Integrand size = 15, antiderivative size = 108 \[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {2 x^2}{21 c^4 \sqrt {\text {sech}(2 \log (c x))}}+\frac {x^6}{7 \sqrt {\text {sech}(2 \log (c x))}}+\frac {\sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{21 c^5 \left (c^4+\frac {1}{x^4}\right ) x \sqrt {\text {sech}(2 \log (c x))}} \] Output:
2/21*x^2/c^4/sech(2*ln(c*x))^(1/2)+1/7*x^6/sech(2*ln(c*x))^(1/2)+1/21*((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)/x/sech(2*ln(c*x))^(1/2)
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 = 0.71 \[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \left (\left (1+c^4 x^4\right )^{3/2}-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-c^4 x^4\right )\right )}{7 c^6} \] Input:
Integrate[x^5/Sqrt[Sech[2*Log[c*x]]],x]
Output:
(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*((1 + c^4*x^4)^(3/2) - Hypergeometric2F1[-1/2, 1/4, 5/4, -(c^4*x^4)]))/(7*c^6)
Time = 0.52 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6085, 6083, 858, 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.
\(\displaystyle \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle \frac {\int \frac {c^5 x^5}{\sqrt {\text {sech}(2 \log (c x))}}d(c x)}{c^6}\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle \frac {\int c^6 \sqrt {1+\frac {1}{c^4 x^4}} x^6d(c x)}{c^7 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {\int \frac {\sqrt {c^4 x^4+1}}{c^8 x^8}d\frac {1}{c x}}{c^7 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {\frac {2}{7} \int \frac {1}{c^4 x^4 \sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}}{c^7 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle -\frac {\frac {2}{7} \left (-\frac {1}{3} \int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}}{c^7 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {\frac {2}{7} \left (-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{6 \sqrt {c^4 x^4+1}}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\sqrt {c^4 x^4+1}}{7 c^7 x^7}}{c^7 x \sqrt {\frac {1}{c^4 x^4}+1} \sqrt {\text {sech}(2 \log (c x))}}\) |
Input:
Int[x^5/Sqrt[Sech[2*Log[c*x]]],x]
Output:
-((-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)/(c^7*Sqrt[1 + 1/(c^4*x^4)]*x*Sqr t[Sech[2*Log[c*x]]]))
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]
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]
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]
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]
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]
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])
Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.20
method | result | size |
risch | \(\frac {x^{2} \left (3 c^{4} x^{4}+2\right ) \sqrt {2}}{42 c^{4} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}-\frac {\sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, x}{21 c^{4} \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\) | \(130\) |
Input:
int(x^5/sech(2*ln(x*c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/42*x^2*(3*c^4*x^4+2)/c^4*2^(1/2)/(c^2*x^2/(c^4*x^4+1))^(1/2)-1/21/c^4/(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)
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.75 \[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=-\frac {4 \, \sqrt {\frac {1}{2}} \sqrt {c^{4}} c \left (-\frac {1}{c^{4}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{c^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - \sqrt {2} {\left (3 \, c^{8} x^{8} + 5 \, c^{4} x^{4} + 2\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{42 \, c^{6}} \] Input:
integrate(x^5/sech(2*log(c*x))^(1/2),x, algorithm="fricas")
Output:
-1/42*(4*sqrt(1/2)*sqrt(c^4)*c*(-1/c^4)^(3/4)*elliptic_f(arcsin((-1/c^4)^( 1/4)/x), -1) - sqrt(2)*(3*c^8*x^8 + 5*c^4*x^4 + 2)*sqrt(c^2*x^2/(c^4*x^4 + 1)))/c^6
\[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^{5}}{\sqrt {\operatorname {sech}{\left (2 \log {\left (c x \right )} \right )}}}\, dx \] Input:
integrate(x**5/sech(2*ln(c*x))**(1/2),x)
Output:
Integral(x**5/sqrt(sech(2*log(c*x))), x)
\[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{5}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \] Input:
integrate(x^5/sech(2*log(c*x))^(1/2),x, algorithm="maxima")
Output:
integrate(x^5/sqrt(sech(2*log(c*x))), x)
\[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int { \frac {x^{5}}{\sqrt {\operatorname {sech}\left (2 \, \log \left (c x\right )\right )}} \,d x } \] Input:
integrate(x^5/sech(2*log(c*x))^(1/2),x, algorithm="giac")
Output:
integrate(x^5/sqrt(sech(2*log(c*x))), x)
Timed out. \[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {x^5}{\sqrt {\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}}} \,d x \] Input:
int(x^5/(1/cosh(2*log(c*x)))^(1/2),x)
Output:
int(x^5/(1/cosh(2*log(c*x)))^(1/2), x)
\[ \int \frac {x^5}{\sqrt {\text {sech}(2 \log (c x))}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x^{5}}{\mathrm {sech}\left (2 \,\mathrm {log}\left (c x \right )\right )}d x \] Input:
int(x^5/sech(2*log(c*x))^(1/2),x)
Output:
int((sqrt(sech(2*log(c*x)))*x**5)/sech(2*log(c*x)),x)