3.1.45 \(\int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx\) [45]

3.1.45.1 Optimal result
3.1.45.2 Mathematica [A] (verified)
3.1.45.3 Rubi [A] (verified)
3.1.45.4 Maple [A] (verified)
3.1.45.5 Fricas [F]
3.1.45.6 Sympy [F]
3.1.45.7 Maxima [F]
3.1.45.8 Giac [F]
3.1.45.9 Mupad [F(-1)]

3.1.45.1 Optimal result

Integrand size = 18, antiderivative size = 94 \[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {a x}{c d}+\frac {b x \text {arctanh}(c x)}{c d}+\frac {(a+b \text {arctanh}(c x)) \log \left (\frac {2}{1+c x}\right )}{c^2 d}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^2 d}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 c^2 d} \]

output
a*x/c/d+b*x*arctanh(c*x)/c/d+(a+b*arctanh(c*x))*ln(2/(c*x+1))/c^2/d+1/2*b* 
ln(-c^2*x^2+1)/c^2/d-1/2*b*polylog(2,1-2/(c*x+1))/c^2/d
 
3.1.45.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80 \[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {2 a c x+2 b \text {arctanh}(c x) \left (c x+\log \left (1+e^{-2 \text {arctanh}(c x)}\right )\right )-2 a \log (1+c x)+b \log \left (1-c^2 x^2\right )-b \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}(c x)}\right )}{2 c^2 d} \]

input
Integrate[(x*(a + b*ArcTanh[c*x]))/(d + c*d*x),x]
 
output
(2*a*c*x + 2*b*ArcTanh[c*x]*(c*x + Log[1 + E^(-2*ArcTanh[c*x])]) - 2*a*Log 
[1 + c*x] + b*Log[1 - c^2*x^2] - b*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(2*c^ 
2*d)
 
3.1.45.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6492, 27, 2009, 6470, 2849, 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.

\(\displaystyle \int \frac {x (a+b \text {arctanh}(c x))}{c d x+d} \, dx\)

\(\Big \downarrow \) 6492

\(\displaystyle \frac {\int (a+b \text {arctanh}(c x))dx}{c d}-\frac {\int \frac {a+b \text {arctanh}(c x)}{d (c x+1)}dx}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a+b \text {arctanh}(c x))dx}{c d}-\frac {\int \frac {a+b \text {arctanh}(c x)}{c x+1}dx}{c d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c d}-\frac {\int \frac {a+b \text {arctanh}(c x)}{c x+1}dx}{c d}\)

\(\Big \downarrow \) 6470

\(\displaystyle \frac {a x+b x \text {arctanh}(c x)+\frac {b \log \left (1-c^2 x^2\right )}{2 c}}{c d}-\frac {b \int \frac {\log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{c}}{c d}\)

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

input
Int[(x*(a + b*ArcTanh[c*x]))/(d + c*d*x),x]
 
output
(a*x + b*x*ArcTanh[c*x] + (b*Log[1 - c^2*x^2])/(2*c))/(c*d) - (-(((a + b*A 
rcTanh[c*x])*Log[2/(1 + c*x)])/c) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/(2*c)) 
/(c*d)
 

3.1.45.3.1 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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 6470
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol 
] :> Simp[(-(a + b*ArcTanh[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c 
*(p/e)   Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^2*x^ 
2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2 
, 0]
 

rule 6492
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + 
(e_.)*(x_)), x_Symbol] :> Simp[f/e   Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x]) 
^p, x], x] - Simp[d*(f/e)   Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d + 
e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 
- e^2, 0] && GtQ[m, 0]
 
3.1.45.4 Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13

method result size
derivativedivides \(\frac {\frac {a \left (c x -\ln \left (c x +1\right )\right )}{d}+\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+c x \,\operatorname {arctanh}\left (c x \right )-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\ln \left (\left (c x -1\right ) \left (c x +1\right )\right )}{2}\right )}{d}}{c^{2}}\) \(106\)
default \(\frac {\frac {a \left (c x -\ln \left (c x +1\right )\right )}{d}+\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+c x \,\operatorname {arctanh}\left (c x \right )-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\ln \left (\left (c x -1\right ) \left (c x +1\right )\right )}{2}\right )}{d}}{c^{2}}\) \(106\)
parts \(\frac {a \left (\frac {x}{c}-\frac {\ln \left (c x +1\right )}{c^{2}}\right )}{d}+\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \ln \left (c x +1\right )+c x \,\operatorname {arctanh}\left (c x \right )-\frac {\left (\ln \left (c x +1\right )-\ln \left (\frac {c x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\operatorname {dilog}\left (\frac {c x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (c x +1\right )^{2}}{4}+\frac {\ln \left (\left (c x -1\right ) \left (c x +1\right )\right )}{2}\right )}{d \,c^{2}}\) \(110\)
risch \(-\frac {b \ln \left (c x +1\right )^{2}}{4 d \,c^{2}}+\frac {b x \ln \left (c x +1\right )}{2 c d}+\frac {b \ln \left (c x +1\right )}{2 c^{2} d}-\frac {b x \ln \left (-c x +1\right )}{2 c d}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 c^{2} d}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 c^{2} d}+\frac {a x}{c d}+\frac {b \ln \left (-c x +1\right )}{2 c^{2} d}-\frac {a \ln \left (-c x -1\right )}{c^{2} d}-\frac {b \operatorname {dilog}\left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 c^{2} d}-\frac {a}{c^{2} d}-\frac {b}{2 d \,c^{2}}\) \(188\)

input
int(x*(a+b*arctanh(c*x))/(c*d*x+d),x,method=_RETURNVERBOSE)
 
output
1/c^2*(a/d*(c*x-ln(c*x+1))+b/d*(-arctanh(c*x)*ln(c*x+1)+c*x*arctanh(c*x)-1 
/2*(ln(c*x+1)-ln(1/2*c*x+1/2))*ln(-1/2*c*x+1/2)+1/2*dilog(1/2*c*x+1/2)+1/4 
*ln(c*x+1)^2+1/2*ln((c*x-1)*(c*x+1))))
 
3.1.45.5 Fricas [F]

\[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{c d x + d} \,d x } \]

input
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="fricas")
 
output
integral((b*x*arctanh(c*x) + a*x)/(c*d*x + d), x)
 
3.1.45.6 Sympy [F]

\[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\frac {\int \frac {a x}{c x + 1}\, dx + \int \frac {b x \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]

input
integrate(x*(a+b*atanh(c*x))/(c*d*x+d),x)
 
output
(Integral(a*x/(c*x + 1), x) + Integral(b*x*atanh(c*x)/(c*x + 1), x))/d
 
3.1.45.7 Maxima [F]

\[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{c d x + d} \,d x } \]

input
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="maxima")
 
output
1/4*(c^2*(2*x/(c^3*d) - log(c*x + 1)/(c^4*d) + log(c*x - 1)/(c^4*d)) + 2*c 
^2*integrate(x^2*log(c*x + 1)/(c^3*d*x^2 - c*d), x) - 4*c*integrate(x*log( 
c*x + 1)/(c^3*d*x^2 - c*d), x) - 2*(c*x - log(c*x + 1))*log(-c*x + 1)/(c^2 
*d) + log(c^3*d*x^2 - c*d)/(c^2*d) - 2*integrate(log(c*x + 1)/(c^3*d*x^2 - 
 c*d), x))*b + a*(x/(c*d) - log(c*x + 1)/(c^2*d))
 
3.1.45.8 Giac [F]

\[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} x}{c d x + d} \,d x } \]

input
integrate(x*(a+b*arctanh(c*x))/(c*d*x+d),x, algorithm="giac")
 
output
integrate((b*arctanh(c*x) + a)*x/(c*d*x + d), x)
 
3.1.45.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x (a+b \text {arctanh}(c x))}{d+c d x} \, dx=\int \frac {x\,\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}{d+c\,d\,x} \,d x \]

input
int((x*(a + b*atanh(c*x)))/(d + c*d*x),x)
 
output
int((x*(a + b*atanh(c*x)))/(d + c*d*x), x)