Integrand size = 16, antiderivative size = 84 \[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\frac {e^{a+b x} \left (1-e^{2 a+2 b x}\right )^n \left (1+e^{2 a+2 b x}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},n,-n,\frac {3}{2},e^{2 a+2 b x},-e^{2 a+2 b x}\right ) \coth ^n(a+b x)}{b} \] Output:
exp(b*x+a)*(1-exp(2*b*x+2*a))^n*AppellF1(1/2,-n,n,3/2,-exp(2*b*x+2*a),exp( 2*b*x+2*a))*coth(b*x+a)^n/b/((1+exp(2*b*x+2*a))^n)
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.26 \[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\frac {e^{a+b x} \left (-1+e^{-2 (a+b x)}\right )^{-n} \left (1+e^{-2 (a+b x)}\right )^{-n} \left (-e^{-4 (a+b x)} \left (-1+e^{2 (a+b x)}\right )^2\right )^n \operatorname {AppellF1}\left (-\frac {1}{2},-n,n,\frac {1}{2},-e^{-2 (a+b x)},e^{-2 (a+b x)}\right ) \coth ^n(a+b x)}{b} \] Input:
Integrate[E^(a + b*x)*Coth[a + b*x]^n,x]
Output:
(E^(a + b*x)*(-((-1 + E^(2*(a + b*x)))^2/E^(4*(a + b*x))))^n*AppellF1[-1/2 , -n, n, 1/2, -E^(-2*(a + b*x)), E^(-2*(a + b*x))]*Coth[a + b*x]^n)/(b*(-1 + E^(-2*(a + b*x)))^n*(1 + E^(-2*(a + b*x)))^n)
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 2050, 334, 333}
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.
\(\displaystyle \int e^{a+b x} \coth ^n(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \left (-\frac {1+e^{2 a+2 b x}}{1-e^{2 a+2 b x}}\right )^nde^{a+b x}}{b}\) |
\(\Big \downarrow \) 2050 |
\(\displaystyle \frac {\int \left (-1-e^{2 a+2 b x}\right )^n \left (1-e^{2 a+2 b x}\right )^{-n}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\left (-e^{2 a+2 b x}-1\right )^n \left (e^{2 a+2 b x}+1\right )^{-n} \int \left (1-e^{2 a+2 b x}\right )^{-n} \left (1+e^{2 a+2 b x}\right )^nde^{a+b x}}{b}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {e^{a+b x} \left (-e^{2 a+2 b x}-1\right )^n \left (e^{2 a+2 b x}+1\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},n,-n,\frac {3}{2},e^{2 a+2 b x},-e^{2 a+2 b x}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Coth[a + b*x]^n,x]
Output:
(E^(a + b*x)*(-1 - E^(2*a + 2*b*x))^n*AppellF1[1/2, n, -n, 3/2, E^(2*a + 2 *b*x), -E^(2*a + 2*b*x)])/(b*(1 + E^(2*a + 2*b*x))^n)
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p _), x_Symbol] :> Int[u*((a*e + b*e*x^n)^p/(c + d*x^n)^p), x] /; FreeQ[{a, b , c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - a*(d/b), 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
\[\int {\mathrm e}^{b x +a} \coth \left (b x +a \right )^{n}d x\]
Input:
int(exp(b*x+a)*coth(b*x+a)^n,x)
Output:
int(exp(b*x+a)*coth(b*x+a)^n,x)
\[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\int { \coth \left (b x + a\right )^{n} e^{\left (b x + a\right )} \,d x } \] Input:
integrate(exp(b*x+a)*coth(b*x+a)^n,x, algorithm="fricas")
Output:
integral(coth(b*x + a)^n*e^(b*x + a), x)
\[ \int e^{a+b x} \coth ^n(a+b x) \, dx=e^{a} \int e^{b x} \coth ^{n}{\left (a + b x \right )}\, dx \] Input:
integrate(exp(b*x+a)*coth(b*x+a)**n,x)
Output:
exp(a)*Integral(exp(b*x)*coth(a + b*x)**n, x)
\[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\int { \coth \left (b x + a\right )^{n} e^{\left (b x + a\right )} \,d x } \] Input:
integrate(exp(b*x+a)*coth(b*x+a)^n,x, algorithm="maxima")
Output:
integrate(coth(b*x + a)^n*e^(b*x + a), x)
\[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\int { \coth \left (b x + a\right )^{n} e^{\left (b x + a\right )} \,d x } \] Input:
integrate(exp(b*x+a)*coth(b*x+a)^n,x, algorithm="giac")
Output:
integrate(coth(b*x + a)^n*e^(b*x + a), x)
Timed out. \[ \int e^{a+b x} \coth ^n(a+b x) \, dx=\int {\mathrm {coth}\left (a+b\,x\right )}^n\,{\mathrm {e}}^{a+b\,x} \,d x \] Input:
int(coth(a + b*x)^n*exp(a + b*x),x)
Output:
int(coth(a + b*x)^n*exp(a + b*x), x)
\[ \int e^{a+b x} \coth ^n(a+b x) \, dx=e^{a} \left (\int e^{b x} \coth \left (b x +a \right )^{n}d x \right ) \] Input:
int(exp(b*x+a)*coth(b*x+a)^n,x)
Output:
e**a*int(e**(b*x)*coth(a + b*x)**n,x)