\(\int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx\) [25]

Optimal result

Integrand size = 18, antiderivative size = 130 \[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=-\frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f} \] Output:

-(a+b*arccoth(d*x+c))*ln(2/(d*x+c+1))/f+(a+b*arccoth(d*x+c))*ln(2*d*(f*x+e 
)/(-c*f+d*e+f)/(d*x+c+1))/f+1/2*b*polylog(2,1-2/(d*x+c+1))/f-1/2*b*polylog 
(2,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+1))/f
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.58 \[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\frac {a \log (e+f x)}{f}+\frac {b \log \left (\frac {f (1-c-d x)}{d e+f-c f}\right ) \log (e+f x)}{2 f}-\frac {b \log \left (-\frac {1-c-d x}{c+d x}\right ) \log (e+f x)}{2 f}-\frac {b \log \left (-\frac {f (1+c+d x)}{d e-f-c f}\right ) \log (e+f x)}{2 f}+\frac {b \log \left (\frac {1+c+d x}{c+d x}\right ) \log (e+f x)}{2 f}-\frac {b \operatorname {PolyLog}\left (2,\frac {d (e+f x)}{d e-f-c f}\right )}{2 f}+\frac {b \operatorname {PolyLog}\left (2,\frac {d (e+f x)}{d e+f-c f}\right )}{2 f} \] Input:

Integrate[(a + b*ArcCoth[c + d*x])/(e + f*x),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6662, 27, 6473, 2849, 2752, 2897}

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.

\(\displaystyle \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx\)

\(\Big \downarrow \) 6662

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

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {a+b \coth ^{-1}(c+d x)}{f (c+d x)-c f+d e}d(c+d x)\)

\(\Big \downarrow \) 6473

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

\(\displaystyle -\frac {b \int \frac {\log \left (\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{1-(c+d x)^2}d(c+d x)}{f}+\frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2 (f (c+d x)-c f+d e)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{2 f}\)

\(\Big \downarrow \) 2897

\(\displaystyle \frac {\left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac {2 (f (c+d x)-c f+d e)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{f}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 (d e-c f+f (c+d x))}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{2 f}\)

Input:

Int[(a + b*ArcCoth[c + d*x])/(e + f*x),x]
 

Output:

-(((a + b*ArcCoth[c + d*x])*Log[2/(1 + c + d*x)])/f) + ((a + b*ArcCoth[c + 
 d*x])*Log[(2*(d*e - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x))]) 
/f + (b*PolyLog[2, 1 - 2/(1 + c + d*x)])/(2*f) - (b*PolyLog[2, 1 - (2*(d*e 
 - c*f + f*(c + d*x)))/((d*e + f - c*f)*(1 + c + d*x))])/(2*f)
 

Defintions of rubi rules used

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 2849

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

rule 2897

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ 
D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && 
PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, 
 x][[2]], Expon[Pq, x]]
 

rule 6473

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> S 
imp[(-(a + b*ArcCoth[c*x]))*(Log[2/(1 + c*x)]/e), x] + (Simp[(a + b*ArcCoth 
[c*x])*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[b*(c/e) 
Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Simp[b*(c/e)   Int[Log[2*c*((d 
 + e*x)/((c*d + e)*(1 + c*x)))]/(1 - c^2*x^2), x], x]) /; FreeQ[{a, b, c, d 
, e}, x] && NeQ[c^2*d^2 - e^2, 0]
 

rule 6662

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* 
ArcCoth[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IG 
tQ[p, 0]
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {a \ln \left (\left (d x +c -1\right ) f -c f +d e +f \right )}{f}-\frac {b \operatorname {dilog}\left (\frac {\left (d x +c -1\right ) f -c f +d e +f}{-c f +d e +f}\right )}{2 f}-\frac {b \ln \left (d x +c -1\right ) \ln \left (\frac {\left (d x +c -1\right ) f -c f +d e +f}{-c f +d e +f}\right )}{2 f}+\frac {b \operatorname {dilog}\left (\frac {\left (d x +c +1\right ) f -c f +d e -f}{-c f +d e -f}\right )}{2 f}+\frac {b \ln \left (d x +c +1\right ) \ln \left (\frac {\left (d x +c +1\right ) f -c f +d e -f}{-c f +d e -f}\right )}{2 f}\)	\(191\)
parts	\(\frac {a \ln \left (f x +e \right )}{f}+\frac {b \ln \left (f \left (d x +c \right )-c f +d e \right ) \operatorname {arccoth}\left (d x +c \right )}{f}+\frac {b \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {f \left (d x +c \right )-f}{c f -d e -f}\right )}{2 f}+\frac {b \operatorname {dilog}\left (\frac {f \left (d x +c \right )-f}{c f -d e -f}\right )}{2 f}-\frac {b \ln \left (f \left (d x +c \right )-c f +d e \right ) \ln \left (\frac {f \left (d x +c \right )+f}{c f -d e +f}\right )}{2 f}-\frac {b \operatorname {dilog}\left (\frac {f \left (d x +c \right )+f}{c f -d e +f}\right )}{2 f}\)	\(192\)
derivativedivides	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}-b d \left (-\frac {\ln \left (c f -d e -f \left (d x +c \right )\right ) \operatorname {arccoth}\left (d x +c \right )}{f}+\frac {-\frac {f \left (\operatorname {dilog}\left (\frac {-f \left (d x +c \right )+f}{-c f +d e +f}\right )+\ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {-f \left (d x +c \right )+f}{-c f +d e +f}\right )\right )}{2}+\frac {f \left (\operatorname {dilog}\left (\frac {-f \left (d x +c \right )-f}{-c f +d e -f}\right )+\ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {-f \left (d x +c \right )-f}{-c f +d e -f}\right )\right )}{2}}{f^{2}}\right )}{d}\)	\(211\)
default	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}-b d \left (-\frac {\ln \left (c f -d e -f \left (d x +c \right )\right ) \operatorname {arccoth}\left (d x +c \right )}{f}+\frac {-\frac {f \left (\operatorname {dilog}\left (\frac {-f \left (d x +c \right )+f}{-c f +d e +f}\right )+\ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {-f \left (d x +c \right )+f}{-c f +d e +f}\right )\right )}{2}+\frac {f \left (\operatorname {dilog}\left (\frac {-f \left (d x +c \right )-f}{-c f +d e -f}\right )+\ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {-f \left (d x +c \right )-f}{-c f +d e -f}\right )\right )}{2}}{f^{2}}\right )}{d}\)	\(211\)

Input:

int((a+b*arccoth(d*x+c))/(f*x+e),x,method=_RETURNVERBOSE)
 

Output:

a*ln((d*x+c-1)*f-c*f+d*e+f)/f-1/2*b*dilog(((d*x+c-1)*f-c*f+d*e+f)/(-c*f+d* 
e+f))/f-1/2*b*ln(d*x+c-1)*ln(((d*x+c-1)*f-c*f+d*e+f)/(-c*f+d*e+f))/f+1/2*b 
*dilog(((d*x+c+1)*f-c*f+d*e-f)/(-c*f+d*e-f))/f+1/2*b*ln(d*x+c+1)*ln(((d*x+ 
c+1)*f-c*f+d*e-f)/(-c*f+d*e-f))/f
 

Fricas [F]

\[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\int { \frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))/(f*x+e),x, algorithm="fricas")
 

Output:

integral((b*arccoth(d*x + c) + a)/(f*x + e), x)
 

Sympy [F]

\[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\int \frac {a + b \operatorname {acoth}{\left (c + d x \right )}}{e + f x}\, dx \] Input:

integrate((a+b*acoth(d*x+c))/(f*x+e),x)
 

Output:

Integral((a + b*acoth(c + d*x))/(e + f*x), x)
 

Maxima [F]

\[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\int { \frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))/(f*x+e),x, algorithm="maxima")
 

Output:

1/2*b*integrate((log(1/(d*x + c) + 1) - log(-1/(d*x + c) + 1))/(f*x + e), 
x) + a*log(f*x + e)/f
 

Giac [F]

\[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\int { \frac {b \operatorname {arcoth}\left (d x + c\right ) + a}{f x + e} \,d x } \] Input:

integrate((a+b*arccoth(d*x+c))/(f*x+e),x, algorithm="giac")
 

Output:

integrate((b*arccoth(d*x + c) + a)/(f*x + e), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\int \frac {a+b\,\mathrm {acoth}\left (c+d\,x\right )}{e+f\,x} \,d x \] Input:

int((a + b*acoth(c + d*x))/(e + f*x),x)
 

Output:

int((a + b*acoth(c + d*x))/(e + f*x), x)
 

Reduce [F]

\[ \int \frac {a+b \coth ^{-1}(c+d x)}{e+f x} \, dx=\frac {\left (\int \frac {\mathit {acoth} \left (d x +c \right )}{f x +e}d x \right ) b f +\mathrm {log}\left (f x +e \right ) a}{f} \] Input:

int((a+b*acoth(d*x+c))/(f*x+e),x)
 

Output:

(int(acoth(c + d*x)/(e + f*x),x)*b*f + log(e + f*x)*a)/f