\(\int \frac {(a+b \text {arctanh}(c \sqrt {x}))^3}{x^2} \, dx\) [206]

Optimal result

Integrand size = 18, antiderivative size = 142 \[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=3 b c^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {3 b c \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{\sqrt {x}}+c^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x}+6 b^2 c^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-3 b^3 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c \sqrt {x}}\right ) \] Output:

3*b*c^2*(a+b*arctanh(c*x^(1/2)))^2-3*b*c*(a+b*arctanh(c*x^(1/2)))^2/x^(1/2 
)+c^2*(a+b*arctanh(c*x^(1/2)))^3-(a+b*arctanh(c*x^(1/2)))^3/x+6*b^2*c^2*(a 
+b*arctanh(c*x^(1/2)))*ln(2-2/(1+c*x^(1/2)))-3*b^3*c^2*polylog(2,-1+2/(1+c 
*x^(1/2)))
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\frac {6 b^2 \left (-1+c \sqrt {x}\right ) \left (a+a c \sqrt {x}+b c \sqrt {x}\right ) \text {arctanh}\left (c \sqrt {x}\right )^2+2 b^3 \left (-1+c^2 x\right ) \text {arctanh}\left (c \sqrt {x}\right )^3-6 b \text {arctanh}\left (c \sqrt {x}\right ) \left (a^2+2 a b c \sqrt {x}-2 b^2 c^2 x \log \left (1-e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )\right )+a \left (-2 a^2-6 a b c \sqrt {x}-3 a b c^2 x \log \left (1-c \sqrt {x}\right )+3 a b c^2 x \log \left (1+c \sqrt {x}\right )+12 b^2 c^2 x \log \left (\frac {c \sqrt {x}}{\sqrt {1-c^2 x}}\right )\right )-6 b^3 c^2 x \operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c \sqrt {x}\right )}\right )}{2 x} \] Input:

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

Output:

(6*b^2*(-1 + c*Sqrt[x])*(a + a*c*Sqrt[x] + b*c*Sqrt[x])*ArcTanh[c*Sqrt[x]] 
^2 + 2*b^3*(-1 + c^2*x)*ArcTanh[c*Sqrt[x]]^3 - 6*b*ArcTanh[c*Sqrt[x]]*(a^2 
 + 2*a*b*c*Sqrt[x] - 2*b^2*c^2*x*Log[1 - E^(-2*ArcTanh[c*Sqrt[x]])]) + a*( 
-2*a^2 - 6*a*b*c*Sqrt[x] - 3*a*b*c^2*x*Log[1 - c*Sqrt[x]] + 3*a*b*c^2*x*Lo 
g[1 + c*Sqrt[x]] + 12*b^2*c^2*x*Log[(c*Sqrt[x])/Sqrt[1 - c^2*x]]) - 6*b^3* 
c^2*x*PolyLog[2, E^(-2*ArcTanh[c*Sqrt[x]])])/(2*x)
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6454, 6452, 6544, 6452, 6510, 6550, 6494, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6510

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

\(\Big \downarrow \) 6550

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

\(\Big \downarrow \) 6494

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

\(\Big \downarrow \) 2897

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

Input:

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

Output:

2*(-1/2*(a + b*ArcTanh[c*Sqrt[x]])^3/x + (3*b*c*(-((a + b*ArcTanh[c*Sqrt[x 
]])^2/Sqrt[x]) + (c*(a + b*ArcTanh[c*Sqrt[x]])^3)/(3*b) + 2*b*c*((a + b*Ar 
cTanh[c*Sqrt[x]])^2/(2*b) + (a + b*ArcTanh[c*Sqrt[x]])*Log[2 - 2/(1 + c*Sq 
rt[x])] - (b*PolyLog[2, -1 + 2/(1 + c*Sqrt[x])])/2)))/2)
 

Defintions of rubi rules used

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 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 6454

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

rule 6494

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x 
_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - 
Simp[b*c*(p/d)   Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6510

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

rule 6544

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

rule 6550

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 15.12 (sec) , antiderivative size = 4700, normalized size of antiderivative = 33.10

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4700\)
default	\(\text {Expression too large to display}\)	\(4700\)
parts	\(\text {Expression too large to display}\)	\(4702\)

Input:

int((a+b*arctanh(c*x^(1/2)))^3/x^2,x,method=_RETURNVERBOSE)
 

Output:

2*c^2*(-1/2*a^3/c^2/x+b^3*(3/8*I*Pi*csgn(I/(1-(1+c*x^(1/2))^2/(c^2*x-1)))* 
csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1-(1+c 
*x^(1/2))^2/(c^2*x-1)))*dilog(1+(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/2*arctan 
h(c*x^(1/2))^2+3/8*I*Pi*csgn(I/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*csgn(I*(1+c* 
x^(1/2))^2/(c^2*x-1)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))^2*arctanh(c*x^(1/2))*l 
n(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*Pi*csgn(I*(1+c*x^(1/2))^2/(c^2*x 
-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))^2*arc 
tanh(c*x^(1/2))*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*Pi*csgn(I/(1-(1 
+c*x^(1/2))^2/(c^2*x-1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^ 
(1/2))^2/(c^2*x-1)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*arctanh(c*x^(1/2))^2-3/8 
*I*Pi*csgn(I/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x- 
1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*polylo 
g(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/8*I*Pi*csgn(I/(1-(1+c*x^(1/2))^2/(c^ 
2*x-1)))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1 
)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2) 
)-3/8*I*Pi*csgn(I/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*csgn(I*(1+c*x^(1/2))^2/(c 
^2*x-1))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1)/(1-(1+c*x^(1/2))^2/(c^2*x-1)))*d 
ilog((1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3/4*I*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x 
+1)^(1/2))*csgn(I*(1+c*x^(1/2))^2/(c^2*x-1))^2*arctanh(c*x^(1/2))*ln(1-(1+ 
c*x^(1/2))/(-c^2*x+1)^(1/2))+3/8*I*Pi*csgn(I*(1+c*x^(1/2))/(-c^2*x+1)^(...
 

Fricas [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )}^{3}}{x^{2}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^(1/2)))^3/x^2,x, algorithm="fricas")
 

Output:

integral((b^3*arctanh(c*sqrt(x))^3 + 3*a*b^2*arctanh(c*sqrt(x))^2 + 3*a^2* 
b*arctanh(c*sqrt(x)) + a^3)/x^2, x)
 

Sympy [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{3}}{x^{2}}\, dx \] Input:

integrate((a+b*atanh(c*x**(1/2)))**3/x**2,x)
 

Output:

Integral((a + b*atanh(c*sqrt(x)))**3/x**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (125) = 250\).

Time = 1.02 (sec) , antiderivative size = 528, normalized size of antiderivative = 3.72 \[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=-3 \, {\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^{3} c^{2} - 3 \, {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b^{3} c^{2} + 3 \, {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b^{3} c^{2} - 3 \, a b^{2} c^{2} \log \left (c \sqrt {x} - 1\right ) - \frac {3}{4} \, {\left ({\left (2 \, c \log \left (c \sqrt {x} - 1\right ) - c \log \left (x\right ) + \frac {2}{\sqrt {x}}\right )} c - \frac {2 \, \log \left (-c \sqrt {x} + 1\right )}{x}\right )} a^{2} b - \frac {a^{3}}{x} + \frac {3}{2} \, {\left (a^{2} b c^{2} - 2 \, a b^{2} c^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - \frac {3}{4} \, {\left (a^{2} b c^{2} - 4 \, a b^{2} c^{2}\right )} \log \left (x\right ) - \frac {12 \, a^{2} b c \sqrt {x} - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )^{3} + {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (-c \sqrt {x} + 1\right )^{3} + 6 \, {\left (b^{3} c \sqrt {x} + a b^{2} - {\left (a b^{2} c^{2} - b^{3} c^{2}\right )} x\right )} \log \left (c \sqrt {x} + 1\right )^{2} + 3 \, {\left (2 \, b^{3} c \sqrt {x} + 2 \, a b^{2} - 2 \, {\left (a b^{2} c^{2} + b^{3} c^{2}\right )} x - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )^{2} + 12 \, {\left (2 \, a b^{2} c \sqrt {x} + a^{2} b\right )} \log \left (c \sqrt {x} + 1\right ) - 3 \, {\left (8 \, a b^{2} c \sqrt {x} - {\left (b^{3} c^{2} x - b^{3}\right )} \log \left (c \sqrt {x} + 1\right )^{2} + 4 \, {\left (b^{3} c \sqrt {x} + a b^{2} - {\left (a b^{2} c^{2} - b^{3} c^{2}\right )} x\right )} \log \left (c \sqrt {x} + 1\right )\right )} \log \left (-c \sqrt {x} + 1\right )}{8 \, x} \] Input:

integrate((a+b*arctanh(c*x^(1/2)))^3/x^2,x, algorithm="maxima")
 

Output:

-3*(log(c*sqrt(x) + 1)*log(-1/2*c*sqrt(x) + 1/2) + dilog(1/2*c*sqrt(x) + 1 
/2))*b^3*c^2 - 3*(log(c*sqrt(x))*log(-c*sqrt(x) + 1) + dilog(-c*sqrt(x) + 
1))*b^3*c^2 + 3*(log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) + 1) 
)*b^3*c^2 - 3*a*b^2*c^2*log(c*sqrt(x) - 1) - 3/4*((2*c*log(c*sqrt(x) - 1) 
- c*log(x) + 2/sqrt(x))*c - 2*log(-c*sqrt(x) + 1)/x)*a^2*b - a^3/x + 3/2*( 
a^2*b*c^2 - 2*a*b^2*c^2)*log(c*sqrt(x) + 1) - 3/4*(a^2*b*c^2 - 4*a*b^2*c^2 
)*log(x) - 1/8*(12*a^2*b*c*sqrt(x) - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1)^ 
3 + (b^3*c^2*x - b^3)*log(-c*sqrt(x) + 1)^3 + 6*(b^3*c*sqrt(x) + a*b^2 - ( 
a*b^2*c^2 - b^3*c^2)*x)*log(c*sqrt(x) + 1)^2 + 3*(2*b^3*c*sqrt(x) + 2*a*b^ 
2 - 2*(a*b^2*c^2 + b^3*c^2)*x - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1))*log( 
-c*sqrt(x) + 1)^2 + 12*(2*a*b^2*c*sqrt(x) + a^2*b)*log(c*sqrt(x) + 1) - 3* 
(8*a*b^2*c*sqrt(x) - (b^3*c^2*x - b^3)*log(c*sqrt(x) + 1)^2 + 4*(b^3*c*sqr 
t(x) + a*b^2 - (a*b^2*c^2 - b^3*c^2)*x)*log(c*sqrt(x) + 1))*log(-c*sqrt(x) 
 + 1))/x
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )}^{3}}{x^{2}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^(1/2)))^3/x^2,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*sqrt(x)) + a)^3/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )\right )}^3}{x^2} \,d x \] Input:

int((a + b*atanh(c*x^(1/2)))^3/x^2,x)
                                                                                    
                                                                                    
 

Output:

int((a + b*atanh(c*x^(1/2)))^3/x^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^3}{x^2} \, dx=\frac {\mathit {atanh} \left (\sqrt {x}\, c \right )^{3} b^{3} c^{2} x -\mathit {atanh} \left (\sqrt {x}\, c \right )^{3} b^{3}-3 \sqrt {x}\, \mathit {atanh} \left (\sqrt {x}\, c \right )^{2} b^{3} c +3 \mathit {atanh} \left (\sqrt {x}\, c \right )^{2} a \,b^{2} c^{2} x -3 \mathit {atanh} \left (\sqrt {x}\, c \right )^{2} a \,b^{2}-6 \sqrt {x}\, \mathit {atanh} \left (\sqrt {x}\, c \right ) a \,b^{2} c +3 \mathit {atanh} \left (\sqrt {x}\, c \right ) a^{2} b \,c^{2} x -3 \mathit {atanh} \left (\sqrt {x}\, c \right ) a^{2} b -6 \mathit {atanh} \left (\sqrt {x}\, c \right ) a \,b^{2} c^{2} x -3 \mathit {atanh} \left (\sqrt {x}\, c \right ) b^{3} c^{2} x +3 \mathit {atanh} \left (\sqrt {x}\, c \right ) b^{3}-3 \sqrt {x}\, a^{2} b c +3 \sqrt {x}\, b^{3} c -3 \left (\int \frac {\mathit {atanh} \left (\sqrt {x}\, c \right )}{c^{2} x^{3}-x^{2}}d x \right ) b^{3} x -6 \,\mathrm {log}\left (\sqrt {x}\, c -1\right ) a \,b^{2} c^{2} x +6 \,\mathrm {log}\left (\sqrt {x}\right ) a \,b^{2} c^{2} x -a^{3}}{x} \] Input:

int((a+b*atanh(c*x^(1/2)))^3/x^2,x)
 

Output:

(atanh(sqrt(x)*c)**3*b**3*c**2*x - atanh(sqrt(x)*c)**3*b**3 - 3*sqrt(x)*at 
anh(sqrt(x)*c)**2*b**3*c + 3*atanh(sqrt(x)*c)**2*a*b**2*c**2*x - 3*atanh(s 
qrt(x)*c)**2*a*b**2 - 6*sqrt(x)*atanh(sqrt(x)*c)*a*b**2*c + 3*atanh(sqrt(x 
)*c)*a**2*b*c**2*x - 3*atanh(sqrt(x)*c)*a**2*b - 6*atanh(sqrt(x)*c)*a*b**2 
*c**2*x - 3*atanh(sqrt(x)*c)*b**3*c**2*x + 3*atanh(sqrt(x)*c)*b**3 - 3*sqr 
t(x)*a**2*b*c + 3*sqrt(x)*b**3*c - 3*int(atanh(sqrt(x)*c)/(c**2*x**3 - x** 
2),x)*b**3*x - 6*log(sqrt(x)*c - 1)*a*b**2*c**2*x + 6*log(sqrt(x))*a*b**2* 
c**2*x - a**3)/x