\(\int \frac {(a+b \text {arctanh}(\frac {c}{x^2}))^2}{x^3} \, dx\) [174]

Optimal result

Integrand size = 16, antiderivative size = 99 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^3} \, dx=-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 c}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 x^2}+\frac {b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \log \left (\frac {2}{1-\frac {c}{x^2}}\right )}{c}+\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {c}{x^2}}\right )}{2 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {a b \left (\frac {c \text {arctanh}\left (\frac {c}{x^2}\right )}{x^2}-\log \left (\frac {1}{\sqrt {1-\frac {c^2}{x^4}}}\right )\right )}{c}-\frac {b^2 \left (\text {arctanh}\left (\frac {c}{x^2}\right ) \left (-\text {arctanh}\left (\frac {c}{x^2}\right )+\frac {c \text {arctanh}\left (\frac {c}{x^2}\right )}{x^2}-2 \log \left (1+e^{-2 \text {arctanh}\left (\frac {c}{x^2}\right )}\right )\right )+\operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (\frac {c}{x^2}\right )}\right )\right )}{2 c} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6454, 6436, 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.

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

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6436

\(\displaystyle \frac {1}{2} \left (2 b c \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{\left (1-\frac {c^2}{x^4}\right ) x^2}d\frac {1}{x^2}-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^2}\right )\)

\(\Big \downarrow \) 6546

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

\(\Big \downarrow \) 6470

\(\displaystyle \frac {1}{2} \left (2 b c \left (\frac {\frac {\log \left (\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{1-\frac {c}{x^2}}\right )}{1-\frac {c^2}{x^4}}d\frac {1}{x^2}}{c}-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{2 b c^2}\right )-\frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^2}\right )\)

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

Input:

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

Output:

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

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 6436

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a 
 + b*ArcTanh[c*x^n])^p, x] - Simp[b*c*n*p   Int[x^n*((a + b*ArcTanh[c*x^n]) 
^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] 
 && (EqQ[n, 1] || EqQ[p, 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 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 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 = 1.60 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(-\frac {\frac {c \,a^{2}}{x^{2}}+b^{2} \left (\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2} \left (\frac {c}{x^{2}}-1\right )+2 \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2}-2 \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )-\operatorname {polylog}\left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )\right )+\frac {2 a b c \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\)	\(134\)
default	\(-\frac {\frac {c \,a^{2}}{x^{2}}+b^{2} \left (\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2} \left (\frac {c}{x^{2}}-1\right )+2 \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2}-2 \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )-\operatorname {polylog}\left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )\right )+\frac {2 a b c \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\)	\(134\)
parts	\(-\frac {a^{2}}{2 x^{2}}-\frac {b^{2} \left (\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2} \left (\frac {c}{x^{2}}-1\right )+2 \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )^{2}-2 \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )-\operatorname {polylog}\left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right )\right )}{2 c}-\frac {a b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{x^{2}}-\frac {a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\)	\(136\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/8*(8*c^3*integrate(log(x)^2/(c*x^7 - c^3*x^3), x) + c^2*(log(x^2 + c)/c^ 
3 + log(x^2 - c)/c^3 - 4*log(x)/c^3) - 8*c^2*integrate(x^2*log(x^2 + c)/(c 
*x^7 - c^3*x^3), x) + 8*c^2*integrate(x^2*log(x)/(c*x^7 - c^3*x^3), x) + 2 
*c*(log(x^2 - c)/c^2 - log(x^2)/c^2 + 1/(c*x^2))*log(-c/x^2 + 1) - c*(log( 
x^2 + c)/c^2 - log(x^2 - c)/c^2) - 8*c*integrate(x^4*log(x)^2/(c*x^7 - c^3 
*x^3), x) - 4*c*integrate(x^4*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 16*c*in 
tegrate(x^4*log(x)/(c*x^7 - c^3*x^3), x) - log(-c/x^2 + 1)^2/x^2 - (x^2*lo 
g(x^2 - c)^2 + 4*x^2*log(x)^2 - 4*x^2*log(x) - 2*(2*x^2*log(x) - x^2)*log( 
x^2 - c) + 2*c)/(c*x^2) - (c*log(x^2 + c)^2 - 2*((x^2 + c)*log(x^2 + c) - 
2*(x^2 + c)*log(x) - c)*log(x^2 - c))/(c*x^2) - 4*integrate(x^6*log(x^2 + 
c)/(c*x^7 - c^3*x^3), x) + 8*integrate(x^6*log(x)/(c*x^7 - c^3*x^3), x))*b 
^2 - 1/2*a*b*(2*c*arctanh(c/x^2)/x^2 + log(-c^2/x^4 + 1))/c - 1/2*a^2/x^2
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )^2}{x^3} \, dx=\frac {-\mathit {atanh} \left (\frac {c}{x^{2}}\right )^{2} b^{2} c -2 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b c +2 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) a b \,x^{2}+4 \left (\int \frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right )}{-x^{5}+c^{2} x}d x \right ) b^{2} c^{2} x^{2}-2 \,\mathrm {log}\left (x^{2}+c \right ) a b \,x^{2}+4 \,\mathrm {log}\left (x \right ) a b \,x^{2}-a^{2} c}{2 c \,x^{2}} \] Input:

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

Output:

( - atanh(c/x**2)**2*b**2*c - 2*atanh(c/x**2)*a*b*c + 2*atanh(c/x**2)*a*b* 
x**2 + 4*int(atanh(c/x**2)/(c**2*x - x**5),x)*b**2*c**2*x**2 - 2*log(c + x 
**2)*a*b*x**2 + 4*log(x)*a*b*x**2 - a**2*c)/(2*c*x**2)