\(\int \frac {(a+b \text {arctanh}(c x^{3/2}))^2}{x^4} \, dx\) [222]

Optimal result

Integrand size = 18, antiderivative size = 96 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=-\frac {2 b c \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )}{3 x^{3/2}}+\frac {1}{3} c^2 \left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2-\frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{3 x^3}+b^2 c^2 \log (x)-\frac {1}{3} b^2 c^2 \log \left (1-c^2 x^3\right ) \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(210\) vs. \(2(96)=192\).

Time = 0.09 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.19 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=-\frac {a^2+2 a b c x^{3/2}+2 b \left (a+b c x^{3/2}\right ) \text {arctanh}\left (c x^{3/2}\right )-b^2 \left (-1+c^2 x^3\right ) \text {arctanh}\left (c x^{3/2}\right )^2+a b c^2 x^3 \log \left (1-\sqrt [3]{c} \sqrt {x}\right )-a b c^2 x^3 \log \left (1+\sqrt [3]{c} \sqrt {x}\right )-3 b^2 c^2 x^3 \log (x)-a b c^2 x^3 \log \left (1-\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right )+a b c^2 x^3 \log \left (1+\sqrt [3]{c} \sqrt {x}+c^{2/3} x\right )+b^2 c^2 x^3 \log \left (1-c^2 x^3\right )}{3 x^3} \] Input:

Integrate[(a + b*ArcTanh[c*x^(3/2)])^2/x^4,x]
 

Output:

-1/3*(a^2 + 2*a*b*c*x^(3/2) + 2*b*(a + b*c*x^(3/2))*ArcTanh[c*x^(3/2)] - b 
^2*(-1 + c^2*x^3)*ArcTanh[c*x^(3/2)]^2 + a*b*c^2*x^3*Log[1 - c^(1/3)*Sqrt[ 
x]] - a*b*c^2*x^3*Log[1 + c^(1/3)*Sqrt[x]] - 3*b^2*c^2*x^3*Log[x] - a*b*c^ 
2*x^3*Log[1 - c^(1/3)*Sqrt[x] + c^(2/3)*x] + a*b*c^2*x^3*Log[1 + c^(1/3)*S 
qrt[x] + c^(2/3)*x] + b^2*c^2*x^3*Log[1 - c^2*x^3])/x^3
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6454, 6452, 6544, 6452, 243, 47, 14, 16, 6510}

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.

\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx\)

\(\Big \downarrow \) 6454

\(\displaystyle \frac {2}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^{9/2}}dx^{3/2}\)

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 6544

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

\(\Big \downarrow \) 6452

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 47

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

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 6510

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

Input:

Int[(a + b*ArcTanh[c*x^(3/2)])^2/x^4,x]
 

Output:

(2*(-1/2*(a + b*ArcTanh[c*x^(3/2)])^2/x^3 + b*c*(-((a + b*ArcTanh[c*x^(3/2 
)])/x^(3/2)) + (c*(a + b*ArcTanh[c*x^(3/2)])^2)/(2*b) + (b*c*(Log[x^3] - L 
og[1 - c^2*x^3]))/2)))/3
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 47

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c 
 - a*d)   Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*x), x 
], x] /; FreeQ[{a, b, c, d}, x]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.24 (sec) , antiderivative size = 3062, normalized size of antiderivative = 31.90

method	result	size
parts	\(\text {Expression too large to display}\)	\(3062\)
derivativedivides	\(\text {Expression too large to display}\)	\(3063\)
default	\(\text {Expression too large to display}\)	\(3063\)

Input:

int((a+b*arctanh(c*x^(3/2)))^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*a^2/x^3+b^2*(-1/3/x^3*arctanh(c*x^(3/2))^2+2*c*(1/6*arctanh(c*x^(3/2) 
)*c*ln(c*x^(3/2)+1)-1/3*arctanh(c*x^(3/2))/x^(3/2)-1/6*arctanh(c*x^(3/2))* 
c*ln(c*x^(3/2)-1)-1/2*c*(c*(Sum(-1/6*(ln(x^(1/2)-_alpha)*ln(c*x^(3/2)-1)-3 
*c*(1/3*ln(x^(1/2)-_alpha)*(ln(1/2*(x^(1/2)+_alpha)/_alpha)+ln((RootOf(_Z^ 
2+_Z*_alpha+_alpha^2,index=1)-x^(1/2)+_alpha)/RootOf(_Z^2+_Z*_alpha+_alpha 
^2,index=1))+ln((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)-x^(1/2)+_alpha)/R 
ootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)))/c+1/3*(dilog(1/2*(x^(1/2)+_alpha) 
/_alpha)+dilog((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=1)-x^(1/2)+_alpha)/Ro 
otOf(_Z^2+_Z*_alpha+_alpha^2,index=1))+dilog((RootOf(_Z^2+_Z*_alpha+_alpha 
^2,index=2)-x^(1/2)+_alpha)/RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)))/c))/ 
c,_alpha=RootOf(_Z^3*c+1))+Sum(1/6*(ln(x^(1/2)-_alpha)*ln(c*x^(3/2)-1)-3*c 
*(1/6/_alpha^2/c*ln(x^(1/2)-_alpha)^2-1/3*_alpha*ln(x^(1/2)-_alpha)*(2*Roo 
tOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^ 
2,index=2)*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x^(1/2)+_alpha) 
/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))+2*RootOf(_Z^2+3*_Z*_alpha+3* 
_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)*ln((RootOf(_ 
Z^2+3*_Z*_alpha+3*_alpha^2,index=2)-x^(1/2)+_alpha)/RootOf(_Z^2+3*_Z*_alph 
a+3*_alpha^2,index=2))+3*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*_alph 
a*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x^(1/2)+_alpha)/RootOf(_ 
Z^2+3*_Z*_alpha+3*_alpha^2,index=1))+6*RootOf(_Z^2+3*_Z*_alpha+3*_alpha...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (80) = 160\).

Time = 0.11 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\frac {24 \, b^{2} c^{2} x^{3} \log \left (\sqrt {x}\right ) + 4 \, {\left (a b - b^{2}\right )} c^{2} x^{3} \log \left (c x^{\frac {3}{2}} + 1\right ) - 4 \, {\left (a b + b^{2}\right )} c^{2} x^{3} \log \left (c x^{\frac {3}{2}} - 1\right ) - 8 \, a b c x^{\frac {3}{2}} + {\left (b^{2} c^{2} x^{3} - b^{2}\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )^{2} - 4 \, a^{2} - 4 \, {\left (b^{2} c x^{\frac {3}{2}} + a b\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )}{12 \, x^{3}} \] Input:

integrate((a+b*arctanh(c*x^(3/2)))^2/x^4,x, algorithm="fricas")
 

Output:

1/12*(24*b^2*c^2*x^3*log(sqrt(x)) + 4*(a*b - b^2)*c^2*x^3*log(c*x^(3/2) + 
1) - 4*(a*b + b^2)*c^2*x^3*log(c*x^(3/2) - 1) - 8*a*b*c*x^(3/2) + (b^2*c^2 
*x^3 - b^2)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^2*x^3 - 1))^2 - 4*a^2 - 4* 
(b^2*c*x^(3/2) + a*b)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^2*x^3 - 1)))/x^3
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\text {Timed out} \] Input:

integrate((a+b*atanh(c*x**(3/2)))**2/x**4,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).

Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.82 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\frac {1}{3} \, {\left ({\left (c \log \left (c x^{\frac {3}{2}} + 1\right ) - c \log \left (c x^{\frac {3}{2}} - 1\right ) - \frac {2}{x^{\frac {3}{2}}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )}{x^{3}}\right )} a b + \frac {1}{12} \, {\left ({\left (2 \, {\left (\log \left (c x^{\frac {3}{2}} - 1\right ) - 2\right )} \log \left (c x^{\frac {3}{2}} + 1\right ) - \log \left (c x^{\frac {3}{2}} + 1\right )^{2} - \log \left (c x^{\frac {3}{2}} - 1\right )^{2} - 4 \, \log \left (c x^{\frac {3}{2}} - 1\right ) + 12 \, \log \left (x\right )\right )} c^{2} + 4 \, {\left (c \log \left (c x^{\frac {3}{2}} + 1\right ) - c \log \left (c x^{\frac {3}{2}} - 1\right ) - \frac {2}{x^{\frac {3}{2}}}\right )} c \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )^{2}}{3 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} \] Input:

integrate((a+b*arctanh(c*x^(3/2)))^2/x^4,x, algorithm="maxima")
 

Output:

1/3*((c*log(c*x^(3/2) + 1) - c*log(c*x^(3/2) - 1) - 2/x^(3/2))*c - 2*arcta 
nh(c*x^(3/2))/x^3)*a*b + 1/12*((2*(log(c*x^(3/2) - 1) - 2)*log(c*x^(3/2) + 
 1) - log(c*x^(3/2) + 1)^2 - log(c*x^(3/2) - 1)^2 - 4*log(c*x^(3/2) - 1) + 
 12*log(x))*c^2 + 4*(c*log(c*x^(3/2) + 1) - c*log(c*x^(3/2) - 1) - 2/x^(3/ 
2))*c*arctanh(c*x^(3/2)))*b^2 - 1/3*b^2*arctanh(c*x^(3/2))^2/x^3 - 1/3*a^2 
/x^3
 

Giac [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{\frac {3}{2}}\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:

integrate((a+b*arctanh(c*x^(3/2)))^2/x^4,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x^(3/2)) + a)^2/x^4, x)
 

Mupad [B] (verification not implemented)

Time = 4.69 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.93 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\frac {2\,b^2\,c^2\,\ln \left (x^{3/2}\right )}{3}-\frac {a^2}{3\,x^3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}-1\right )}{3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}+1\right )}{3}+\frac {b^2\,c^2\,{\ln \left (c\,x^{3/2}+1\right )}^2}{12}+\frac {b^2\,c^2\,{\ln \left (1-c\,x^{3/2}\right )}^2}{12}-\frac {b^2\,{\ln \left (c\,x^{3/2}+1\right )}^2}{12\,x^3}-\frac {b^2\,{\ln \left (1-c\,x^{3/2}\right )}^2}{12\,x^3}-\frac {a\,b\,c^2\,\ln \left (c\,x^{3/2}-1\right )}{3}+\frac {a\,b\,c^2\,\ln \left (c\,x^{3/2}+1\right )}{3}-\frac {2\,a\,b\,c}{3\,x^{3/2}}-\frac {a\,b\,\ln \left (c\,x^{3/2}+1\right )}{3\,x^3}+\frac {a\,b\,\ln \left (1-c\,x^{3/2}\right )}{3\,x^3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}+1\right )\,\ln \left (1-c\,x^{3/2}\right )}{6}-\frac {b^2\,c\,\ln \left (c\,x^{3/2}+1\right )}{3\,x^{3/2}}+\frac {b^2\,c\,\ln \left (1-c\,x^{3/2}\right )}{3\,x^{3/2}}+\frac {b^2\,\ln \left (c\,x^{3/2}+1\right )\,\ln \left (1-c\,x^{3/2}\right )}{6\,x^3} \] Input:

int((a + b*atanh(c*x^(3/2)))^2/x^4,x)
 

Output:

(2*b^2*c^2*log(x^(3/2)))/3 - a^2/(3*x^3) - (b^2*c^2*log(c*x^(3/2) - 1))/3 
- (b^2*c^2*log(c*x^(3/2) + 1))/3 + (b^2*c^2*log(c*x^(3/2) + 1)^2)/12 + (b^ 
2*c^2*log(1 - c*x^(3/2))^2)/12 - (b^2*log(c*x^(3/2) + 1)^2)/(12*x^3) - (b^ 
2*log(1 - c*x^(3/2))^2)/(12*x^3) - (a*b*c^2*log(c*x^(3/2) - 1))/3 + (a*b*c 
^2*log(c*x^(3/2) + 1))/3 - (2*a*b*c)/(3*x^(3/2)) - (a*b*log(c*x^(3/2) + 1) 
)/(3*x^3) + (a*b*log(1 - c*x^(3/2)))/(3*x^3) - (b^2*c^2*log(c*x^(3/2) + 1) 
*log(1 - c*x^(3/2)))/6 - (b^2*c*log(c*x^(3/2) + 1))/(3*x^(3/2)) + (b^2*c*l 
og(1 - c*x^(3/2)))/(3*x^(3/2)) + (b^2*log(c*x^(3/2) + 1)*log(1 - c*x^(3/2) 
))/(6*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.77 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^{3/2}\right )\right )^2}{x^4} \, dx=\frac {\mathit {atanh} \left (\sqrt {x}\, c x \right )^{2} b^{2} c^{2} x^{3}-\mathit {atanh} \left (\sqrt {x}\, c x \right )^{2} b^{2}-2 \sqrt {x}\, \mathit {atanh} \left (\sqrt {x}\, c x \right ) b^{2} c x +2 \mathit {atanh} \left (\sqrt {x}\, c x \right ) a b \,c^{2} x^{3}-2 \mathit {atanh} \left (\sqrt {x}\, c x \right ) a b +2 \mathit {atanh} \left (\sqrt {x}\, c x \right ) b^{2} c^{2} x^{3}-2 \sqrt {x}\, a b c x -2 \,\mathrm {log}\left (c^{\frac {2}{3}} x -\sqrt {x}\, c^{\frac {1}{3}}+1\right ) b^{2} c^{2} x^{3}-2 \,\mathrm {log}\left (\sqrt {x}\, c^{\frac {2}{3}}+c^{\frac {1}{3}}\right ) b^{2} c^{2} x^{3}+6 \,\mathrm {log}\left (\sqrt {x}\right ) b^{2} c^{2} x^{3}-a^{2}}{3 x^{3}} \] Input:

int((a+b*atanh(c*x^(3/2)))^2/x^4,x)
 

Output:

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