Integrand size = 21, antiderivative size = 107 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=-\frac {a e x}{d^2}+\frac {b e n x}{d^2}-\frac {b n x^2}{4 d}-\frac {b e x \log \left (c x^n\right )}{d^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^3}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3} \]
-a*e*x/d^2+b*e*n*x/d^2-1/4*b*n*x^2/d-b*e*x*ln(c*x^n)/d^2+1/2*x^2*(a+b*ln(c *x^n))/d+e^2*(a+b*ln(c*x^n))*ln(1+d*x/e)/d^3+b*e^2*n*polylog(2,-d*x/e)/d^3
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {-4 a d e x+4 b d e n x+2 a d^2 x^2-b d^2 n x^2+4 a e^2 \log \left (1+\frac {d x}{e}\right )+2 b \log \left (c x^n\right ) \left (d x (-2 e+d x)+2 e^2 \log \left (1+\frac {d x}{e}\right )\right )+4 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{4 d^3} \]
(-4*a*d*e*x + 4*b*d*e*n*x + 2*a*d^2*x^2 - b*d^2*n*x^2 + 4*a*e^2*Log[1 + (d *x)/e] + 2*b*Log[c*x^n]*(d*x*(-2*e + d*x) + 2*e^2*Log[1 + (d*x)/e]) + 4*b* e^2*n*PolyLog[2, -((d*x)/e)])/(4*d^3)
Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2005, 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.
\(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{d x+e}dx\) |
\(\Big \downarrow \) 2793 |
\(\displaystyle \int \left (\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{d^2 (d x+e)}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{d^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{d}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d}-\frac {a e x}{d^2}-\frac {b e x \log \left (c x^n\right )}{d^2}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3}+\frac {b e n x}{d^2}-\frac {b n x^2}{4 d}\) |
-((a*e*x)/d^2) + (b*e*n*x)/d^2 - (b*n*x^2)/(4*d) - (b*e*x*Log[c*x^n])/d^2 + (x^2*(a + b*Log[c*x^n]))/(2*d) + (e^2*(a + b*Log[c*x^n])*Log[1 + (d*x)/e ])/d^3 + (b*e^2*n*PolyLog[2, -((d*x)/e)])/d^3
3.4.31.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
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 = 0.13 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.18
method | result | size |
risch | \(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 d}-\frac {b \ln \left (x^{n}\right ) e x}{d^{2}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (d x +e \right )}{d^{3}}-\frac {b n \,e^{2} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{3}}-\frac {b n \,x^{2}}{4 d}+\frac {b e n x}{d^{2}}+\frac {5 b n \,e^{2}}{4 d^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} d \,x^{2}-e x}{d^{2}}+\frac {e^{2} \ln \left (d x +e \right )}{d^{3}}\right )\) | \(233\) |
1/2*b*ln(x^n)/d*x^2-b*ln(x^n)/d^2*e*x+b*ln(x^n)*e^2/d^3*ln(d*x+e)-b*n*e^2/ d^3*ln(d*x+e)*ln(-d*x/e)-b*n*e^2/d^3*dilog(-d*x/e)-1/4*b*n*x^2/d+b*e*n*x/d ^2+5/4*b*n*e^2/d^3+(-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*(1/2*d*x^2-e*x)+e^2/d^3*ln(d*x+e) )
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{d + \frac {e}{x}} \,d x } \]
Time = 53.55 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.04 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{2}}{2 d} + \frac {a e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e x}{d^{2}} - \frac {b n x^{2}}{4 d} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 d} - \frac {b e^{2} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b e n x}{d^{2}} - \frac {b e x \log {\left (c x^{n} \right )}}{d^{2}} \]
a*x**2/(2*d) + a*e**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e)/d, True))/d **2 - a*e*x/d**2 - b*n*x**2/(4*d) + b*x**2*log(c*x**n)/(2*d) - b*e**2*n*Pi ecewise((x/e, Eq(d, 0)), (Piecewise((-polylog(2, d*x*exp_polar(I*pi)/e), ( Abs(x) < 1) & (1/Abs(x) < 1)), (log(e)*log(x) - polylog(2, d*x*exp_polar(I *pi)/e), Abs(x) < 1), (-log(e)*log(1/x) - polylog(2, d*x*exp_polar(I*pi)/e ), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijer g(((1, 1), ()), ((), (0, 0)), x)*log(e) - polylog(2, d*x*exp_polar(I*pi)/e ), True))/d, True))/d**2 + b*e**2*Piecewise((x/e, Eq(d, 0)), (log(d*x + e) /d, True))*log(c*x**n)/d**2 + b*e*n*x/d**2 - b*e*x*log(c*x**n)/d**2
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{d + \frac {e}{x}} \,d x } \]
1/2*a*(2*e^2*log(d*x + e)/d^3 + (d*x^2 - 2*e*x)/d^2) + b*integrate((x^2*lo g(c) + x^2*log(x^n))/(d*x + e), x)
\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{d + \frac {e}{x}} \,d x } \]
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{d+\frac {e}{x}} \,d x \]