\(\int x^2 \tanh (a+2 \log (x)) \, dx\) [147]

Optimal result

Integrand size = 11, antiderivative size = 151 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {x^3}{3}+\frac {e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}} \]

[Out]

1/3*x^3-1/2*arctan(-1+exp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)-1/2*arctan(1+exp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2
^(1/2)-1/4*ln(1+exp(a)*x^2-exp(1/2*a)*x*2^(1/2))/exp(3/2*a)*2^(1/2)+1/4*ln(1+exp(a)*x^2+exp(1/2*a)*x*2^(1/2))/
exp(3/2*a)*2^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {5656, 470, 303, 1176, 631, 210, 1179, 642} \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \arctan \left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+\frac {x^3}{3} \]

[In]

Int[x^2*Tanh[a + 2*Log[x]],x]

[Out]

x^3/3 + ArcTan[1 - Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^((3*a)/2)) - ArcTan[1 + Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^((3*a)/
2)) - Log[1 - Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*Sqrt[2]*E^((3*a)/2)) + Log[1 + Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*S
qrt[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 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 )}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}-2 \int \frac {x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}+e^{-a} \int \frac {1-e^a x^2}{1+e^{2 a} x^4} \, dx-e^{-a} \int \frac {1+e^a x^2}{1+e^{2 a} x^4} \, dx \\ & = \frac {x^3}{3}-\frac {1}{2} e^{-2 a} \int \frac {1}{e^{-a}-\sqrt {2} e^{-a/2} x+x^2} \, dx-\frac {1}{2} e^{-2 a} \int \frac {1}{e^{-a}+\sqrt {2} e^{-a/2} x+x^2} \, dx-\frac {e^{-3 a/2} \int \frac {\sqrt {2} e^{-a/2}+2 x}{-e^{-a}-\sqrt {2} e^{-a/2} x-x^2} \, dx}{2 \sqrt {2}}-\frac {e^{-3 a/2} \int \frac {\sqrt {2} e^{-a/2}-2 x}{-e^{-a}+\sqrt {2} e^{-a/2} x-x^2} \, dx}{2 \sqrt {2}} \\ & = \frac {x^3}{3}-\frac {e^{-3 a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}-\frac {e^{-3 a/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{-3 a/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}} \\ & = \frac {x^3}{3}+\frac {e^{-3 a/2} \arctan \left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \arctan \left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-3 a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}+\frac {e^{-3 a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.42 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {1}{6} \left (2 x^3+3 \text {RootSum}\left [\cosh (a)-\sinh (a)+\cosh (a) \text {$\#$1}^4+\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)-\log (x-\text {$\#$1})}{\text {$\#$1}}\&\right ] (\cosh (2 a)-\sinh (2 a))\right ) \]

[In]

Integrate[x^2*Tanh[a + 2*Log[x]],x]

[Out]

(2*x^3 + 3*RootSum[Cosh[a] - Sinh[a] + Cosh[a]*#1^4 + Sinh[a]*#1^4 & , (Log[x] - Log[x - #1])/#1 & ]*(Cosh[2*a
] - Sinh[2*a]))/6

Maple [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.25

method	result	size
risch	\(\frac {x^{3}}{3}-\frac {{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{2}\)	\(37\)

[In]

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

[Out]

1/3*x^3-1/2*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 = 113, normalized size of antiderivative = 0.75 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {1}{2} \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) + \frac {1}{2} i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) - \frac {1}{2} i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-i \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) + \frac {1}{2} \, \left (-e^{\left (-6 \, a\right )}\right )^{\frac {1}{4}} \log \left (-\left (-e^{\left (-6 \, a\right )}\right )^{\frac {3}{4}} e^{\left (4 \, a\right )} + x\right ) \]

[In]

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

[Out]

1/3*x^3 - 1/2*(-e^(-6*a))^(1/4)*log((-e^(-6*a))^(3/4)*e^(4*a) + x) + 1/2*I*(-e^(-6*a))^(1/4)*log(I*(-e^(-6*a))
^(3/4)*e^(4*a) + x) - 1/2*I*(-e^(-6*a))^(1/4)*log(-I*(-e^(-6*a))^(3/4)*e^(4*a) + x) + 1/2*(-e^(-6*a))^(1/4)*lo
g(-(-e^(-6*a))^(3/4)*e^(4*a) + x)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*tanh(a + 2*log(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) \]

[In]

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

[Out]

1/3*x^3 - 1/2*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) - 1/2*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) + 1/4*sqrt(2)*e^(-3/2*a)*log(x^2*e^a + s
qrt(2)*x*e^(1/2*a) + 1) - 1/4*sqrt(2)*e^(-3/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {1}{3} \, x^{3} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {3}{2} \, a\right )} + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {3}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) \]

[In]

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

[Out]

1/3*x^3 - 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(-3/2*a) - 1/2*sqrt(2)*arctan
(-1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2*a))*e^(-3/2*a) + 1/4*sqrt(2)*e^(-3/2*a)*log(sqrt(2)*x*e^(-1/2*
a) + x^2 + e^(-a)) - 1/4*sqrt(2)*e^(-3/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a))

Mupad [B] (verification not implemented)

Time = 1.72 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.31 \[ \int x^2 \tanh (a+2 \log (x)) \, dx=\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}-\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{3/4}}+\frac {x^3}{3} \]

[In]

int(x^2*tanh(a + 2*log(x)),x)

[Out]

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