3.181 \(\int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx\)

Optimal. Leaf size=66 \[ \frac {1}{2} x \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {1}{2} c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {csch}^{-1}\left (c^2 x^2\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x)) \]

[Out]

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*arccsch(c^2*x^2)*sech(2*ln(c*x))^(3/2)

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Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5551, 5549, 335, 275, 288, 215} \[ \frac {1}{2} x \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {1}{2} c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {csch}^{-1}\left (c^2 x^2\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[2*Log[c*x]]^(3/2)/x^4,x]

[Out]

((c^4 + x^(-4))*x*Sech[2*Log[c*x]]^(3/2))/2 - (c^6*(1 + 1/(c^4*x^4))^(3/2)*x^3*ArcCsch[c^2*x^2]*Sech[2*Log[c*x
]]^(3/2))/2

Rule 215

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]

Rule 275

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 288

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

Rule 335

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] && IntegerQ[m]

Rule 5549

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/(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 5551

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*} \int \frac {\text {sech}^{\frac {3}{2}}(2 \log (c x))}{x^4} \, dx &=c^3 \operatorname {Subst}\left (\int \frac {\text {sech}^{\frac {3}{2}}(2 \log (x))}{x^4} \, dx,x,c x\right )\\ &=\left (c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {1}{x^4}\right )^{3/2} x^7} \, dx,x,c x\right )\\ &=-\left (\left (c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \operatorname {Subst}\left (\int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx,x,\frac {1}{c x}\right )\right )\\ &=-\left (\frac {1}{2} \left (c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^{3/2}} \, dx,x,\frac {1}{c^2 x^2}\right )\right )\\ &=\frac {1}{2} \left (c^4+\frac {1}{x^4}\right ) x \text {sech}^{\frac {3}{2}}(2 \log (c x))-\frac {1}{2} \left (c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\frac {1}{c^2 x^2}\right )\\ &=\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))\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 51, normalized size = 0.77 \[ \frac {\sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{c^4 x^4+1}} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};c^4 x^4+1\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[2*Log[c*x]]^(3/2)/x^4,x]

[Out]

(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

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fricas [A]  time = 0.44, size = 93, normalized size = 1.41 \[ \frac {\sqrt {2} c^{3} x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} c^{2}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="fricas")

[Out]

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)*sq
rt(c^2*x^2/(c^4*x^4 + 1))*c^2)/x

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {sech}\left (2 \ln \left (c x \right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(2*ln(c*x))^(3/2)/x^4,x)

[Out]

int(sech(2*ln(c*x))^(3/2)/x^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2*log(c*x))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sech(2*log(c*x))^(3/2)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cosh(2*log(c*x)))^(3/2)/x^4,x)

[Out]

int((1/cosh(2*log(c*x)))^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2*ln(c*x))**(3/2)/x**4,x)

[Out]

Integral(sech(2*log(c*x))**(3/2)/x**4, x)

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