\(\int \tanh (a+2 \log (x)) \, dx\) [154]

Optimal result

Integrand size = 7, antiderivative size = 104 \[ \int \tanh (a+2 \log (x)) \, dx=x+\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{\sqrt {2}} \] Output:

x-1/2*arctan(-1+2^(1/2)*exp(1/2*a)*x)*2^(1/2)/exp(1/2*a)-1/2*arctan(1+2^(1 
/2)*exp(1/2*a)*x)*2^(1/2)/exp(1/2*a)-1/2*arctanh(2^(1/2)*exp(1/2*a)*x/(1+e 
xp(a)*x^2))*2^(1/2)/exp(1/2*a)
 

Mathematica [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.56 \[ \int \tanh (a+2 \log (x)) \, dx=x+\frac {1}{2} \text {RootSum}\left [\cosh (a)-\sinh (a)+\cosh (a) \text {$\#$1}^4+\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)-\log (x-\text {$\#$1})}{\text {$\#$1}^3}\&\right ] (\cosh (2 a)-\sinh (2 a)) \] Input:

Integrate[Tanh[a + 2*Log[x]],x]
 

Output:

x + (RootSum[Cosh[a] - Sinh[a] + Cosh[a]*#1^4 + Sinh[a]*#1^4 & , (Log[x] - 
 Log[x - #1])/#1^3 & ]*(Cosh[2*a] - Sinh[2*a]))/2
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.52, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.429, Rules used = {6067, 913, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \tanh (a+2 \log (x)) \, dx\)

\(\Big \downarrow \) 6067

\(\displaystyle \int \frac {e^{2 a} x^4-1}{e^{2 a} x^4+1}dx\)

\(\Big \downarrow \) 913

\(\displaystyle x-2 \int \frac {1}{e^{2 a} x^4+1}dx\)

\(\Big \downarrow \) 755

\(\displaystyle x-2 \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \int \frac {e^a x^2+1}{e^{2 a} x^4+1}dx\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle x-2 \left (\frac {1}{2} \left (\frac {1}{2} e^{-a} \int \frac {1}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx+\frac {1}{2} e^{-a} \int \frac {1}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle x-2 \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {1}{-\left (1-\sqrt {2} e^{a/2} x\right )^2-1}d\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \int \frac {1}{-\left (\sqrt {2} e^{a/2} x+1\right )^2-1}d\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle x-2 \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle x-2 \left (\frac {1}{2} \left (-\frac {e^{-a/2} \int -\frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}-\frac {e^{-a/2} \int -\frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle x-2 \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {e^{-a/2} \int \frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle x-2 \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {1}{2} e^{-a/2} \int \frac {\sqrt {2} x+e^{-a/2}}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle x-2 \left (\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )\)

Input:

Int[Tanh[a + 2*Log[x]],x]
 

Output:

x - 2*((-(ArcTan[1 - Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^(a/2))) + ArcTan[1 + Sq 
rt[2]*E^(a/2)*x]/(Sqrt[2]*E^(a/2)))/2 + (-1/2*Log[1 - Sqrt[2]*E^(a/2)*x + 
E^a*x^2]/(Sqrt[2]*E^(a/2)) + Log[1 + Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*Sqrt[ 
2]*E^(a/2)))/2)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 755

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

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 6067

Int[Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(-1 + E^(2* 
a*d)*x^(2*b*d))^p/(1 + E^(2*a*d)*x^(2*b*d))^p, x] /; FreeQ[{a, b, d, p}, x]
 

Maple [C] (verified)

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

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.32

method	result	size
risch	\(x -\frac {{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{2}\)	\(33\)

Input:

int(tanh(a+2*ln(x)),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

x-1/2*exp(-2*a)*sum(1/_R^3*ln(x-_R),_R=RootOf(exp(2*a)*_Z^4+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \tanh (a+2 \log (x)) \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + x \] Input:

integrate(tanh(a+2*log(x)),x, algorithm="fricas")
 

Output:

-1/2*sqrt(2)*arctan(sqrt(2)*x*e^(1/2*a) + 1)*e^(-1/2*a) - 1/2*sqrt(2)*arct 
an(sqrt(2)*x*e^(1/2*a) - 1)*e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^ 
a + sqrt(2)*x*e^(1/2*a) + 1) + 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^a - sqrt(2 
)*x*e^(1/2*a) + 1) + x
 

Sympy [F]

\[ \int \tanh (a+2 \log (x)) \, dx=\int \tanh {\left (a + 2 \log {\left (x \right )} \right )}\, dx \] Input:

integrate(tanh(a+2*ln(x)),x)
 

Output:

Integral(tanh(a + 2*log(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \tanh (a+2 \log (x)) \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + x \] Input:

integrate(tanh(a+2*log(x)),x, algorithm="maxima")
 

Output:

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a + sqrt(2)*e^(1/2*a))*e^(-1/2*a))* 
e^(-1/2*a) - 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a - sqrt(2)*e^(1/2*a))* 
e^(-1/2*a))*e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^a + sqrt(2)*x*e^ 
(1/2*a) + 1) + 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 
1) + x
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.14 \[ \int \tanh (a+2 \log (x)) \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + x \] Input:

integrate(tanh(a+2*log(x)),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(- 
1/2*a) - 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2 
*a))*e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + 
e^(-a)) + 1/4*sqrt(2)*e^(-1/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) 
 + x
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 2.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.42 \[ \int \tanh (a+2 \log (x)) \, dx=x-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}-\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}} \] Input:

int(tanh(a + 2*log(x)),x)
 

Output:

x - atan(x*(-exp(2*a))^(1/4))/(-exp(2*a))^(1/4) - atanh(x*(-exp(2*a))^(1/4 
))/(-exp(2*a))^(1/4)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.38 \[ \int \tanh (a+2 \log (x)) \, dx=\frac {2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )-2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )+e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )-e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )+4 e^{a} x}{4 e^{a}} \] Input:

int(tanh(a+2*log(x)),x)
 

Output:

(2*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt(2))) 
 - 2*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2) 
)) + e**(a/2)*sqrt(2)*log( - e**(a/2)*sqrt(2)*x + e**a*x**2 + 1) - e**(a/2 
)*sqrt(2)*log(e**(a/2)*sqrt(2)*x + e**a*x**2 + 1) + 4*e**a*x)/(4*e**a)