Integrand size = 13, antiderivative size = 214 \[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {12}{5 \left (c^4+\frac {1}{x^4}\right ) \left (c^2+\frac {1}{x^2}\right ) x^4 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {6}{5 \left (c^4+\frac {1}{x^4}\right ) x^2 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^2}{5 \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {12 c \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}(c x)|\frac {1}{2}\right )}{5 \left (c^4+\frac {1}{x^4}\right )^2 x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {6 c \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{5 \left (c^4+\frac {1}{x^4}\right )^2 x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]
-12/5/(c^4+1/x^4)/(c^2+1/x^2)/x^4/sech(2*ln(c*x))^(3/2)+6/5/(c^4+1/x^4)/x^ 2/sech(2*ln(c*x))^(3/2)+1/5*x^2/sech(2*ln(c*x))^(3/2)+12/5*c*(c^2+1/x^2)*( cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot(c*x))*EllipticE(sin(2*arccot(c*x) ),1/2*2^(1/2))*((c^4+1/x^4)/(c^2+1/x^2)^2)^(1/2)/(c^4+1/x^4)^2/x^3/sech(2* ln(c*x))^(3/2)-6/5*c*(c^2+1/x^2)*(cos(2*arccot(c*x))^2)^(1/2)/cos(2*arccot (c*x))*EllipticF(sin(2*arccot(c*x)),1/2*2^(1/2))*((c^4+1/x^4)/(c^2+1/x^2)^ 2)^(1/2)/(c^4+1/x^4)^2/x^3/sech(2*ln(c*x))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.30 \[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},-c^4 x^4\right )}{2 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{1+c^4 x^4}} \sqrt {1+c^4 x^4}} \]
-1/2*Hypergeometric2F1[-3/2, -1/4, 3/4, -(c^4*x^4)]/(Sqrt[2]*c^2*Sqrt[(c^2 *x^2)/(1 + c^4*x^4)]*Sqrt[1 + c^4*x^4])
Time = 0.39 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6085, 6083, 858, 809, 809, 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.
| \(\displaystyle \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) |
| \(\Big \downarrow \) 6085 |
| \(\displaystyle \frac {\int \frac {c x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^2}\) |
| \(\Big \downarrow \) 6083 |
| \(\displaystyle \frac {\int c^4 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^4d(c x)}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 858 |
| \(\displaystyle -\frac {\int \frac {\left (c^4 x^4+1\right )^{3/2}}{c^6 x^6}d\frac {1}{c x}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 809 |
| \(\displaystyle -\frac {\frac {6}{5} \int \frac {\sqrt {c^4 x^4+1}}{c^2 x^2}d\frac {1}{c x}-\frac {\left (c^4 x^4+1\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 809 |
| \(\displaystyle -\frac {\frac {6}{5} \left (2 \int \frac {c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{c x}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle -\frac {\frac {6}{5} \left (2 \left (\int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\int \frac {1-c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\right )-\frac {\sqrt {c^4 x^4+1}}{c x}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle -\frac {\frac {6}{5} \left (2 \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\int \frac {1-c^2 x^2}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}\right )-\frac {\sqrt {c^4 x^4+1}}{c x}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle -\frac {\frac {6}{5} \left (2 \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{2 \sqrt {c^4 x^4+1}}-\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} E\left (2 \arctan \left (\frac {1}{c x}\right )|\frac {1}{2}\right )}{\sqrt {c^4 x^4+1}}+\frac {\sqrt {c^4 x^4+1}}{c x \left (c^2 x^2+1\right )}\right )-\frac {\sqrt {c^4 x^4+1}}{c x}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
-((-1/5*(1 + c^4*x^4)^(3/2)/(c^5*x^5) + (6*(-(Sqrt[1 + c^4*x^4]/(c*x)) + 2 *(Sqrt[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*(1 + 1/(c^4*x^4))^(3/2)*x^3*S ech[2*Log[c*x]]^(3/2)))
3.2.76.3.1 Defintions of rubi rules used
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[(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]
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[((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]
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 = 0.68 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.74
| method | result | size |
| risch | \(\frac {\left (c^{8} x^{8}-4 c^{4} x^{4}-5\right ) \sqrt {2}}{20 \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {3 i \sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i c^{2}}, i\right )\right ) \sqrt {2}\, x}{5 \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\) | \(159\) |
1/20*(c^8*x^8-4*c^4*x^4-5)/(c^4*x^4+1)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^( 1/2)+3/5*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)*(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)
\[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]
\[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]