Integrand size = 38, antiderivative size = 27 \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {g (c+d x)}{a+b x}\right )}{b c-a d} \] Output:
polylog(2,g*(d*x+c)/(b*x+a))/(-a*d+b*c)
Leaf count is larger than twice the leaf count of optimal. \(375\) vs. \(2(27)=54\).
Time = 0.10 (sec) , antiderivative size = 375, normalized size of antiderivative = 13.89 \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {-\log ^2\left (\frac {a}{b}+x\right )+2 \log \left (\frac {a}{b}+x\right ) \log (a+b x)-2 \log \left (\frac {a-c g}{b-d g}+x\right ) \log (a+b x)+2 \log \left (\frac {a-c g}{b-d g}+x\right ) \log \left (\frac {(b-d g) (a+b x)}{(b c-a d) g}\right )-2 \log \left (\frac {a}{b}+x\right ) \log (c+d x)+2 \log \left (\frac {a-c g}{b-d g}+x\right ) \log (c+d x)+2 \log \left (\frac {a}{b}+x\right ) \log \left (\frac {b (c+d x)}{b c-a d}\right )-2 \log \left (\frac {a-c g}{b-d g}+x\right ) \log \left (\frac {(b-d g) (c+d x)}{b c-a d}\right )+2 \log (a+b x) \log \left (\frac {a-c g+b x-d g x}{a+b x}\right )-2 \log (c+d x) \log \left (\frac {a-c g+b x-d g x}{a+b x}\right )+2 \operatorname {PolyLog}\left (2,\frac {d (a+b x)}{-b c+a d}\right )+2 \operatorname {PolyLog}\left (2,-\frac {b (a-c g+b x-d g x)}{(b c-a d) g}\right )-2 \operatorname {PolyLog}\left (2,-\frac {d (-a+c g-b x+d g x)}{-b c+a d}\right )}{2 b c-2 a d} \] Input:
Integrate[Log[(a - c*g + b*x - d*g*x)/(a + b*x)]/((a + b*x)*(c + d*x)),x]
Output:
(-Log[a/b + x]^2 + 2*Log[a/b + x]*Log[a + b*x] - 2*Log[(a - c*g)/(b - d*g) + x]*Log[a + b*x] + 2*Log[(a - c*g)/(b - d*g) + x]*Log[((b - d*g)*(a + b* x))/((b*c - a*d)*g)] - 2*Log[a/b + x]*Log[c + d*x] + 2*Log[(a - c*g)/(b - d*g) + x]*Log[c + d*x] + 2*Log[a/b + x]*Log[(b*(c + d*x))/(b*c - a*d)] - 2 *Log[(a - c*g)/(b - d*g) + x]*Log[((b - d*g)*(c + d*x))/(b*c - a*d)] + 2*L og[a + b*x]*Log[(a - c*g + b*x - d*g*x)/(a + b*x)] - 2*Log[c + d*x]*Log[(a - c*g + b*x - d*g*x)/(a + b*x)] + 2*PolyLog[2, (d*(a + b*x))/(-(b*c) + a* d)] + 2*PolyLog[2, -((b*(a - c*g + b*x - d*g*x))/((b*c - a*d)*g))] - 2*Pol yLog[2, -((d*(-a + c*g - b*x + d*g*x))/(-(b*c) + a*d))])/(2*b*c - 2*a*d)
Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2971, 2965, 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.
\(\displaystyle \int \frac {\log \left (\frac {a+b x-c g-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx\) |
\(\Big \downarrow \) 2971 |
\(\displaystyle \int \frac {\log \left (\frac {a+x (b-d g)-c g}{a+b x}\right )}{(a+b x) (c+d x)}dx\) |
\(\Big \downarrow \) 2965 |
\(\displaystyle \int \frac {\log \left (\frac {a+x (b-d g)-c g}{a+b x}\right )}{-\frac {(b c-a d) (a+x (b-d g)-c g)}{a+b x}-a d+b c}d\frac {a+x (b-d g)-c g}{a+b x}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,1-\frac {a-c g+(b-d g) x}{a+b x}\right )}{b c-a d}\) |
Input:
Int[Log[(a - c*g + b*x - d*g*x)/(a + b*x)]/((a + b*x)*(c + d*x)),x]
Output:
PolyLog[2, 1 - (a - c*g + (b - d*g)*x)/(a + b*x)]/(b*c - a*d)
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]
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]
Int[((A_.) + Log[(e_.)*((u_)/(v_))^(n_.)]*(B_.))^(p_.)*(w_)^(m_.)*(y_)^(q_. ), x_Symbol] :> Int[ExpandToSum[w, x]^m*ExpandToSum[y, x]^q*(A + B*Log[e*(E xpandToSum[u, x]/ExpandToSum[v, x])^n])^p, x] /; FreeQ[{e, A, B, m, n, p, q }, x] && LinearQ[{u, v, w, y}, x] && !LinearMatchQ[{u, v, w, y}, x]
Time = 2.69 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70
method | result | size |
derivativedivides | \(-\frac {\operatorname {dilog}\left (\frac {a d g -b c g}{b \left (b x +a \right )}+\frac {-d g +b}{b}\right )}{d a -b c}\) | \(46\) |
default | \(-\frac {\operatorname {dilog}\left (\frac {a d g -b c g}{b \left (b x +a \right )}+\frac {-d g +b}{b}\right )}{d a -b c}\) | \(46\) |
risch | \(-\frac {\operatorname {dilog}\left (\frac {a d g -b c g}{b \left (b x +a \right )}+\frac {-d g +b}{b}\right )}{d a -b c}\) | \(46\) |
parts | \(\frac {\ln \left (\frac {-x d g +b x -c g +a}{b x +a}\right ) \ln \left (d x +c \right )}{d a -b c}-\frac {\ln \left (\frac {-x d g +b x -c g +a}{b x +a}\right ) \ln \left (b x +a \right )}{d a -b c}-\frac {g \left (-\frac {\left (\frac {\operatorname {dilog}\left (\frac {\left (-d g +b \right ) \left (b x +a \right )+a d g -b c g}{a d g -b c g}\right )}{-d g +b}+\frac {\ln \left (b x +a \right ) \ln \left (\frac {\left (-d g +b \right ) \left (b x +a \right )+a d g -b c g}{a d g -b c g}\right )}{-d g +b}\right ) b \left (-d g +b \right )}{g \left (d a -b c \right )}+\frac {\ln \left (b x +a \right )^{2} b}{2 g \left (d a -b c \right )}\right )}{b}+\frac {g \left (-\frac {\left (\frac {\operatorname {dilog}\left (\frac {\left (-d g +b \right ) \left (d x +c \right )+d a -b c}{d a -b c}\right )}{-d g +b}+\frac {\ln \left (d x +c \right ) \ln \left (\frac {\left (-d g +b \right ) \left (d x +c \right )+d a -b c}{d a -b c}\right )}{-d g +b}\right ) d \left (-d g +b \right )}{\left (d a -b c \right ) g}+\frac {\left (\frac {\operatorname {dilog}\left (\frac {d a -b c +b \left (d x +c \right )}{d a -b c}\right )}{b}+\frac {\ln \left (d x +c \right ) \ln \left (\frac {d a -b c +b \left (d x +c \right )}{d a -b c}\right )}{b}\right ) d b}{\left (d a -b c \right ) g}\right )}{d}\) | \(435\) |
Input:
int(ln((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE )
Output:
-dilog((a*d*g-b*c*g)/b/(b*x+a)+(-d*g+b)/b)/(a*d-b*c)
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {{\rm Li}_2\left (\frac {c g + {\left (d g - b\right )} x - a}{b x + a} + 1\right )}{b c - a d} \] Input:
integrate(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="fr icas")
Output:
dilog((c*g + (d*g - b)*x - a)/(b*x + a) + 1)/(b*c - a*d)
Timed out. \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(ln((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (26) = 52\).
Time = 0.05 (sec) , antiderivative size = 343, normalized size of antiderivative = 12.70 \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx={\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (-\frac {d g x + c g - b x - a}{b x + a}\right ) + \frac {\log \left (b x + a\right )^{2} - 2 \, \log \left (b x + a\right ) \log \left (d x + c\right )}{2 \, {\left (b c - a d\right )}} - \frac {\log \left (b x + a\right ) \log \left (\frac {{\left (d g - b\right )} a + {\left (b d g - b^{2}\right )} x}{b c g - a d g} + 1\right ) + {\rm Li}_2\left (-\frac {{\left (d g - b\right )} a + {\left (b d g - b^{2}\right )} x}{b c g - a d g}\right )}{b c - a d} + \frac {\log \left (d x + c\right ) \log \left (\frac {c d g - b c + {\left (d^{2} g - b d\right )} x}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {c d g - b c + {\left (d^{2} g - b d\right )} x}{b c - a d}\right )}{b c - a d} + \frac {\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )}{b c - a d} \] Input:
integrate(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="ma xima")
Output:
(log(b*x + a)/(b*c - a*d) - log(d*x + c)/(b*c - a*d))*log(-(d*g*x + c*g - b*x - a)/(b*x + a)) + 1/2*(log(b*x + a)^2 - 2*log(b*x + a)*log(d*x + c))/( b*c - a*d) - (log(b*x + a)*log(((d*g - b)*a + (b*d*g - b^2)*x)/(b*c*g - a* d*g) + 1) + dilog(-((d*g - b)*a + (b*d*g - b^2)*x)/(b*c*g - a*d*g)))/(b*c - a*d) + (log(d*x + c)*log((c*d*g - b*c + (d^2*g - b*d)*x)/(b*c - a*d) + 1 ) + dilog(-(c*d*g - b*c + (d^2*g - b*d)*x)/(b*c - a*d)))/(b*c - a*d) + (lo g(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 - a*d)
\[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int { \frac {\log \left (-\frac {d g x + c g - b x - a}{b x + a}\right )}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \] Input:
integrate(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x, algorithm="gi ac")
Output:
integrate(log(-(d*g*x + c*g - b*x - a)/(b*x + a))/((b*x + a)*(d*x + c)), x )
Timed out. \[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\int \frac {\ln \left (\frac {a-c\,g+b\,x-d\,g\,x}{a+b\,x}\right )}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \] Input:
int(log((a - c*g + b*x - d*g*x)/(a + b*x))/((a + b*x)*(c + d*x)),x)
Output:
int(log((a - c*g + b*x - d*g*x)/(a + b*x))/((a + b*x)*(c + d*x)), x)
\[ \int \frac {\log \left (\frac {a-c g+b x-d g x}{a+b x}\right )}{(a+b x) (c+d x)} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (\frac {-d g x +b x -c g +a}{b x +a}\right )}{-b \,d^{2} g \,x^{3}-a \,d^{2} g \,x^{2}+b^{2} d \,x^{3}-2 b c d g \,x^{2}+2 a b d \,x^{2}-2 a c d g x +b^{2} c \,x^{2}-b \,c^{2} g x +a^{2} d x +2 a b c x -a \,c^{2} g +a^{2} c}d x \right ) a^{2} d^{2} g -4 \left (\int \frac {\mathrm {log}\left (\frac {-d g x +b x -c g +a}{b x +a}\right )}{-b \,d^{2} g \,x^{3}-a \,d^{2} g \,x^{2}+b^{2} d \,x^{3}-2 b c d g \,x^{2}+2 a b d \,x^{2}-2 a c d g x +b^{2} c \,x^{2}-b \,c^{2} g x +a^{2} d x +2 a b c x -a \,c^{2} g +a^{2} c}d x \right ) a b c d g +2 \left (\int \frac {\mathrm {log}\left (\frac {-d g x +b x -c g +a}{b x +a}\right )}{-b \,d^{2} g \,x^{3}-a \,d^{2} g \,x^{2}+b^{2} d \,x^{3}-2 b c d g \,x^{2}+2 a b d \,x^{2}-2 a c d g x +b^{2} c \,x^{2}-b \,c^{2} g x +a^{2} d x +2 a b c x -a \,c^{2} g +a^{2} c}d x \right ) b^{2} c^{2} g -\mathrm {log}\left (\frac {-d g x +b x -c g +a}{b x +a}\right )^{2} b +\mathrm {log}\left (\frac {-d g x +b x -c g +a}{b x +a}\right )^{2} d g}{2 d g \left (a d -b c \right )} \] Input:
int(log((-d*g*x+b*x-c*g+a)/(b*x+a))/(b*x+a)/(d*x+c),x)
Output:
(2*int(log((a + b*x - c*g - d*g*x)/(a + b*x))/(a**2*c + a**2*d*x + 2*a*b*c *x + 2*a*b*d*x**2 - a*c**2*g - 2*a*c*d*g*x - a*d**2*g*x**2 + b**2*c*x**2 + b**2*d*x**3 - b*c**2*g*x - 2*b*c*d*g*x**2 - b*d**2*g*x**3),x)*a**2*d**2*g - 4*int(log((a + b*x - c*g - d*g*x)/(a + b*x))/(a**2*c + a**2*d*x + 2*a*b *c*x + 2*a*b*d*x**2 - a*c**2*g - 2*a*c*d*g*x - a*d**2*g*x**2 + b**2*c*x**2 + b**2*d*x**3 - b*c**2*g*x - 2*b*c*d*g*x**2 - b*d**2*g*x**3),x)*a*b*c*d*g + 2*int(log((a + b*x - c*g - d*g*x)/(a + b*x))/(a**2*c + a**2*d*x + 2*a*b *c*x + 2*a*b*d*x**2 - a*c**2*g - 2*a*c*d*g*x - a*d**2*g*x**2 + b**2*c*x**2 + b**2*d*x**3 - b*c**2*g*x - 2*b*c*d*g*x**2 - b*d**2*g*x**3),x)*b**2*c**2 *g - log((a + b*x - c*g - d*g*x)/(a + b*x))**2*b + log((a + b*x - c*g - d* g*x)/(a + b*x))**2*d*g)/(2*d*g*(a*d - b*c))