\(\int x^2 \tanh ^2(a+2 \log (x)) \, dx\) [166]

Optimal result

Integrand size = 13, antiderivative size = 134 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}+\frac {3 e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}+\frac {3 e^{-3 a/2} \text {arctanh}\left (\frac {\sqrt {2} e^{a/2} x}{1+e^a x^2}\right )}{2 \sqrt {2}} \] Output:

1/3*x^3+x^3/(1+exp(2*a)*x^4)-3/4*arctan(-1+2^(1/2)*exp(1/2*a)*x)*2^(1/2)/e 
xp(3/2*a)-3/4*arctan(1+2^(1/2)*exp(1/2*a)*x)*2^(1/2)/exp(3/2*a)+3/4*arctan 
h(2^(1/2)*exp(1/2*a)*x/(1+exp(a)*x^2))*2^(1/2)/exp(3/2*a)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{12} \left (4 x^3+\frac {12 x^3}{1+e^{2 a} x^4}+9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}-e^{-a} x\right )+9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}-e^{-a} x\right )-9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}+e^{-a} x\right )-9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}+e^{-a} x\right )\right ) \] Input:

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

Output:

(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
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.56, 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

\(\displaystyle \frac {x^3}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} e^{-a} \left (\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {1}{3} e^{4 a} x^3\right )\)

Input:

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

Output:

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)
 

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 826

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]]))
 

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 = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.40

method	result	size
risch	\(\frac {x^{3}}{3}+\frac {x^{3}}{1+{\mathrm e}^{2 a} x^{4}}-\frac {3 \,{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{4}\)	\(53\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.26 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {8 \, x^{7} e^{\left (3 \, a\right )} + 32 \, x^{3} e^{a} - 18 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {1}{2} \, a\right )} - 18 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {1}{2} \, a\right )} + 9 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - 9 \, \sqrt {2} {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right )}{24 \, {\left (x^{4} e^{\left (3 \, a\right )} + e^{a}\right )}} \] Input:

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

Output:

1/24*(8*x^7*e^(3*a) + 32*x^3*e^a - 18*sqrt(2)*(x^4*e^(2*a) + 1)*arctan(sqr 
t(2)*x*e^(1/2*a) + 1)*e^(-1/2*a) - 18*sqrt(2)*(x^4*e^(2*a) + 1)*arctan(sqr 
t(2)*x*e^(1/2*a) - 1)*e^(-1/2*a) + 9*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) - 9*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))/(x^4*e^(3*a) + e^a)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \] Input:

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

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \] Input:

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

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.50 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{{\mathrm {e}}^{2\,a}\,x^4+1}+\frac {3\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}} \] Input:

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

Output:

x^3/(x^4*exp(2*a) + 1) + (3*atan(x*(-exp(2*a))^(1/4)))/(2*(-exp(2*a))^(3/4 
)) + (atan(x*(-exp(2*a))^(1/4)*1i)*3i)/(2*(-exp(2*a))^(3/4)) + x^3/3
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.37 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {18 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}+18 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )-18 e^{\frac {5 a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}-18 e^{\frac {a}{2}} \sqrt {2}\, \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )-9 e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{4}+9 e^{\frac {5 a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{4}-9 e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )+9 e^{\frac {a}{2}} \sqrt {2}\, \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right )+8 e^{4 a} x^{7}+32 e^{2 a} x^{3}}{24 e^{2 a} \left (e^{2 a} x^{4}+1\right )} \] Input:

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

Output:

(18*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt 
(2)))*x**4 + 18*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a 
/2)*sqrt(2))) - 18*e**((5*a)/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2*e**a*x) 
/(e**(a/2)*sqrt(2)))*x**4 - 18*e**(a/2)*sqrt(2)*atan((e**(a/2)*sqrt(2) + 2 
*e**a*x)/(e**(a/2)*sqrt(2))) - 9*e**((5*a)/2)*sqrt(2)*log( - e**(a/2)*sqrt 
(2)*x + e**a*x**2 + 1)*x**4 + 9*e**((5*a)/2)*sqrt(2)*log(e**(a/2)*sqrt(2)* 
x + e**a*x**2 + 1)*x**4 - 9*e**(a/2)*sqrt(2)*log( - e**(a/2)*sqrt(2)*x + e 
**a*x**2 + 1) + 9*e**(a/2)*sqrt(2)*log(e**(a/2)*sqrt(2)*x + e**a*x**2 + 1) 
 + 8*e**(4*a)*x**7 + 32*e**(2*a)*x**3)/(24*e**(2*a)*(e**(2*a)*x**4 + 1))