\(\int x^2 \tanh ^2(a+2 \log (x)) \, dx\) [154]

Optimal result

Integrand size = 13, antiderivative size = 173 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}+\frac {3 e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}}+\frac {3 e^{-3 a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{4 \sqrt {2}} \]

[Out]

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+exp(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+e
xp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {5656, 474, 470, 303, 1176, 631, 210, 1179, 642} \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {3 e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {3 e^{-3 a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}+\frac {3 e^{-3 a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{4 \sqrt {2}}+\frac {x^3}{e^{2 a} x^4+1}+\frac {x^3}{3} \]

[In]

Int[x^2*Tanh[a + 2*Log[x]]^2,x]

[Out]

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

Rule 210

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 470

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] - Dist[(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 474

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] + Dist[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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 5656

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 \left (-1+e^{2 a} x^4\right )^2}{\left (1+e^{2 a} x^4\right )^2} \, dx \\ & = \frac {x^3}{1+e^{2 a} x^4}-\frac {1}{4} e^{-4 a} \int \frac {x^2 \left (8 e^{4 a}-4 e^{6 a} x^4\right )}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}-3 \int \frac {x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}+\frac {1}{2} \left (3 e^{-a}\right ) \int \frac {1-e^a x^2}{1+e^{2 a} x^4} \, dx-\frac {1}{2} \left (3 e^{-a}\right ) \int \frac {1+e^a x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}-\frac {1}{4} \left (3 e^{-2 a}\right ) \int \frac {1}{e^{-a}-\sqrt {2} e^{-a/2} x+x^2} \, dx-\frac {1}{4} \left (3 e^{-2 a}\right ) \int \frac {1}{e^{-a}+\sqrt {2} e^{-a/2} x+x^2} \, dx-\frac {\left (3 e^{-3 a/2}\right ) \int \frac {\sqrt {2} e^{-a/2}+2 x}{-e^{-a}-\sqrt {2} e^{-a/2} x-x^2} \, dx}{4 \sqrt {2}}-\frac {\left (3 e^{-3 a/2}\right ) \int \frac {\sqrt {2} e^{-a/2}-2 x}{-e^{-a}+\sqrt {2} e^{-a/2} x-x^2} \, dx}{4 \sqrt {2}} \\ & = \frac {x^3}{3}+\frac {x^3}{1+e^{2 a} x^4}-\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}}-\frac {\left (3 e^{-3 a/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}}+\frac {\left (3 e^{-3 a/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^{a/2} x\right )}{2 \sqrt {2}} \\ & = \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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.01 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{12} \left (4 x^3+\frac {12 x^3}{1+e^{2 a} x^4}+9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}-e^{-a} x\right )+9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}-e^{-a} x\right )-9 (-1)^{3/4} e^{-3 a/2} \log \left (\sqrt [4]{-1} e^{-3 a/2}+e^{-a} x\right )-9 \sqrt [4]{-1} e^{-3 a/2} \log \left ((-1)^{3/4} e^{-3 a/2}+e^{-a} x\right )\right ) \]

[In]

Integrate[x^2*Tanh[a + 2*Log[x]]^2,x]

[Out]

(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

Maple [C] (verified)

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

[In]

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

[Out]

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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {4 \, x^{7} e^{\left (2 \, a\right )} + 16 \, x^{3} - 9 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) - 9 \, {\left (-i \, x^{4} e^{\left (2 \, a\right )} - i\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) - 9 \, {\left (i \, x^{4} e^{\left (2 \, a\right )} + i\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) + 9 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right )}{12 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )}} \]

[In]

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

[Out]

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)*log(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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]

[In]

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

[Out]

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)*ar
ctan(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 + s
qrt(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)

none

Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.80 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {3}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {3}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {3}{8} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {x^{3}}{x^{4} e^{\left (2 \, a\right )} + 1} \]

[In]

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

[Out]

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 = 1.76 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.39 \[ \int x^2 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^3}{{\mathrm {e}}^{2\,a}\,x^4+1}+\frac {3\,\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3}+\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}} \]

[In]

int(x^2*tanh(a + 2*log(x))^2,x)

[Out]

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)*3
i)/(2*(-exp(2*a))^(3/4)) + x^3/3