\(\int \frac {x^3}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) [160]

Optimal result

Integrand size = 15, antiderivative size = 203 \[ \int \frac {x^3}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {2}{5 c^4 \sqrt {\text {sech}(2 \log (c x))}}-\frac {2}{5 c^4 \left (c^2+\frac {1}{x^2}\right ) x^2 \sqrt {\text {sech}(2 \log (c x))}}+\frac {x^4}{5 \sqrt {\text {sech}(2 \log (c x))}}+\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 c^3 \left (c^4+\frac {1}{x^4}\right ) x \sqrt {\text {sech}(2 \log (c x))}}-\frac {\sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{5 c^3 \left (c^4+\frac {1}{x^4}\right ) x \sqrt {\text {sech}(2 \log (c x))}} \] Output:

2/5/c^4/sech(2*ln(c*x))^(1/2)-2/5/c^4/(c^2+1/x^2)/x^2/sech(2*ln(c*x))^(1/2 
)+1/5*x^4/sech(2*ln(c*x))^(1/2)+2/5*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)*(c^2 
+1/x^2)*EllipticE(sin(2*arccot(c*x)),1/2*2^(1/2))/c^3/(c^4+1/x^4)/x/sech(2 
*ln(c*x))^(1/2)-1/5*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)*(c^2+1/x^2)*InverseJ 
acobiAM(2*arccot(c*x),1/2*2^(1/2))/c^3/(c^4+1/x^4)/x/sech(2*ln(c*x))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.32 \[ \int \frac {x^3}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {\left (\frac {c^2 x^2}{1+c^4 x^4}\right )^{3/2} \left (1+c^4 x^4\right )^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-c^4 x^4\right )}{3 \sqrt {2} c^4} \] Input:

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

Output:

(((c^2*x^2)/(1 + c^4*x^4))^(3/2)*(1 + c^4*x^4)^(3/2)*Hypergeometric2F1[-1/ 
2, 3/4, 7/4, -(c^4*x^4)])/(3*Sqrt[2]*c^4)
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6085, 6083, 858, 809, 847, 834, 761, 1510}

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.

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

\(\Big \downarrow \) 6085

\(\displaystyle \frac {\int \frac {c^3 x^3}{\sqrt {\text {sech}(2 \log (c x))}}d(c x)}{c^4}\)

\(\Big \downarrow \) 6083

\(\displaystyle \frac {\int c^4 \sqrt {1+\frac {1}{c^4 x^4}} x^4d(c x)}{c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {\int \frac {\sqrt {c^4 x^4+1}}{c^6 x^6}d\frac {1}{c x}}{c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\)

\(\Big \downarrow \) 809

\(\displaystyle -\frac {\frac {2}{5} \int \frac {1}{c^2 x^2 \sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{5 c^5 x^5}}{c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\)

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 761

\(\displaystyle -\frac {\frac {2}{5} \left (-\int \frac {1-c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}+\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\frac {\sqrt {c^4 x^4+1}}{c x}\right )-\frac {\sqrt {c^4 x^4+1}}{5 c^5 x^5}}{c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\)

\(\Big \downarrow \) 1510

\(\displaystyle -\frac {\frac {2}{5} \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} E\left (2 \arctan \left (\frac {1}{c x}\right )|\frac {1}{2}\right )}{\sqrt {c^4 x^4+1}}-\frac {\sqrt {c^4 x^4+1}}{c x}+\frac {\sqrt {c^4 x^4+1}}{c x \left (c^2 x^2+1\right )}\right )-\frac {\sqrt {c^4 x^4+1}}{5 c^5 x^5}}{c^5 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\)

Input:

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

Output:

-((-1/5*Sqrt[1 + c^4*x^4]/(c^5*x^5) + (2*(-(Sqrt[1 + c^4*x^4]/(c*x)) + Sqr 
t[1 + c^4*x^4]/(c*x*(1 + c^2*x^2)) - ((1 + c^2*x^2)*Sqrt[(1 + c^4*x^4)/(1 
+ c^2*x^2)^2]*EllipticE[2*ArcTan[1/(c*x)], 1/2])/Sqrt[1 + c^4*x^4] + ((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])/(2*Sqrt[1 + c^4*x^4])))/5)/(c^5*Sqrt[1 + 1/(c^4*x^4)]*x*Sqrt[Sech[2 
*Log[c*x]]]))
 

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 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

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 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

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 = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {\sqrt {2}\, x^{4}}{10 \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {i \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i c^{2}}, i\right )\right ) \sqrt {2}\, x}{5 \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\)	\(134\)

Input:

int(x^3/sech(2*ln(x*c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/10*2^(1/2)*x^4/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/5*I/(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)/c^2*(EllipticF(x*(I*c^2)^(1/2), 
I)-EllipticE(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 = 120, normalized size of antiderivative = 0.59 \[ \int \frac {x^3}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {4 \, \sqrt {\frac {1}{2}} \sqrt {c^{4}} c x^{2} \left (-\frac {1}{c^{4}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{c^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 4 \, \sqrt {\frac {1}{2}} \sqrt {c^{4}} c x^{2} \left (-\frac {1}{c^{4}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{c^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {2} {\left (c^{8} x^{8} + 3 \, c^{4} x^{4} + 2\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{10 \, c^{6} x^{2}} \] Input:

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="fricas")
 

Output:

1/10*(4*sqrt(1/2)*sqrt(c^4)*c*x^2*(-1/c^4)^(3/4)*elliptic_e(arcsin((-1/c^4 
)^(1/4)/x), -1) - 4*sqrt(1/2)*sqrt(c^4)*c*x^2*(-1/c^4)^(3/4)*elliptic_f(ar 
csin((-1/c^4)^(1/4)/x), -1) + sqrt(2)*(c^8*x^8 + 3*c^4*x^4 + 2)*sqrt(c^2*x 
^2/(c^4*x^4 + 1)))/(c^6*x^2)
 

Sympy [F]

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

integrate(x**3/sech(2*ln(c*x))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^3/sech(2*log(c*x))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly exception caught Unable to 
 convert to real %%{poly1[1.0000000000000000000000000000000,0.000000000000 
000000000
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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