\(\int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [170]

Optimal result

Integrand size = 15, antiderivative size = 141 \[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {4}{77 c^4 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {6 x^4}{77 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^8}{11 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{77 c^5 \left (c^4+\frac {1}{x^4}\right )^2 x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \] Output:

4/77/c^4/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+6/77*x^4/(c^4+1/x^4)/sech(2*ln( 
c*x))^(3/2)+1/11*x^8/sech(2*ln(c*x))^(3/2)+2/77*((c^4+1/x^4)/(c^2+1/x^2)^2 
)^(1/2)*(c^2+1/x^2)*InverseJacobiAM(2*arccot(c*x),1/2*2^(1/2))/c^5/(c^4+1/ 
x^4)^2/x^3/sech(2*ln(c*x))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.55 \[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \left (\left (1+c^4 x^4\right )^{5/2}-\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-c^4 x^4\right )\right )}{22 c^8} \] Input:

Integrate[x^7/Sech[2*Log[c*x]]^(3/2),x]
 

Output:

(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*((1 + c^4*x^4)^(5/2) - 
Hypergeometric2F1[-3/2, 1/4, 5/4, -(c^4*x^4)]))/(22*c^8)
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.26, 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, 809, 809, 847, 761}

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.

\(\displaystyle \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

\(\Big \downarrow \) 6085

\(\displaystyle \frac {\int \frac {c^7 x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^8}\)

\(\Big \downarrow \) 6083

\(\displaystyle \frac {\int c^{10} \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^{10}d(c x)}{c^{11} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 809

\(\displaystyle -\frac {\frac {6}{11} \int \frac {\sqrt {c^4 x^4+1}}{c^8 x^8}d\frac {1}{c x}-\frac {\left (c^4 x^4+1\right )^{3/2}}{11 c^{11} x^{11}}}{c^{11} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 809

\(\displaystyle -\frac {\frac {6}{11} \left (\frac {2}{7} \int \frac {1}{c^4 x^4 \sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{11 c^{11} x^{11}}}{c^{11} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 847

\(\displaystyle -\frac {\frac {6}{11} \left (\frac {2}{7} \left (-\frac {1}{3} \int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{11 c^{11} x^{11}}}{c^{11} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 761

\(\displaystyle -\frac {\frac {6}{11} \left (\frac {2}{7} \left (-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{6 \sqrt {c^4 x^4+1}}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{11 c^{11} x^{11}}}{c^{11} x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

Int[x^7/Sech[2*Log[c*x]]^(3/2),x]
 

Output:

-((-1/11*(1 + c^4*x^4)^(3/2)/(c^11*x^11) + (6*(-1/7*Sqrt[1 + c^4*x^4]/(c^7 
*x^7) + (2*(-1/3*Sqrt[1 + c^4*x^4]/(c^3*x^3) - ((1 + c^2*x^2)*Sqrt[(1 + c^ 
4*x^4)/(1 + c^2*x^2)^2]*EllipticF[2*ArcTan[1/(c*x)], 1/2])/(6*Sqrt[1 + c^4 
*x^4])))/7))/11)/(c^11*(1 + 1/(c^4*x^4))^(3/2)*x^3*Sech[2*Log[c*x]]^(3/2)) 
)
 

Defintions of rubi rules used

rule 761

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)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 809

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] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(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] && IntB 
inomialQ[a, b, c, n, m, p, x]
 

rule 847

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] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {x^{2} \left (7 c^{8} x^{8}+13 c^{4} x^{4}+4\right ) \sqrt {2}}{308 c^{6} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}-\frac {\sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, x}{77 c^{6} \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\)	\(138\)

Input:

int(x^7/sech(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/308*x^2*(7*c^8*x^8+13*c^4*x^4+4)/c^6*2^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2) 
-1/77/c^6/(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)*x/(c^2*x^2/(c^4*x^4+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.63 \[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {8 \, \sqrt {\frac {1}{2}} \sqrt {c^{4}} c \left (-\frac {1}{c^{4}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{c^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - \sqrt {2} {\left (7 \, c^{12} x^{12} + 20 \, c^{8} x^{8} + 17 \, c^{4} x^{4} + 4\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{308 \, c^{8}} \] Input:

integrate(x^7/sech(2*log(c*x))^(3/2),x, algorithm="fricas")
 

Output:

-1/308*(8*sqrt(1/2)*sqrt(c^4)*c*(-1/c^4)^(3/4)*elliptic_f(arcsin((-1/c^4)^ 
(1/4)/x), -1) - sqrt(2)*(7*c^12*x^12 + 20*c^8*x^8 + 17*c^4*x^4 + 4)*sqrt(c 
^2*x^2/(c^4*x^4 + 1)))/c^8
 

Sympy [F]

\[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{7}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

integrate(x**7/sech(2*ln(c*x))**(3/2),x)
 

Output:

Integral(x**7/sech(2*log(c*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{7}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^7/sech(2*log(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^7/sech(2*log(c*x))^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

integrate(x^7/sech(2*log(c*x))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^7}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

int(x^7/(1/cosh(2*log(c*x)))^(3/2),x)
 

Output:

int(x^7/(1/cosh(2*log(c*x)))^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {x^7}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x^{7}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

int(x^7/sech(2*log(c*x))^(3/2),x)
 

Output:

int((sqrt(sech(2*log(c*x)))*x**7)/sech(2*log(c*x))**2,x)