Integrand size = 15, antiderivative size = 79 \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\frac {(e x)^{1+m}}{e (1+m)}+\frac {(e x)^{1+m}}{e \left (1+e^{2 a} x^4\right )}-\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-e^{2 a} x^4\right )}{e} \]
(e*x)^(1+m)/e/(1+m)+(e*x)^(1+m)/e/(1+exp(2*a)*x^4)-(e*x)^(1+m)*hypergeom([ 1, 1/4+1/4*m],[5/4+1/4*m],-exp(2*a)*x^4)/e
Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=-\frac {x (e x)^m \left (-1+4 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-x^4 (\cosh (2 a)+\sinh (2 a))\right )-4 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{4},\frac {5+m}{4},-x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{1+m} \]
-((x*(e*x)^m*(-1 + 4*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, -(x^4*(Cos h[2*a] + Sinh[2*a]))] - 4*Hypergeometric2F1[2, (1 + m)/4, (5 + m)/4, -(x^4 *(Cosh[2*a] + Sinh[2*a]))]))/(1 + m))
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6071, 963, 27, 959, 888}
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.
\(\displaystyle \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx\) |
\(\Big \downarrow \) 6071 |
\(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right )^2 (e x)^m}{\left (e^{2 a} x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 963 |
\(\displaystyle \frac {(e x)^{m+1}}{e \left (e^{2 a} x^4+1\right )}-\frac {1}{4} e^{-4 a} \int \frac {4 (e x)^m \left (e^{4 a} m-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(e x)^{m+1}}{e \left (e^{2 a} x^4+1\right )}-e^{-4 a} \int \frac {(e x)^m \left (e^{4 a} m-e^{6 a} x^4\right )}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(e x)^{m+1}}{e \left (e^{2 a} x^4+1\right )}-e^{-4 a} \left (e^{4 a} (m+1) \int \frac {(e x)^m}{e^{2 a} x^4+1}dx-\frac {e^{4 a} (e x)^{m+1}}{e (m+1)}\right )\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(e x)^{m+1}}{e \left (e^{2 a} x^4+1\right )}-e^{-4 a} \left (\frac {e^{4 a} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-e^{2 a} x^4\right )}{e}-\frac {e^{4 a} (e x)^{m+1}}{e (m+1)}\right )\) |
(e*x)^(1 + m)/(e*(1 + E^(2*a)*x^4)) - (-((E^(4*a)*(e*x)^(1 + m))/(e*(1 + m ))) + (E^(4*a)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, -( E^(2*a)*x^4)])/e)/E^(4*a)
3.2.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
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]
\[\int \left (e x \right )^{m} \tanh \left (a +2 \ln \left (x \right )\right )^{2}d x\]
\[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right )^{2} \,d x } \]
\[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int \left (e x\right )^{m} \tanh ^{2}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \]
\[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right )^{2} \,d x } \]
\[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right )^{2} \,d x } \]
Timed out. \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int {\mathrm {tanh}\left (a+2\,\ln \left (x\right )\right )}^2\,{\left (e\,x\right )}^m \,d x \]