\(\int x^4 \tanh ^2(a+2 \log (x)) \, dx\) [165]

Optimal result

Integrand size = 13, antiderivative size = 146 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=-4 e^{-2 a} x+\frac {x^5}{5}-\frac {e^{-2 a} x}{1+e^{2 a} x^4}-\frac {5 e^{-5 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}+\frac {5 e^{-5 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}+\frac {5 e^{-5 a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{2 \sqrt {2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.31 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=-4 e^{-2 a} x+\frac {x^5}{5}-\frac {e^{-2 a} x}{1+e^{2 a} x^4}-\frac {5}{4} \sqrt [4]{-1} e^{-5 a/2} \log \left (\sqrt [4]{-1} e^{-5 a/2}-e^{-2 a} x\right )-\frac {5}{4} (-1)^{3/4} e^{-5 a/2} \log \left ((-1)^{3/4} e^{-5 a/2}-e^{-2 a} x\right )+\frac {5}{4} \sqrt [4]{-1} e^{-5 a/2} \log \left (\sqrt [4]{-1} e^{-5 a/2}+e^{-2 a} x\right )+\frac {5}{4} (-1)^{3/4} e^{-5 a/2} \log \left ((-1)^{3/4} e^{-5 a/2}+e^{-2 a} x\right ) \] Input:

Integrate[x^4*Tanh[a + 2*Log[x]]^2,x]
 

Output:

(-4*x)/E^(2*a) + x^5/5 - x/(E^(2*a)*(1 + E^(2*a)*x^4)) - (5*(-1)^(1/4)*Log 
[(-1)^(1/4)/E^((5*a)/2) - x/E^(2*a)])/(4*E^((5*a)/2)) - (5*(-1)^(3/4)*Log[ 
(-1)^(3/4)/E^((5*a)/2) - x/E^(2*a)])/(4*E^((5*a)/2)) + (5*(-1)^(1/4)*Log[( 
-1)^(1/4)/E^((5*a)/2) + x/E^(2*a)])/(4*E^((5*a)/2)) + (5*(-1)^(3/4)*Log[(- 
1)^(3/4)/E^((5*a)/2) + x/E^(2*a)])/(4*E^((5*a)/2))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.47, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6071, 963, 27, 959, 843, 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.

\(\displaystyle \int x^4 \tanh ^2(a+2 \log (x)) \, dx\)

\(\Big \downarrow \) 6071

\(\displaystyle \int \frac {x^4 \left (e^{2 a} x^4-1\right )^2}{\left (e^{2 a} x^4+1\right )^2}dx\)

\(\Big \downarrow \) 963

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-\frac {1}{4} e^{-4 a} \int \frac {4 x^4 \left (4 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \int \frac {x^4 \left (4 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \int \frac {x^4}{e^{2 a} x^4+1}dx-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \int \frac {1}{e^{2 a} x^4+1}dx\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \int \frac {e^a x^2+1}{e^{2 a} x^4+1}dx\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \left (\frac {1}{2} e^{-a} \int \frac {1}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx+\frac {1}{2} e^{-a} \int \frac {1}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {1}{-\left (1-\sqrt {2} e^{a/2} x\right )^2-1}d\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \int \frac {1}{-\left (\sqrt {2} e^{a/2} x+1\right )^2-1}d\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \int \frac {1-e^a x^2}{e^{2 a} x^4+1}dx+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \left (-\frac {e^{-a/2} \int -\frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}-\frac {e^{-a/2} \int -\frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {e^{-a/2} \int \frac {\sqrt {2} \left (\sqrt {2} x+e^{-a/2}\right )}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{x^2-\sqrt {2} e^{-a/2} x+e^{-a}}dx}{2 \sqrt {2}}+\frac {1}{2} e^{-a/2} \int \frac {\sqrt {2} x+e^{-a/2}}{x^2+\sqrt {2} e^{-a/2} x+e^{-a}}dx\right )+\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {x^5}{e^{2 a} x^4+1}-e^{-4 a} \left (5 e^{4 a} \left (e^{-2 a} x-e^{-2 a} \left (\frac {1}{2} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )\right )-\frac {1}{5} e^{4 a} x^5\right )\)

Input:

Int[x^4*Tanh[a + 2*Log[x]]^2,x]
 

Output:

x^5/(1 + E^(2*a)*x^4) - (-1/5*(E^(4*a)*x^5) + 5*E^(4*a)*(x/E^(2*a) - ((-(A 
rcTan[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 + (-1/2*Log[1 - Sqrt[2]*E^(a/2)*x + E^a*x^2]/(S 
qrt[2]*E^(a/2)) + Log[1 + Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*Sqrt[2]*E^(a/2)) 
)/2)/E^(2*a)))/E^(4*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 755

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]]))
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 959

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]
 

rule 963

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]
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

rule 6071

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]
 

Maple [C] (verified)

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

Time = 0.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {{\mathrm e}^{2 a} {\mathrm e}^{-2 a} x^{5}}{5}-4 \,{\mathrm e}^{-2 a} x -\frac {x \,{\mathrm e}^{-2 a}}{1+{\mathrm e}^{2 a} x^{4}}+\frac {5 \,{\mathrm e}^{-4 a} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4}\)	\(71\)

Input:

int(x^4*tanh(a+2*ln(x))^2,x,method=_RETURNVERBOSE)
 

Output:

1/5*exp(2*a)*exp(-2*a)*x^5-4*exp(-2*a)*x-x*exp(-2*a)/(1+exp(2*a)*x^4)+5/4* 
exp(-4*a)*sum(1/_R^3*ln(x-_R),_R=RootOf(exp(2*a)*_Z^4+1))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.21 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\frac {8 \, x^{9} e^{\left (4 \, a\right )} - 152 \, x^{5} e^{\left (2 \, a\right )} + 50 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {1}{2} \, a\right )} + 50 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {1}{2} \, a\right )} + 25 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - 25 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - 200 \, x}{40 \, {\left (x^{4} e^{\left (4 \, a\right )} + e^{\left (2 \, a\right )}\right )}} \] Input:

integrate(x^4*tanh(a+2*log(x))^2,x, algorithm="fricas")
 

Output:

1/40*(8*x^9*e^(4*a) - 152*x^5*e^(2*a) + 50*sqrt(2)*(x^4*e^(2*a) + 1)*arcta 
n(sqrt(2)*x*e^(1/2*a) + 1)*e^(-1/2*a) + 50*sqrt(2)*(x^4*e^(2*a) + 1)*arcta 
n(sqrt(2)*x*e^(1/2*a) - 1)*e^(-1/2*a) + 25*sqrt(2)*(x^4*e^(2*a) + 1)*e^(-1 
/2*a)*log(x^2*e^a + sqrt(2)*x*e^(1/2*a) + 1) - 25*sqrt(2)*(x^4*e^(2*a) + 1 
)*e^(-1/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1) - 200*x)/(x^4*e^(4*a) 
+ e^(2*a))
 

Sympy [F]

\[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\int x^{4} \tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \] Input:

integrate(x**4*tanh(a+2*ln(x))**2,x)
 

Output:

Integral(x**4*tanh(a + 2*log(x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.13 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{5} \, {\left (x^{5} e^{\left (2 \, a\right )} - 20 \, x\right )} e^{\left (-2 \, a\right )} + \frac {5}{8} \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} + 2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} + \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right )\right )} e^{\left (-2 \, a\right )} - \frac {x}{x^{4} e^{\left (4 \, a\right )} + e^{\left (2 \, a\right )}} \] Input:

integrate(x^4*tanh(a+2*log(x))^2,x, algorithm="maxima")
 

Output:

1/5*(x^5*e^(2*a) - 20*x)*e^(-2*a) + 5/8*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x 
*e^a + sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) + 2*sqrt(2)*arctan(1/2*sq 
rt(2)*(2*x*e^a - sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) + sqrt(2)*e^(-1 
/2*a)*log(x^2*e^a + sqrt(2)*x*e^(1/2*a) + 1) - sqrt(2)*e^(-1/2*a)*log(x^2* 
e^a - sqrt(2)*x*e^(1/2*a) + 1))*e^(-2*a) - x/(x^4*e^(4*a) + e^(2*a))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.09 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\frac {5}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {5}{2} \, a\right )} + \frac {5}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {5}{2} \, a\right )} + \frac {5}{8} \, \sqrt {2} e^{\left (-\frac {5}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {5}{8} \, \sqrt {2} e^{\left (-\frac {5}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{5} \, {\left (x^{5} e^{\left (20 \, a\right )} - 20 \, x e^{\left (18 \, a\right )}\right )} e^{\left (-20 \, a\right )} - \frac {x e^{\left (-2 \, a\right )}}{x^{4} e^{\left (2 \, a\right )} + 1} \] Input:

integrate(x^4*tanh(a+2*log(x))^2,x, algorithm="giac")
 

Output:

5/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(-5 
/2*a) + 5/4*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2* 
a))*e^(-5/2*a) + 5/8*sqrt(2)*e^(-5/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e 
^(-a)) - 5/8*sqrt(2)*e^(-5/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) 
+ 1/5*(x^5*e^(20*a) - 20*x*e^(18*a))*e^(-20*a) - x*e^(-2*a)/(x^4*e^(2*a) + 
 1)
 

Mupad [B] (verification not implemented)

Time = 2.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.52 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^5}{5}-\frac {5\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{5/4}}-\frac {x}{{\mathrm {e}}^{4\,a}\,x^4+{\mathrm {e}}^{2\,a}}-4\,x\,{\mathrm {e}}^{-2\,a}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{5/4}} \] Input:

int(x^4*tanh(a + 2*log(x))^2,x)
 

Output:

(atan(x*(-exp(2*a))^(1/4)*1i)*5i)/(2*(-exp(2*a))^(5/4)) - (5*atan(x*(-exp( 
2*a))^(1/4)))/(2*(-exp(2*a))^(5/4)) - x/(exp(2*a) + x^4*exp(4*a)) + x^5/5 
- 4*x*exp(-2*a)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.21 \[ \int x^4 \tanh ^2(a+2 \log (x)) \, dx=\frac {-50 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}-50 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )+50 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}+50 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )-25 e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{4}+25 e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{4}-25 e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )+25 e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )+8 e^{5 a} x^{9}-152 e^{3 a} x^{5}-200 e^{a} x}{40 e^{3 a} \left (e^{2 a} x^{4}+1\right )} \] Input:

int(x^4*tanh(a+2*log(x))^2,x)
 

Output:

( - 50*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*s 
qrt(2)))*x**4 - 50*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e* 
*(a/2)*sqrt(2))) + 50*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e**a 
*x)/(e**(a/2)*sqrt(2)))*x**4 + 50*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) 
+ 2*e**a*x)/(e**(a/2)*sqrt(2))) - 25*e**((5*a)/2)*sqrt(2)*log( - e**(a/2)* 
sqrt(2)*x + e**a*x**2 + 1)*x**4 + 25*e**((5*a)/2)*sqrt(2)*log(e**(a/2)*sqr 
t(2)*x + e**a*x**2 + 1)*x**4 - 25*e**(a/2)*sqrt(2)*log( - e**(a/2)*sqrt(2) 
*x + e**a*x**2 + 1) + 25*e**(a/2)*sqrt(2)*log(e**(a/2)*sqrt(2)*x + e**a*x* 
*2 + 1) + 8*e**(5*a)*x**9 - 152*e**(3*a)*x**5 - 200*e**a*x)/(40*e**(3*a)*( 
e**(2*a)*x**4 + 1))