3.162 \(\int \frac{x}{\sqrt{\text{sech}(2 \log (c x))}} \, dx\)

Optimal. Leaf size=87 \[ \frac{x^2}{3 \sqrt{\text{sech}(2 \log (c x))}}-\frac{\sqrt{\frac{c^4+\frac{1}{x^4}}{\left (c^2+\frac{1}{x^2}\right )^2}} \left (c^2+\frac{1}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}(c x),\frac{1}{2}\right )}{3 c x \left (c^4+\frac{1}{x^4}\right ) \sqrt{\text{sech}(2 \log (c x))}} \]

[Out]

x^2/(3*Sqrt[Sech[2*Log[c*x]]]) - (Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(-2))*EllipticF[2*ArcCot[c*x]
, 1/2])/(3*c*(c^4 + x^(-4))*x*Sqrt[Sech[2*Log[c*x]]])

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Rubi [A]  time = 0.0595054, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5551, 5549, 335, 277, 220} \[ \frac{x^2}{3 \sqrt{\text{sech}(2 \log (c x))}}-\frac{\sqrt{\frac{c^4+\frac{1}{x^4}}{\left (c^2+\frac{1}{x^2}\right )^2}} \left (c^2+\frac{1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac{1}{2}\right )}{3 c x \left (c^4+\frac{1}{x^4}\right ) \sqrt{\text{sech}(2 \log (c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(3*Sqrt[Sech[2*Log[c*x]]]) - (Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(-2))*EllipticF[2*ArcCot[c*x]
, 1/2])/(3*c*(c^4 + x^(-4))*x*Sqrt[Sech[2*Log[c*x]]])

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])

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 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 277

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{x}{\sqrt{\text{sech}(2 \log (c x))}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{\sqrt{\text{sech}(2 \log (x))}} \, dx,x,c x\right )}{c^2}\\ &=\frac{\operatorname{Subst}\left (\int \sqrt{1+\frac{1}{x^4}} x^2 \, dx,x,c x\right )}{c^3 \sqrt{1+\frac{1}{c^4 x^4}} x \sqrt{\text{sech}(2 \log (c x))}}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1+x^4}}{x^4} \, dx,x,\frac{1}{c x}\right )}{c^3 \sqrt{1+\frac{1}{c^4 x^4}} x \sqrt{\text{sech}(2 \log (c x))}}\\ &=\frac{x^2}{3 \sqrt{\text{sech}(2 \log (c x))}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^4}} \, dx,x,\frac{1}{c x}\right )}{3 c^3 \sqrt{1+\frac{1}{c^4 x^4}} x \sqrt{\text{sech}(2 \log (c x))}}\\ &=\frac{x^2}{3 \sqrt{\text{sech}(2 \log (c x))}}-\frac{\sqrt{\frac{c^4+\frac{1}{x^4}}{\left (c^2+\frac{1}{x^2}\right )^2}} \left (c^2+\frac{1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac{1}{2}\right )}{3 c \left (c^4+\frac{1}{x^4}\right ) x \sqrt{\text{sech}(2 \log (c x))}}\\ \end{align*}

Mathematica [C]  time = 0.0964022, size = 58, normalized size = 0.67 \[ \frac{\sqrt{c^4 x^4+1} \sqrt{\frac{c^2 x^2}{2 c^4 x^4+2}} \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-c^4 x^4\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*Hypergeometric2F1[-1/2, 1/4, 5/4, -(c^4*x^4)])/c^2

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Maple [C]  time = 0.035, size = 114, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}\sqrt{2}}{6}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}}{{c}^{4}{x}^{4}+1}}}}}}+{\frac{\sqrt{2}x}{3\,{c}^{4}{x}^{4}+3}\sqrt{1-i{c}^{2}{x}^{2}}\sqrt{1+i{c}^{2}{x}^{2}}{\it EllipticF} \left ( x\sqrt{i{c}^{2}},i \right ){\frac{1}{\sqrt{i{c}^{2}}}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}}{{c}^{4}{x}^{4}+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^2*2^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/3/(I*c^2)^(1/2)*(1-I*c^2*x^2)^(1/2)*(1+I*c^2*x^2)^(1/2)/(c^4*x^4
+1)*EllipticF(x*(I*c^2)^(1/2),I)*2^(1/2)*x/(c^2*x^2/(c^4*x^4+1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\operatorname{sech}\left (2 \, \log \left (c x\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\sqrt{\operatorname{sech}\left (2 \, \log \left (c x\right )\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/sqrt(sech(2*log(c*x))), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\operatorname{sech}{\left (2 \log{\left (c x \right )} \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(sech(2*log(c*x))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\operatorname{sech}\left (2 \, \log \left (c x\right )\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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