\(\int \frac {x (a+b \log (c x^n))}{d+\frac {e}{x}} \, dx\) [330]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=-\frac {a e x}{d^2}+\frac {b e n x}{d^2}-\frac {b n x^2}{4 d}-\frac {b e x \log \left (c x^n\right )}{d^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d x}{e}\right )}{d^3}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3} \] Output:

-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
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {-4 a d e x+4 b d e n x+2 a d^2 x^2-b d^2 n x^2+4 a e^2 \log \left (1+\frac {d x}{e}\right )+2 b \log \left (c x^n\right ) \left (d x (-2 e+d x)+2 e^2 \log \left (1+\frac {d x}{e}\right )\right )+4 b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{4 d^3} \] Input:

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

Output:

(-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)
 

Rubi [A] (verified)

Time = 0.37 (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

\(\displaystyle \frac {e^2 \log \left (\frac {d x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^3}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d}-\frac {a e x}{d^2}-\frac {b e x \log \left (c x^n\right )}{d^2}+\frac {b e^2 n \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d^3}+\frac {b e n x}{d^2}-\frac {b n x^2}{4 d}\)

Input:

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

Output:

-((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
 

Defintions of rubi rules used

rule 2005

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2793

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]))
 

Maple [C] (warning: unable to verify)

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

Time = 0.26 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.18

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 d}-\frac {b \ln \left (x^{n}\right ) e x}{d^{2}}+\frac {b \ln \left (x^{n}\right ) e^{2} \ln \left (d x +e \right )}{d^{3}}-\frac {b n \,e^{2} \ln \left (d x +e \right ) \ln \left (-\frac {d x}{e}\right )}{d^{3}}-\frac {b n \,e^{2} \operatorname {dilog}\left (-\frac {d x}{e}\right )}{d^{3}}-\frac {b n \,x^{2}}{4 d}+\frac {b e n x}{d^{2}}+\frac {5 b n \,e^{2}}{4 d^{3}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} d \,x^{2}-e x}{d^{2}}+\frac {e^{2} \ln \left (d x +e \right )}{d^{3}}\right )\)	\(233\)

Input:

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

Output:

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*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^2*(1/2*d*x^2-e*x)+e^2/d^3*ln(d*x+e))
 

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 52.59 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.04 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {a x^{2}}{2 d} + \frac {a e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} - \frac {a e x}{d^{2}} - \frac {b n x^{2}}{4 d} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 d} - \frac {b e^{2} n \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\begin {cases} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (e \right )} \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (e \right )} \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (e \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (e \right )} - \operatorname {Li}_{2}\left (\frac {d x e^{i \pi }}{e}\right ) & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right )}{d^{2}} + \frac {b e^{2} \left (\begin {cases} \frac {x}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d x + e \right )}}{d} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{d^{2}} + \frac {b e n x}{d^{2}} - \frac {b e x \log {\left (c x^{n} \right )}}{d^{2}} \] Input:

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

Output:

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
 

Maxima [F]

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

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

Output:

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)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

Reduce [F]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+\frac {e}{x}} \, dx=\frac {-4 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{d \,x^{2}+e x}d x \right ) b \,e^{3} n +4 \,\mathrm {log}\left (d x +e \right ) a \,e^{2} n +2 \mathrm {log}\left (x^{n} c \right )^{2} b \,e^{2}+2 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} n \,x^{2}-4 \,\mathrm {log}\left (x^{n} c \right ) b d e n x +2 a \,d^{2} n \,x^{2}-4 a d e n x -b \,d^{2} n^{2} x^{2}+4 b d e \,n^{2} x}{4 d^{3} n} \] Input:

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

Output:

( - 4*int(log(x**n*c)/(d*x**2 + e*x),x)*b*e**3*n + 4*log(d*x + e)*a*e**2*n 
 + 2*log(x**n*c)**2*b*e**2 + 2*log(x**n*c)*b*d**2*n*x**2 - 4*log(x**n*c)*b 
*d*e*n*x + 2*a*d**2*n*x**2 - 4*a*d*e*n*x - b*d**2*n**2*x**2 + 4*b*d*e*n**2 
*x)/(4*d**3*n)