\(\int \frac {\tanh (a+2 \log (x))}{x^3} \, dx\) [152]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\frac {1}{2 x^2}+e^a \arctan \left (e^a x^2\right ) \]

[Out]

1/2/x^2+exp(a)*arctan(exp(a)*x^2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5656, 464, 281, 209} \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=e^a \arctan \left (e^a x^2\right )+\frac {1}{2 x^2} \]

[In]

Int[Tanh[a + 2*Log[x]]/x^3,x]

[Out]

1/(2*x^2) + E^a*ArcTan[E^a*x^2]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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 {-1+e^{2 a} x^4}{x^3 \left (1+e^{2 a} x^4\right )} \, dx \\ & = \frac {1}{2 x^2}+\left (2 e^{2 a}\right ) \int \frac {x}{1+e^{2 a} x^4} \, dx \\ & = \frac {1}{2 x^2}+e^{2 a} \text {Subst}\left (\int \frac {1}{1+e^{2 a} x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2 x^2}+e^a \arctan \left (e^a x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\frac {1}{2 x^2}-\arctan \left (\frac {\cosh (a)-\sinh (a)}{x^2}\right ) \cosh (a)-\arctan \left (\frac {\cosh (a)-\sinh (a)}{x^2}\right ) \sinh (a) \]

[In]

Integrate[Tanh[a + 2*Log[x]]/x^3,x]

[Out]

1/(2*x^2) - ArcTan[(Cosh[a] - Sinh[a])/x^2]*Cosh[a] - ArcTan[(Cosh[a] - Sinh[a])/x^2]*Sinh[a]

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.20

method	result	size
risch	\(\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{2 a}+\textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (4 \,{\mathrm e}^{2 a}+5 \textit {\_R}^{2}\right ) x^{2}-\textit {\_R} \right )\right )}{2}\)	\(44\)

[In]

int(tanh(a+2*ln(x))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\frac {2 \, x^{2} \arctan \left (x^{2} e^{a}\right ) e^{a} + 1}{2 \, x^{2}} \]

[In]

integrate(tanh(a+2*log(x))/x^3,x, algorithm="fricas")

[Out]

1/2*(2*x^2*arctan(x^2*e^a)*e^a + 1)/x^2

Sympy [F]

\[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\int \frac {\tanh {\left (a + 2 \log {\left (x \right )} \right )}}{x^{3}}\, dx \]

[In]

integrate(tanh(a+2*ln(x))/x**3,x)

[Out]

Integral(tanh(a + 2*log(x))/x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=-\arctan \left (\frac {e^{\left (-a\right )}}{x^{2}}\right ) e^{a} + \frac {1}{2 \, x^{2}} \]

[In]

integrate(tanh(a+2*log(x))/x^3,x, algorithm="maxima")

[Out]

-arctan(e^(-a)/x^2)*e^a + 1/2/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\arctan \left (x^{2} e^{a}\right ) e^{a} + \frac {1}{2 \, x^{2}} \]

[In]

integrate(tanh(a+2*log(x))/x^3,x, algorithm="giac")

[Out]

arctan(x^2*e^a)*e^a + 1/2/x^2

Mupad [B] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {\tanh (a+2 \log (x))}{x^3} \, dx=\mathrm {atan}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )\,\sqrt {{\mathrm {e}}^{2\,a}}+\frac {1}{2\,x^2} \]

[In]

int(tanh(a + 2*log(x))/x^3,x)

[Out]

atan(x^2*exp(2*a)^(1/2))*exp(2*a)^(1/2) + 1/(2*x^2)