Integrand size = 16, antiderivative size = 109 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\frac {8 e^{a+2 d+3 b x}}{3 b \left (1-e^{2 d+2 b x}\right )^3}-\frac {2 e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )^2}+\frac {e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )}+\frac {e^{a-d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
8/3*exp(3*b*x+a+2*d)/b/(1-exp(2*b*x+2*d))^3-2*exp(b*x+a)/b/(1-exp(2*b*x+2* d))^2+exp(b*x+a)/b/(1-exp(2*b*x+2*d))+exp(a-d)*arctanh(exp(b*x+d))/b
Time = 0.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.50 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\frac {e^a \left (3 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) \cosh (d)-3 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) \sinh (d)-\frac {8 e^{b x} (\cosh (d)-\sinh (d))^3}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^3}-\frac {14 e^{b x} (\cosh (d)-\sinh (d))^2}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}+\frac {3 e^{b x} (-\cosh (d)+\sinh (d))}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}\right )}{3 b} \] Input:
Integrate[E^(a + b*x)*Csch[d + b*x]^4,x]
Output:
(E^a*(3*ArcTanh[E^(b*x)*(Cosh[d] + Sinh[d])]*Cosh[d] - 3*ArcTanh[E^(b*x)*( Cosh[d] + Sinh[d])]*Sinh[d] - (8*E^(b*x)*(Cosh[d] - Sinh[d])^3)/((-1 + E^( 2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^3 - (14*E^(b*x)*(Cosh[d] - Sinh [d])^2)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^2 + (3*E^(b*x )*(-Cosh[d] + Sinh[d]))/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d ])))/(3*b)
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2720, 27, 252, 252, 215, 219}
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.
\(\displaystyle \int e^{a+b x} \text {csch}^4(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {16 e^{a+4 b x}}{\left (1-e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {16 e^a \int \frac {e^{4 b x}}{\left (1-e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 e^a \left (\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}-\frac {1}{2} \int \frac {e^{2 b x}}{\left (1-e^{2 b x}\right )^3}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (1-e^{2 b x}\right )^2}de^{b x}-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{1-e^{2 b x}}de^{b x}+\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}\right )-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {16 e^a \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \text {arctanh}\left (e^{b x}\right )+\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}\right )-\frac {e^{b x}}{4 \left (1-e^{2 b x}\right )^2}\right )+\frac {e^{3 b x}}{6 \left (1-e^{2 b x}\right )^3}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Csch[d + b*x]^4,x]
Output:
(16*E^a*(E^(3*b*x)/(6*(1 - E^(2*b*x))^3) + (-1/4*E^(b*x)/(1 - E^(2*b*x))^2 + (E^(b*x)/(2*(1 - E^(2*b*x))) + ArcTanh[E^(b*x)]/2)/4)/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_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
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 = 0.97 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13
method | result | size |
risch | \(-\frac {\left (-3 \,{\mathrm e}^{4 b x +4 a +4 d}-8 \,{\mathrm e}^{2 b x +4 a +2 d}+3 \,{\mathrm e}^{4 a}\right ) {\mathrm e}^{b x +3 a}}{3 \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{3} b}+\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}-\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(123\) |
Input:
int(exp(b*x+a)*csch(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
-1/3/(-exp(2*b*x+2*a+2*d)+exp(2*a))^3/b*(-3*exp(4*b*x+4*a+4*d)-8*exp(2*b*x +4*a+2*d)+3*exp(4*a))*exp(b*x+3*a)+1/2*ln(exp(b*x+a)+exp(a-d))/b*exp(a-d)- 1/2*ln(exp(b*x+a)-exp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 1466 vs. \(2 (94) = 188\).
Time = 0.09 (sec) , antiderivative size = 1466, normalized size of antiderivative = 13.45 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^4,x, algorithm="fricas")
Output:
-1/6*(6*cosh(b*x + d)^5*cosh(-a + d) + 6*(cosh(-a + d) - sinh(-a + d))*sin h(b*x + d)^5 + 30*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d) )*sinh(b*x + d)^4 + 16*cosh(b*x + d)^3*cosh(-a + d) + 4*(15*cosh(b*x + d)^ 2*cosh(-a + d) - (15*cosh(b*x + d)^2 + 4)*sinh(-a + d) + 4*cosh(-a + d))*s inh(b*x + d)^3 + 12*(5*cosh(b*x + d)^3*cosh(-a + d) + 4*cosh(b*x + d)*cosh (-a + d) - (5*cosh(b*x + d)^3 + 4*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^2 - 6*cosh(b*x + d)*cosh(-a + d) - 3*(cosh(b*x + d)^6*cosh(-a + d) + (c osh(-a + d) - sinh(-a + d))*sinh(b*x + d)^6 + 6*(cosh(b*x + d)*cosh(-a + d ) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^5 - 3*cosh(b*x + d)^4*cosh(- a + d) + 3*(5*cosh(b*x + d)^2*cosh(-a + d) - (5*cosh(b*x + d)^2 - 1)*sinh( -a + d) - cosh(-a + d))*sinh(b*x + d)^4 + 4*(5*cosh(b*x + d)^3*cosh(-a + d ) - 3*cosh(b*x + d)*cosh(-a + d) - (5*cosh(b*x + d)^3 - 3*cosh(b*x + d))*s inh(-a + d))*sinh(b*x + d)^3 + 3*cosh(b*x + d)^2*cosh(-a + d) + 3*(5*cosh( b*x + d)^4*cosh(-a + d) - 6*cosh(b*x + d)^2*cosh(-a + d) - (5*cosh(b*x + d )^4 - 6*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*sinh(b*x + d)^2 + 6*(cosh(b*x + d)^5*cosh(-a + d) - 2*cosh(b*x + d)^3*cosh(-a + d) + cosh( b*x + d)*cosh(-a + d) - (cosh(b*x + d)^5 - 2*cosh(b*x + d)^3 + cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^6 - 3*cosh(b*x + d)^4 + 3 *cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*log(cosh(b*x + d) + sin h(b*x + d) + 1) + 3*(cosh(b*x + d)^6*cosh(-a + d) + (cosh(-a + d) - sin...
\[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=e^{a} \int e^{b x} \operatorname {csch}^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*csch(b*x+d)**4,x)
Output:
exp(a)*Integral(exp(b*x)*csch(b*x + d)**4, x)
Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{2 \, b} - \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{2 \, b} - \frac {3 \, e^{\left (5 \, b x + 7 \, a + 4 \, d\right )} + 8 \, e^{\left (3 \, b x + 7 \, a + 2 \, d\right )} - 3 \, e^{\left (b x + 7 \, a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a + 6 \, d\right )} - 3 \, e^{\left (4 \, b x + 6 \, a + 4 \, d\right )} + 3 \, e^{\left (2 \, b x + 6 \, a + 2 \, d\right )} - e^{\left (6 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^4,x, algorithm="maxima")
Output:
1/2*e^(a - d)*log(e^(b*x + a + d) + e^a)/b - 1/2*e^(a - d)*log(e^(b*x + a + d) - e^a)/b - 1/3*(3*e^(5*b*x + 7*a + 4*d) + 8*e^(3*b*x + 7*a + 2*d) - 3 *e^(b*x + 7*a))/(b*(e^(6*b*x + 6*a + 6*d) - 3*e^(4*b*x + 6*a + 4*d) + 3*e^ (2*b*x + 6*a + 2*d) - e^(6*a)))
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\frac {1}{6} \, {\left (\frac {3 \, e^{\left (-5 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right )}{b} - \frac {3 \, e^{\left (-5 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right )}{b} - \frac {2 \, {\left (3 \, e^{\left (5 \, b x + 4 \, d\right )} + 8 \, e^{\left (3 \, b x + 2 \, d\right )} - 3 \, e^{\left (b x\right )}\right )} e^{\left (-4 \, d\right )}}{b {\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{3}}\right )} e^{\left (a + 4 \, d\right )} \] Input:
integrate(exp(b*x+a)*csch(b*x+d)^4,x, algorithm="giac")
Output:
1/6*(3*e^(-5*d)*log(e^(b*x + d) + 1)/b - 3*e^(-5*d)*log(abs(e^(b*x + d) - 1))/b - 2*(3*e^(5*b*x + 4*d) + 8*e^(3*b*x + 2*d) - 3*e^(b*x))*e^(-4*d)/(b* (e^(2*b*x + 2*d) - 1)^3))*e^(a + 4*d)
Timed out. \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {sinh}\left (d+b\,x\right )}^4} \,d x \] Input:
int(exp(a + b*x)/sinh(d + b*x)^4,x)
Output:
int(exp(a + b*x)/sinh(d + b*x)^4, x)
Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.21 \[ \int e^{a+b x} \text {csch}^4(d+b x) \, dx=\frac {e^{a} \left (-3 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}-1\right )+3 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}+1\right )-6 e^{5 b x +5 d}+9 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )-9 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-16 e^{3 b x +3 d}-9 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )+9 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )+6 e^{b x +d}+3 \,\mathrm {log}\left (e^{b x +d}-1\right )-3 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{6 e^{d} b \left (e^{6 b x +6 d}-3 e^{4 b x +4 d}+3 e^{2 b x +2 d}-1\right )} \] Input:
int(exp(b*x+a)*csch(b*x+d)^4,x)
Output:
(e**a*( - 3*e**(6*b*x + 6*d)*log(e**(b*x + d) - 1) + 3*e**(6*b*x + 6*d)*lo g(e**(b*x + d) + 1) - 6*e**(5*b*x + 5*d) + 9*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) - 9*e**(4*b*x + 4*d)*log(e**(b*x + d) + 1) - 16*e**(3*b*x + 3*d) - 9*e**(2*b*x + 2*d)*log(e**(b*x + d) - 1) + 9*e**(2*b*x + 2*d)*log(e**(b *x + d) + 1) + 6*e**(b*x + d) + 3*log(e**(b*x + d) - 1) - 3*log(e**(b*x + d) + 1)))/(6*e**d*b*(e**(6*b*x + 6*d) - 3*e**(4*b*x + 4*d) + 3*e**(2*b*x + 2*d) - 1))