\(\int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx\) [158]

Optimal result

Integrand size = 13, antiderivative size = 190 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}+\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}}+\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}} \]

[Out]

-1/x/(1+exp(2*a)*x^4)-2*exp(2*a)*x^3/(1+exp(2*a)*x^4)-1/4*exp(1/2*a)*arctan(-1+exp(1/2*a)*x*2^(1/2))*2^(1/2)-1
/4*exp(1/2*a)*arctan(1+exp(1/2*a)*x*2^(1/2))*2^(1/2)-1/8*exp(1/2*a)*ln(1+exp(a)*x^2-exp(1/2*a)*x*2^(1/2))*2^(1
/2)+1/8*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 = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {5656, 473, 468, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {e^{a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}+\frac {e^{a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1} \]

[In]

Int[Tanh[a + 2*Log[x]]^2/x^2,x]

[Out]

-(1/(x*(1 + E^(2*a)*x^4))) - (2*E^(2*a)*x^3)/(1 + E^(2*a)*x^4) + (E^(a/2)*ArcTan[1 - Sqrt[2]*E^(a/2)*x])/(2*Sq
rt[2]) - (E^(a/2)*ArcTan[1 + Sqrt[2]*E^(a/2)*x])/(2*Sqrt[2]) - (E^(a/2)*Log[1 - Sqrt[2]*E^(a/2)*x + E^a*x^2])/
(4*Sqrt[2]) + (E^(a/2)*Log[1 + Sqrt[2]*E^(a/2)*x + E^a*x^2])/(4*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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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 {\left (-1+e^{2 a} x^4\right )^2}{x^2 \left (1+e^{2 a} x^4\right )^2} \, dx \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}+\int \frac {x^2 \left (-7 e^{2 a}+e^{4 a} x^4\right )}{\left (1+e^{2 a} x^4\right )^2} \, dx \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}-e^{2 a} \int \frac {x^2}{1+e^{2 a} x^4} \, dx \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}+\frac {1}{2} e^a \int \frac {1-e^a x^2}{1+e^{2 a} x^4} \, dx-\frac {1}{2} e^a \int \frac {1+e^a x^2}{1+e^{2 a} x^4} \, dx \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}-\frac {1}{4} \int \frac {1}{e^{-a}-\sqrt {2} e^{-a/2} x+x^2} \, dx-\frac {1}{4} \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}{4 \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}{4 \sqrt {2}} \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}-\frac {e^{a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}}+\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \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 )}{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 )}{2 \sqrt {2}} \\ & = -\frac {1}{x \left (1+e^{2 a} x^4\right )}-\frac {2 e^{2 a} x^3}{1+e^{2 a} x^4}+\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}}+\frac {e^{a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {1}{4} \left (-\frac {4}{x}-\frac {4}{\frac {e^{-2 a}}{x^3}+x}+(-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}-e^{a/2} x\right )}{x^4}\right )+\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}-e^{a/2} x\right )}{x^4}\right )-(-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}+e^{a/2} x\right )}{x^4}\right )-\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}+e^{a/2} x\right )}{x^4}\right )\right ) \]

[In]

Integrate[Tanh[a + 2*Log[x]]^2/x^2,x]

[Out]

(-4/x - 4/(1/(E^(2*a)*x^3) + x) + (-1)^(3/4)*E^(a/2)*Log[((-1)^(1/4) - E^(a/2)*x)/(E^(2*a)*x^4)] + (-1)^(1/4)*
E^(a/2)*Log[((-1)^(3/4) - E^(a/2)*x)/(E^(2*a)*x^4)] - (-1)^(3/4)*E^(a/2)*Log[((-1)^(1/4) + E^(a/2)*x)/(E^(2*a)
*x^4)] - (-1)^(1/4)*E^(a/2)*Log[((-1)^(3/4) + E^(a/2)*x)/(E^(2*a)*x^4)])/4

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.34

method	result	size
risch	\(\frac {-2 \,{\mathrm e}^{2 a} x^{4}-1}{x \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+{\mathrm e}^{2 a}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4}+4 \,{\mathrm e}^{2 a}\right ) x +\textit {\_R}^{3}\right )\right )}{4}\)	\(64\)

[In]

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

[Out]

(-2*exp(2*a)*x^4-1)/x/(1+exp(2*a)*x^4)+1/4*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 = 180, normalized size of antiderivative = 0.95 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {8 \, x^{4} e^{\left (2 \, a\right )} + {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} + \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) - {\left (i \, x^{5} e^{\left (2 \, a\right )} + i \, x\right )} \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} + i \, \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) - {\left (-i \, x^{5} e^{\left (2 \, a\right )} - i \, x\right )} \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} - i \, \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) - {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} \left (-e^{\left (2 \, a\right )}\right )^{\frac {1}{4}} \log \left (x e^{\left (2 \, a\right )} - \left (-e^{\left (2 \, a\right )}\right )^{\frac {3}{4}}\right ) + 4}{4 \, {\left (x^{5} e^{\left (2 \, a\right )} + x\right )}} \]

[In]

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

[Out]

-1/4*(8*x^4*e^(2*a) + (x^5*e^(2*a) + x)*(-e^(2*a))^(1/4)*log(x*e^(2*a) + (-e^(2*a))^(3/4)) - (I*x^5*e^(2*a) +
I*x)*(-e^(2*a))^(1/4)*log(x*e^(2*a) + I*(-e^(2*a))^(3/4)) - (-I*x^5*e^(2*a) - I*x)*(-e^(2*a))^(1/4)*log(x*e^(2
*a) - I*(-e^(2*a))^(3/4)) - (x^5*e^(2*a) + x)*(-e^(2*a))^(1/4)*log(x*e^(2*a) - (-e^(2*a))^(3/4)) + 4)/(x^5*e^(
2*a) + x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.77 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{x} - \frac {e^{\left (2 \, a\right )}}{x {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \]

[In]

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

[Out]

1/4*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)*arctan(-1/2*sqrt(
2)*(sqrt(2)*e^(1/2*a) - 2/x)*e^(-1/2*a))*e^(1/2*a) + 1/8*sqrt(2)*e^(1/2*a)*log(sqrt(2)*e^(1/2*a)/x + 1/x^2 + e
^a) - 1/8*sqrt(2)*e^(1/2*a)*log(-sqrt(2)*e^(1/2*a)/x + 1/x^2 + e^a) - 1/x - e^(2*a)/(x*(1/x^4 + e^(2*a)))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.75 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {2 \, x^{4} e^{\left (2 \, a\right )} + 1}{x^{5} e^{\left (2 \, a\right )} + x} \]

[In]

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

[Out]

-1/4*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)*arctan(-1/2*sqrt
(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2*a))*e^(1/2*a) + 1/8*sqrt(2)*e^(1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e
^(-a)) - 1/8*sqrt(2)*e^(1/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) - (2*x^4*e^(2*a) + 1)/(x^5*e^(2*a) +
x)

Mupad [B] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.36 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {2\,{\mathrm {e}}^{2\,a}\,x^4+1}{{\mathrm {e}}^{2\,a}\,x^5+x} \]

[In]

int(tanh(a + 2*log(x))^2/x^2,x)

[Out]

(atanh(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(1/4))/2 - (atan(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(1/4))/2 - (2*x^4*ex
p(2*a) + 1)/(x + x^5*exp(2*a))