Integrand size = 13, antiderivative size = 55 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=-\frac {1}{6 x^6}+\frac {2 e^{2 a}}{x^2}+\frac {e^{4 a} x^2}{1+e^{2 a} x^4}+3 e^{3 a} \arctan \left (e^a x^2\right ) \] Output:
-1/6/x^6+2*exp(2*a)/x^2+exp(4*a)*x^2/(1+exp(2*a)*x^4)+3*exp(3*a)*arctan(ex p(a)*x^2)
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.04 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=-\frac {1}{6 x^6}+\frac {2 e^{2 a}}{x^2}+\frac {e^{4 a} x^2}{1+e^{2 a} x^4}-3 e^{3 a} \arctan \left (\frac {e^{-a}}{x^2}\right ) \] Input:
Integrate[Tanh[a + 2*Log[x]]^2/x^7,x]
Output:
-1/6*1/x^6 + (2*E^(2*a))/x^2 + (E^(4*a)*x^2)/(1 + E^(2*a)*x^4) - 3*E^(3*a) *ArcTan[1/(E^a*x^2)]
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.40, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {6071, 962, 27, 957, 807, 264, 216}
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.
\(\displaystyle \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx\) |
\(\Big \downarrow \) 6071 |
\(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right )^2}{x^7 \left (e^{2 a} x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 962 |
\(\displaystyle \frac {1}{6} \int -\frac {2 \left (11 e^{2 a}-3 e^{4 a} x^4\right )}{x^3 \left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} \int \frac {11 e^{2 a}-3 e^{4 a} x^4}{x^3 \left (e^{2 a} x^4+1\right )^2}dx-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {1}{3} \left (-18 e^{2 a} \int \frac {1}{x^3 \left (e^{2 a} x^4+1\right )}dx-\frac {7 e^{2 a}}{2 x^2 \left (e^{2 a} x^4+1\right )}\right )-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \int \frac {1}{x^4 \left (e^{2 a} x^4+1\right )}dx^2-\frac {7 e^{2 a}}{2 x^2 \left (e^{2 a} x^4+1\right )}\right )-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (-e^{2 a} \int \frac {1}{e^{2 a} x^4+1}dx^2-\frac {1}{x^2}\right )-\frac {7 e^{2 a}}{2 x^2 \left (e^{2 a} x^4+1\right )}\right )-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (-9 e^{2 a} \left (-e^a \arctan \left (e^a x^2\right )-\frac {1}{x^2}\right )-\frac {7 e^{2 a}}{2 x^2 \left (e^{2 a} x^4+1\right )}\right )-\frac {1}{6 x^6 \left (e^{2 a} x^4+1\right )}\) |
Input:
Int[Tanh[a + 2*Log[x]]^2/x^7,x]
Output:
-1/6*1/(x^6*(1 + E^(2*a)*x^4)) + ((-7*E^(2*a))/(2*x^2*(1 + E^(2*a)*x^4)) - 9*E^(2*a)*(-x^(-2) - E^a*ArcTan[E^a*x^2]))/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^2, x_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1)) ), x] - Simp[1/(a*e^n*(m + 1)) Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b*c^2 *n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; Fre eQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.44
method | result | size |
risch | \(\frac {3 \,{\mathrm e}^{4 a} x^{8}+\frac {11 \,{\mathrm e}^{2 a} x^{4}}{6}-\frac {1}{6}}{x^{6} \left (1+{\mathrm e}^{2 a} x^{4}\right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left ({\mathrm e}^{6 a}+\textit {\_Z}^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (4 \,{\mathrm e}^{6 a}+5 \textit {\_R}^{2}\right ) x^{2}-{\mathrm e}^{2 a} \textit {\_R} \right )\right )}{2}\) | \(79\) |
Input:
int(tanh(a+2*ln(x))^2/x^7,x,method=_RETURNVERBOSE)
Output:
(3*exp(4*a)*x^8+11/6*exp(2*a)*x^4-1/6)/x^6/(1+exp(2*a)*x^4)+3/2*sum(_R*ln( (4*exp(6*a)+5*_R^2)*x^2-exp(2*a)*_R),_R=RootOf(exp(6*a)+_Z^2))
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.13 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=\frac {18 \, x^{8} e^{\left (4 \, a\right )} + 11 \, x^{4} e^{\left (2 \, a\right )} + 18 \, {\left (x^{10} e^{\left (5 \, a\right )} + x^{6} e^{\left (3 \, a\right )}\right )} \arctan \left (x^{2} e^{a}\right ) - 1}{6 \, {\left (x^{10} e^{\left (2 \, a\right )} + x^{6}\right )}} \] Input:
integrate(tanh(a+2*log(x))^2/x^7,x, algorithm="fricas")
Output:
1/6*(18*x^8*e^(4*a) + 11*x^4*e^(2*a) + 18*(x^10*e^(5*a) + x^6*e^(3*a))*arc tan(x^2*e^a) - 1)/(x^10*e^(2*a) + x^6)
\[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=\int \frac {\tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}}{x^{7}}\, dx \] Input:
integrate(tanh(a+2*ln(x))**2/x**7,x)
Output:
Integral(tanh(a + 2*log(x))**2/x**7, x)
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=-3 \, \arctan \left (\frac {e^{\left (-a\right )}}{x^{2}}\right ) e^{\left (3 \, a\right )} + \frac {2 \, e^{\left (2 \, a\right )}}{x^{2}} + \frac {e^{\left (4 \, a\right )}}{x^{2} {\left (\frac {1}{x^{4}} + e^{\left (2 \, a\right )}\right )}} - \frac {1}{6 \, x^{6}} \] Input:
integrate(tanh(a+2*log(x))^2/x^7,x, algorithm="maxima")
Output:
-3*arctan(e^(-a)/x^2)*e^(3*a) + 2*e^(2*a)/x^2 + e^(4*a)/(x^2*(1/x^4 + e^(2 *a))) - 1/6/x^6
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=\frac {x^{2} e^{\left (4 \, a\right )}}{x^{4} e^{\left (2 \, a\right )} + 1} + 3 \, \arctan \left (x^{2} e^{a}\right ) e^{\left (3 \, a\right )} + \frac {12 \, x^{4} e^{\left (2 \, a\right )} - 1}{6 \, x^{6}} \] Input:
integrate(tanh(a+2*log(x))^2/x^7,x, algorithm="giac")
Output:
x^2*e^(4*a)/(x^4*e^(2*a) + 1) + 3*arctan(x^2*e^a)*e^(3*a) + 1/6*(12*x^4*e^ (2*a) - 1)/x^6
Time = 2.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=3\,\mathrm {atan}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )\,{\left ({\mathrm {e}}^{2\,a}\right )}^{3/2}+\frac {3\,{\mathrm {e}}^{4\,a}\,x^8+\frac {11\,{\mathrm {e}}^{2\,a}\,x^4}{6}-\frac {1}{6}}{{\mathrm {e}}^{2\,a}\,x^{10}+x^6} \] Input:
int(tanh(a + 2*log(x))^2/x^7,x)
Output:
3*atan(x^2*exp(2*a)^(1/2))*exp(2*a)^(3/2) + ((11*x^4*exp(2*a))/6 + 3*x^8*e xp(4*a) - 1/6)/(x^10*exp(2*a) + x^6)
Time = 0.25 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.49 \[ \int \frac {\tanh ^2(a+2 \log (x))}{x^7} \, dx=\frac {-18 e^{5 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{10}-18 e^{3 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{6}-18 e^{5 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{10}-18 e^{3 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{6}+18 e^{4 a} x^{8}+11 e^{2 a} x^{4}-1}{6 x^{6} \left (e^{2 a} x^{4}+1\right )} \] Input:
int(tanh(a+2*log(x))^2/x^7,x)
Output:
( - 18*e**(5*a)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt(2)))*x** 10 - 18*e**(3*a)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt(2)))*x* *6 - 18*e**(5*a)*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2)))*x* *10 - 18*e**(3*a)*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2)))*x **6 + 18*e**(4*a)*x**8 + 11*e**(2*a)*x**4 - 1)/(6*x**6*(e**(2*a)*x**4 + 1) )