\(\int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [169]

Optimal result

Integrand size = 15, antiderivative size = 122 \[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {x}{32 c^4 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^5}{16 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^9}{12 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{c^4 x^4}}\right )}{32 c^{12} \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \] Output:

1/32*x/c^4/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+1/16*x^5/(c^4+1/x^4)/sech(2*l 
n(c*x))^(3/2)+1/12*x^9/sech(2*ln(c*x))^(3/2)-1/32*arctanh((1+1/c^4/x^4)^(1 
/2))/c^12/(1+1/c^4/x^4)^(3/2)/x^3/sech(2*ln(c*x))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {c^3 x^3 \sqrt {1+c^4 x^4} \left (3+14 c^4 x^4+8 c^8 x^8\right )-3 c x \text {arcsinh}\left (c^2 x^2\right )}{192 \sqrt {2} c^9 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \] Input:

Integrate[x^8/Sech[2*Log[c*x]]^(3/2),x]
 

Output:

(c^3*x^3*Sqrt[1 + c^4*x^4]*(3 + 14*c^4*x^4 + 8*c^8*x^8) - 3*c*x*ArcSinh[c^ 
2*x^2])/(192*Sqrt[2]*c^9*Sqrt[(c^2*x^2)/(1 + c^4*x^4)]*Sqrt[1 + c^4*x^4])
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6085, 6083, 798, 51, 51, 52, 73, 220}

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.

\(\displaystyle \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

\(\Big \downarrow \) 6085

\(\displaystyle \frac {\int \frac {c^8 x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^9}\)

\(\Big \downarrow \) 6083

\(\displaystyle \frac {\int c^{11} \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^{11}d(c x)}{c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 798

\(\displaystyle -\frac {\int \frac {\left (1+\frac {1}{c^4 x^4}\right )^{3/2}}{c^4 x^4}d\frac {1}{c^4 x^4}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {\frac {1}{2} \int \frac {\sqrt {1+\frac {1}{c^4 x^4}}}{c^3 x^3}d\frac {1}{c^4 x^4}-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{c^2 \sqrt {1+\frac {1}{c^4 x^4}} x^2}d\frac {1}{c^4 x^4}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{2 c^2 x^2}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {1}{c \sqrt {1+\frac {1}{c^4 x^4}} x}d\frac {1}{c^4 x^4}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{2 c^2 x^2}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (-\int \frac {1}{c^2 x^2-1}d\sqrt {1+\frac {1}{c^4 x^4}}-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{2 c^2 x^2}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (\text {arctanh}\left (\sqrt {\frac {1}{c^4 x^4}+1}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{c x}\right )-\frac {\sqrt {\frac {1}{c^4 x^4}+1}}{2 c^2 x^2}\right )-\frac {\left (\frac {1}{c^4 x^4}+1\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

Int[x^8/Sech[2*Log[c*x]]^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 51

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] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 220

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

rule 798

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

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99

method	result	size
risch	\(\frac {x^{3} \left (8 c^{8} x^{8}+14 c^{4} x^{4}+3\right ) \sqrt {2}}{384 c^{6} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}-\frac {\ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}+1}\right ) \sqrt {2}\, x}{128 c^{6} \sqrt {c^{4}}\, \sqrt {c^{4} x^{4}+1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\)	\(121\)

Input:

int(x^8/sech(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/384*x^3*(8*c^8*x^8+14*c^4*x^4+3)/c^6*2^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2) 
-1/128/c^6*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4+1)^(1/2))/(c^4)^(1/2)*2^(1/2)*x 
/(c^4*x^4+1)^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {2 \, \sqrt {2} {\left (8 \, c^{13} x^{13} + 22 \, c^{9} x^{9} + 17 \, c^{5} x^{5} + 3 \, c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} + 3 \, \sqrt {2} \log \left (-2 \, c^{4} x^{4} + 2 \, {\left (c^{5} x^{5} + c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}} - 1\right )}{768 \, c^{9}} \] Input:

integrate(x^8/sech(2*log(c*x))^(3/2),x, algorithm="fricas")
 

Output:

1/768*(2*sqrt(2)*(8*c^13*x^13 + 22*c^9*x^9 + 17*c^5*x^5 + 3*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^9
 

Sympy [F]

\[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{8}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

integrate(x**8/sech(2*ln(c*x))**(3/2),x)
 

Output:

Integral(x**8/sech(2*log(c*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{8}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^8/sech(2*log(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^8/sech(2*log(c*x))^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

integrate(x^8/sech(2*log(c*x))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^8}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

int(x^8/(1/cosh(2*log(c*x)))^(3/2),x)
 

Output:

int(x^8/(1/cosh(2*log(c*x)))^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^8}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x^{8}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

int(x^8/sech(2*log(c*x))^(3/2),x)
 

Output:

int((sqrt(sech(2*log(c*x)))*x**8)/sech(2*log(c*x))**2,x)