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

Optimal. Leaf size=203 \[ \frac {2}{5 c^4 \sqrt {\text {sech}(2 \log (c x))}}-\frac {2}{5 c^4 x^2 \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))}}-\frac {\sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 c^3 x \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}+\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 c^3 x \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}+\frac {x^4}{5 \sqrt {\text {sech}(2 \log (c x))}} \]

[Out]

2/5/c^4/sech(2*ln(c*x))^(1/2)-2/5/c^4/(c^2+1/x^2)/x^2/sech(2*ln(c*x))^(1/2)+1/5*x^4/sech(2*ln(c*x))^(1/2)+2/5*
(c^2+1/x^2)*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot(c*x))*EllipticE(sin(2*arccot(c*x)),1/2*2^(1/2))*((c^4+1/
x^4)/(c^2+1/x^2)^2)^(1/2)/c^3/(c^4+1/x^4)/x/sech(2*ln(c*x))^(1/2)-1/5*(c^2+1/x^2)*(cos(2*arccot(c*x))^2)^(1/2)
/cos(2*arccot(c*x))*EllipticF(sin(2*arccot(c*x)),1/2*2^(1/2))*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)/c^3/(c^4+1/x^4
)/x/sech(2*ln(c*x))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5551, 5549, 335, 277, 325, 305, 220, 1196} \[ -\frac {2}{5 c^4 x^2 \left (c^2+\frac {1}{x^2}\right ) \sqrt {\text {sech}(2 \log (c x))}}-\frac {\sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 c^3 x \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}+\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 c^3 x \left (c^4+\frac {1}{x^4}\right ) \sqrt {\text {sech}(2 \log (c x))}}+\frac {2}{5 c^4 \sqrt {\text {sech}(2 \log (c x))}}+\frac {x^4}{5 \sqrt {\text {sech}(2 \log (c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(5*c^4*Sqrt[Sech[2*Log[c*x]]]) - 2/(5*c^4*(c^2 + x^(-2))*x^2*Sqrt[Sech[2*Log[c*x]]]) + x^4/(5*Sqrt[Sech[2*Lo
g[c*x]]]) + (2*Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(-2))*EllipticE[2*ArcCot[c*x], 1/2])/(5*c^3*(c^4
+ x^(-4))*x*Sqrt[Sech[2*Log[c*x]]]) - (Sqrt[(c^4 + x^(-4))/(c^2 + x^(-2))^2]*(c^2 + x^(-2))*EllipticF[2*ArcCo
t[c*x], 1/2])/(5*c^3*(c^4 + x^(-4))*x*Sqrt[Sech[2*Log[c*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]

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 305

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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

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Mathematica [C]  time = 0.12, size = 65, normalized size = 0.32 \[ \frac {\left (\frac {c^2 x^2}{c^4 x^4+1}\right )^{3/2} \left (c^4 x^4+1\right )^{3/2} \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-c^4 x^4\right )}{3 \sqrt {2} c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((c^2*x^2)/(1 + c^4*x^4))^(3/2)*(1 + c^4*x^4)^(3/2)*Hypergeometric2F1[-1/2, 3/4, 7/4, -(c^4*x^4)])/(3*Sqrt[2]
*c^4)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.23, size = 134, normalized size = 0.66 \[ \frac {x^{4} \sqrt {2}}{10 \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {i \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \left (\EllipticF \left (x \sqrt {i c^{2}}, i\right )-\EllipticE \left (x \sqrt {i c^{2}}, i\right )\right ) \sqrt {2}\, x}{5 \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/sech(2*ln(c*x))^(1/2),x)

[Out]

1/10*x^4*2^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/5*I/(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)/c^2*(EllipticF(x*(I*c^2)^(1/2),I)-EllipticE(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/sech(2*ln(c*x))**(1/2),x)

[Out]

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

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