\(\int \frac {\tanh (a+2 \log (x))}{x^2} \, dx\) [155]

Optimal result

Integrand size = 11, antiderivative size = 106 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {1}{x}-\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{\sqrt {2}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {2-x \text {RootSum}\left [\cosh (a)+\sinh (a)+\cosh (a) \text {$\#$1}^4-\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)+\log \left (\frac {1}{x}-\text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ] (\cosh (a)+\sinh (a))^2}{2 x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.65, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {6071, 955, 826, 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.

\(\displaystyle \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx\)

\(\Big \downarrow \) 6071

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]
 

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 6071

Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] 
 :> Int[(e*x)^m*((-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, e, m, p}, x]
 

Maple [C] (verified)

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

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.40

method	result	size
risch	\(\frac {1}{x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+{\mathrm e}^{2 a}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4}+4 \,{\mathrm e}^{2 a}\right ) x -\textit {\_R}^{3}\right )\right )}{2}\)	\(42\)

Input:

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

Output:

1/x+1/2*sum(_R*ln((5*_R^4+4*exp(2*a))*x-_R^3),_R=RootOf(_Z^4+exp(2*a)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.01 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {2 \, \sqrt {2} x \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (\frac {1}{2} \, a\right )} + 2 \, \sqrt {2} x \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (\frac {1}{2} \, a\right )} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + 4}{4 \, x} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.18 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) + \frac {1}{x} \] Input:

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

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.14 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{x} \] Input:

integrate(tanh(a+2*log(x))/x^2,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)) + 1/ 
x
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 2.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.42 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}-\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}+\frac {1}{x} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.33 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {-2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x +2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x +e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x -e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x +4}{4 x} \] Input:

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

Output:

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