\(\int (e x)^m \coth (a+2 \log (x)) \, dx\) [165]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 13, antiderivative size = 59 \[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\frac {(e x)^{1+m}}{e (1+m)}-\frac {2 (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},e^{2 a} x^4\right )}{e (1+m)} \]

[Out]

(e*x)^(1+m)/e/(1+m)-2*(e*x)^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],exp(2*a)*x^4)/e/(1+m)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, 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 = {5657, 470, 371} \[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\frac {(e x)^{m+1}}{e (m+1)}-\frac {2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},e^{2 a} x^4\right )}{e (m+1)} \]

[In]

Int[(e*x)^m*Coth[a + 2*Log[x]],x]

[Out]

(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))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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] - Dist[(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]

Rule 5657

Int[Coth[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), 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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(e x)^m \left (-1-e^{2 a} x^4\right )}{1-e^{2 a} x^4} \, dx \\ & = \frac {(e x)^{1+m}}{e (1+m)}-2 \int \frac {(e x)^m}{1-e^{2 a} x^4} \, 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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int (e x)^m \coth (a+2 \log (x)) \, dx=-\frac {x (e x)^m \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{1+m} \]

[In]

Integrate[(e*x)^m*Coth[a + 2*Log[x]],x]

[Out]

-((x*(e*x)^m*(-1 + 2*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, x^4*(Cosh[2*a] + Sinh[2*a])]))/(1 + m))

Maple [F]

\[\int \left (e x \right )^{m} \coth \left (a +2 \ln \left (x \right )\right )d x\]

[In]

int((e*x)^m*coth(a+2*ln(x)),x)

[Out]

int((e*x)^m*coth(a+2*ln(x)),x)

Fricas [F]

\[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \coth \left (a + 2 \, \log \left (x\right )\right ) \,d x } \]

[In]

integrate((e*x)^m*coth(a+2*log(x)),x, algorithm="fricas")

[Out]

integral((e*x)^m*coth(a + 2*log(x)), x)

Sympy [F]

\[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\int \left (e x\right )^{m} \coth {\left (a + 2 \log {\left (x \right )} \right )}\, dx \]

[In]

integrate((e*x)**m*coth(a+2*ln(x)),x)

[Out]

Integral((e*x)**m*coth(a + 2*log(x)), x)

Maxima [F]

\[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \coth \left (a + 2 \, \log \left (x\right )\right ) \,d x } \]

[In]

integrate((e*x)^m*coth(a+2*log(x)),x, algorithm="maxima")

[Out]

integrate((e*x)^m*coth(a + 2*log(x)), x)

Giac [F]

\[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\int { \left (e x\right )^{m} \coth \left (a + 2 \, \log \left (x\right )\right ) \,d x } \]

[In]

integrate((e*x)^m*coth(a+2*log(x)),x, algorithm="giac")

[Out]

integrate((e*x)^m*coth(a + 2*log(x)), x)

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \coth (a+2 \log (x)) \, dx=\int \mathrm {coth}\left (a+2\,\ln \left (x\right )\right )\,{\left (e\,x\right )}^m \,d x \]

[In]

int(coth(a + 2*log(x))*(e*x)^m,x)

[Out]

int(coth(a + 2*log(x))*(e*x)^m, x)