Integrand size = 13, antiderivative size = 60 \[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\frac {(e x)^{1+m}}{e (1+m)}-\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-e^{2 a} x^4\right )}{e (1+m)} \] Output:
(e*x)^(1+m)/e/(1+m)-2*(e*x)^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-ex p(2*a)*x^4)/e/(1+m)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=-\frac {x (e x)^m \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{1+m} \] Input:
Integrate[(e*x)^m*Tanh[a + 2*Log[x]],x]
Output:
-((x*(e*x)^m*(-1 + 2*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, -(x^4*(Cos h[2*a] + Sinh[2*a]))]))/(1 + m))
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6071, 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.
\(\displaystyle \int (e x)^m \tanh (a+2 \log (x)) \, dx\) |
\(\Big \downarrow \) 6071 |
\(\displaystyle \int \frac {\left (e^{2 a} x^4-1\right ) (e x)^m}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(e x)^{m+1}}{e (m+1)}-2 \int \frac {(e x)^m}{e^{2 a} x^4+1}dx\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {(e x)^{m+1}}{e (m+1)}-\frac {2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-e^{2 a} x^4\right )}{e (m+1)}\) |
Input:
Int[(e*x)^m*Tanh[a + 2*Log[x]],x]
Output:
(e*x)^(1 + m)/(e*(1 + m)) - (2*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/ 4, (5 + m)/4, -(E^(2*a)*x^4)])/(e*(1 + m))
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_.)*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]
\[\int \left (e x \right )^{m} \tanh \left (a +2 \ln \left (x \right )\right )d x\]
Input:
int((e*x)^m*tanh(a+2*ln(x)),x)
Output:
int((e*x)^m*tanh(a+2*ln(x)),x)
\[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right ) \,d x } \] Input:
integrate((e*x)^m*tanh(a+2*log(x)),x, algorithm="fricas")
Output:
integral((e*x)^m*tanh(a + 2*log(x)), x)
\[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\int \left (e x\right )^{m} \tanh {\left (a + 2 \log {\left (x \right )} \right )}\, dx \] Input:
integrate((e*x)**m*tanh(a+2*ln(x)),x)
Output:
Integral((e*x)**m*tanh(a + 2*log(x)), x)
\[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right ) \,d x } \] Input:
integrate((e*x)^m*tanh(a+2*log(x)),x, algorithm="maxima")
Output:
integrate((e*x)^m*tanh(a + 2*log(x)), x)
\[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \tanh \left (a + 2 \, \log \left (x\right )\right ) \,d x } \] Input:
integrate((e*x)^m*tanh(a+2*log(x)),x, algorithm="giac")
Output:
integrate((e*x)^m*tanh(a + 2*log(x)), x)
Timed out. \[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\int \mathrm {tanh}\left (a+2\,\ln \left (x\right )\right )\,{\left (e\,x\right )}^m \,d x \] Input:
int(tanh(a + 2*log(x))*(e*x)^m,x)
Output:
int(tanh(a + 2*log(x))*(e*x)^m, x)
\[ \int (e x)^m \tanh (a+2 \log (x)) \, dx=\frac {e^{m} \left (x^{m} x -2 \left (\int \frac {x^{m}}{e^{2 a} x^{4}+1}d x \right ) m -2 \left (\int \frac {x^{m}}{e^{2 a} x^{4}+1}d x \right )\right )}{m +1} \] Input:
int((e*x)^m*tanh(a+2*log(x)),x)
Output:
(e**m*(x**m*x - 2*int(x**m/(e**(2*a)*x**4 + 1),x)*m - 2*int(x**m/(e**(2*a) *x**4 + 1),x)))/(m + 1)