\(\int \frac {1}{x \sqrt {\tanh (a+b \log (c x^n))}} \, dx\) [207]

Optimal result

Integrand size = 19, antiderivative size = 47 \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {\arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}+\frac {\text {arctanh}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \] Output:

arctan(tanh(a+b*ln(c*x^n))^(1/2))/b/n+arctanh(tanh(a+b*ln(c*x^n))^(1/2))/b 
/n
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {\arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}+\frac {\text {arctanh}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \] Input:

Integrate[1/(x*Sqrt[Tanh[a + b*Log[c*x^n]]]),x]
 

Output:

ArcTan[Sqrt[Tanh[a + b*Log[c*x^n]]]]/(b*n) + ArcTanh[Sqrt[Tanh[a + b*Log[c 
*x^n]]]]/(b*n)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3039, 3042, 3957, 25, 266, 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.

\(\displaystyle \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx\)

\(\Big \downarrow \) 3039

\(\displaystyle \frac {\int \frac {1}{\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sqrt {-i \tan \left (i a+i b \log \left (c x^n\right )\right )}}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {1}{1-\tanh \left (a+b \log \left (c x^n\right )\right )}d\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}+\frac {1}{2} \arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{b n}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (\frac {1}{2} \arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )\right )}{b n}\)

Input:

Int[1/(x*Sqrt[Tanh[a + b*Log[c*x^n]]]),x]
 

Output:

(2*(ArcTan[Sqrt[Tanh[a + b*Log[c*x^n]]]]/2 + ArcTanh[Sqrt[Tanh[a + b*Log[c 
*x^n]]]]/2))/(b*n)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

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 3039

Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst 
[[3]]   Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /;  !FalseQ[lst]] /; 
NonsumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+\arctan \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )}{n b}\)	\(37\)
default	\(\frac {\operatorname {arctanh}\left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )+\arctan \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )}{n b}\)	\(37\)

Input:

int(1/x/tanh(a+b*ln(c*x^n))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/n/b*(arctanh(tanh(a+b*ln(c*x^n))^(1/2))+arctan(tanh(a+b*ln(c*x^n))^(1/2) 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (43) = 86\).

Time = 0.11 (sec) , antiderivative size = 305, normalized size of antiderivative = 6.49 \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 \, \arctan \left (-\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 1\right )} \sqrt {\frac {\sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}}\right ) - \log \left (-\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 1\right )} \sqrt {\frac {\sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}}\right )}{2 \, b n} \] Input:

integrate(1/x/tanh(a+b*log(c*x^n))^(1/2),x, algorithm="fricas")
 

Output:

1/2*(2*arctan(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*l 
og(c) + a)*sinh(b*n*log(x) + b*log(c) + a) - sinh(b*n*log(x) + b*log(c) + 
a)^2 + (cosh(b*n*log(x) + b*log(c) + a)^2 + 2*cosh(b*n*log(x) + b*log(c) + 
 a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 + 
1)*sqrt(sinh(b*n*log(x) + b*log(c) + a)/cosh(b*n*log(x) + b*log(c) + a))) 
- log(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*log(c) + 
a)*sinh(b*n*log(x) + b*log(c) + a) - sinh(b*n*log(x) + b*log(c) + a)^2 + ( 
cosh(b*n*log(x) + b*log(c) + a)^2 + 2*cosh(b*n*log(x) + b*log(c) + a)*sinh 
(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 + 1)*sqrt( 
sinh(b*n*log(x) + b*log(c) + a)/cosh(b*n*log(x) + b*log(c) + a))))/(b*n)
 

Sympy [A] (verification not implemented)

Time = 1.91 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=- \frac {\log {\left (\sqrt {\tanh {\left (a + b \log {\left (c x^{n} \right )} \right )}} - 1 \right )}}{2 b n} + \frac {\log {\left (\sqrt {\tanh {\left (a + b \log {\left (c x^{n} \right )} \right )}} + 1 \right )}}{2 b n} + \frac {\operatorname {atan}{\left (\sqrt {\tanh {\left (a + b \log {\left (c x^{n} \right )} \right )}} \right )}}{b n} \] Input:

integrate(1/x/tanh(a+b*ln(c*x**n))**(1/2),x)
 

Output:

-log(sqrt(tanh(a + b*log(c*x**n))) - 1)/(2*b*n) + log(sqrt(tanh(a + b*log( 
c*x**n))) + 1)/(2*b*n) + atan(sqrt(tanh(a + b*log(c*x**n))))/(b*n)
 

Maxima [F]

\[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int { \frac {1}{x \sqrt {\tanh \left (b \log \left (c x^{n}\right ) + a\right )}} \,d x } \] Input:

integrate(1/x/tanh(a+b*log(c*x^n))^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(x*sqrt(tanh(b*log(c*x^n) + a))), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\text {Timed out} \] Input:

integrate(1/x/tanh(a+b*log(c*x^n))^(1/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 2.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {\mathrm {atan}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )+\mathrm {atanh}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n} \] Input:

int(1/(x*tanh(a + b*log(c*x^n))^(1/2)),x)
 

Output:

(atan(tanh(a + b*log(c*x^n))^(1/2)) + atanh(tanh(a + b*log(c*x^n))^(1/2))) 
/(b*n)
 

Reduce [F]

\[ \int \frac {1}{x \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\int \frac {\sqrt {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}}{\tanh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right ) x}d x \] Input:

int(1/x/tanh(a+b*log(c*x^n))^(1/2),x)
 

Output:

int(sqrt(tanh(log(x**n*c)*b + a))/(tanh(log(x**n*c)*b + a)*x),x)