Integrand size = 15, antiderivative size = 92 \[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 x}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{8 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {3 \text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{16 c^8 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]
3/16*x/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+1/8*x^5/sech(2*ln(c*x))^(3/2)+3/1 6*arctanh((1+1/c^4/x^4)^(1/2))/c^8/(1+1/c^4/x^4)^(3/2)/x^3/sech(2*ln(c*x)) ^(3/2)
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98 \[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {c^3 x^3 \sqrt {1+c^4 x^4} \left (5+2 c^4 x^4\right )+3 c x \text {arcsinh}\left (c^2 x^2\right )}{32 \sqrt {2} c^5 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \]
(c^3*x^3*Sqrt[1 + c^4*x^4]*(5 + 2*c^4*x^4) + 3*c*x*ArcSinh[c^2*x^2])/(32*S qrt[2]*c^5*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*Sqrt[1 + c^4*x^4])
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6085, 6083, 798, 51, 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.
\(\displaystyle \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle \frac {\int \frac {c^4 x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^5}\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle \frac {\int c^7 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^7d(c x)}{c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {\int \frac {\left (1+\frac {1}{c^4 x^4}\right )^{3/2}}{c^3 x^3}d\frac {1}{c^4 x^4}}{4 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {3}{4} \int \frac {\sqrt {1+\frac {1}{c^4 x^4}}}{c^2 x^2}d\frac {1}{c^4 x^4}-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{2 c^2 x^2}}{4 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{c \sqrt {1+\frac {1}{c^4 x^4}} x}d\frac {1}{c^4 x^4}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{2 c^2 x^2}}{4 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {3}{4} \left (\int \frac {1}{c^2 x^2-1}d\sqrt {1+\frac {1}{c^4 x^4}}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{2 c^2 x^2}}{4 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {\frac {3}{4} \left (-\text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{2 c^2 x^2}}{4 c^8 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
-1/4*(-1/2*(1 + 1/(c^4*x^4))^(3/2)/(c^2*x^2) + (3*(-(Sqrt[1 + 1/(c^4*x^4)] /(c*x)) - ArcTanh[Sqrt[1 + 1/(c^4*x^4)]]))/4)/(c^8*(1 + 1/(c^4*x^4))^(3/2) *x^3*Sech[2*Log[c*x]]^(3/2))
3.2.73.3.1 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 = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.23
method | result | size |
risch | \(\frac {x^{3} \left (2 c^{4} x^{4}+5\right ) \sqrt {2}}{64 c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {3 \ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{64 \sqrt {c^{4}}\, c^{2} \sqrt {c^{4} x^{4}+1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\) | \(113\) |
1/64*x^3*(2*c^4*x^4+5)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^(1/2)+3/64*ln(c^4 *x^2/(c^4)^(1/2)+(c^4*x^4+1)^(1/2))/(c^4)^(1/2)*2^(1/2)/c^2*x/(c^4*x^4+1)^ (1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)
Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {2 \, \sqrt {2} {\left (2 \, c^{9} x^{9} + 7 \, c^{5} x^{5} + 5 \, c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} + 3 \, \sqrt {2} \log \left (-2 \, c^{4} x^{4} - 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right )}{128 \, c^{5}} \]
1/128*(2*sqrt(2)*(2*c^9*x^9 + 7*c^5*x^5 + 5*c*x)*sqrt(c^2*x^2/(c^4*x^4 + 1 )) + 3*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^5
\[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{4}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]
\[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{4}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^4}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^4}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]