Integrand size = 16, antiderivative size = 87 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=-\frac {4 e^{a-d}}{b \left (1-e^{2 d+2 b x}\right )^4}+\frac {32 e^{a-d}}{3 b \left (1-e^{2 d+2 b x}\right )^3}-\frac {8 e^{a-d}}{b \left (1-e^{2 d+2 b x}\right )^2} \] Output:
-4*exp(a-d)/b/(1-exp(2*b*x+2*d))^4+32/3*exp(a-d)/b/(1-exp(2*b*x+2*d))^3-8* exp(a-d)/b/(1-exp(2*b*x+2*d))^2
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=-\frac {4 e^a \left (-4 e^{2 b x}+\left (1+6 e^{4 b x}\right ) \cosh (2 d)+\left (-1+6 e^{4 b x}\right ) \sinh (2 d)\right ) (\cosh (3 d)-\sinh (3 d))}{3 b \left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^4} \] Input:
Integrate[E^(a + b*x)*Csch[d + b*x]^5,x]
Output:
(-4*E^a*(-4*E^(2*b*x) + (1 + 6*E^(4*b*x))*Cosh[2*d] + (-1 + 6*E^(4*b*x))*S inh[2*d])*(Cosh[3*d] - Sinh[3*d]))/(3*b*((-1 + E^(2*b*x))*Cosh[d] + (1 + E ^(2*b*x))*Sinh[d])^4)
Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2720, 27, 243, 53, 2009}
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.
\(\displaystyle \int e^{a+b x} \text {csch}^5(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {32 e^{a+5 b x}}{\left (1-e^{2 b x}\right )^5}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {32 e^a \int \frac {e^{5 b x}}{\left (1-e^{2 b x}\right )^5}de^{b x}}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {16 e^a \int \frac {e^{2 b x}}{\left (1-e^{2 b x}\right )^5}de^{2 b x}}{b}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {16 e^a \int \left (-\frac {1}{\left (-1+e^{2 b x}\right )^3}-\frac {2}{\left (-1+e^{2 b x}\right )^4}-\frac {1}{\left (-1+e^{2 b x}\right )^5}\right )de^{2 b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 e^a \left (\frac {1}{2 \left (1-e^{2 b x}\right )^2}-\frac {2}{3 \left (1-e^{2 b x}\right )^3}+\frac {1}{4 \left (1-e^{2 b x}\right )^4}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Csch[d + b*x]^5,x]
Output:
(-16*E^a*(1/(4*(1 - E^(2*b*x))^4) - 2/(3*(1 - E^(2*b*x))^3) + 1/(2*(1 - E^ (2*b*x))^2)))/b
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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]]]
Time = 1.35 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {4 \left (6 \,{\mathrm e}^{4 b x +4 a +4 d}-4 \,{\mathrm e}^{2 b x +4 a +2 d}+{\mathrm e}^{4 a}\right ) {\mathrm e}^{5 a -d}}{3 \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{4} b}\) | \(68\) |
parallelrisch | \(\frac {{\mathrm e}^{b x +a} \left (3 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}-3 \coth \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+2 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-2 \coth \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-22 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+22 \coth \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}-38 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )+38 \coth \left (\frac {b x}{2}+\frac {d}{2}\right )\right )}{192 b}\) | \(113\) |
Input:
int(exp(b*x+a)*csch(b*x+d)^5,x,method=_RETURNVERBOSE)
Output:
-4/3/(-exp(2*b*x+2*a+2*d)+exp(2*a))^4/b*(6*exp(4*b*x+4*a+4*d)-4*exp(2*b*x+ 4*a+2*d)+exp(4*a))*exp(5*a-d)
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (73) = 146\).
Time = 0.08 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.48 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=-\frac {4 \, {\left (7 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + 7 \, {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} + 10 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (7 \, \cosh \left (b x + d\right )^{2} - 4\right )} \sinh \left (-a + d\right ) - 4 \, \cosh \left (-a + d\right )\right )}}{3 \, {\left (b \cosh \left (b x + d\right )^{6} + 6 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right )^{5} + b \sinh \left (b x + d\right )^{6} - 4 \, b \cosh \left (b x + d\right )^{4} + {\left (15 \, b \cosh \left (b x + d\right )^{2} - 4 \, b\right )} \sinh \left (b x + d\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + d\right )^{3} - 4 \, b \cosh \left (b x + d\right )\right )} \sinh \left (b x + d\right )^{3} + 7 \, b \cosh \left (b x + d\right )^{2} + {\left (15 \, b \cosh \left (b x + d\right )^{4} - 24 \, b \cosh \left (b x + d\right )^{2} + 7 \, b\right )} \sinh \left (b x + d\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + d\right )^{5} - 8 \, b \cosh \left (b x + d\right )^{3} + 5 \, b \cosh \left (b x + d\right )\right )} \sinh \left (b x + d\right ) - 4 \, b\right )}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^5,x, algorithm="fricas")
Output:
-4/3*(7*cosh(b*x + d)^2*cosh(-a + d) + 7*(cosh(-a + d) - sinh(-a + d))*sin h(b*x + d)^2 + 10*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d) )*sinh(b*x + d) - (7*cosh(b*x + d)^2 - 4)*sinh(-a + d) - 4*cosh(-a + d))/( b*cosh(b*x + d)^6 + 6*b*cosh(b*x + d)*sinh(b*x + d)^5 + b*sinh(b*x + d)^6 - 4*b*cosh(b*x + d)^4 + (15*b*cosh(b*x + d)^2 - 4*b)*sinh(b*x + d)^4 + 4*( 5*b*cosh(b*x + d)^3 - 4*b*cosh(b*x + d))*sinh(b*x + d)^3 + 7*b*cosh(b*x + d)^2 + (15*b*cosh(b*x + d)^4 - 24*b*cosh(b*x + d)^2 + 7*b)*sinh(b*x + d)^2 + 2*(3*b*cosh(b*x + d)^5 - 8*b*cosh(b*x + d)^3 + 5*b*cosh(b*x + d))*sinh( b*x + d) - 4*b)
\[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=e^{a} \int e^{b x} \operatorname {csch}^{5}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*csch(b*x+d)**5,x)
Output:
exp(a)*Integral(exp(b*x)*csch(b*x + d)**5, x)
Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (73) = 146\).
Time = 0.04 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.68 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=-\frac {8 \, e^{\left (4 \, b x + 9 \, a + 4 \, d\right )}}{b {\left (e^{\left (8 \, b x + 8 \, a + 9 \, d\right )} - 4 \, e^{\left (6 \, b x + 8 \, a + 7 \, d\right )} + 6 \, e^{\left (4 \, b x + 8 \, a + 5 \, d\right )} - 4 \, e^{\left (2 \, b x + 8 \, a + 3 \, d\right )} + e^{\left (8 \, a + d\right )}\right )}} + \frac {16 \, e^{\left (2 \, b x + 9 \, a + 2 \, d\right )}}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a + 9 \, d\right )} - 4 \, e^{\left (6 \, b x + 8 \, a + 7 \, d\right )} + 6 \, e^{\left (4 \, b x + 8 \, a + 5 \, d\right )} - 4 \, e^{\left (2 \, b x + 8 \, a + 3 \, d\right )} + e^{\left (8 \, a + d\right )}\right )}} - \frac {4 \, e^{\left (9 \, a\right )}}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a + 9 \, d\right )} - 4 \, e^{\left (6 \, b x + 8 \, a + 7 \, d\right )} + 6 \, e^{\left (4 \, b x + 8 \, a + 5 \, d\right )} - 4 \, e^{\left (2 \, b x + 8 \, a + 3 \, d\right )} + e^{\left (8 \, a + d\right )}\right )}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^5,x, algorithm="maxima")
Output:
-8*e^(4*b*x + 9*a + 4*d)/(b*(e^(8*b*x + 8*a + 9*d) - 4*e^(6*b*x + 8*a + 7* d) + 6*e^(4*b*x + 8*a + 5*d) - 4*e^(2*b*x + 8*a + 3*d) + e^(8*a + d))) + 1 6/3*e^(2*b*x + 9*a + 2*d)/(b*(e^(8*b*x + 8*a + 9*d) - 4*e^(6*b*x + 8*a + 7 *d) + 6*e^(4*b*x + 8*a + 5*d) - 4*e^(2*b*x + 8*a + 3*d) + e^(8*a + d))) - 4/3*e^(9*a)/(b*(e^(8*b*x + 8*a + 9*d) - 4*e^(6*b*x + 8*a + 7*d) + 6*e^(4*b *x + 8*a + 5*d) - 4*e^(2*b*x + 8*a + 3*d) + e^(8*a + d)))
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.55 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=-\frac {4 \, {\left (6 \, e^{\left (4 \, b x + 4 \, d\right )} - 4 \, e^{\left (2 \, b x + 2 \, d\right )} + 1\right )} e^{\left (a - d\right )}}{3 \, b {\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{4}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^5,x, algorithm="giac")
Output:
-4/3*(6*e^(4*b*x + 4*d) - 4*e^(2*b*x + 2*d) + 1)*e^(a - d)/(b*(e^(2*b*x + 2*d) - 1)^4)
Timed out. \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {sinh}\left (d+b\,x\right )}^5} \,d x \] Input:
int(exp(a + b*x)/sinh(d + b*x)^5,x)
Output:
int(exp(a + b*x)/sinh(d + b*x)^5, x)
Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int e^{a+b x} \text {csch}^5(d+b x) \, dx=\frac {4 e^{a} \left (-6 e^{4 b x +4 d}+4 e^{2 b x +2 d}-1\right )}{3 e^{d} b \left (e^{8 b x +8 d}-4 e^{6 b x +6 d}+6 e^{4 b x +4 d}-4 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*csch(b*x+d)^5,x)
Output:
(4*e**a*( - 6*e**(4*b*x + 4*d) + 4*e**(2*b*x + 2*d) - 1))/(3*e**d*b*(e**(8 *b*x + 8*d) - 4*e**(6*b*x + 6*d) + 6*e**(4*b*x + 4*d) - 4*e**(2*b*x + 2*d) + 1))