Integrand size = 15, antiderivative size = 111 \[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {2}{7 \left (c^4+\frac {1}{x^4}\right ) \text {sech}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^4}{7 \text {sech}^{\frac {3}{2}}(2 \log (c x))}-\frac {2 \sqrt {\frac {c^4+\frac {1}{x^4}}{\left (c^2+\frac {1}{x^2}\right )^2}} \left (c^2+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}(c x),\frac {1}{2}\right )}{7 c \left (c^4+\frac {1}{x^4}\right )^2 x^3 \text {sech}^{\frac {3}{2}}(2 \log (c x))} \]
2/7/(c^4+1/x^4)/sech(2*ln(c*x))^(3/2)+1/7*x^4/sech(2*ln(c*x))^(3/2)-2/7*(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/(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 = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.55 \[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {\sqrt {1+c^4 x^4} \sqrt {\frac {c^2 x^2}{2+2 c^4 x^4}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-c^4 x^4\right )}{2 c^4} \]
(Sqrt[1 + c^4*x^4]*Sqrt[(c^2*x^2)/(2 + 2*c^4*x^4)]*Hypergeometric2F1[-3/2, 1/4, 5/4, -(c^4*x^4)])/(2*c^4)
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.34, 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, 809, 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.
\(\displaystyle \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx\) |
\(\Big \downarrow \) 6085 |
\(\displaystyle \frac {\int \frac {c^3 x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^4}\) |
\(\Big \downarrow \) 6083 |
\(\displaystyle \frac {\int c^6 \left (1+\frac {1}{c^4 x^4}\right )^{3/2} x^6d(c x)}{c^7 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {\int \frac {\left (c^4 x^4+1\right )^{3/2}}{c^8 x^8}d\frac {1}{c x}}{c^7 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {\frac {6}{7} \int \frac {\sqrt {c^4 x^4+1}}{c^4 x^4}d\frac {1}{c x}-\frac {\left (c^4 x^4+1\right )^{3/2}}{7 c^7 x^7}}{c^7 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle -\frac {\frac {6}{7} \left (\frac {2}{3} \int \frac {1}{\sqrt {c^4 x^4+1}}d\frac {1}{c x}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{7 c^7 x^7}}{c^7 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle -\frac {\frac {6}{7} \left (\frac {\left (c^2 x^2+1\right ) \sqrt {\frac {c^4 x^4+1}{\left (c^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{c x}\right ),\frac {1}{2}\right )}{3 \sqrt {c^4 x^4+1}}-\frac {\sqrt {c^4 x^4+1}}{3 c^3 x^3}\right )-\frac {\left (c^4 x^4+1\right )^{3/2}}{7 c^7 x^7}}{c^7 x^3 \left (\frac {1}{c^4 x^4}+1\right )^{3/2} \text {sech}^{\frac {3}{2}}(2 \log (c x))}\) |
-((-1/7*(1 + c^4*x^4)^(3/2)/(c^7*x^7) + (6*(-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])/(3*Sqrt[1 + c^4*x^4])))/7)/(c^7*(1 + 1/(c^4*x^4))^(3/2)*x ^3*Sech[2*Log[c*x]]^(3/2)))
3.2.74.3.1 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_)^(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 = 0.59 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {x^{2} \left (c^{4} x^{4}+3\right ) \sqrt {2}}{28 c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}+\frac {\sqrt {-i c^{2} x^{2}+1}\, \sqrt {i c^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i c^{2}}, i\right ) \sqrt {2}\, x}{7 \sqrt {i c^{2}}\, \left (c^{4} x^{4}+1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}+1}}}\) | \(129\) |
1/28*x^2*(c^4*x^4+3)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4+1))^(1/2)+1/7/(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)/c^2*x/(c^2*x^2/(c^4*x^4+1))^(1/2)
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.71 \[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {4 \, \sqrt {2} \sqrt {c^{4}} c \left (-\frac {1}{c^{4}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{c^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {2} {\left (c^{8} x^{8} + 4 \, c^{4} x^{4} + 3\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} + 1}}}{28 \, c^{4}} \]
1/28*(4*sqrt(2)*sqrt(c^4)*c*(-1/c^4)^(3/4)*elliptic_f(arcsin((-1/c^4)^(1/4 )/x), -1) + sqrt(2)*(c^8*x^8 + 4*c^4*x^4 + 3)*sqrt(c^2*x^2/(c^4*x^4 + 1))) /c^4
\[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{3}}{\operatorname {sech}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]
\[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{3}}{\operatorname {sech}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^3}{\text {sech}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^3}{{\left (\frac {1}{\mathrm {cosh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]