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

Optimal. Leaf size=92 \[ \frac{3 x}{16 \left (c^4+\frac{1}{x^4}\right ) \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{3 \tanh ^{-1}\left (\sqrt{\frac{1}{c^4 x^4}+1}\right )}{16 c^8 x^3 \left (\frac{1}{c^4 x^4}+1\right )^{3/2} \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{x^5}{8 \text{sech}^{\frac{3}{2}}(2 \log (c x))} \]

[Out]

(3*x)/(16*(c^4 + x^(-4))*Sech[2*Log[c*x]]^(3/2)) + x^5/(8*Sech[2*Log[c*x]]^(3/2)) + (3*ArcTanh[Sqrt[1 + 1/(c^4
*x^4)]])/(16*c^8*(1 + 1/(c^4*x^4))^(3/2)*x^3*Sech[2*Log[c*x]]^(3/2))

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Rubi [A]  time = 0.0658133, antiderivative size = 92, 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, 266, 47, 63, 207} \[ \frac{3 x}{16 \left (c^4+\frac{1}{x^4}\right ) \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{3 \tanh ^{-1}\left (\sqrt{\frac{1}{c^4 x^4}+1}\right )}{16 c^8 x^3 \left (\frac{1}{c^4 x^4}+1\right )^{3/2} \text{sech}^{\frac{3}{2}}(2 \log (c x))}+\frac{x^5}{8 \text{sech}^{\frac{3}{2}}(2 \log (c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/(16*(c^4 + x^(-4))*Sech[2*Log[c*x]]^(3/2)) + x^5/(8*Sech[2*Log[c*x]]^(3/2)) + (3*ArcTanh[Sqrt[1 + 1/(c^4
*x^4)]])/(16*c^8*(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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 47

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] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^4}{\text{sech}^{\frac{3}{2}}(2 \log (c x))} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\text{sech}^{\frac{3}{2}}(2 \log (x))} \, dx,x,c x\right )}{c^5}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}\right )^{3/2} x^7 \, dx,x,c x\right )}{c^8 \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{(1+x)^{3/2}}{x^3} \, dx,x,\frac{1}{c^4 x^4}\right )}{4 c^8 \left (1+\frac{1}{c^4 x^4}\right )^{3/2} x^3 \text{sech}^{\frac{3}{2}}(2 \log (c x))}\\ &=\frac{x^5}{8 \text{sech}^{\frac{3}{2}}(2 \log (c x))}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{x^2} \, dx,x,\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))}\\ &=\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 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,\frac{1}{c^4 x^4}\right )}{32 c^8 \left (1+\frac{1}{c^4 x^4}\right )^{3/2} x^3 \text{sech}^{\frac{3}{2}}(2 \log (c x))}\\ &=\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 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\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))}\\ &=\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 \tanh ^{-1}\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))}\\ \end{align*}

Mathematica [A]  time = 0.167401, size = 90, normalized size = 0.98 \[ \frac{c^3 x^3 \sqrt{c^4 x^4+1} \left (2 c^4 x^4+5\right )+3 c x \sinh ^{-1}\left (c^2 x^2\right )}{32 \sqrt{2} c^5 \sqrt{\frac{c^2 x^2}{c^4 x^4+1}} \sqrt{c^4 x^4+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 113, normalized size = 1.2 \begin{align*}{\frac{{x}^{3} \left ( 2\,{c}^{4}{x}^{4}+5 \right ) \sqrt{2}}{64\,{c}^{2}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}}{{c}^{4}{x}^{4}+1}}}}}}+{\frac{3\,\sqrt{2}x}{64\,{c}^{2}}\ln \left ({{c}^{4}{x}^{2}{\frac{1}{\sqrt{{c}^{4}}}}}+\sqrt{{c}^{4}{x}^{4}+1} \right ){\frac{1}{\sqrt{{c}^{4}}}}{\frac{1}{\sqrt{{c}^{4}{x}^{4}+1}}}{\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^4/sech(2*ln(c*x))^(3/2),x)

[Out]

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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^4/sech(2*log(c*x))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.10366, size = 220, normalized size = 2.39 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\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**4/sech(2*ln(c*x))**(3/2),x)

[Out]

Integral(x**4/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^{4}}{\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^4/sech(2*log(c*x))^(3/2),x, algorithm="giac")

[Out]

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