\(\int \frac {a+b \text {arctanh}(c \sqrt {x})}{1-c^2 x} \, dx\) [40]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{b c^2}+\frac {2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^2}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{c^2} \] Output:

-(a+b*arctanh(c*x^(1/2)))^2/b/c^2+2*(a+b*arctanh(c*x^(1/2)))*ln(2/(1-c*x^( 
1/2)))/c^2+b*polylog(2,1-2/(1-c*x^(1/2)))/c^2
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=-\frac {a \log \left (1-c^2 x\right )}{c^2}-\frac {b \left (-\text {arctanh}\left (c \sqrt {x}\right ) \left (\text {arctanh}\left (c \sqrt {x}\right )+2 \log \left (1+e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )\right )}{c^2} \] Input:

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(1 - c^2*x),x]
 

Output:

-((a*Log[1 - c^2*x])/c^2) - (b*(-(ArcTanh[c*Sqrt[x]]*(ArcTanh[c*Sqrt[x]] + 
 2*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])])) + PolyLog[2, -E^(-2*ArcTanh[c*Sqrt 
[x]])]))/c^2
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6484, 6546, 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.

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

\(\Big \downarrow \) 6484

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

\(\Big \downarrow \) 6546

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

\(\Big \downarrow \) 6470

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

Input:

Int[(a + b*ArcTanh[c*Sqrt[x]])/(1 - c^2*x),x]
 

Output:

2*(-1/2*(a + b*ArcTanh[c*Sqrt[x]])^2/(b*c^2) + (((a + b*ArcTanh[c*Sqrt[x]] 
)*Log[2/(1 - c*Sqrt[x])])/c + (b*PolyLog[2, 1 - 2/(1 - c*Sqrt[x])])/(2*c)) 
/c)
 

Defintions of rubi rules used

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 6484

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] 
 :> With[{k = Denominator[n]}, Simp[k   Subst[Int[x^(k - 1)*((a + b*ArcTanh 
[c*x^(k*n)])/(d + e*x^k)), x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e}, x 
] && FractionQ[n]
 

rule 6546

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.79

method	result	size
parts	\(-\frac {a \ln \left (c^{2} x -1\right )}{c^{2}}-\frac {2 b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )}{c^{2}}\)	\(140\)
derivativedivides	\(-\frac {2 \left (a \left (\frac {\ln \left (c \sqrt {x}-1\right )}{2}+\frac {\ln \left (1+c \sqrt {x}\right )}{2}\right )+b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )\right )}{c^{2}}\)	\(150\)
default	\(-\frac {2 \left (a \left (\frac {\ln \left (c \sqrt {x}-1\right )}{2}+\frac {\ln \left (1+c \sqrt {x}\right )}{2}\right )+b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )\right )}{c^{2}}\)	\(150\)

Input:

int((a+b*arctanh(c*x^(1/2)))/(-c^2*x+1),x,method=_RETURNVERBOSE)
 

Output:

-a*ln(c^2*x-1)/c^2-2*b/c^2*(1/2*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)+1/2*arc 
tanh(c*x^(1/2))*ln(1+c*x^(1/2))-1/2*dilog(1/2*c*x^(1/2)+1/2)-1/4*ln(c*x^(1 
/2)-1)*ln(1/2*c*x^(1/2)+1/2)+1/8*ln(c*x^(1/2)-1)^2-1/8*ln(1+c*x^(1/2))^2+1 
/4*(ln(1+c*x^(1/2))-ln(1/2*c*x^(1/2)+1/2))*ln(-1/2*c*x^(1/2)+1/2))
 

Fricas [F]

\[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=\int { -\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{c^{2} x - 1} \,d x } \] Input:

integrate((a+b*arctanh(c*x^(1/2)))/(-c^2*x+1),x, algorithm="fricas")
 

Output:

integral(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x - 1), x)
 

Sympy [F]

\[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=- \int \frac {a}{c^{2} x - 1}\, dx - \int \frac {b \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x - 1}\, dx \] Input:

integrate((a+b*atanh(c*x**(1/2)))/(-c**2*x+1),x)
 

Output:

-Integral(a/(c**2*x - 1), x) - Integral(b*atanh(c*sqrt(x))/(c**2*x - 1), x 
)
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.29 \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=-\frac {{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b}{c^{2}} - \frac {a \log \left (c^{2} x - 1\right )}{c^{2}} - \frac {b \log \left (c \sqrt {x} + 1\right )^{2} - 2 \, b \log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x} + 1\right ) - b \log \left (-c \sqrt {x} + 1\right )^{2}}{4 \, c^{2}} \] Input:

integrate((a+b*arctanh(c*x^(1/2)))/(-c^2*x+1),x, algorithm="maxima")
 

Output:

-(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1/2 
))*b/c^2 - a*log(c^2*x - 1)/c^2 - 1/4*(b*log(c*sqrt(x) + 1)^2 - 2*b*log(c* 
sqrt(x) + 1)*log(-c*sqrt(x) + 1) - b*log(-c*sqrt(x) + 1)^2)/c^2
 

Giac [F]

\[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=\int { -\frac {b \operatorname {artanh}\left (c \sqrt {x}\right ) + a}{c^{2} x - 1} \,d x } \] Input:

integrate((a+b*arctanh(c*x^(1/2)))/(-c^2*x+1),x, algorithm="giac")
 

Output:

integrate(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=\int -\frac {a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{c^2\,x-1} \,d x \] Input:

int(-(a + b*atanh(c*x^(1/2)))/(c^2*x - 1),x)
 

Output:

int(-(a + b*atanh(c*x^(1/2)))/(c^2*x - 1), x)
 

Reduce [F]

\[ \int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x} \, dx=\frac {-\left (\int \frac {\mathit {atanh} \left (\sqrt {x}\, c \right )}{c^{2} x -1}d x \right ) b \,c^{2}-\mathrm {log}\left (c^{2} x -1\right ) a}{c^{2}} \] Input:

int((a+b*atanh(c*x^(1/2)))/(-c^2*x+1),x)
 

Output:

( - (int(atanh(sqrt(x)*c)/(c**2*x - 1),x)*b*c**2 + log(c**2*x - 1)*a))/c** 
2