Integrand size = 13, antiderivative size = 138 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {1}{x}-\frac {e^{2 a} x^3}{1+e^{2 a} x^4}+\frac {e^{a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {e^{a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}+\frac {e^{a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{2 \sqrt {2}} \] Output:
-1/x-exp(2*a)*x^3/(1+exp(2*a)*x^4)-1/4*exp(1/2*a)*arctan(-1+2^(1/2)*exp(1/ 2*a)*x)*2^(1/2)-1/4*exp(1/2*a)*arctan(1+2^(1/2)*exp(1/2*a)*x)*2^(1/2)+1/4* exp(1/2*a)*arctanh(2^(1/2)*exp(1/2*a)*x/(1+exp(a)*x^2))*2^(1/2)
Time = 0.54 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.31 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {1}{4} \left (-\frac {4}{x}-\frac {4}{\frac {e^{-2 a}}{x^3}+x}+(-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}-e^{a/2} x\right )}{x^4}\right )+\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}-e^{a/2} x\right )}{x^4}\right )-(-1)^{3/4} e^{a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}+e^{a/2} x\right )}{x^4}\right )-\sqrt [4]{-1} e^{a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}+e^{a/2} x\right )}{x^4}\right )\right ) \] Input:
Integrate[Tanh[a + 2*Log[x]]^2/x^2,x]
Output:
(-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
Time = 0.50 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.54, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {6071, 962, 25, 957, 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.
| \(\displaystyle \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx\) |
| \(\Big \downarrow \) 6071 |
| \(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right )^2}{x^2 \left (e^{2 a} x^4+1\right )^2}dx\) |
| \(\Big \downarrow \) 962 |
| \(\displaystyle \int -\frac {x^2 \left (7 e^{2 a}-e^{4 a} x^4\right )}{\left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{x \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {x^2 \left (7 e^{2 a}-e^{4 a} x^4\right )}{\left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{x \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle -e^{2 a} \int \frac {x^2}{e^{2 a} x^4+1}dx-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \int \frac {e^a x^2+1}{e^{2 a} x^4+1}dx-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {1}{2} e^{-a} \int \frac {1}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx+\frac {1}{2} e^{-a} \int \frac {1}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {1}{-\left (1-\sqrt {2} e^{a/2} x\right )^2-1}d\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \int \frac {1}{-\left (\sqrt {2} e^{a/2} x+1\right )^2-1}d\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (-\frac {e^{-a/2} \int -\frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}-\frac {e^{-a/2} \int -\frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {e^{-a/2} \int \frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {1}{2} e^{-a/2} \int \frac {\sqrt {2} x+e^{-a/2}}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -e^{2 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {1}{x \left (e^{2 a} x^4+1\right )}-\frac {2 e^{2 a} x^3}{e^{2 a} x^4+1}\) |
Input:
Int[Tanh[a + 2*Log[x]]^2/x^2,x]
Output:
-(1/(x*(1 + E^(2*a)*x^4))) - (2*E^(2*a)*x^3)/(1 + E^(2*a)*x^4) - 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 - 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*E^a))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]]))
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] - Simp[(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] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
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] - Simp[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] /; Fre eQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]
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]
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]
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]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.46
| method | result | size |
| risch | \(\frac {-2 \,{\mathrm e}^{2 a} x^{4}-1}{x \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+{\mathrm e}^{2 a}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4}+4 \,{\mathrm e}^{2 a}\right ) x +\textit {\_R}^{3}\right )\right )}{4}\) | \(64\) |
Input:
int(tanh(a+2*ln(x))^2/x^2,x,method=_RETURNVERBOSE)
Output:
(-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)))
Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.17 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {16 \, x^{4} e^{\left (2 \, a\right )} + 2 \, \sqrt {2} {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (\frac {1}{2} \, a\right )} + 2 \, \sqrt {2} {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (\frac {1}{2} \, a\right )} - \sqrt {2} {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \sqrt {2} {\left (x^{5} e^{\left (2 \, a\right )} + x\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + 8}{8 \, {\left (x^{5} e^{\left (2 \, a\right )} + x\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^2,x, algorithm="fricas")
Output:
-1/8*(16*x^4*e^(2*a) + 2*sqrt(2)*(x^5*e^(2*a) + x)*arctan(sqrt(2)*x*e^(1/2 *a) + 1)*e^(1/2*a) + 2*sqrt(2)*(x^5*e^(2*a) + x)*arctan(sqrt(2)*x*e^(1/2*a ) - 1)*e^(1/2*a) - sqrt(2)*(x^5*e^(2*a) + x)*e^(1/2*a)*log(x^2*e^a + sqrt( 2)*x*e^(1/2*a) + 1) + sqrt(2)*(x^5*e^(2*a) + x)*e^(1/2*a)*log(x^2*e^a - sq rt(2)*x*e^(1/2*a) + 1) + 8)/(x^5*e^(2*a) + x)
\[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\int \frac {\tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}}{x^{2}}\, dx \] Input:
integrate(tanh(a+2*ln(x))**2/x**2,x)
Output:
Integral(tanh(a + 2*log(x))**2/x**2, x)
Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.06 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) - \frac {1}{x} - \frac {e^{\left (2 \, a\right )}}{x {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^2,x, algorithm="maxima")
Output:
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)))
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {1}{2} \, a\right )} + \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{8} \, \sqrt {2} e^{\left (\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {2 \, x^{4} e^{\left (2 \, a\right )} + 1}{x^{5} e^{\left (2 \, a\right )} + x} \] Input:
integrate(tanh(a+2*log(x))^2/x^2,x, algorithm="giac")
Output:
-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)
Time = 2.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.49 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}{2}-\frac {2\,{\mathrm {e}}^{2\,a}\,x^4+1}{{\mathrm {e}}^{2\,a}\,x^5+x} \] Input:
int(tanh(a + 2*log(x))^2/x^2,x)
Output:
(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*exp(2*a) + 1)/(x + x^5*exp(2*a))
Time = 0.28 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.22 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^2} \, dx=\frac {2 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{5}+2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x -2 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{5}-2 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x -e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{5}+e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{5}-e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x +e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x -16 e^{2 a} x^{4}-8}{8 x \left (e^{2 a} x^{4}+1\right )} \] Input:
int(tanh(a+2*log(x))^2/x^2,x)
Output:
(2*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt( 2)))*x**5 + 2*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2 )*sqrt(2)))*x - 2*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/ (e**(a/2)*sqrt(2)))*x**5 - 2*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e **a*x)/(e**(a/2)*sqrt(2)))*x - e**((5*a)/2)*sqrt(2)*log( - e**(a/2)*sqrt(2 )*x + e**a*x**2 + 1)*x**5 + e**((5*a)/2)*sqrt(2)*log(e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**5 - 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 - 1 6*e**(2*a)*x**4 - 8)/(8*x*(e**(2*a)*x**4 + 1))