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

Optimal. Leaf size=88 \[ \frac{1}{2 x \left (c^4+\frac{1}{x^4}\right ) \text{sech}^{\frac{3}{2}}(2 \log (c x))}-\frac{\text{csch}^{-1}\left (c^2 x^2\right )}{2 c^6 x^3 \left (\frac{1}{c^4 x^4}+1\right )^{3/2} \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{x^3}{6 \text{sech}^{\frac{3}{2}}(2 \log (c x))} \]

[Out]

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

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Rubi [A]  time = 0.0670216, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5551, 5549, 335, 275, 277, 215} \[ \frac{1}{2 x \left (c^4+\frac{1}{x^4}\right ) \text{sech}^{\frac{3}{2}}(2 \log (c x))}-\frac{\text{csch}^{-1}\left (c^2 x^2\right )}{2 c^6 x^3 \left (\frac{1}{c^4 x^4}+1\right )^{3/2} \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{x^3}{6 \text{sech}^{\frac{3}{2}}(2 \log (c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.168229, size = 88, normalized size = 1. \[ \frac{x \left (\sqrt{c^4 x^4+1} \left (c^4 x^4+4\right )-3 \tanh ^{-1}\left (\sqrt{c^4 x^4+1}\right )\right )}{12 \sqrt{2} c^2 \sqrt{\frac{c^2 x^2}{c^4 x^4+1}} \sqrt{c^4 x^4+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[1 + c^4*x^4]*(4 + c^4*x^4) - 3*ArcTanh[Sqrt[1 + c^4*x^4]]))/(12*Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(1 + c^4*x
^4)]*Sqrt[1 + c^4*x^4])

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \left ({\rm sech} \left (2\,\ln \left ( cx \right ) \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{sech}\left (2 \, \log \left (c x\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.1002, size = 234, normalized size = 2.66 \begin{align*} \frac{3 \, \sqrt{2} c 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}{\left (c^{8} x^{8} + 5 \, c^{4} x^{4} + 4\right )} \sqrt{\frac{c^{2} x^{2}}{c^{4} x^{4} + 1}}}{48 \, c^{4} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*sqrt(2)*c*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)*(
c^8*x^8 + 5*c^4*x^4 + 4)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/(c^4*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{sech}^{\frac{3}{2}}{\left (2 \log{\left (c x \right )} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{sech}\left (2 \, \log \left (c x\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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