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

Optimal. Leaf size=111 \[ \frac {2}{7 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{7 c x^3 \left (c^4+\frac {1}{x^4}\right )^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

[Out]

2/7/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+1/7*x^4/sech(2*ln(c*x))^(3/2)-2/7*(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/(c^4+1/x^4
)^2/x^3/sech(2*ln(c*x))^(3/2)

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Rubi [A]  time = 0.08, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5551, 5549, 335, 277, 220} \[ \frac {2}{7 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{7 c x^3 \left (c^4+\frac {1}{x^4}\right )^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2/(7*(c^4 + x^(-4))*Sech[2*Log[c*x]]^(3/2)) + x^4/(7*Sech[2*Log[c*x]]^(3/2)) - (2*Sqrt[(c^4 + x^(-4))/(c^2 + x
^(-2))^2]*(c^2 + x^(-2))*EllipticF[2*ArcCot[c*x], 1/2])/(7*c*(c^4 + x^(-4))^2*x^3*Sech[2*Log[c*x]]^(3/2))

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 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 {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (x))} \, dx,x,c x\right )}{c^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {1}{x^4}\right )^{3/2} x^6 \, dx,x,c x\right )}{c^7 \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^8} \, dx,x,\frac {1}{c x}\right )}{c^7 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {6 \operatorname {Subst}\left (\int \frac {\sqrt {1+x^4}}{x^4} \, dx,x,\frac {1}{c x}\right )}{7 c^7 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {2}{7 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\frac {1}{c x}\right )}{7 c^7 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ &=\frac {2}{7 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(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 ) F\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{7 c \left (c^4+\frac {1}{x^4}\right )^2 x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 61, normalized size = 0.55 \[ \frac {\sqrt {c^4 x^4+1} \sqrt {\frac {c^2 x^2}{2 c^4 x^4+2}} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-c^4 x^4\right )}{2 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*Hypergeometric2F1[-3/2, 1/4, 5/4, -(c^4*x^4)])/(2*c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.22, size = 129, normalized size = 1.16 \[ \frac {x^{2} \left (c^{4} x^{4}+3\right ) \sqrt {2}}{28 c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {\sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \EllipticF \left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, x}{7 \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/28*x^2*(c^4*x^4+3)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/7/(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)/c^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 \[ \int \frac {x^{3}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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