Integrand size = 15, antiderivative size = 66 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\frac {1}{2} \left (c^4+\frac {1}{x^4}\right ) x \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {1}{2} c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{-1}\left (c^2 x^2\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x)) \] Output:
1/2*(c^4+1/x^4)*x*sech(2*ln(c*x))^(3/2)-1/2*c^6*(1+1/c^4/x^4)^(3/2)*x^3*ar ccsch(c^2*x^2)*sech(2*ln(c*x))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.77 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\frac {\sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+c^4 x^4\right )}{x} \] Input:
Integrate[Sech[2*Log[c*x]]^(3/2)/x^4,x]
Output:
(Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*Hypergeometric2F1[-1/2, 1, 1/2, 1 + c^4*x^4])/x
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6085, 6083, 858, 807, 252, 222}
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.
\(\displaystyle \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle c^3 \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{c^4 x^4}d(c x)\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \int \frac {1}{c^7 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^7}d(c x)\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \int \frac {c^5 x^5}{\left (c^4 x^4+1\right )^{3/2}}d\frac {1}{c x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {1}{2} c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \int \frac {c^2 x^2}{\left (c^2 x^2+1\right )^{3/2}}d\left (c^2 x^2\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {1}{2} c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x)) \left (\int \frac {1}{\sqrt {c^2 x^2+1}}d\left (c^2 x^2\right )-\frac {c^2 x^2}{\sqrt {c^2 x^2+1}}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {1}{2} c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \left (\text {arcsinh}\left (c^2 x^2\right )-\frac {c^2 x^2}{\sqrt {c^2 x^2+1}}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))\) |
Input:
Int[Sech[2*Log[c*x]]^(3/2)/x^4,x]
Output:
-1/2*(c^6*(1 + 1/(c^4*x^4))^(3/2)*x^3*(-((c^2*x^2)/Sqrt[1 + c^2*x^2]) + Ar cSinh[c^2*x^2])*Sech[2*Log[c*x]]^(3/2))
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
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])
\[\int \frac {\operatorname {sech}\left (2 \ln \left (x c \right )\right )^{\frac {3}{2}}}{x^{4}}d x\]
Input:
int(sech(2*ln(x*c))^(3/2)/x^4,x)
Output:
int(sech(2*ln(x*c))^(3/2)/x^4,x)
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41 \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\frac {\sqrt {2} c^{3} x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} c^{2}}{2 \, x} \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="fricas")
Output:
1/2*(sqrt(2)*c^3*x*log((c^5*x^5 + 2*c*x - 2*(c^4*x^4 + 1)*sqrt(c^2*x^2/(c^ 4*x^4 + 1)))/(c*x^5)) + 2*sqrt(2)*sqrt(c^2*x^2/(c^4*x^4 + 1))*c^2)/x
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\int \frac {\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}{x^{4}}\, dx \] Input:
integrate(sech(2*ln(c*x))**(3/2)/x**4,x)
Output:
Integral(sech(2*log(c*x))**(3/2)/x**4, x)
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\int { \frac {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}{x^{4}} \,d x } \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="maxima")
Output:
integrate(sech(2*log(c*x))^(3/2)/x^4, x)
Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\text {Timed out} \] Input:
integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\int \frac {{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}}{x^4} \,d x \] Input:
int((1/cosh(2*log(c*x)))^(3/2)/x^4,x)
Output:
int((1/cosh(2*log(c*x)))^(3/2)/x^4, x)
\[ \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, \mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}{x^{4}}d x \] Input:
int(sech(2*log(c*x))^(3/2)/x^4,x)
Output:
int((sqrt(sech(2*log(c*x)))*sech(2*log(c*x)))/x**4,x)