\(\int \frac {a+b \text {arctanh}(\frac {c}{x^2})}{x^6} \, dx\) [170]

Optimal result

Integrand size = 14, antiderivative size = 65 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=-\frac {2 b}{15 c x^3}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}}-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}+\frac {b \text {arctanh}\left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}} \] Output:

-2/15*b/c/x^3+1/5*b*arctan(x/c^(1/2))/c^(5/2)-1/5*(a+b*arctanh(c/x^2))/x^5 
+1/5*b*arctanh(x/c^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {2 b}{15 c x^3}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{5/2}}-\frac {b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}-\frac {b \log \left (\sqrt {c}-x\right )}{10 c^{5/2}}+\frac {b \log \left (\sqrt {c}+x\right )}{10 c^{5/2}} \] Input:

Integrate[(a + b*ArcTanh[c/x^2])/x^6,x]
 

Output:

-1/5*a/x^5 - (2*b)/(15*c*x^3) + (b*ArcTan[x/Sqrt[c]])/(5*c^(5/2)) - (b*Arc 
Tanh[c/x^2])/(5*x^5) - (b*Log[Sqrt[c] - x])/(10*c^(5/2)) + (b*Log[Sqrt[c] 
+ x])/(10*c^(5/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6452, 795, 847, 756, 216, 219}

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.

\(\displaystyle \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx\)

\(\Big \downarrow \) 6452

\(\displaystyle -\frac {2}{5} b c \int \frac {1}{\left (1-\frac {c^2}{x^4}\right ) x^8}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}\)

\(\Big \downarrow \) 795

\(\displaystyle -\frac {2}{5} b c \int \frac {1}{x^4 \left (x^4-c^2\right )}dx-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {2}{5} b c \left (\frac {\int \frac {1}{x^4-c^2}dx}{c^2}+\frac {1}{3 c^2 x^3}\right )-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {2}{5} b c \left (\frac {-\frac {\int \frac {1}{c-x^2}dx}{2 c}-\frac {\int \frac {1}{x^2+c}dx}{2 c}}{c^2}+\frac {1}{3 c^2 x^3}\right )-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2}{5} b c \left (\frac {-\frac {\int \frac {1}{c-x^2}dx}{2 c}-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{2 c^{3/2}}}{c^2}+\frac {1}{3 c^2 x^3}\right )-\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{5 x^5}-\frac {2}{5} b c \left (\frac {-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{2 c^{3/2}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right )}{2 c^{3/2}}}{c^2}+\frac {1}{3 c^2 x^3}\right )\)

Input:

Int[(a + b*ArcTanh[c/x^2])/x^6,x]
 

Output:

-1/5*(a + b*ArcTanh[c/x^2])/x^5 - (2*b*c*(1/(3*c^2*x^3) + (-1/2*ArcTan[x/S 
qrt[c]]/c^(3/2) - ArcTanh[x/Sqrt[c]]/(2*c^(3/2)))/c^2))/5
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 795

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* 
(b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
 

rule 847

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85

method	result	size
parts	\(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}\)	\(55\)
derivativedivides	\(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}-\frac {b \arctan \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}\)	\(57\)
default	\(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5 x^{5}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}-\frac {b \arctan \left (\frac {\sqrt {c}}{x}\right )}{5 c^{\frac {5}{2}}}\)	\(57\)
risch	\(-\frac {b \ln \left (x^{2}+c \right )}{10 x^{5}}-\frac {4 a -2 i b \pi -i b \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}+2 i b \pi {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}+i b \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )-i b \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )+i b \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}-i b \pi \,\operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}+i b \pi \,\operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}-i b \pi {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}-i b \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{20 x^{5}}+\frac {b \ln \left (-x^{2}+c \right )}{10 x^{5}}-\frac {2 b}{15 c \,x^{3}}+\frac {b \,\operatorname {arctanh}\left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctan \left (\frac {x}{\sqrt {c}}\right )}{5 c^{\frac {5}{2}}}\)	\(320\)

Input:

int((a+b*arctanh(c/x^2))/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/5*a/x^5-1/5*b/x^5*arctanh(c/x^2)-2/15*b/c/x^3+1/5*b*arctan(x/c^(1/2))/c 
^(5/2)+1/5*b/c^(5/2)*arctanh(1/x*c^(1/2))
 

Fricas [A] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (49) = 98\).

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 3.02 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=\left [\frac {6 \, b \sqrt {c} x^{5} \arctan \left (\frac {x}{\sqrt {c}}\right ) + 3 \, b \sqrt {c} x^{5} \log \left (\frac {x^{2} + 2 \, \sqrt {c} x + c}{x^{2} - c}\right ) - 4 \, b c^{2} x^{2} - 3 \, b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) - 6 \, a c^{3}}{30 \, c^{3} x^{5}}, -\frac {6 \, b \sqrt {-c} x^{5} \arctan \left (\frac {\sqrt {-c} x}{c}\right ) + 3 \, b \sqrt {-c} x^{5} \log \left (\frac {x^{2} - 2 \, \sqrt {-c} x - c}{x^{2} + c}\right ) + 4 \, b c^{2} x^{2} + 3 \, b c^{3} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + 6 \, a c^{3}}{30 \, c^{3} x^{5}}\right ] \] Input:

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="fricas")
 

Output:

[1/30*(6*b*sqrt(c)*x^5*arctan(x/sqrt(c)) + 3*b*sqrt(c)*x^5*log((x^2 + 2*sq 
rt(c)*x + c)/(x^2 - c)) - 4*b*c^2*x^2 - 3*b*c^3*log((x^2 + c)/(x^2 - c)) - 
 6*a*c^3)/(c^3*x^5), -1/30*(6*b*sqrt(-c)*x^5*arctan(sqrt(-c)*x/c) + 3*b*sq 
rt(-c)*x^5*log((x^2 - 2*sqrt(-c)*x - c)/(x^2 + c)) + 4*b*c^2*x^2 + 3*b*c^3 
*log((x^2 + c)/(x^2 - c)) + 6*a*c^3)/(c^3*x^5)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 994 vs. \(2 (60) = 120\).

Time = 6.72 (sec) , antiderivative size = 994, normalized size of antiderivative = 15.29 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx =\text {Too large to display} \] Input:

integrate((a+b*atanh(c/x**2))/x**6,x)
 

Output:

Piecewise((-a/(5*x**5), Eq(c, 0)), (-(a - oo*b)/(5*x**5), Eq(c, -x**2)), ( 
-(a + oo*b)/(5*x**5), Eq(c, x**2)), (6*a*c**13*sqrt(-c)/(-30*c**13*x**5*sq 
rt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*a*c**11*x**4*sqrt(-c)/(-30*c**13*x**5 
*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 6*b*c**(21/2)*x**5*sqrt(-c)*log(-sqr 
t(c) + x)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 3*b*c**(21/ 
2)*x**5*sqrt(-c)*log(x - sqrt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x** 
9*sqrt(-c)) - 3*b*c**(21/2)*x**5*sqrt(-c)*log(x + sqrt(-c))/(-30*c**13*x** 
5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) + 6*b*c**(21/2)*x**5*sqrt(-c)*atanh(c 
/x**2)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**(17/2)* 
x**9*sqrt(-c)*log(-sqrt(c) + x)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*s 
qrt(-c)) + 3*b*c**(17/2)*x**9*sqrt(-c)*log(x - sqrt(-c))/(-30*c**13*x**5*s 
qrt(-c) + 30*c**11*x**9*sqrt(-c)) + 3*b*c**(17/2)*x**9*sqrt(-c)*log(x + sq 
rt(-c))/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**(17/2) 
*x**9*sqrt(-c)*atanh(c/x**2)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt 
(-c)) + 6*b*c**13*sqrt(-c)*atanh(c/x**2)/(-30*c**13*x**5*sqrt(-c) + 30*c** 
11*x**9*sqrt(-c)) + 4*b*c**12*x**2*sqrt(-c)/(-30*c**13*x**5*sqrt(-c) + 30* 
c**11*x**9*sqrt(-c)) - 3*b*c**11*x**5*log(x - sqrt(-c))/(-30*c**13*x**5*sq 
rt(-c) + 30*c**11*x**9*sqrt(-c)) + 3*b*c**11*x**5*log(x + sqrt(-c))/(-30*c 
**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 6*b*c**11*x**4*sqrt(-c)*ata 
nh(c/x**2)/(-30*c**13*x**5*sqrt(-c) + 30*c**11*x**9*sqrt(-c)) - 4*b*c**...
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=\frac {1}{30} \, {\left (c {\left (\frac {6 \, \arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {7}{2}}} - \frac {3 \, \log \left (\frac {x - \sqrt {c}}{x + \sqrt {c}}\right )}{c^{\frac {7}{2}}} - \frac {4}{c^{2} x^{3}}\right )} - \frac {6 \, \operatorname {artanh}\left (\frac {c}{x^{2}}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \] Input:

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="maxima")
 

Output:

1/30*(c*(6*arctan(x/sqrt(c))/c^(7/2) - 3*log((x - sqrt(c))/(x + sqrt(c)))/ 
c^(7/2) - 4/(c^2*x^3)) - 6*arctanh(c/x^2)/x^5)*b - 1/5*a/x^5
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=-\frac {1}{5} \, b {\left (\frac {\arctan \left (\frac {x}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} - \frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} - \frac {b \log \left (\frac {x^{2} + c}{x^{2} - c}\right )}{10 \, x^{5}} - \frac {2 \, b x^{2} + 3 \, a c}{15 \, c x^{5}} \] Input:

integrate((a+b*arctanh(c/x^2))/x^6,x, algorithm="giac")
 

Output:

-1/5*b*(arctan(x/sqrt(-c))/(sqrt(-c)*c^2) - arctan(x/sqrt(c))/c^(5/2)) - 1 
/10*b*log((x^2 + c)/(x^2 - c))/x^5 - 1/15*(2*b*x^2 + 3*a*c)/(c*x^5)
 

Mupad [B] (verification not implemented)

Time = 3.92 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=\frac {b\,\mathrm {atan}\left (\frac {x}{\sqrt {c}}\right )}{5\,c^{5/2}}-\frac {2\,b}{15\,c\,x^3}-\frac {a}{5\,x^5}-\frac {b\,\ln \left (x^2+c\right )}{10\,x^5}+\frac {b\,\ln \left (x^2-c\right )}{10\,x^5}-\frac {b\,\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{5\,c^{5/2}} \] Input:

int((a + b*atanh(c/x^2))/x^6,x)
 

Output:

(b*atan(x/c^(1/2)))/(5*c^(5/2)) - (2*b)/(15*c*x^3) - a/(5*x^5) - (b*atan(( 
x*1i)/c^(1/2))*1i)/(5*c^(5/2)) - (b*log(c + x^2))/(10*x^5) + (b*log(x^2 - 
c))/(10*x^5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.45 \[ \int \frac {a+b \text {arctanh}\left (\frac {c}{x^2}\right )}{x^6} \, dx=\frac {6 \sqrt {c}\, \mathit {atan} \left (\frac {x}{\sqrt {c}}\right ) b \,x^{5}-6 \sqrt {c}\, \mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,x^{5}-6 \mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,c^{3}-6 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}-x \right ) b \,x^{5}+3 \sqrt {c}\, \mathrm {log}\left (x^{2}+c \right ) b \,x^{5}-6 a \,c^{3}-4 b \,c^{2} x^{2}}{30 c^{3} x^{5}} \] Input:

int((a+b*atanh(c/x^2))/x^6,x)
 

Output:

(6*sqrt(c)*atan(x/sqrt(c))*b*x**5 - 6*sqrt(c)*atanh(c/x**2)*b*x**5 - 6*ata 
nh(c/x**2)*b*c**3 - 6*sqrt(c)*log(sqrt(c) - x)*b*x**5 + 3*sqrt(c)*log(c + 
x**2)*b*x**5 - 6*a*c**3 - 4*b*c**2*x**2)/(30*c**3*x**5)