\(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^2} \, dx\) [49]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(e f-d g) (f+g x)}-\frac {2 b e n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g (e f-d g)}-\frac {2 b^2 e n^2 \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g (e f-d g)} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^2} \, dx=\frac {-\left (\left (a+b \log \left (c (d+e x)^n\right )\right ) \left (a g (d+e x)+b g (d+e x) \log \left (c (d+e x)^n\right )-2 b e n (f+g x) \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right )+2 b^2 e n^2 (f+g x) \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{g (-e f+d g) (f+g x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2844, 2841, 2840, 2838}

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.

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

\(\Big \downarrow \) 2844

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

\(\Big \downarrow \) 2841

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

\(\Big \downarrow \) 2840

\(\displaystyle \frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x) (e f-d g)}-\frac {2 b e n \left (\frac {\log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {b n \int \frac {\log \left (\frac {g (d+e x)}{e f-d g}+1\right )}{d+e x}d(d+e x)}{g}\right )}{e f-d g}\)

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 2840

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_ 
Symbol] :> Simp[1/g   Subst[Int[(a + b*Log[1 + c*e*(x/g)])/x, x], x, f + g* 
x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g + c 
*(e*f - d*g), 0]
 

rule 2841

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_ 
)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x 
)^n])/g), x] - Simp[b*e*(n/g)   Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]
 

rule 2844

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. 
)*(x_))^2, x_Symbol] :> Simp[(d + e*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f 
- d*g)*(f + g*x))), x] - Simp[b*e*n*(p/(e*f - d*g))   Int[(a + b*Log[c*(d + 
 e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] & 
& NeQ[e*f - d*g, 0] && GtQ[p, 0]
 

Maple [C] (warning: unable to verify)

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

Time = 1.85 (sec) , antiderivative size = 537, normalized size of antiderivative = 4.07

method	result	size
risch	\(-\frac {b^{2} \ln \left (\left (e x +d \right )^{n}\right )^{2}}{\left (g x +f \right ) g}+\frac {2 b^{2} n e \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g x +f \right )}{g \left (d g -e f \right )}-\frac {2 b^{2} n e \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (e x +d \right )}{g \left (d g -e f \right )}-\frac {2 b^{2} n^{2} e \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g \left (d g -e f \right )}-\frac {2 b^{2} n^{2} e \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 \left (d g -e f \right )}+\frac {b^{2} n^{2} e \ln \left (e x +d \right )^{2}}{g \left (d g -e f \right )}+\left (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}-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 )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+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 )+2 a \right ) b \left (-\frac {\ln \left (\left (e x +d \right )^{n}\right )}{\left (g x +f \right ) g}+\frac {n e \left (\frac {\ln \left (g x +f \right )}{d g -e f}-\frac {\ln \left (e x +d \right )}{d g -e f}\right )}{g}\right )-\frac {{\left (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}-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 )-i b \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}+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 )+2 a \right )}^{2}}{4 \left (g x +f \right ) g}\)	\(537\)

Input:

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

Output:

-b^2*ln((e*x+d)^n)^2/(g*x+f)/g+2*b^2/g*n*e*ln((e*x+d)^n)/(d*g-e*f)*ln(g*x+ 
f)-2*b^2/g*n*e*ln((e*x+d)^n)/(d*g-e*f)*ln(e*x+d)-2*b^2/g*n^2*e/(d*g-e*f)*d 
ilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))-2*b^2/g*n^2*e/(d*g-e*f)*ln(g*x+f)*ln(( 
(g*x+f)*e+d*g-e*f)/(d*g-e*f))+b^2/g*n^2*e/(d*g-e*f)*ln(e*x+d)^2+(I*b*Pi*cs 
gn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e 
*x+d)^n)*csgn(I*c)-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n) 
^2*csgn(I*c)+2*b*ln(c)+2*a)*b*(-ln((e*x+d)^n)/(g*x+f)/g+1/g*n*e*(1/(d*g-e* 
f)*ln(g*x+f)-1/(d*g-e*f)*ln(e*x+d)))-1/4*(I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I* 
c*(e*x+d)^n)^2-I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-I*b* 
Pi*csgn(I*c*(e*x+d)^n)^3+I*b*Pi*csgn(I*c*(e*x+d)^n)^2*csgn(I*c)+2*b*ln(c)+ 
2*a)^2/(g*x+f)/g
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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