Integrand size = 13, antiderivative size = 173 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}+\frac {3 e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}}+\frac {3 e^{-3 a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}} \]
1/3*x^3+x^3/(1+exp(2*a)*x^4)-3/4*arctan(-1+exp(1/2*a)*x*2^(1/2))/exp(3/2*a )*2^(1/2)-3/4*arctan(1+exp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)-3/8*ln(1+e xp(a)*x^2-exp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)+3/8*ln(1+exp(a)*x^2+exp (1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)
Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.01 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{12} \left (4 x^3+\frac {12 x^3}{1+e^{2 a} x^4}+9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}-e^{-a} x\right )+9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}-e^{-a} x\right )-9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}+e^{-a} x\right )-9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}+e^{-a} x\right )\right ) \]
(4*x^3 + (12*x^3)/(1 + E^(2*a)*x^4) + (9*(-1)^(3/4)*Log[(-1)^(1/4)/E^((3*a )/2) - x/E^a])/E^((3*a)/2) + (9*(-1)^(1/4)*Log[(-1)^(3/4)/E^((3*a)/2) - x/ E^a])/E^((3*a)/2) - (9*(-1)^(3/4)*Log[(-1)^(1/4)/E^((3*a)/2) + x/E^a])/E^( (3*a)/2) - (9*(-1)^(1/4)*Log[(-1)^(3/4)/E^((3*a)/2) + x/E^a])/E^((3*a)/2)) /12
Time = 0.50 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.21, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {6071, 963, 27, 959, 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.
| \(\displaystyle \int x^2 \tanh ^2(a+2 \log (x)) \, dx\) |
| \(\Big \downarrow \) 6071 |
| \(\displaystyle \int \frac {x^2 \left (e^{2 a} x^4-1\right )^2}{\left (e^{2 a} x^4+1\right )^2}dx\) |
| \(\Big \downarrow \) 963 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-\frac {1}{4} e^{-4 a} \int \frac {4 x^2 \left (2 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \int \frac {x^2 \left (2 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \int \frac {x^2}{e^{2 a} x^4+1}dx-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \int \frac {e^a x^2+1}{e^{2 a} x^4+1}dx-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {1}{2} e^{-a} \int \frac {1}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx+\frac {1}{2} e^{-a} \int \frac {1}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {1}{-\left (1-\sqrt {2} e^{a/2} x\right )^2-1}d\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \int \frac {1}{-\left (\sqrt {2} e^{a/2} x+1\right )^2-1}d\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (-\frac {e^{-a/2} \int -\frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}-\frac {e^{-a/2} \int -\frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {e^{-a/2} \int \frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {1}{2} e^{-a/2} \int \frac {\sqrt {2} x+e^{-a/2}}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {1}{3} e^{4 a} x^3\right )\) |
x^3/(1 + E^(2*a)*x^4) - (-1/3*(E^(4*a)*x^3) + 3*E^(4*a)*((-(ArcTan[1 - Sqr t[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)))/E^(4*a)
3.2.54.3.1 Defintions of rubi rules used
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[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1) /(a*b^2*e*n*(p + 1))), x] + Simp[1/(a*b^2*n*(p + 1)) Int[(e*x)^m*(a + b*x ^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]
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 = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.31
| method | result | size |
| risch | \(\frac {x^{3}}{3}+\frac {x^{3}}{1+{\mathrm e}^{2 a} x^{4}}-\frac {3 \,{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{4}\) | \(53\) |
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {4 \, x^{7} e^{\left (2 \, a\right )} + 16 \, x^{3} - 9 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) - 9 \, {\left (-i \, x^{4} e^{\left (2 \, a\right )} - i\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) - 9 \, {\left (i \, x^{4} e^{\left (2 \, a\right )} + i\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) + 9 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right )}{12 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )}} \]
1/12*(4*x^7*e^(2*a) + 16*x^3 - 9*(x^4*e^(2*a) + 1)*(-e^(-6*a))^(1/4)*log(( -e^(-6*a))^(3/4)*e^(4*a) + x) - 9*(-I*x^4*e^(2*a) - I)*(-e^(-6*a))^(1/4)*l og(I*(-e^(-6*a))^(3/4)*e^(4*a) + x) - 9*(I*x^4*e^(2*a) + I)*(-e^(-6*a))^(1 /4)*log(-I*(-e^(-6*a))^(3/4)*e^(4*a) + x) + 9*(x^4*e^(2*a) + 1)*(-e^(-6*a) )^(1/4)*log(-(-e^(-6*a))^(3/4)*e^(4*a) + x))/(x^4*e^(2*a) + 1)
\[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\int x^{2} \tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]
1/3*x^3 - 3/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a + sqrt(2)*e^(1/2*a))*e^( -1/2*a))*e^(-3/2*a) - 3/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a - sqrt(2)*e^ (1/2*a))*e^(-1/2*a))*e^(-3/2*a) + 3/8*sqrt(2)*e^(-3/2*a)*log(x^2*e^a + sqr t(2)*x*e^(1/2*a) + 1) - 3/8*sqrt(2)*e^(-3/2*a)*log(x^2*e^a - sqrt(2)*x*e^( 1/2*a) + 1) + x^3/(x^4*e^(2*a) + 1)
Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.80 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]
1/3*x^3 - 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^3/(x^4*e^(2*a) + 1)
Time = 1.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.39 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{{\mathrm {e}}^{2\,a}\,x^4+1}+\frac {3\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}} \]