Integrand size = 18, antiderivative size = 73 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {arctanh}\left (e^{a+b x}\right )}{b} \]
-2*exp(3*b*x+3*a)/b/(1-exp(2*b*x+2*a))^2+3*exp(b*x+a)/b/(1-exp(2*b*x+2*a)) -3*arctanh(exp(b*x+a))/b
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=\frac {3 e^{a+b x}-5 e^{3 (a+b x)}-3 \left (-1+e^{2 (a+b x)}\right )^2 \text {arctanh}\left (e^{a+b x}\right )}{b \left (-1+e^{2 (a+b x)}\right )^2} \]
(3*E^(a + b*x) - 5*E^(3*(a + b*x)) - 3*(-1 + E^(2*(a + b*x)))^2*ArcTanh[E^ (a + b*x)])/(b*(-1 + E^(2*(a + b*x)))^2)
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2720, 27, 252, 252, 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.
\(\displaystyle \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {8 e^{4 a+4 b x}}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8 \int \frac {e^{4 a+4 b x}}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {8 \left (\frac {e^{3 a+3 b x}}{4 \left (1-e^{2 a+2 b x}\right )^2}-\frac {3}{4} \int \frac {e^{2 a+2 b x}}{\left (1-e^{2 a+2 b x}\right )^2}de^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {8 \left (\frac {e^{3 a+3 b x}}{4 \left (1-e^{2 a+2 b x}\right )^2}-\frac {3}{4} \left (\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )}-\frac {1}{2} \int \frac {1}{1-e^{2 a+2 b x}}de^{a+b x}\right )\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {8 \left (\frac {e^{3 a+3 b x}}{4 \left (1-e^{2 a+2 b x}\right )^2}-\frac {3}{4} \left (\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )}-\frac {1}{2} \text {arctanh}\left (e^{a+b x}\right )\right )\right )}{b}\) |
(-8*(E^(3*a + 3*b*x)/(4*(1 - E^(2*a + 2*b*x))^2) - (3*(E^(a + b*x)/(2*(1 - E^(2*a + 2*b*x))) - ArcTanh[E^(a + b*x)]/2))/4))/b
3.9.77.3.1 Defintions of rubi rules used
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)^(-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 = 0.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {{\mathrm e}^{b x +a} \left (5 \,{\mathrm e}^{2 b x +2 a}-3\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}\) | \(67\) |
-exp(b*x+a)*(5*exp(2*b*x+2*a)-3)/b/(exp(2*b*x+2*a)-1)^2-3/2/b*ln(exp(b*x+a )+1)+3/2/b*ln(exp(b*x+a)-1)
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (64) = 128\).
Time = 0.25 (sec) , antiderivative size = 388, normalized size of antiderivative = 5.32 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=-\frac {10 \, \cosh \left (b x + a\right )^{3} + 30 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 10 \, \sinh \left (b x + a\right )^{3} + 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 6 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 6 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]
-1/2*(10*cosh(b*x + a)^3 + 30*cosh(b*x + a)*sinh(b*x + a)^2 + 10*sinh(b*x + a)^3 + 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4* (cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + s inh(b*x + a) + 1) - 3*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b *x + a) + sinh(b*x + a) - 1) + 6*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 6 *cosh(b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b *sinh(b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh( b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)
\[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=e^{2 a} \int e^{2 b x} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]
Time = 0.18 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.21 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=-\frac {3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac {5 \, e^{\left (-b x - a\right )} - 3 \, e^{\left (-3 \, b x - 3 \, a\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}} \]
-3/2*log(e^(-b*x - a) + 1)/b + 3/2*log(e^(-b*x - a) - 1)/b + (5*e^(-b*x - a) - 3*e^(-3*b*x - 3*a))/(b*(2*e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a) - 1))
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (5 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} + 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \]
-1/2*(2*(5*e^(3*b*x + 3*a) - 3*e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^2 + 3*lo g(e^(b*x + a) + 1) - 3*log(abs(e^(b*x + a) - 1)))/b
Time = 2.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int e^{2 (a+b x)} \text {csch}^3(a+b x) \, dx=-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{3\,a+3\,b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]