\(\int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [175]

Optimal result

Integrand size = 15, antiderivative size = 88 \[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {1}{2 \left (c^4+\frac {1}{x^4}\right ) x \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^3}{6 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {\text {csch}^{-1}\left (c^2 x^2\right )}{2 c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \] Output:

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)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {x \left (\sqrt {1+c^4 x^4} \left (4+c^4 x^4\right )-3 \text {arctanh}\left (\sqrt {1+c^4 x^4}\right )\right )}{12 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6085, 6083, 858, 807, 247, 247, 222}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6085

\(\displaystyle \frac {\int \frac {c^2 x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^3}\)

\(\Big \downarrow \) 6083

\(\displaystyle \frac {\int c^5 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^5d(c x)}{c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {\int \frac {\left (c^4 x^4+1\right )^{3/2}}{c^7 x^7}d\frac {1}{c x}}{c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 807

\(\displaystyle -\frac {\int \frac {\left (c^2 x^2+1\right )^{3/2}}{c^4 x^4}d\left (c^2 x^2\right )}{2 c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 247

\(\displaystyle -\frac {\int \frac {\sqrt {c^2 x^2+1}}{c^2 x^2}d\left (c^2 x^2\right )-\frac {\left (c^2 x^2+1\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 247

\(\displaystyle -\frac {\int \frac {1}{\sqrt {c^2 x^2+1}}d\left (c^2 x^2\right )-c^2 x^2 \sqrt {c^2 x^2+1}-\frac {\left (c^2 x^2+1\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 222

\(\displaystyle -\frac {\text {arcsinh}\left (c^2 x^2\right )-c^2 x^2 \sqrt {c^2 x^2+1}-\frac {\left (c^2 x^2+1\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 222

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 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[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 858

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] && Int 
egerQ[m]
 

rule 6083

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] 
 :> Simp[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*(1/(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 6085

Int[((e_.)*(x_))^(m_.)*Sech[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p 
_.), x_Symbol] :> Simp[(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])
 

Maple [F]

\[\int \frac {x^{2}}{\operatorname {sech}\left (2 \ln \left (x c \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c x \log \left (\frac {c^{5} x^{5} + 2 \, c x - 2 \, {\left (c^{4} x^{4} + 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{c x^{5}}\right ) + 2 \, \sqrt {2} {\left (c^{8} x^{8} + 5 \, c^{4} x^{4} + 4\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{48 \, c^{4} x} \] Input:

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

Output:

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)
 

Sympy [F]

\[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{2}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{2}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^2}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^2}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x^{2}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

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

Output:

int((sqrt(sech(2*log(c*x)))*x**2)/sech(2*log(c*x))**2,x)