\(\int \frac {\tanh (a+2 \log (x))}{x^2} \, dx\) [151]

Optimal result

Integrand size = 11, antiderivative size = 147 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {1}{x}-\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}} \]

[Out]

1/x+1/2*exp(1/2*a)*arctan(-1+exp(1/2*a)*x*2^(1/2))*2^(1/2)+1/2*exp(1/2*a)*arctan(1+exp(1/2*a)*x*2^(1/2))*2^(1/
2)+1/4*exp(1/2*a)*ln(1+exp(a)*x^2-exp(1/2*a)*x*2^(1/2))*2^(1/2)-1/4*exp(1/2*a)*ln(1+exp(a)*x^2+exp(1/2*a)*x*2^
(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {5656, 464, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=-\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}+\frac {e^{a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {1}{x} \]

[In]

Int[Tanh[a + 2*Log[x]]/x^2,x]

[Out]

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

Rule 210

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 303

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

Rule 464

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] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

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

Rule 5656

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+e^{2 a} x^4}{x^2 \left (1+e^{2 a} x^4\right )} \, dx \\ & = \frac {1}{x}+\left (2 e^{2 a}\right ) \int \frac {x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {1}{x}-e^a \int \frac {1-e^a x^2}{1+e^{2 a} x^4} \, dx+e^a \int \frac {1+e^a x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {1}{x}+\frac {1}{2} \int \frac {1}{e^{-a}-\sqrt {2} e^{-a/2} x+x^2} \, dx+\frac {1}{2} \int \frac {1}{e^{-a}+\sqrt {2} e^{-a/2} x+x^2} \, dx+\frac {e^{a/2} \int \frac {\sqrt {2} e^{-a/2}+2 x}{-e^{-a}-\sqrt {2} e^{-a/2} x-x^2} \, dx}{2 \sqrt {2}}+\frac {e^{a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{-e^{-a}+\sqrt {2} e^{-a/2} x-x^2} \, dx}{2 \sqrt {2}} \\ & = \frac {1}{x}+\frac {e^{a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}+\frac {e^{a/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{a/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}} \\ & = \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} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.40 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {2-x \text {RootSum}\left [\cosh (a)+\sinh (a)+\cosh (a) \text {$\#$1}^4-\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)+\log \left (\frac {1}{x}-\text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ] (\cosh (a)+\sinh (a))^2}{2 x} \]

[In]

Integrate[Tanh[a + 2*Log[x]]/x^2,x]

[Out]

(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)

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.29

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

[In]

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

[Out]

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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {x \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} + \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) - i \, x \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} + i \, \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) + i \, x \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} - i \, \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) - x \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} - \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) + 2}{2 \, x} \]

[In]

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

[Out]

1/2*(x*(-e^(2*a))^(1/4)*log(x*e^(2*a) + (-e^(2*a))^(3/4)) - I*x*(-e^(2*a))^(1/4)*log(x*e^(2*a) + I*(-e^(2*a))^
(3/4)) + I*x*(-e^(2*a))^(1/4)*log(x*e^(2*a) - I*(-e^(2*a))^(3/4)) - x*(-e^(2*a))^(1/4)*log(x*e^(2*a) - (-e^(2*
a))^(3/4)) + 2)/x

Sympy [F]

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

[In]

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

[Out]

Integral(tanh(a + 2*log(x))/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.85 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) + \frac {1}{x} \]

[In]

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

[Out]

-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)

none

Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{x} \]

[In]

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

[Out]

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 = 1.73 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.31 \[ \int \frac {\tanh (a+2 \log (x))}{x^2} \, dx=\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}-\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}+\frac {1}{x} \]

[In]

int(tanh(a + 2*log(x))/x^2,x)

[Out]

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