\(\int \frac {\log (\frac {c+d x}{a+b x})}{(a+b x) ((a-c) h+(b-d) h x)} \, dx\) [255]

Optimal result

Integrand size = 40, antiderivative size = 33 \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx=-\frac {\operatorname {PolyLog}\left (2,1-\frac {c+d x}{a+b x}\right )}{(b c-a d) h} \] Output:

-polylog(2,1-(d*x+c)/(b*x+a))/(-a*d+b*c)/h
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(298\) vs. \(2(33)=66\).

Time = 0.24 (sec) , antiderivative size = 298, normalized size of antiderivative = 9.03 \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx=\frac {-\log ^2\left (\frac {-b c+a d}{d (a+b x)}\right )+2 \log \left (\frac {(b-d) (a+b x)}{b c-a d}\right ) \log (a-c+b x-d x)-2 \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )-2 \log (a-c+b x-d x) \log \left (\frac {(b-d) (c+d x)}{b c-a d}\right )+2 \log \left (\frac {-b c+a d}{d (a+b x)}\right ) \log \left (\frac {c+d x}{a+b x}\right )+2 \log (a-c+b x-d x) \log \left (\frac {c+d x}{a+b x}\right )+2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )+2 \operatorname {PolyLog}\left (2,-\frac {b (a-c+b x-d x)}{b c-a d}\right )-2 \operatorname {PolyLog}\left (2,-\frac {d (-a+c-b x+d x)}{-b c+a d}\right )}{(2 b c-2 a d) h} \] Input:

Integrate[Log[(c + d*x)/(a + b*x)]/((a + b*x)*((a - c)*h + (b - d)*h*x)),x 
]
 

Output:

(-Log[(-(b*c) + a*d)/(d*(a + b*x))]^2 + 2*Log[((b - d)*(a + b*x))/(b*c - a 
*d)]*Log[a - c + b*x - d*x] - 2*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(b*( 
c + d*x))/(b*c - a*d)] - 2*Log[a - c + b*x - d*x]*Log[((b - d)*(c + d*x))/ 
(b*c - a*d)] + 2*Log[(-(b*c) + a*d)/(d*(a + b*x))]*Log[(c + d*x)/(a + b*x) 
] + 2*Log[a - c + b*x - d*x]*Log[(c + d*x)/(a + b*x)] + 2*PolyLog[2, (d*(a 
 + b*x))/(-(b*c) + a*d)] + 2*PolyLog[2, -((b*(a - c + b*x - d*x))/(b*c - a 
*d))] - 2*PolyLog[2, -((d*(-a + c - b*x + d*x))/(-(b*c) + a*d))])/((2*b*c 
- 2*a*d)*h)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2965, 25, 27, 2752}

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.

\(\displaystyle \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) (h (a-c)+h x (b-d))} \, dx\)

\(\Big \downarrow \) 2965

\(\displaystyle \int -\frac {\log \left (\frac {c+d x}{a+b x}\right )}{h \left (-\frac {(c+d x) (b c-a d)}{a+b x}-a d+b c\right )}d\frac {c+d x}{a+b x}\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{h \left (b c-a d-\frac {(b c-a d) (c+d x)}{a+b x}\right )}d\frac {c+d x}{a+b x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{b c-a d-\frac {(b c-a d) (c+d x)}{a+b x}}d\frac {c+d x}{a+b x}}{h}\)

\(\Big \downarrow \) 2752

\(\displaystyle -\frac {\operatorname {PolyLog}\left (2,1-\frac {c+d x}{a+b x}\right )}{h (b c-a d)}\)

Input:

Int[Log[(c + d*x)/(a + b*x)]/((a + b*x)*((a - c)*h + (b - d)*h*x)),x]
 

Output:

-(PolyLog[2, 1 - (c + d*x)/(a + b*x)]/((b*c - a*d)*h))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2965

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*( 
B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol 
] :> Simp[(b*c - a*d)^(q + 1)*(i/d)^q   Subst[Int[(b*f - a*g - (d*f - c*g)* 
x)^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d* 
x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && IntegersQ[m, q] && IGtQ[p, 0] && EqQ[d*h - c*i, 0]
 

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {\operatorname {dilog}\left (-\frac {d a -b c}{b \left (b x +a \right )}+\frac {d}{b}\right )}{h \left (d a -b c \right )}\)	\(42\)
default	\(\frac {\operatorname {dilog}\left (-\frac {d a -b c}{b \left (b x +a \right )}+\frac {d}{b}\right )}{h \left (d a -b c \right )}\)	\(42\)
risch	\(\frac {\operatorname {dilog}\left (-\frac {d a -b c}{b \left (b x +a \right )}+\frac {d}{b}\right )}{h \left (d a -b c \right )}\)	\(42\)
parts	\(-\frac {\ln \left (\frac {d x +c}{b x +a}\right ) \ln \left (b x -d x +a -c \right ) b}{h \left (d a -b c \right ) \left (b -d \right )}+\frac {\ln \left (\frac {d x +c}{b x +a}\right ) \ln \left (b x -d x +a -c \right ) d}{h \left (d a -b c \right ) \left (b -d \right )}+\frac {\ln \left (\frac {d x +c}{b x +a}\right ) \ln \left (b x +a \right )}{h \left (d a -b c \right )}-\frac {\frac {\operatorname {dilog}\left (\frac {-d a +b c +d \left (b x +a \right )}{-d a +b c}\right )}{d a -b c}+\frac {\ln \left (b x +a \right ) \ln \left (\frac {-d a +b c +d \left (b x +a \right )}{-d a +b c}\right )}{d a -b c}-\frac {\ln \left (b x +a \right )^{2}}{2 \left (d a -b c \right )}-\frac {\frac {\left (\frac {\operatorname {dilog}\left (\frac {d \left (a -c +\left (b -d \right ) x \right )-d a +b c}{-d a +b c}\right )}{d}+\frac {\ln \left (a -c +\left (b -d \right ) x \right ) \ln \left (\frac {d \left (a -c +\left (b -d \right ) x \right )-d a +b c}{-d a +b c}\right )}{d}\right ) d \left (b -d \right )}{d a -b c}-\frac {\left (\frac {\operatorname {dilog}\left (\frac {\left (a -c +\left (b -d \right ) x \right ) b -d a +b c}{-d a +b c}\right )}{b}+\frac {\ln \left (a -c +\left (b -d \right ) x \right ) \ln \left (\frac {\left (a -c +\left (b -d \right ) x \right ) b -d a +b c}{-d a +b c}\right )}{b}\right ) b \left (b -d \right )}{d a -b c}}{b -d}}{h}\)	\(468\)

Input:

int(ln((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x,method=_RETURNVERBOS 
E)
 

Output:

dilog(-1/b*(a*d-b*c)/(b*x+a)+d/b)/h/(a*d-b*c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx=-\frac {{\rm Li}_2\left (-\frac {d x + c}{b x + a} + 1\right )}{{\left (b c - a d\right )} h} \] Input:

integrate(log((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x, algorithm="f 
ricas")
 

Output:

-dilog(-(d*x + c)/(b*x + a) + 1)/((b*c - a*d)*h)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx=\text {Timed out} \] Input:

integrate(ln((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (32) = 64\).

Time = 0.05 (sec) , antiderivative size = 357, normalized size of antiderivative = 10.82 \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx={\left (\frac {\log \left (-{\left (b - d\right )} x - a + c\right )}{{\left (b c - a d\right )} h} - \frac {\log \left (b x + a\right )}{{\left (b c - a d\right )} h}\right )} \log \left (\frac {d x + c}{b x + a}\right ) + \frac {2 \, \log \left (-{\left (b - d\right )} x - a + c\right ) \log \left (b x + a\right ) - \log \left (b x + a\right )^{2}}{2 \, {\left (b c h - a d h\right )}} + \frac {\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )}{b c h - a d h} - \frac {\log \left (b x + a\right ) \log \left (-\frac {a {\left (b - d\right )} + {\left (b^{2} - b d\right )} x}{b c - a d} + 1\right ) + {\rm Li}_2\left (\frac {a {\left (b - d\right )} + {\left (b^{2} - b d\right )} x}{b c - a d}\right )}{b c h - a d h} - \frac {\log \left (-{\left (b - d\right )} x - a + c\right ) \log \left (\frac {a d - c d + {\left (b d - d^{2}\right )} x}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {a d - c d + {\left (b d - d^{2}\right )} x}{b c - a d}\right )}{b c h - a d h} \] Input:

integrate(log((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x, algorithm="m 
axima")
 

Output:

(log(-(b - d)*x - a + c)/((b*c - a*d)*h) - log(b*x + a)/((b*c - a*d)*h))*l 
og((d*x + c)/(b*x + a)) + 1/2*(2*log(-(b - d)*x - a + c)*log(b*x + a) - lo 
g(b*x + a)^2)/(b*c*h - a*d*h) + (log(b*x + a)*log((b*d*x + a*d)/(b*c - a*d 
) + 1) + dilog(-(b*d*x + a*d)/(b*c - a*d)))/(b*c*h - a*d*h) - (log(b*x + a 
)*log(-(a*(b - d) + (b^2 - b*d)*x)/(b*c - a*d) + 1) + dilog((a*(b - d) + ( 
b^2 - b*d)*x)/(b*c - a*d)))/(b*c*h - a*d*h) - (log(-(b - d)*x - a + c)*log 
((a*d - c*d + (b*d - d^2)*x)/(b*c - a*d) + 1) + dilog(-(a*d - c*d + (b*d - 
 d^2)*x)/(b*c - a*d)))/(b*c*h - a*d*h)
 

Giac [F]

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

integrate(log((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x, algorithm="g 
iac")
 

Output:

integrate(log((d*x + c)/(b*x + a))/(((b - d)*h*x + (a - c)*h)*(b*x + a)), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx=\int \frac {\ln \left (\frac {c+d\,x}{a+b\,x}\right )}{\left (h\,\left (a-c\right )+h\,x\,\left (b-d\right )\right )\,\left (a+b\,x\right )} \,d x \] Input:

int(log((c + d*x)/(a + b*x))/((h*(a - c) + h*x*(b - d))*(a + b*x)),x)
 

Output:

int(log((c + d*x)/(a + b*x))/((h*(a - c) + h*x*(b - d))*(a + b*x)), x)
 

Reduce [F]

\[ \int \frac {\log \left (\frac {c+d x}{a+b x}\right )}{(a+b x) ((a-c) h+(b-d) h x)} \, dx =\text {Too large to display} \] Input:

int(log((d*x+c)/(b*x+a))/(b*x+a)/((a-c)*h+(b-d)*h*x),x)
 

Output:

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