\(\int \frac {\tanh ^{\frac {5}{2}}(a+b \log (c x^n))}{x} \, dx\) [204]

Optimal result

Integrand size = 19, antiderivative size = 73 \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {\arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}+\frac {\text {arctanh}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {2 \tanh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b n} \] 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-2/3*tanh(a+b*ln(c*x^n))^(3/2)/b/n
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85 \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {\arctan \left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )-\text {arctanh}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )+\frac {2}{3} \tanh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{b n} \] Input:

Integrate[Tanh[a + b*Log[c*x^n]]^(5/2)/x,x]
 

Output:

-((ArcTan[Sqrt[Tanh[a + b*Log[c*x^n]]]] - ArcTanh[Sqrt[Tanh[a + b*Log[c*x^ 
n]]]] + (2*Tanh[a + b*Log[c*x^n]]^(3/2))/3)/(b*n))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3039, 3042, 3954, 3042, 3957, 25, 266, 827, 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.

\(\displaystyle \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx\)

\(\Big \downarrow \) 3039

\(\displaystyle \frac {\int \tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (-i \tan \left (i a+i b \log \left (c x^n\right )\right )\right )^{5/2}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {2 \tanh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{3 b}+\int \sqrt {-i \tan \left (i a+i b \log \left (c x^n\right )\right )}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[Tanh[a + b*Log[c*x^n]]^(5/2)/x,x]
 

Output:

((2*(-1/2*ArcTan[Sqrt[Tanh[a + b*Log[c*x^n]]]] + ArcTanh[Sqrt[Tanh[a + b*L 
og[c*x^n]]]]/2))/b - (2*Tanh[a + b*Log[c*x^n]]^(3/2))/(3*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 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   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 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

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 = 1.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {-\frac {2 {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}}{3}-\frac {\ln \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}-1\right )}{2}+\frac {\ln \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}+1\right )}{2}-\arctan \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )}{n b}\)	\(76\)
default	\(\frac {-\frac {2 {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}}}{3}-\frac {\ln \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}-1\right )}{2}+\frac {\ln \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}+1\right )}{2}-\arctan \left (\sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )}{n b}\)	\(76\)

Input:

int(tanh(a+b*ln(c*x^n))^(5/2)/x,x,method=_RETURNVERBOSE)
 

Output:

1/n/b*(-2/3*tanh(a+b*ln(c*x^n))^(3/2)-1/2*ln(tanh(a+b*ln(c*x^n))^(1/2)-1)+ 
1/2*ln(tanh(a+b*ln(c*x^n))^(1/2)+1)-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. 625 vs. \(2 (65) = 130\).

Time = 0.12 (sec) , antiderivative size = 625, normalized size of antiderivative = 8.56 \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Too large to display} \] Input:

integrate(tanh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="fricas")
 

Output:

-1/6*(6*(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)*arctan(-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))) + 4 
*cosh(b*n*log(x) + b*log(c) + a)^2 + 3*(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)*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*cos 
h(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*lo 
g(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))) + 8*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log( 
x) + b*log(c) + a) + 4*sinh(b*n*log(x) + b*log(c) + a)^2 + 4*(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)) + 4)/(b*n*cosh(b*n*lo 
g(x) + b*log(c) + a)^2 + 2*b*n*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate(tanh(a+b*ln(c*x**n))**(5/2)/x,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {\tanh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}}{x} \,d x } \] Input:

integrate(tanh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="maxima")
 

Output:

integrate(tanh(b*log(c*x^n) + a)^(5/2)/x, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate(tanh(a+b*log(c*x^n))^(5/2)/x,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\mathrm {atanh}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {\mathrm {atan}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {2\,{\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2}}{3\,b\,n} \] Input:

int(tanh(a + b*log(c*x^n))^(5/2)/x,x)
 

Output:

atanh(tanh(a + b*log(c*x^n))^(1/2))/(b*n) - atan(tanh(a + b*log(c*x^n))^(1 
/2))/(b*n) - (2*tanh(a + b*log(c*x^n))^(3/2))/(3*b*n)
 

Reduce [F]

\[ \int \frac {\tanh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int \frac {\sqrt {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}\, {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}^{2}}{x}d x \] Input:

int(tanh(a+b*log(c*x^n))^(5/2)/x,x)
 

Output:

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