Integrand size = 15, antiderivative size = 67 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x^3}{4 \sqrt {\text {sech}(2 \log (c x))}}+\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{4 c^4 \sqrt {1+\frac {1}{c^4 x^4}} x \sqrt {\text {sech}(2 \log (c x))}} \]
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)
Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {x \left (c^2 x^2 \sqrt {1+c^4 x^4}+\text {arcsinh}\left (c^2 x^2\right )\right )}{4 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \]
(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])
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6085, 6083, 798, 51, 73, 220}
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.
\(\displaystyle \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle \frac {\int \frac {c^2 x^2}{\sqrt {\text {sech}(2 \log (c x))}}d(c x)}{c^3}\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle \frac {\int c^3 \sqrt {1+\frac {1}{c^4 x^4}} x^3d(c x)}{c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {\int \frac {\sqrt {1+\frac {1}{c^4 x^4}}}{c^2 x^2}d\frac {1}{c^4 x^4}}{4 c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {1}{c \sqrt {1+\frac {1}{c^4 x^4}} x}d\frac {1}{c^4 x^4}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}}{4 c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\int \frac {1}{c^2 x^2-1}d\sqrt {1+\frac {1}{c^4 x^4}}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}}{4 c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {-\text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}}{4 c^4 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
-1/4*(-(Sqrt[1 + 1/(c^4*x^4)]/(c*x)) - ArcTanh[Sqrt[1 + 1/(c^4*x^4)]])/(c^ 4*Sqrt[1 + 1/(c^4*x^4)]*x*Sqrt[Sech[2*Log[c*x]]])
3.2.61.3.1 Defintions of rubi rules used
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[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]
Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p _.), x_Symbol] :> Simp[(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])
Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.45
method | result | size |
risch | \(\frac {\sqrt {2}\, x^{3}}{8 \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {\ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{8 \sqrt {c^{4}}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}\, \sqrt {c^{4} x^{4}+1}}\) | \(97\) |
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)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34 \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {2 \, \sqrt {2} {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} + \sqrt {2} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right )}{16 \, c^{3}} \]
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
\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^{2}}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \]
\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{2}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]
\[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{2}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^2}{\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}} \,d x \]