\(\int \frac {x^2 (a+b \log (c (d+e x)^n))}{f+g x} \, dx\) [243]

Optimal result

Integrand size = 25, antiderivative size = 181 \[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=-\frac {a f x}{g^2}+\frac {b f n x}{g^2}+\frac {b d n x}{2 e g}-\frac {b n x^2}{4 g}-\frac {b d^2 n \log (d+e x)}{2 e^2 g}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}+\frac {b f^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^3} \] Output:

-a*f*x/g^2+b*f*n*x/g^2+1/2*b*d*n*x/e/g-1/4*b*n*x^2/g-1/2*b*d^2*n*ln(e*x+d) 
/e^2/g-b*f*(e*x+d)*ln(c*(e*x+d)^n)/e/g^2+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g+f 
^2*(a+b*ln(c*(e*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/g^3+b*f^2*n*polylog(2,-g 
*(e*x+d)/(-d*g+e*f))/g^3
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=-\frac {a f x}{g^2}+\frac {b f n x}{g^2}+\frac {b n \left (\frac {2 d x}{e}-x^2-\frac {2 d^2 \log (d+e x)}{e^2}\right )}{4 g}-\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e g^2}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}+\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^3}+\frac {b f^2 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^3} \] Input:

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

Output:

-((a*f*x)/g^2) + (b*f*n*x)/g^2 + (b*n*((2*d*x)/e - x^2 - (2*d^2*Log[d + e* 
x])/e^2))/(4*g) - (b*f*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^2) + (x^2*(a + b 
*Log[c*(d + e*x)^n]))/(2*g) + (f^2*(a + b*Log[c*(d + e*x)^n])*Log[(e*(f + 
g*x))/(e*f - d*g)])/g^3 + (b*f^2*n*PolyLog[2, -((g*(d + e*x))/(e*f - d*g)) 
])/g^3
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((a*f*x)/g^2) + (b*f*n*x)/g^2 + (b*d*n*x)/(2*e*g) - (b*n*x^2)/(4*g) - (b* 
d^2*n*Log[d + e*x])/(2*e^2*g) - (b*f*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^2) 
 + (x^2*(a + b*Log[c*(d + e*x)^n]))/(2*g) + (f^2*(a + b*Log[c*(d + e*x)^n] 
)*Log[(e*(f + g*x))/(e*f - d*g)])/g^3 + (b*f^2*n*PolyLog[2, -((g*(d + e*x) 
)/(e*f - d*g))])/g^3
 

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 = 1.62 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.14

method	result	size
risch	\(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{2}}{2 g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f x}{g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{2} \ln \left (g x +f \right )}{g^{3}}-\frac {b n \,f^{2} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}-\frac {b n \,f^{2} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{3}}-\frac {b n \,x^{2}}{4 g}+\frac {b f n x}{g^{2}}+\frac {5 b \,f^{2} n}{4 g^{3}}+\frac {b d n x}{2 e g}+\frac {b d f n}{2 e \,g^{2}}-\frac {b n \,d^{2} \ln \left (\left (g x +f \right ) e +d g -e f \right )}{2 e^{2} g}-\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right ) f}{e \,g^{2}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} g \,x^{2}-f x}{g^{2}}+\frac {f^{2} \ln \left (g x +f \right )}{g^{3}}\right )\)	\(388\)

Input:

int(x^2*(a+b*ln(c*(e*x+d)^n))/(g*x+f),x,method=_RETURNVERBOSE)
 

Output:

1/2*b*ln((e*x+d)^n)/g*x^2-b*ln((e*x+d)^n)/g^2*f*x+b*ln((e*x+d)^n)*f^2/g^3* 
ln(g*x+f)-b*n/g^3*f^2*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-b*n/g^3*f^2*ln( 
g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-1/4*b*n*x^2/g+b*f*n*x/g^2+5/4*b*f 
^2*n/g^3+1/2*b*d*n*x/e/g+1/2*b*d*f*n/e/g^2-1/2*b/e^2*n/g*d^2*ln((g*x+f)*e+ 
d*g-e*f)-b/e*n/g^2*d*ln((g*x+f)*e+d*g-e*f)*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)*cs 
gn(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)*(1/g^2*(1/2*g*x^2-f*x)+f^2/g^3*ln(g*x+f))
 

Fricas [F]

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

integrate(x^2*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x^2*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="maxima")
 

Output:

1/2*a*(2*f^2*log(g*x + f)/g^3 + (g*x^2 - 2*f*x)/g^2) + b*integrate((x^2*lo 
g((e*x + d)^n) + x^2*log(c))/(g*x + f), x)
 

Giac [F]

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

integrate(x^2*(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^2/(g*x + f), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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