\(\int \frac {x (a+b \text {arctanh}(c \sqrt {x}))}{1-c^2 x} \, dx\) [39]

Optimal result

Integrand size = 24, antiderivative size = 120 \[ \int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x} \, dx=-\frac {b \sqrt {x}}{c^3}+\frac {b \text {arctanh}\left (c \sqrt {x}\right )}{c^4}-\frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{c^2}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{b c^4}+\frac {2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^4}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{c^4} \] Output:

-b*x^(1/2)/c^3+b*arctanh(c*x^(1/2))/c^4-x*(a+b*arctanh(c*x^(1/2)))/c^2-(a+ 
b*arctanh(c*x^(1/2)))^2/b/c^4+2*(a+b*arctanh(c*x^(1/2)))*ln(2/(1-c*x^(1/2) 
))/c^4+b*polylog(2,1-2/(1-c*x^(1/2)))/c^4
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.80 \[ \int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x} \, dx=-\frac {b c \sqrt {x}+a c^2 x-b \text {arctanh}\left (c \sqrt {x}\right )^2+b \text {arctanh}\left (c \sqrt {x}\right ) \left (-1+c^2 x-2 \log \left (1+e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )\right )+a \log \left (1-c^2 x\right )+b \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )}{c^4} \] Input:

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

Output:

-((b*c*Sqrt[x] + a*c^2*x - b*ArcTanh[c*Sqrt[x]]^2 + b*ArcTanh[c*Sqrt[x]]*( 
-1 + c^2*x - 2*Log[1 + E^(-2*ArcTanh[c*Sqrt[x]])]) + a*Log[1 - c^2*x] + b* 
PolyLog[2, -E^(-2*ArcTanh[c*Sqrt[x]])])/c^4)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {7267, 6542, 6452, 262, 219, 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.

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

\(\Big \downarrow \) 7267

\(\displaystyle 2 \int \frac {x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}\)

\(\Big \downarrow \) 6542

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {\int \frac {\sqrt {x} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{2} x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {\sqrt {x}}{c^2}\right )}{c^2}\right )\)

\(\Big \downarrow \) 6546

\(\displaystyle 2 \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c \sqrt {x}}d\sqrt {x}}{c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {\sqrt {x}}{c^2}\right )}{c^2}\right )\)

\(\Big \downarrow \) 6470

\(\displaystyle 2 \left (\frac {\frac {\frac {\log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{1-c \sqrt {x}}\right )}{1-c^2 x}d\sqrt {x}}{c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {\sqrt {x}}{c^2}\right )}{c^2}\right )\)

\(\Big \downarrow \) 2849

\(\displaystyle 2 \left (\frac {\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c \sqrt {x}}\right )}{1-\frac {2}{1-c \sqrt {x}}}d\frac {1}{1-c \sqrt {x}}}{c}+\frac {\log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{c}}{c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {\sqrt {x}}{c^2}\right )}{c^2}\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle 2 \left (\frac {\frac {\frac {\log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{2 c}}{c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c \sqrt {x}\right )}{c^3}-\frac {\sqrt {x}}{c^2}\right )}{c^2}\right )\)

Input:

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

Output:

2*(-(((x*(a + b*ArcTanh[c*Sqrt[x]]))/2 - (b*c*(-(Sqrt[x]/c^2) + ArcTanh[c* 
Sqrt[x]]/c^3))/2)/c^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)/c^2)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, 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 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

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 6542

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

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]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.54

method	result	size
parts	\(-\frac {a x}{c^{2}}-\frac {a \ln \left (c^{2} x -1\right )}{c^{4}}-\frac {2 b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{2} x}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {c \sqrt {x}}{2}+\frac {\ln \left (c \sqrt {x}-1\right )}{4}-\frac {\ln \left (1+c \sqrt {x}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )}{c^{4}}\)	\(185\)
derivativedivides	\(-\frac {2 \left (a \left (\frac {c^{2} x}{2}+\frac {\ln \left (c \sqrt {x}-1\right )}{2}+\frac {\ln \left (1+c \sqrt {x}\right )}{2}\right )+b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{2} x}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {c \sqrt {x}}{2}+\frac {\ln \left (c \sqrt {x}-1\right )}{4}-\frac {\ln \left (1+c \sqrt {x}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )\right )}{c^{4}}\)	\(194\)
default	\(-\frac {2 \left (a \left (\frac {c^{2} x}{2}+\frac {\ln \left (c \sqrt {x}-1\right )}{2}+\frac {\ln \left (1+c \sqrt {x}\right )}{2}\right )+b \left (\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{2} x}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {c \sqrt {x}}{2}+\frac {\ln \left (c \sqrt {x}-1\right )}{4}-\frac {\ln \left (1+c \sqrt {x}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{8}+\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}\right )\right )}{c^{4}}\)	\(194\)

Input:

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

Output:

-a/c^2*x-a/c^4*ln(c^2*x-1)-2*b/c^4*(1/2*arctanh(c*x^(1/2))*c^2*x+1/2*arcta 
nh(c*x^(1/2))*ln(c*x^(1/2)-1)+1/2*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))+1/2*c 
*x^(1/2)+1/4*ln(c*x^(1/2)-1)-1/4*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]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.38 \[ \int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x} \, dx=-a {\left (\frac {x}{c^{2}} + \frac {\log \left (c^{2} x - 1\right )}{c^{4}}\right )} - \frac {{\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b}{c^{4}} + \frac {b \log \left (c \sqrt {x} + 1\right )}{2 \, c^{4}} - \frac {b \log \left (c \sqrt {x} - 1\right )}{2 \, c^{4}} - \frac {2 \, b c^{2} x \log \left (c \sqrt {x} + 1\right ) + b \log \left (c \sqrt {x} + 1\right )^{2} - b \log \left (-c \sqrt {x} + 1\right )^{2} + 4 \, b c \sqrt {x} - 2 \, {\left (b c^{2} x + b \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )}{4 \, c^{4}} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x} \, dx=\frac {-\left (\int \frac {\mathit {atanh} \left (\sqrt {x}\, c \right ) x}{c^{2} x -1}d x \right ) b \,c^{4}-\mathrm {log}\left (c^{2} x -1\right ) a -a \,c^{2} x}{c^{4}} \] Input:

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

Output:

( - (int((atanh(sqrt(x)*c)*x)/(c**2*x - 1),x)*b*c**4 + log(c**2*x - 1)*a + 
 a*c**2*x))/c**4