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+1/x^4)/x/sech(2*ln(c*x))^(3/2)+1/6*x^3/sech(2*ln(c*x))^(3/2)-1/2*arccsch(c^2*x^2)/c^6/(1+1/c^4/x^4)^(
3/2)/x^3/sech(2*ln(c*x))^(3/2)

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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]

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 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^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*}

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Mathematica [A]  time = 0.17, size = 88, normalized size = 1.00 \[ \frac {x \left (\sqrt {c^4 x^4+1} \left (c^4 x^4+4\right )-3 \tanh ^{-1}\left (\sqrt {c^4 x^4+1}\right )\right )}{12 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{c^4 x^4+1}} \sqrt {c^4 x^4+1}} \]

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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fricas [A]  time = 0.42, size = 109, normalized size = 1.24 \[ \frac {3 \, \sqrt {2} c x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} {\left (c^{8} x^{8} + 5 \, c^{4} x^{4} + 4\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{48 \, c^{4} x} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]U
nable to cancel step at 0 of 1/2/c^6*c^4*(1/2*ln(sqrt(c^4*t_nostep^4+1)-1)-1/2*ln(sqrt(c^4*t_nostep^4+1)+1)+sq
rt(c^4*t_nostep^4+1))-1/2/c^6*c^4*(-1/2*ln(sqrt(c^4*t_nostep^4+1)-1)+1/2*ln(sqrt(c^4*t_nostep^4+1)+1)-sqrt(c^4
*t_nostep^4+1))Unable to divide, perhaps due to rounding error%%%{1,[4,4,1,0]%%%}+%%%{1,[0,0,1,0]%%%} / %%%{1,
[0,2,0,1]%%%} Error: Bad Argument Value

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\mathrm {sech}\left (2 \ln \left (c x \right )\right )^{\frac {3}{2}}}\, dx \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**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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