Integrand size = 18, antiderivative size = 88 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {2 e^{2 a+d+3 b x}}{b \left (1-e^{2 d+2 b x}\right )^2}+\frac {3 e^{2 a-d+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {3 e^{2 a-2 d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
-2*exp(3*b*x+2*a+d)/b/(1-exp(2*b*x+2*d))^2+3*exp(b*x+2*a-d)/b/(1-exp(2*b*x +2*d))-3*exp(2*a-2*d)*arctanh(exp(b*x+d))/b
Leaf count is larger than twice the leaf count of optimal. \(239\) vs. \(2(88)=176\).
Time = 0.44 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.72 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=\frac {e^{2 a} \left (-3 \cosh (2 d) \log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )+3 \cosh (2 d) \log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )-\frac {4 e^{b x} (\cosh (d)-\sinh (d))^3}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {10 e^{b x} (\cosh (d)-\sinh (d))^2}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}+3 \log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right ) \sinh (2 d)-3 \log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right ) \sinh (2 d)\right )}{2 b} \] Input:
Integrate[E^(2*(a + b*x))*Csch[d + b*x]^3,x]
Output:
(E^(2*a)*(-3*Cosh[2*d]*Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[d /2]] + 3*Cosh[2*d]*Log[(-1 + E^(b*x))*Cosh[d/2] + (1 + E^(b*x))*Sinh[d/2]] - (4*E^(b*x)*(Cosh[d] - Sinh[d])^3)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2 *b*x))*Sinh[d])^2 - (10*E^(b*x)*(Cosh[d] - Sinh[d])^2)/((-1 + E^(2*b*x))*C osh[d] + (1 + E^(2*b*x))*Sinh[d]) + 3*Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[d/2]]*Sinh[2*d] - 3*Log[(-1 + E^(b*x))*Cosh[d/2] + (1 + E^(b *x))*Sinh[d/2]]*Sinh[2*d]))/(2*b)
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78, 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.
\(\displaystyle \int e^{2 (a+b x)} \text {csch}^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {8 e^{2 a+4 b x}}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {8 e^{2 a} \int \frac {e^{4 b x}}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {8 e^{2 a} \left (\frac {e^{3 b x}}{4 \left (1-e^{2 b x}\right )^2}-\frac {3}{4} \int \frac {e^{2 b x}}{\left (1-e^{2 b x}\right )^2}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {8 e^{2 a} \left (\frac {e^{3 b x}}{4 \left (1-e^{2 b x}\right )^2}-\frac {3}{4} \left (\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}-\frac {1}{2} \int \frac {1}{1-e^{2 b x}}de^{b x}\right )\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {8 e^{2 a} \left (\frac {e^{3 b x}}{4 \left (1-e^{2 b x}\right )^2}-\frac {3}{4} \left (\frac {e^{b x}}{2 \left (1-e^{2 b x}\right )}-\frac {1}{2} \text {arctanh}\left (e^{b x}\right )\right )\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Csch[d + b*x]^3,x]
Output:
(-8*E^(2*a)*(E^(3*b*x)/(4*(1 - E^(2*b*x))^2) - (3*(E^(b*x)/(2*(1 - E^(2*b* x))) - ArcTanh[E^(b*x)]/2))/4))/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)^(-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 = 0.51 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {\left (-5 \,{\mathrm e}^{2 b x +2 a +2 d}+3 \,{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +4 a -d}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}\) | \(115\) |
Input:
int(exp(2*b*x+2*a)*csch(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/(-exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(-5*exp(2*b*x+2*a+2*d)+3*exp(2*a))*ex p(b*x+4*a-d)-3/2*ln(exp(b*x+a)+exp(a-d))/b*exp(2*a-2*d)+3/2*ln(exp(b*x+a)- exp(a-d))/b*exp(2*a-2*d)
Leaf count of result is larger than twice the leaf count of optimal. 883 vs. \(2 (78) = 156\).
Time = 0.10 (sec) , antiderivative size = 883, normalized size of antiderivative = 10.03 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^3,x, algorithm="fricas")
Output:
-1/2*(10*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 10*(cosh(-2*a + 2*d) - sinh(-2 *a + 2*d))*sinh(b*x + d)^3 + 30*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - 6*cosh(b*x + d)*cosh(-2*a + 2*d) + 3*(cosh(b*x + d)^4*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2 *d))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*s inh(-2*a + 2*d))*sinh(b*x + d)^3 - 2*cosh(b*x + d)^2*cosh(-2*a + 2*d) + 2* (3*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (3*cosh(b*x + d)^2 - 1)*sinh(-2*a + 2*d) - cosh(-2*a + 2*d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-2*a + 2*d) - cosh(b*x + d)*cosh(-2*a + 2*d) - (cosh(b*x + d)^3 - cosh(b*x + d))* sinh(-2*a + 2*d))*sinh(b*x + d) - (cosh(b*x + d)^4 - 2*cosh(b*x + d)^2 + 1 )*sinh(-2*a + 2*d) + cosh(-2*a + 2*d))*log(cosh(b*x + d) + sinh(b*x + d) + 1) - 3*(cosh(b*x + d)^4*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d )*sinh(-2*a + 2*d))*sinh(b*x + d)^3 - 2*cosh(b*x + d)^2*cosh(-2*a + 2*d) + 2*(3*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (3*cosh(b*x + d)^2 - 1)*sinh(-2*a + 2*d) - cosh(-2*a + 2*d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-2*a + 2*d) - cosh(b*x + d)*cosh(-2*a + 2*d) - (cosh(b*x + d)^3 - cosh(b*x + d ))*sinh(-2*a + 2*d))*sinh(b*x + d) - (cosh(b*x + d)^4 - 2*cosh(b*x + d)^2 + 1)*sinh(-2*a + 2*d) + cosh(-2*a + 2*d))*log(cosh(b*x + d) + sinh(b*x + d ) - 1) + 6*(5*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (5*cosh(b*x + d)^2 - 1...
\[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=e^{2 a} \int e^{2 b x} \operatorname {csch}^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)**3,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*csch(b*x + d)**3, x)
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {3 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{2 \, b} + \frac {3 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{2 \, b} + \frac {{\left (5 \, e^{\left (-b x - d\right )} - 3 \, e^{\left (-3 \, b x - 3 \, d\right )}\right )} e^{\left (2 \, a - 2 \, d\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, d\right )} - e^{\left (-4 \, b x - 4 \, d\right )} - 1\right )}} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^3,x, algorithm="maxima")
Output:
-3/2*e^(2*a - 2*d)*log(e^(-b*x - d) + 1)/b + 3/2*e^(2*a - 2*d)*log(e^(-b*x - d) - 1)/b + (5*e^(-b*x - d) - 3*e^(-3*b*x - 3*d))*e^(2*a - 2*d)/(b*(2*e ^(-2*b*x - 2*d) - e^(-4*b*x - 4*d) - 1))
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {{\left (3 \, e^{\left (-2 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right ) - 3 \, e^{\left (-2 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right ) + \frac {2 \, {\left (5 \, e^{\left (3 \, b x + 3 \, d\right )} - 3 \, e^{\left (b x + d\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{2}}\right )} e^{\left (2 \, a\right )}}{2 \, b} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^3,x, algorithm="giac")
Output:
-1/2*(3*e^(-2*d)*log(e^(b*x + d) + 1) - 3*e^(-2*d)*log(abs(e^(b*x + d) - 1 )) + 2*(5*e^(3*b*x + 3*d) - 3*e^(b*x + d))*e^(-2*d)/(e^(2*b*x + 2*d) - 1)^ 2)*e^(2*a)/b
Time = 2.97 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=-\frac {3\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {3\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{2\,a+d+3\,b\,x}}{b\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}-2\,{\mathrm {e}}^{2\,d+2\,b\,x}+1\right )} \] Input:
int(exp(2*a + 2*b*x)/sinh(d + b*x)^3,x)
Output:
- (3*exp(2*a - d + b*x))/(b*(exp(2*d + 2*b*x) - 1)) - (3*exp(4*a - 4*d)^(1 /2)*atan((exp(2*a)*exp(-d)*exp(b*x)*(-b^2)^(1/2))/(b*(exp(4*a)*exp(-4*d))^ (1/2))))/(-b^2)^(1/2) - (2*exp(2*a + d + 3*b*x))/(b*(exp(4*d + 4*b*x) - 2* exp(2*d + 2*b*x) + 1))
Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.01 \[ \int e^{2 (a+b x)} \text {csch}^3(d+b x) \, dx=\frac {e^{2 a} \left (3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )-3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-10 e^{3 b x +3 d}-6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )+6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )+6 e^{b x +d}+3 \,\mathrm {log}\left (e^{b x +d}-1\right )-3 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{2 e^{2 d} b \left (e^{4 b x +4 d}-2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*csch(b*x+d)^3,x)
Output:
(e**(2*a)*(3*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) - 3*e**(4*b*x + 4*d)*l og(e**(b*x + d) + 1) - 10*e**(3*b*x + 3*d) - 6*e**(2*b*x + 2*d)*log(e**(b* x + d) - 1) + 6*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)))/(2*e**(2*d)*b*(e**(4*b *x + 4*d) - 2*e**(2*b*x + 2*d) + 1))