\(\int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) [161]

Optimal result

Integrand size = 15, antiderivative size = 67 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x^3}{4 \sqrt {\text {sech}(2 \log (c x))}}+\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}} \]

[Out]

1/4*x^3/sech(2*ln(c*x))^(1/2)+1/4*arctanh((1+1/c^4/x^4)^(1/2))/c^4/x/(1+1/c^4/x^4)^(1/2)/sech(2*ln(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5670, 5668, 272, 43, 65, 213} \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {\text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )}{4 c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}+\frac {x^3}{4 \sqrt {\text {sech}(2 \log (c x))}} \]

[In]

Int[x^2/Sqrt[Sech[2*Log[c*x]]],x]

[Out]

x^3/(4*Sqrt[Sech[2*Log[c*x]]]) + ArcTanh[Sqrt[1 + 1/(c^4*x^4)]]/(4*c^4*Sqrt[1 + 1/(c^4*x^4)]*x*Sqrt[Sech[2*Log
[c*x]]])

Rule 43

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 213

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

Rule 272

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

Rule 5668

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[Sech[d*(a + b*Log[x])]^p*(
(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p/x^((-b)*d*p)), Int[(e*x)^m*(1/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p)), x]
, x] /; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]

Rule 5670

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Dist[(e*x)^(m + 1
)/(e*n*(c*x^n)^((m + 1)/n)), Subst[Int[x^((m + 1)/n - 1)*Sech[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[
{a, b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x \left (c^2 x^2 \sqrt {1+c^4 x^4}+\text {arcsinh}\left (c^2 x^2\right )\right )}{4 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \]

[In]

Integrate[x^2/Sqrt[Sech[2*Log[c*x]]],x]

[Out]

(x*(c^2*x^2*Sqrt[1 + c^4*x^4] + ArcSinh[c^2*x^2]))/(4*Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*Sqrt[1 + c^4*x
^4])

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.45

method	result	size
risch	\(\frac {\sqrt {2}\, x^{3}}{8 \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {\ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{8 \sqrt {c^{4}}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}\, \sqrt {c^{4} x^{4}+1}}\)	\(97\)

[In]

int(x^2/sech(2*ln(c*x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*2^(1/2)*x^3/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/8*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4+1)^(1/2))/(c^4)^(1/2)*2^(1/2)*
x/(c^2*x^2/(c^4*x^4+1))^(1/2)/(c^4*x^4+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {2 \, \sqrt {2} {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} + \sqrt {2} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right )}{16 \, c^{3}} \]

[In]

integrate(x^2/sech(2*log(c*x))^(1/2),x, algorithm="fricas")

[Out]

1/16*(2*sqrt(2)*(c^5*x^5 + c*x)*sqrt(c^2*x^2/(c^4*x^4 + 1)) + sqrt(2)*log(-2*c^4*x^4 - 2*(c^5*x^5 + c*x)*sqrt(
c^2*x^2/(c^4*x^4 + 1)) - 1))/c^3

Sympy [F]

\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^{2}}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \]

[In]

integrate(x**2/sech(2*ln(c*x))**(1/2),x)

[Out]

Integral(x**2/sqrt(sech(2*log(c*x))), x)

Maxima [F]

\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{2}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]

[In]

integrate(x^2/sech(2*log(c*x))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(sech(2*log(c*x))), x)

Giac [F]

\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{2}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]

[In]

integrate(x^2/sech(2*log(c*x))^(1/2),x, algorithm="giac")

[Out]

integrate(x^2/sqrt(sech(2*log(c*x))), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^2}{\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}} \,d x \]

[In]

int(x^2/(1/cosh(2*log(c*x)))^(1/2),x)

[Out]

int(x^2/(1/cosh(2*log(c*x)))^(1/2), x)