\(\int \frac {x (a+b \log (c (d+e x)^n))}{f+g x} \, dx\) [244]

Optimal result

Integrand size = 23, antiderivative size = 104 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {a x}{g}-\frac {b n x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^2}-\frac {b f n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^2} \] Output:

a*x/g-b*n*x/g+b*(e*x+d)*ln(c*(e*x+d)^n)/e/g-f*(a+b*ln(c*(e*x+d)^n))*ln(e*( 
g*x+f)/(-d*g+e*f))/g^2-b*f*n*polylog(2,-g*(e*x+d)/(-d*g+e*f))/g^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {a g x-b g n x+\frac {b g (d+e x) \log \left (c (d+e x)^n\right )}{e}-f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-b f n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2863, 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.

\(\displaystyle \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx\)

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g (f+g x)}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(a*x)/g - (b*n*x)/g + (b*(d + e*x)*Log[c*(d + e*x)^n])/(e*g) - (f*(a + b*L 
og[c*(d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^2 - (b*f*n*PolyLog[2, 
 -((g*(d + e*x))/(e*f - d*g))])/g^2
 

Defintions of rubi rules used

rule 2009

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

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

Maple [C] (warning: unable to verify)

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

Time = 1.44 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.73

method	result	size
risch	\(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x}{g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \ln \left (g x +f \right )}{g^{2}}-\frac {b n x}{g}-\frac {b f n}{g^{2}}+\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right )}{e g}+\frac {b n f \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}+\frac {b n f \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x}{g}-\frac {f \ln \left (g x +f \right )}{g^{2}}\right )\)	\(284\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(a + b*log(c*(d + e*x)**n))/(f + g*x), x)
 

Maxima [F]

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

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

Output:

a*(x/g - f*log(g*x + f)/g^2) + b*integrate((x*log((e*x + d)^n) + x*log(c)) 
/(g*x + f), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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