Integrand size = 13, antiderivative size = 64 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=-\frac {1}{8 x^8}+\frac {e^{2 a}}{x^4}+\frac {e^{4 a}}{1+e^{2 a} x^4}+8 e^{4 a} \log (x)-2 e^{4 a} \log \left (1+e^{2 a} x^4\right ) \] Output:
-1/8/x^8+exp(2*a)/x^4+exp(4*a)/(1+exp(2*a)*x^4)+8*exp(4*a)*ln(x)-2*exp(4*a )*ln(1+exp(2*a)*x^4)
Time = 0.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.83 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=-\frac {1}{8 x^8}+\frac {\cosh (2 a)}{x^4}+8 \cosh (4 a) \log (x)-2 \cosh (4 a) \log \left (\left (1+x^4\right ) \cosh (a)+\left (-1+x^4\right ) \sinh (a)\right )+\frac {\sinh (2 a)}{x^4}+\frac {\cosh (3 a)+\sinh (3 a)}{\left (1+x^4\right ) \cosh (a)+\left (-1+x^4\right ) \sinh (a)}+8 \log (x) \sinh (4 a)-2 \log \left (\left (1+x^4\right ) \cosh (a)+\left (-1+x^4\right ) \sinh (a)\right ) \sinh (4 a) \] Input:
Integrate[Tanh[a + 2*Log[x]]^2/x^9,x]
Output:
-1/8*1/x^8 + Cosh[2*a]/x^4 + 8*Cosh[4*a]*Log[x] - 2*Cosh[4*a]*Log[(1 + x^4 )*Cosh[a] + (-1 + x^4)*Sinh[a]] + Sinh[2*a]/x^4 + (Cosh[3*a] + Sinh[3*a])/ ((1 + x^4)*Cosh[a] + (-1 + x^4)*Sinh[a]) + 8*Log[x]*Sinh[4*a] - 2*Log[(1 + x^4)*Cosh[a] + (-1 + x^4)*Sinh[a]]*Sinh[4*a]
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6071, 948, 99, 2009}
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.
\(\displaystyle \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx\) |
\(\Big \downarrow \) 6071 |
\(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right )^2}{x^9 \left (e^{2 a} x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{4} \int \frac {\left (1-e^{2 a} x^4\right )^2}{x^{12} \left (e^{2 a} x^4+1\right )^2}dx^4\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{4} \int \left (\frac {8 e^{4 a}}{x^4}-\frac {4 e^{2 a}}{x^8}+\frac {1}{x^{12}}-\frac {8 e^{6 a}}{e^{2 a} x^4+1}-\frac {4 e^{6 a}}{\left (e^{2 a} x^4+1\right )^2}\right )dx^4\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {4 e^{4 a}}{e^{2 a} x^4+1}+\frac {4 e^{2 a}}{x^4}+8 e^{4 a} \log \left (x^4\right )-8 e^{4 a} \log \left (e^{2 a} x^4+1\right )-\frac {1}{2 x^8}\right )\) |
Input:
Int[Tanh[a + 2*Log[x]]^2/x^9,x]
Output:
(-1/2*1/x^8 + (4*E^(2*a))/x^4 + (4*E^(4*a))/(1 + E^(2*a)*x^4) + 8*E^(4*a)* Log[x^4] - 8*E^(4*a)*Log[1 + E^(2*a)*x^4])/4
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((-1 + E^(2*a*d)*x^(2*b*d))^p/(1 + E^(2*a*d)*x^(2*b*d))^p), x] /; FreeQ[{a, b, d, e, m, p}, x]
Time = 0.63 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{4 a} x^{8}+\frac {7 \,{\mathrm e}^{2 a} x^{4}}{8}-\frac {1}{8}}{x^{8} \left (1+{\mathrm e}^{2 a} x^{4}\right )}+8 \,{\mathrm e}^{4 a} \ln \left (x \right )-2 \,{\mathrm e}^{4 a} \ln \left (1+{\mathrm e}^{2 a} x^{4}\right )\) | \(63\) |
Input:
int(tanh(a+2*ln(x))^2/x^9,x,method=_RETURNVERBOSE)
Output:
(2*exp(4*a)*x^8+7/8*exp(2*a)*x^4-1/8)/x^8/(1+exp(2*a)*x^4)+8*exp(4*a)*ln(x )-2*exp(4*a)*ln(1+exp(2*a)*x^4)
Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.36 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=\frac {16 \, x^{8} e^{\left (4 \, a\right )} + 7 \, x^{4} e^{\left (2 \, a\right )} - 16 \, {\left (x^{12} e^{\left (6 \, a\right )} + x^{8} e^{\left (4 \, a\right )}\right )} \log \left (x^{4} e^{\left (2 \, a\right )} + 1\right ) + 64 \, {\left (x^{12} e^{\left (6 \, a\right )} + x^{8} e^{\left (4 \, a\right )}\right )} \log \left (x\right ) - 1}{8 \, {\left (x^{12} e^{\left (2 \, a\right )} + x^{8}\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^9,x, algorithm="fricas")
Output:
1/8*(16*x^8*e^(4*a) + 7*x^4*e^(2*a) - 16*(x^12*e^(6*a) + x^8*e^(4*a))*log( x^4*e^(2*a) + 1) + 64*(x^12*e^(6*a) + x^8*e^(4*a))*log(x) - 1)/(x^12*e^(2* a) + x^8)
\[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=\int \frac {\tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}}{x^{9}}\, dx \] Input:
integrate(tanh(a+2*ln(x))**2/x**9,x)
Output:
Integral(tanh(a + 2*log(x))**2/x**9, x)
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.70 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=-2 \, e^{\left (4 \, a\right )} \log \left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right ) - \frac {e^{\left (6 \, a\right )}}{\frac {1}{x^{4}} + e^{\left (2 \, a\right )}} + \frac {e^{\left (2 \, a\right )}}{x^{4}} - \frac {1}{8 \, x^{8}} \] Input:
integrate(tanh(a+2*log(x))^2/x^9,x, algorithm="maxima")
Output:
-2*e^(4*a)*log(1/x^4 + e^(2*a)) - e^(6*a)/(1/x^4 + e^(2*a)) + e^(2*a)/x^4 - 1/8/x^8
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=-2 \, e^{\left (4 \, a\right )} \log \left (x^{4} e^{\left (2 \, a\right )} + 1\right ) + 2 \, e^{\left (4 \, a\right )} \log \left (x^{4}\right ) + \frac {2 \, x^{4} e^{\left (6 \, a\right )} + 3 \, e^{\left (4 \, a\right )}}{x^{4} e^{\left (2 \, a\right )} + 1} - \frac {24 \, x^{8} e^{\left (4 \, a\right )} - 8 \, x^{4} e^{\left (2 \, a\right )} + 1}{8 \, x^{8}} \] Input:
integrate(tanh(a+2*log(x))^2/x^9,x, algorithm="giac")
Output:
-2*e^(4*a)*log(x^4*e^(2*a) + 1) + 2*e^(4*a)*log(x^4) + (2*x^4*e^(6*a) + 3* e^(4*a))/(x^4*e^(2*a) + 1) - 1/8*(24*x^8*e^(4*a) - 8*x^4*e^(2*a) + 1)/x^8
Time = 2.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=8\,{\mathrm {e}}^{4\,a}\,\ln \left (x\right )-2\,{\mathrm {e}}^{4\,a}\,\ln \left (x^4+{\mathrm {e}}^{-2\,a}\right )+\frac {2\,{\mathrm {e}}^{4\,a}\,x^8+\frac {7\,{\mathrm {e}}^{2\,a}\,x^4}{8}-\frac {1}{8}}{{\mathrm {e}}^{2\,a}\,x^{12}+x^8} \] Input:
int(tanh(a + 2*log(x))^2/x^9,x)
Output:
8*exp(4*a)*log(x) - 2*exp(4*a)*log(exp(-2*a) + x^4) + ((7*x^4*exp(2*a))/8 + 2*x^8*exp(4*a) - 1/8)/(x^12*exp(2*a) + x^8)
Time = 0.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.84 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^9} \, dx=\frac {-16 e^{6 a} \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{12}-16 e^{6 a} \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{12}+64 e^{6 a} \mathrm {log}\left (x \right ) x^{12}-16 e^{6 a} x^{12}-16 e^{4 a} \mathrm {log}\left (-e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{8}-16 e^{4 a} \mathrm {log}\left (e^{\frac {a}{2}} \sqrt {2}\, x +e^{a} x^{2}+1\right ) x^{8}+64 e^{4 a} \mathrm {log}\left (x \right ) x^{8}+7 e^{2 a} x^{4}-1}{8 x^{8} \left (e^{2 a} x^{4}+1\right )} \] Input:
int(tanh(a+2*log(x))^2/x^9,x)
Output:
( - 16*e**(6*a)*log( - e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**12 - 16*e**( 6*a)*log(e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**12 + 64*e**(6*a)*log(x)*x* *12 - 16*e**(6*a)*x**12 - 16*e**(4*a)*log( - e**(a/2)*sqrt(2)*x + e**a*x** 2 + 1)*x**8 - 16*e**(4*a)*log(e**(a/2)*sqrt(2)*x + e**a*x**2 + 1)*x**8 + 6 4*e**(4*a)*log(x)*x**8 + 7*e**(2*a)*x**4 - 1)/(8*x**8*(e**(2*a)*x**4 + 1))