Integrand size = 13, antiderivative size = 56 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=-2 e^{-2 a} x^2+\frac {x^6}{6}-\frac {e^{-2 a} x^2}{1+e^{2 a} x^4}+3 e^{-3 a} \arctan \left (e^a x^2\right ) \] Output:
-2*x^2/exp(2*a)+1/6*x^6-x^2/exp(2*a)/(1+exp(2*a)*x^4)+3*arctan(exp(a)*x^2) /exp(3*a)
Time = 0.33 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{6} x^2 \left (-12 e^{-2 a}+x^4-\frac {6}{e^{2 a}+e^{4 a} x^4}\right )+3 e^{-3 a} \arctan \left (e^a x^2\right ) \] Input:
Integrate[x^5*Tanh[a + 2*Log[x]]^2,x]
Output:
(x^2*(-12/E^(2*a) + x^4 - 6/(E^(2*a) + E^(4*a)*x^4)))/6 + (3*ArcTan[E^a*x^ 2])/E^(3*a)
Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {6071, 963, 27, 959, 807, 262, 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.
\(\displaystyle \int x^5 \tanh ^2(a+2 \log (x)) \, dx\) |
\(\Big \downarrow \) 6071 |
\(\displaystyle \int \frac {x^5 \left (e^{2 a} x^4-1\right )^2}{\left (e^{2 a} x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 963 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-\frac {1}{4} e^{-4 a} \int \frac {4 x^5 \left (5 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-e^{-4 a} \int \frac {x^5 \left (5 e^{4 a}-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-e^{-4 a} \left (6 e^{4 a} \int \frac {x^5}{e^{2 a} x^4+1}dx-\frac {1}{6} e^{4 a} x^6\right )\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \int \frac {x^4}{e^{2 a} x^4+1}dx^2-\frac {1}{6} e^{4 a} x^6\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (e^{-2 a} x^2-e^{-2 a} \int \frac {1}{e^{2 a} x^4+1}dx^2\right )-\frac {1}{6} e^{4 a} x^6\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {x^6}{e^{2 a} x^4+1}-e^{-4 a} \left (3 e^{4 a} \left (e^{-2 a} x^2-e^{-3 a} \arctan \left (e^a x^2\right )\right )-\frac {1}{6} e^{4 a} x^6\right )\) |
Input:
Int[x^5*Tanh[a + 2*Log[x]]^2,x]
Output:
x^6/(1 + E^(2*a)*x^4) - (-1/6*(E^(4*a)*x^6) + 3*E^(4*a)*(x^2/E^(2*a) - Arc Tan[E^a*x^2]/E^(3*a)))/E^(4*a)
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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1) /(a*b^2*e*n*(p + 1))), x] + Simp[1/(a*b^2*n*(p + 1)) Int[(e*x)^m*(a + b*x ^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]
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 complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.38
method | result | size |
risch | \(\frac {x^{6}}{6}-2 \,{\mathrm e}^{-2 a} x^{2}+\frac {8 \,{\mathrm e}^{-3 a}}{3}-\frac {x^{2} {\mathrm e}^{-2 a}}{1+{\mathrm e}^{2 a} x^{4}}+\frac {3 i {\mathrm e}^{-3 a} \ln \left ({\mathrm e}^{a} x^{2}+i\right )}{2}-\frac {3 i {\mathrm e}^{-3 a} \ln \left ({\mathrm e}^{a} x^{2}-i\right )}{2}\) | \(77\) |
Input:
int(x^5*tanh(a+2*ln(x))^2,x,method=_RETURNVERBOSE)
Output:
1/6*x^6-2/exp(a)^2*x^2+8/3/exp(a)^3-x^2/exp(a)^2/(exp(a)^2*x^4+1)+3/2*I/ex p(a)^3*ln(exp(a)*x^2+I)-3/2*I/exp(a)^3*ln(exp(a)*x^2-I)
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\frac {x^{10} e^{\left (5 \, a\right )} - 11 \, x^{6} e^{\left (3 \, a\right )} - 18 \, x^{2} e^{a} + 18 \, {\left (x^{4} e^{\left (2 \, a\right )} + 1\right )} \arctan \left (x^{2} e^{a}\right )}{6 \, {\left (x^{4} e^{\left (5 \, a\right )} + e^{\left (3 \, a\right )}\right )}} \] Input:
integrate(x^5*tanh(a+2*log(x))^2,x, algorithm="fricas")
Output:
1/6*(x^10*e^(5*a) - 11*x^6*e^(3*a) - 18*x^2*e^a + 18*(x^4*e^(2*a) + 1)*arc tan(x^2*e^a))/(x^4*e^(5*a) + e^(3*a))
\[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\int x^{5} \tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \] Input:
integrate(x**5*tanh(a+2*ln(x))**2,x)
Output:
Integral(x**5*tanh(a + 2*log(x))**2, x)
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\frac {1}{6} \, {\left (x^{6} e^{\left (2 \, a\right )} - 12 \, x^{2}\right )} e^{\left (-2 \, a\right )} + 3 \, \arctan \left (x^{2} e^{a}\right ) e^{\left (-3 \, a\right )} - \frac {x^{2}}{x^{4} e^{\left (4 \, a\right )} + e^{\left (2 \, a\right )}} \] Input:
integrate(x^5*tanh(a+2*log(x))^2,x, algorithm="maxima")
Output:
1/6*(x^6*e^(2*a) - 12*x^2)*e^(-2*a) + 3*arctan(x^2*e^a)*e^(-3*a) - x^2/(x^ 4*e^(4*a) + e^(2*a))
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=-\frac {x^{2} e^{\left (-2 \, a\right )}}{x^{4} e^{\left (2 \, a\right )} + 1} + 3 \, \arctan \left (x^{2} e^{a}\right ) e^{\left (-3 \, a\right )} + \frac {1}{6} \, {\left (x^{6} e^{\left (12 \, a\right )} - 12 \, x^{2} e^{\left (10 \, a\right )}\right )} e^{\left (-12 \, a\right )} \] Input:
integrate(x^5*tanh(a+2*log(x))^2,x, algorithm="giac")
Output:
-x^2*e^(-2*a)/(x^4*e^(2*a) + 1) + 3*arctan(x^2*e^a)*e^(-3*a) + 1/6*(x^6*e^ (12*a) - 12*x^2*e^(10*a))*e^(-12*a)
Time = 2.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\frac {3\,\mathrm {atan}\left (x^2\,\sqrt {{\mathrm {e}}^{2\,a}}\right )}{{\left ({\mathrm {e}}^{2\,a}\right )}^{3/2}}-2\,x^2\,{\mathrm {e}}^{-2\,a}-\frac {x^2}{{\mathrm {e}}^{4\,a}\,x^4+{\mathrm {e}}^{2\,a}}+\frac {x^6}{6} \] Input:
int(x^5*tanh(a + 2*log(x))^2,x)
Output:
(3*atan(x^2*exp(2*a)^(1/2)))/exp(2*a)^(3/2) - 2*x^2*exp(-2*a) - x^2/(exp(2 *a) + x^4*exp(4*a)) + x^6/6
Time = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 3.32 \[ \int x^5 \tanh ^2(a+2 \log (x)) \, dx=\frac {-18 e^{2 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}-18 \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}-2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )-18 e^{2 a} \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right ) x^{4}-18 \mathit {atan} \left (\frac {e^{\frac {a}{2}} \sqrt {2}+2 e^{a} x}{e^{\frac {a}{2}} \sqrt {2}}\right )+e^{5 a} x^{10}-11 e^{3 a} x^{6}-18 e^{a} x^{2}}{6 e^{3 a} \left (e^{2 a} x^{4}+1\right )} \] Input:
int(x^5*tanh(a+2*log(x))^2,x)
Output:
( - 18*e**(2*a)*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt(2)))*x** 4 - 18*atan((e**(a/2)*sqrt(2) - 2*e**a*x)/(e**(a/2)*sqrt(2))) - 18*e**(2*a )*atan((e**(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2)))*x**4 - 18*atan((e **(a/2)*sqrt(2) + 2*e**a*x)/(e**(a/2)*sqrt(2))) + e**(5*a)*x**10 - 11*e**( 3*a)*x**6 - 18*e**a*x**2)/(6*e**(3*a)*(e**(2*a)*x**4 + 1))