Integrand size = 13, antiderivative size = 138 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=-\frac {1}{3 x^3}-\frac {e^{2 a} x}{1+e^{2 a} x^4}+\frac {3 e^{3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{3 a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{2 \sqrt {2}} \] Output:
-1/3/x^3-exp(2*a)*x/(1+exp(2*a)*x^4)-3/4*exp(3/2*a)*arctan(-1+2^(1/2)*exp( 1/2*a)*x)*2^(1/2)-3/4*exp(3/2*a)*arctan(1+2^(1/2)*exp(1/2*a)*x)*2^(1/2)-3/ 4*exp(3/2*a)*arctanh(2^(1/2)*exp(1/2*a)*x/(1+exp(a)*x^2))*2^(1/2)
Time = 0.49 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.37 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=\frac {1}{12} \left (-\frac {4}{x^3}-\frac {12 e^{2 a} x}{1+e^{2 a} x^4}+9 \sqrt [4]{-1} e^{3 a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}-e^{a/2} x\right )}{x^4}\right )+9 (-1)^{3/4} e^{3 a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}-e^{a/2} x\right )}{x^4}\right )-9 \sqrt [4]{-1} e^{3 a/2} \log \left (\frac {e^{-2 a} \left (\sqrt [4]{-1}+e^{a/2} x\right )}{x^4}\right )-9 (-1)^{3/4} e^{3 a/2} \log \left (\frac {e^{-2 a} \left ((-1)^{3/4}+e^{a/2} x\right )}{x^4}\right )\right ) \] Input:
Integrate[Tanh[a + 2*Log[x]]^2/x^4,x]
Output:
(-4/x^3 - (12*E^(2*a)*x)/(1 + E^(2*a)*x^4) + 9*(-1)^(1/4)*E^((3*a)/2)*Log[ ((-1)^(1/4) - E^(a/2)*x)/(E^(2*a)*x^4)] + 9*(-1)^(3/4)*E^((3*a)/2)*Log[((- 1)^(3/4) - E^(a/2)*x)/(E^(2*a)*x^4)] - 9*(-1)^(1/4)*E^((3*a)/2)*Log[((-1)^ (1/4) + E^(a/2)*x)/(E^(2*a)*x^4)] - 9*(-1)^(3/4)*E^((3*a)/2)*Log[((-1)^(3/ 4) + E^(a/2)*x)/(E^(2*a)*x^4)])/12
Time = 0.50 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.51, 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, 910, 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.
| \(\displaystyle \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx\) |
| \(\Big \downarrow \) 6071 |
| \(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right )^2}{x^4 \left (e^{2 a} x^4+1\right )^2}dx\) |
| \(\Big \downarrow \) 962 |
| \(\displaystyle \frac {1}{3} \int -\frac {13 e^{2 a}-3 e^{4 a} x^4}{\left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{3} \int \frac {13 e^{2 a}-3 e^{4 a} x^4}{\left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 910 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \int \frac {1}{e^{2 a} x^4+1}dx-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \int \frac {e^a x^2+1}{e^{2 a} x^4+1}dx\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \left (\frac {1}{2} e^{-a} \int \frac {1}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx+\frac {1}{2} e^{-a} \int \frac {1}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {1}{-\left (1-\sqrt {2} e^{a/2} x\right )^2-1}d\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \int \frac {1}{-\left (\sqrt {2} e^{a/2} x+1\right )^2-1}d\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \left (-\frac {e^{-a/2} \int -\frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}-\frac {e^{-a/2} \int -\frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {e^{-a/2} \int \frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {1}{2} e^{-a/2} \int \frac {\sqrt {2} x+e^{-a/2}}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {4 e^{2 a} x}{e^{2 a} x^4+1}\right )-\frac {1}{3 x^3 \left (e^{2 a} x^4+1\right )}\) |
Input:
Int[Tanh[a + 2*Log[x]]^2/x^4,x]
Output:
-1/3*1/(x^3*(1 + E^(2*a)*x^4)) + ((-4*E^(2*a)*x)/(1 + E^(2*a)*x^4) - 9*E^( 2*a)*((-(ArcTan[1 - Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^(a/2))) + ArcTan[1 + Sqr t[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))/3
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[((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]]))
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
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 = 0.32 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.49
| method | result | size |
| risch | \(\frac {-\frac {4 \,{\mathrm e}^{2 a} x^{4}}{3}-\frac {1}{3}}{x^{3} \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{6 a}+\textit {\_Z}^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (-4 \,{\mathrm e}^{6 a}-5 \textit {\_R}^{4}\right ) x -\textit {\_R} \,{\mathrm e}^{4 a}\right )\right )}{4}\) | \(68\) |
Input:
int(tanh(a+2*ln(x))^2/x^4,x,method=_RETURNVERBOSE)
Output:
(-4/3*exp(2*a)*x^4-1/3)/x^3/(1+exp(2*a)*x^4)+3/4*sum(_R*ln((-4*exp(6*a)-5* _R^4)*x-_R*exp(4*a)),_R=RootOf(exp(6*a)+_Z^4))
Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.33 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=-\frac {32 \, x^{4} e^{\left (2 \, a\right )} + 18 \, \sqrt {2} {\left (x^{7} e^{\left (3 \, a\right )} + x^{3} e^{a}\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (\frac {1}{2} \, a\right )} + 18 \, \sqrt {2} {\left (x^{7} e^{\left (3 \, a\right )} + x^{3} e^{a}\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (\frac {1}{2} \, a\right )} + 9 \, \sqrt {2} {\left (x^{7} e^{\left (3 \, a\right )} + x^{3} e^{a}\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - 9 \, \sqrt {2} {\left (x^{7} e^{\left (3 \, a\right )} + x^{3} e^{a}\right )} e^{\left (\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + 8}{24 \, {\left (x^{7} e^{\left (2 \, a\right )} + x^{3}\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^4,x, algorithm="fricas")
Output:
-1/24*(32*x^4*e^(2*a) + 18*sqrt(2)*(x^7*e^(3*a) + x^3*e^a)*arctan(sqrt(2)* x*e^(1/2*a) + 1)*e^(1/2*a) + 18*sqrt(2)*(x^7*e^(3*a) + x^3*e^a)*arctan(sqr t(2)*x*e^(1/2*a) - 1)*e^(1/2*a) + 9*sqrt(2)*(x^7*e^(3*a) + x^3*e^a)*e^(1/2 *a)*log(x^2*e^a + sqrt(2)*x*e^(1/2*a) + 1) - 9*sqrt(2)*(x^7*e^(3*a) + x^3* e^a)*e^(1/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1) + 8)/(x^7*e^(2*a) + x^3)
\[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=\int \frac {\tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}}{x^{4}}\, dx \] Input:
integrate(tanh(a+2*ln(x))**2/x**4,x)
Output:
Integral(tanh(a + 2*log(x))**2/x**4, x)
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=\frac {3}{8} \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} + \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (\frac {1}{2} \, a\right )} - \frac {2}{x}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right ) + \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\frac {\sqrt {2} e^{\left (\frac {1}{2} \, a\right )}}{x} + \frac {1}{x^{2}} + e^{a}\right )\right )} e^{\left (2 \, a\right )} - \frac {1}{3 \, x^{3}} - \frac {e^{\left (2 \, a\right )}}{x^{3} {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^4,x, algorithm="maxima")
Output:
3/8*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(1/2*a) + 2/x)*e^(-1/2*a))*e^ (-1/2*a) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*e^(1/2*a) - 2/x)*e^(-1/2 *a))*e^(-1/2*a) - sqrt(2)*e^(-1/2*a)*log(sqrt(2)*e^(1/2*a)/x + 1/x^2 + e^a ) + sqrt(2)*e^(-1/2*a)*log(-sqrt(2)*e^(1/2*a)/x + 1/x^2 + e^a))*e^(2*a) - 1/3/x^3 - e^(2*a)/(x^3*(1/x^4 + e^(2*a)))
Time = 0.13 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=-\frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (\frac {3}{2} \, a\right )} - \frac {3}{8} \, \sqrt {2} e^{\left (\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {3}{8} \, \sqrt {2} e^{\left (\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {x e^{\left (2 \, a\right )}}{x^{4} e^{\left (2 \, a\right )} + 1} - \frac {1}{3 \, x^{3}} \] Input:
integrate(tanh(a+2*log(x))^2/x^4,x, algorithm="giac")
Output:
-3/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(3 /2*a) - 3/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2* a))*e^(3/2*a) - 3/8*sqrt(2)*e^(3/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e^( -a)) + 3/8*sqrt(2)*e^(3/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) - x *e^(2*a)/(x^4*e^(2*a) + 1) - 1/3/x^3
Time = 2.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.51 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=\frac {3\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}{2}-\frac {\frac {4\,{\mathrm {e}}^{2\,a}\,x^4}{3}+\frac {1}{3}}{{\mathrm {e}}^{2\,a}\,x^7+x^3}+\frac {3\,\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}{2} \] Input:
int(tanh(a + 2*log(x))^2/x^4,x)
Output:
(3*atan(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(3/4))/2 - ((4*x^4*exp(2*a))/3 + 1/3)/(x^7*exp(2*a) + x^3) + (3*atanh(x*(-exp(2*a))^(1/4))*(-exp(2*a))^(3/4 ))/2
Time = 0.31 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.29 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^4} \, dx=\frac {18 e^{\frac {7 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{7}+18 e^{\frac {3 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{3}-18 e^{\frac {7 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{7}-18 e^{\frac {3 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{3}+9 e^{\frac {7 a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{7}-9 e^{\frac {7 a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{7}+9 e^{\frac {3 a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{3}-9 e^{\frac {3 a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{3}-32 e^{2 a} x^{4}-8}{24 x^{3} \left (e^{2 a} x^{4}+1\right )} \] Input:
int(tanh(a+2*log(x))^2/x^4,x)
Output:
(18*e**((7*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt (2)))*x**7 + 18*e**((3*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e **(a/2)*sqrt(2)))*x**3 - 18*e**((7*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2)))*x**7 - 18*e**((3*a)/2)*sqrt(2)*atan((e**(a/2 )*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2)))*x**3 + 9*e**((7*a)/2)*sqrt(2)*lo g( - e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**7 - 9*e**((7*a)/2)*sqrt(2)*log (e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**7 + 9*e**((3*a)/2)*sqrt(2)*log( - e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**3 - 9*e**((3*a)/2)*sqrt(2)*log(e**( a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**3 - 32*e**(2*a)*x**4 - 8)/(24*x**3*(e** (2*a)*x**4 + 1))