Integrand size = 15, antiderivative size = 91 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {1}{2 \left (c^4-\frac {1}{x^4}\right ) x \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^3}{6 \text {csch}^{\frac {3}{2}}(2 \log (c x))}-\frac {\csc ^{-1}\left (c^2 x^2\right )}{2 c^6 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))} \]
-1/2/(c^4-1/x^4)/x/csch(2*ln(c*x))^(3/2)+1/6*x^3/csch(2*ln(c*x))^(3/2)-1/2 *arccsc(c^2*x^2)/c^6/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c*x))^(3/2)
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {x \left (\left (-4+c^4 x^4\right ) \sqrt {-1+c^4 x^4}+3 \arctan \left (\sqrt {-1+c^4 x^4}\right )\right )}{12 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}} \sqrt {-1+c^4 x^4}} \]
(x*((-4 + c^4*x^4)*Sqrt[-1 + c^4*x^4] + 3*ArcTan[Sqrt[-1 + c^4*x^4]]))/(12 *Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(-1 + c^4*x^4)]*Sqrt[-1 + c^4*x^4])
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6086, 6084, 858, 807, 247, 247, 223}
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.
\(\displaystyle \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\) |
\(\Big \downarrow \) 6086 |
\(\displaystyle \frac {\int \frac {c^2 x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^3}\) |
\(\Big \downarrow \) 6084 |
\(\displaystyle \frac {\int c^5 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^5d(c x)}{c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {\int \frac {\left (1-c^4 x^4\right )^{3/2}}{c^7 x^7}d\frac {1}{c x}}{c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {\int \frac {\left (1-c^2 x^2\right )^{3/2}}{c^4 x^4}d\left (c^2 x^2\right )}{2 c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -\frac {-\int \frac {\sqrt {1-c^2 x^2}}{c^2 x^2}d\left (c^2 x^2\right )-\frac {\left (1-c^2 x^2\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}d\left (c^2 x^2\right )+c^2 x^2 \sqrt {1-c^2 x^2}-\frac {\left (1-c^2 x^2\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {\arcsin \left (c^2 x^2\right )+c^2 x^2 \sqrt {1-c^2 x^2}-\frac {\left (1-c^2 x^2\right )^{3/2}}{3 c^3 x^3}}{2 c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\) |
-1/2*(c^2*x^2*Sqrt[1 - c^2*x^2] - (1 - c^2*x^2)^(3/2)/(3*c^3*x^3) + ArcSin [c^2*x^2])/(c^6*(1 - 1/(c^4*x^4))^(3/2)*x^3*Csch[2*Log[c*x]]^(3/2))
3.2.49.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[Csch[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[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m _.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n)) Subst[Int[ x^((m + 1)/n - 1)*Csch[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])
\[\int \frac {x^{2}}{\operatorname {csch}\left (2 \ln \left (c x \right )\right )^{\frac {3}{2}}}d x\]
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c x \arctan \left (\frac {{\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{c x}\right ) + \sqrt {2} {\left (c^{8} x^{8} - 5 \, c^{4} x^{4} + 4\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{24 \, c^{4} x} \]
1/24*(3*sqrt(2)*c*x*arctan((c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1))/(c*x) ) + sqrt(2)*(c^8*x^8 - 5*c^4*x^4 + 4)*sqrt(c^2*x^2/(c^4*x^4 - 1)))/(c^4*x)
\[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{2}}{\operatorname {csch}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \]
\[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{2}}{\operatorname {csch}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^2}{{\left (\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \]