/d)/e^2-b*d*n*poly log(2,-e*x/d)/e^2
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {a e x-b e n x-a d \log \left (1+\frac {e x}{d}\right )+b \log \left (c x^n\right ) \left (e x-d \log \left (1+\frac {e x}{d}\right )\right )-b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2} \] Input:
Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x),x]
Output:
(a*e*x - b*e*n*x - a*d*Log[1 + (e*x)/d] + b*Log[c*x^n]*(e*x - d*Log[1 + (e *x)/d]) - b*d*n*PolyLog[2, -((e*x)/d)])/e^2
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2793, 2009}
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.
| \(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx\) |
| \(\Big \downarrow \) 2793 |
| \(\displaystyle \int \left (\frac {a+b \log \left (c x^n\right )}{e}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{e (d+e x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d \log \left (\frac {e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {a x}{e}+\frac {b x \log \left (c x^n\right )}{e}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{e^2}-\frac {b n x}{e}\) |
Input:
Int[(x*(a + b*Log[c*x^n]))/(d + e*x),x]
Output:
(a*x)/e - (b*n*x)/e + (b*x*Log[c*x^n])/e - (d*(a + b*Log[c*x^n])*Log[1 + ( e*x)/d])/e^2 - (b*d*n*PolyLog[2, -((e*x)/d)])/e^2
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && Integer Q[r]))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.42 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.72
| method | result | size |
| risch | \(\frac {b \ln \left (x^{n}\right ) x}{e}-\frac {b \ln \left (x^{n}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {b n x}{e}-\frac {b n d}{e^{2}}+\frac {b n d \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{2}}+\frac {b n d \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\) | \(188\) |
Input:
int(x*(a+b*ln(c*x^n))/(e*x+d),x,method=_RETURNVERBOSE)
Output:
b*ln(x^n)/e*x-b*ln(x^n)*d/e^2*ln(e*x+d)-b*n*x/e-b*n*d/e^2+b*n*d/e^2*ln(e*x +d)*ln(-e*x/d)+b*n*d/e^2*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)*(x/e-d/e^2*ln(e*x+d))
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x+d),x, algorithm="fricas")
Output:
integral((b*x*log(c*x^n) + a*x)/(e*x + d), x)
Time = 9.78 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.36 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=- \frac {a d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e} + \frac {a x}{e} + \frac {b d n \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \operatorname {Li}_{2}\left (\frac {e x e^{i \pi }}{d}\right ) & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e} - \frac {b d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e} - \frac {b n x}{e} + \frac {b x \log {\left (c x^{n} \right )}}{e} \] Input:
integrate(x*(a+b*ln(c*x**n))/(e*x+d),x)
Output:
-a*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e + a*x/e + b*d*n* Piecewise((x/d, Eq(e, 0)), (Piecewise((-polylog(2, e*x*exp_polar(I*pi)/d), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x*exp_polar (I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi) /d), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meij erg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*pi) /d), True))/e, True))/e - b*d*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/e - b*n*x/e + b*x*log(c*x**n)/e
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x+d),x, algorithm="maxima")
Output:
a*(x/e - d*log(e*x + d)/e^2) + b*integrate((x*log(c) + x*log(x^n))/(e*x + d), x)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{e x + d} \,d x } \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x+d),x, algorithm="giac")
Output:
integrate((b*log(c*x^n) + a)*x/(e*x + d), x)
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+e\,x} \,d x \] Input:
int((x*(a + b*log(c*x^n)))/(d + e*x),x)
Output:
int((x*(a + b*log(c*x^n)))/(d + e*x), x)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e \,x^{2}+d x}d x \right ) b \,d^{2} n -2 \,\mathrm {log}\left (e x +d \right ) a d n -\mathrm {log}\left (x^{n} c \right )^{2} b d +2 \,\mathrm {log}\left (x^{n} c \right ) b e n x +2 a e n x -2 b e \,n^{2} x}{2 e^{2} n} \] Input:
int(x*(a+b*log(c*x^n))/(e*x+d),x)
Output:
(2*int(log(x**n*c)/(d*x + e*x**2),x)*b*d**2*n - 2*log(d + e*x)*a*d*n - log (x**n*c)**2*b*d + 2*log(x**n*c)*b*e*n*x + 2*a*e*n*x - 2*b*e*n**2*x)/(2*e** 2*n)