Integrand size = 26, antiderivative size = 204 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=-\frac {4 e^{2 a-2 d}}{b \left (1-e^{2 d+2 b x}\right )^4}+\frac {40 e^{2 a-2 d}}{3 b \left (1-e^{2 d+2 b x}\right )^3}-\frac {12 e^{2 a-2 d}}{b \left (1-e^{2 d+2 b x}\right )^2}-\frac {4 e^{2 a-2 d}}{b \left (1-e^{2 d+2 b x}\right )}+\frac {2 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )^2}-\frac {6 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )}+\frac {2 e^{2 a-2 d} \text {arctanh}\left (e^{2 d+2 b x}\right )}{b} \] Output:
-4*exp(2*a-2*d)/b/(1-exp(2*b*x+2*d))^4+40/3*exp(2*a-2*d)/b/(1-exp(2*b*x+2* d))^3-12*exp(2*a-2*d)/b/(1-exp(2*b*x+2*d))^2-4*exp(2*a-2*d)/b/(1-exp(2*b*x +2*d))+2*exp(2*a-2*d)/b/(1+exp(2*b*x+2*d))^2-6*exp(2*a-2*d)/b/(1+exp(2*b*x +2*d))+2*exp(2*a-2*d)*arctanh(exp(2*b*x+2*d))/b
Time = 0.46 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.68 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=\frac {256 e^{2 a-2 d} \left (-\frac {1}{64 \left (1-e^{2 (d+b x)}\right )^4}+\frac {5}{96 \left (1-e^{2 (d+b x)}\right )^3}-\frac {3}{64 \left (1-e^{2 (d+b x)}\right )^2}-\frac {1}{64 \left (1-e^{2 (d+b x)}\right )}+\frac {1}{128 \left (1+e^{2 (d+b x)}\right )^2}-\frac {3}{128 \left (1+e^{2 (d+b x)}\right )}+\frac {1}{128} \text {arctanh}\left (e^{2 (d+b x)}\right )\right )}{b} \] Input:
Integrate[E^(2*(a + b*x))*Csch[d + b*x]^5*Sech[d + b*x]^3,x]
Output:
(256*E^(2*a - 2*d)*(-1/64*1/(1 - E^(2*(d + b*x)))^4 + 5/(96*(1 - E^(2*(d + b*x)))^3) - 3/(64*(1 - E^(2*(d + b*x)))^2) - 1/(64*(1 - E^(2*(d + b*x)))) + 1/(128*(1 + E^(2*(d + b*x)))^2) - 3/(128*(1 + E^(2*(d + b*x)))) + ArcTa nh[E^(2*(d + b*x))]/128))/b
Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.56, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2720, 27, 354, 99, 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.
| \(\displaystyle \int e^{2 (a+b x)} \text {csch}^5(b x+d) \text {sech}^3(b x+d) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\int -\frac {256 e^{2 a+9 b x}}{\left (1-e^{2 b x}\right )^5 \left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {256 e^{2 a} \int \frac {e^{9 b x}}{\left (1-e^{2 b x}\right )^5 \left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {128 e^{2 a} \int \frac {e^{4 b x}}{\left (1-e^{2 b x}\right )^5 \left (1+e^{2 b x}\right )^3}de^{2 b x}}{b}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {128 e^{2 a} \int \left (-\frac {3}{64 \left (1+e^{2 b x}\right )^2}+\frac {1}{32 \left (1+e^{2 b x}\right )^3}+\frac {1}{64 \left (-1+e^{2 b x}\right )}+\frac {1}{32 \left (-1+e^{2 b x}\right )^2}-\frac {3}{16 \left (-1+e^{2 b x}\right )^3}-\frac {5}{16 \left (-1+e^{2 b x}\right )^4}-\frac {1}{8 \left (-1+e^{2 b x}\right )^5}\right )de^{2 b x}}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {128 e^{2 a} \left (-\frac {1}{64} \text {arctanh}\left (e^{2 b x}\right )+\frac {1}{32 \left (1-e^{2 b x}\right )}+\frac {3}{64 \left (e^{2 b x}+1\right )}+\frac {3}{32 \left (1-e^{2 b x}\right )^2}-\frac {1}{64 \left (e^{2 b x}+1\right )^2}-\frac {5}{48 \left (1-e^{2 b x}\right )^3}+\frac {1}{32 \left (1-e^{2 b x}\right )^4}\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Csch[d + b*x]^5*Sech[d + b*x]^3,x]
Output:
(-128*E^(2*a)*(1/(32*(1 - E^(2*b*x))^4) - 5/(48*(1 - E^(2*b*x))^3) + 3/(32 *(1 - E^(2*b*x))^2) + 1/(32*(1 - E^(2*b*x))) - 1/(64*(1 + E^(2*b*x))^2) + 3/(64*(1 + E^(2*b*x))) - ArcTanh[E^(2*b*x)]/64))/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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(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 = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.97
\[-\frac {2 \left (3 \,{\mathrm e}^{10 b x +10 a +10 d}-6 \,{\mathrm e}^{8 b x +10 a +8 d}+62 \,{\mathrm e}^{6 b x +10 a +6 d}-22 \,{\mathrm e}^{4 b x +10 a +4 d}-29 \,{\mathrm e}^{2 b x +10 a +2 d}+16 \,{\mathrm e}^{10 a}\right ) {\mathrm e}^{4 a -2 d}}{3 \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{4} \left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}-\frac {\ln \left ({\mathrm e}^{2 b x +2 a}-{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}\]
Input:
int(exp(2*b*x+2*a)*csch(b*x+d)^5*sech(b*x+d)^3,x)
Output:
-2/3/(-exp(2*b*x+2*a+2*d)+exp(2*a))^4/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(3 *exp(10*b*x+10*a+10*d)-6*exp(8*b*x+10*a+8*d)+62*exp(6*b*x+10*a+6*d)-22*exp (4*b*x+10*a+4*d)-29*exp(2*b*x+10*a+2*d)+16*exp(10*a))*exp(4*a-2*d)+ln(exp( 2*b*x+2*a)+exp(2*a-2*d))/b*exp(2*a-2*d)-ln(exp(2*b*x+2*a)-exp(2*a-2*d))/b* exp(2*a-2*d)
Leaf count of result is larger than twice the leaf count of optimal. 4726 vs. \(2 (180) = 360\).
Time = 0.13 (sec) , antiderivative size = 4726, normalized size of antiderivative = 23.17 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^5*sech(b*x+d)^3,x, algorithm="fricas" )
Output:
Too large to include
\[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=e^{2 a} \int e^{2 b x} \operatorname {csch}^{5}{\left (b x + d \right )} \operatorname {sech}^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)**5*sech(b*x+d)**3,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*csch(b*x + d)**5*sech(b*x + d)**3, x)
Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.03 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=-\frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{b} - \frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{b} + \frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}{b} + \frac {2 \, {\left (35 \, e^{\left (-2 \, b x - 2 \, d\right )} + 10 \, e^{\left (-4 \, b x - 4 \, d\right )} - 2 \, e^{\left (-6 \, b x - 6 \, d\right )} - 6 \, e^{\left (-8 \, b x - 8 \, d\right )} + 3 \, e^{\left (-10 \, b x - 10 \, d\right )} - 16\right )} e^{\left (2 \, a - 2 \, d\right )}}{3 \, b {\left (2 \, e^{\left (-2 \, b x - 2 \, d\right )} + e^{\left (-4 \, b x - 4 \, d\right )} - 4 \, e^{\left (-6 \, b x - 6 \, d\right )} + e^{\left (-8 \, b x - 8 \, d\right )} + 2 \, e^{\left (-10 \, b x - 10 \, d\right )} - e^{\left (-12 \, b x - 12 \, d\right )} - 1\right )}} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^5*sech(b*x+d)^3,x, algorithm="maxima" )
Output:
-e^(2*a - 2*d)*log(e^(-b*x - d) + 1)/b - e^(2*a - 2*d)*log(e^(-b*x - d) - 1)/b + e^(2*a - 2*d)*log(e^(-2*b*x - 2*d) + 1)/b + 2/3*(35*e^(-2*b*x - 2*d ) + 10*e^(-4*b*x - 4*d) - 2*e^(-6*b*x - 6*d) - 6*e^(-8*b*x - 8*d) + 3*e^(- 10*b*x - 10*d) - 16)*e^(2*a - 2*d)/(b*(2*e^(-2*b*x - 2*d) + e^(-4*b*x - 4* d) - 4*e^(-6*b*x - 6*d) + e^(-8*b*x - 8*d) + 2*e^(-10*b*x - 10*d) - e^(-12 *b*x - 12*d) - 1))
Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.75 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=\frac {{\left (12 \, e^{\left (-2 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right ) - 12 \, e^{\left (-2 \, d\right )} \log \left ({\left | e^{\left (2 \, b x + 2 \, d\right )} - 1 \right |}\right ) - \frac {6 \, {\left (3 \, e^{\left (4 \, b x + 4 \, d\right )} + 18 \, e^{\left (2 \, b x + 2 \, d\right )} + 11\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}} + \frac {{\left (25 \, e^{\left (8 \, b x + 8 \, d\right )} - 52 \, e^{\left (6 \, b x + 6 \, d\right )} - 138 \, e^{\left (4 \, b x + 4 \, d\right )} + 172 \, e^{\left (2 \, b x + 2 \, d\right )} - 55\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{4}}\right )} e^{\left (2 \, a\right )}}{12 \, b} \] Input:
integrate(exp(2*b*x+2*a)*csch(b*x+d)^5*sech(b*x+d)^3,x, algorithm="giac")
Output:
1/12*(12*e^(-2*d)*log(e^(2*b*x + 2*d) + 1) - 12*e^(-2*d)*log(abs(e^(2*b*x + 2*d) - 1)) - 6*(3*e^(4*b*x + 4*d) + 18*e^(2*b*x + 2*d) + 11)*e^(-2*d)/(e ^(2*b*x + 2*d) + 1)^2 + (25*e^(8*b*x + 8*d) - 52*e^(6*b*x + 6*d) - 138*e^( 4*b*x + 4*d) + 172*e^(2*b*x + 2*d) - 55)*e^(-2*d)/(e^(2*b*x + 2*d) - 1)^4) *e^(2*a)/b
Time = 0.29 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.40 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=\frac {2\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}-\frac {12\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}-2\,{\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {40\,{\mathrm {e}}^{2\,a-2\,d}}{3\,b\,\left (3\,{\mathrm {e}}^{2\,d+2\,b\,x}-3\,{\mathrm {e}}^{4\,d+4\,b\,x}+{\mathrm {e}}^{6\,d+6\,b\,x}-1\right )}-\frac {4\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left (6\,{\mathrm {e}}^{4\,d+4\,b\,x}-4\,{\mathrm {e}}^{2\,d+2\,b\,x}-4\,{\mathrm {e}}^{6\,d+6\,b\,x}+{\mathrm {e}}^{8\,d+8\,b\,x}+1\right )}+\frac {2\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{\sqrt {-b^2}}+\frac {4\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {6\,{\mathrm {e}}^{2\,a-2\,d}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )} \] Input:
int(exp(2*a + 2*b*x)/(cosh(d + b*x)^3*sinh(d + b*x)^5),x)
Output:
(2*exp(2*a - 2*d))/(b*(2*exp(2*d + 2*b*x) + exp(4*d + 4*b*x) + 1)) - (12*e xp(2*a - 2*d))/(b*(exp(4*d + 4*b*x) - 2*exp(2*d + 2*b*x) + 1)) - (40*exp(2 *a - 2*d))/(3*b*(3*exp(2*d + 2*b*x) - 3*exp(4*d + 4*b*x) + exp(6*d + 6*b*x ) - 1)) - (4*exp(2*a - 2*d))/(b*(6*exp(4*d + 4*b*x) - 4*exp(2*d + 2*b*x) - 4*exp(6*d + 6*b*x) + exp(8*d + 8*b*x) + 1)) + (2*exp(4*a - 4*d)^(1/2)*ata n((exp(2*a)*exp(2*b*x)*(-b^2)^(1/2))/(b*(exp(4*a)*exp(-4*d))^(1/2))))/(-b^ 2)^(1/2) + (4*exp(2*a - 2*d))/(b*(exp(2*d + 2*b*x) - 1)) - (6*exp(2*a - 2* d))/(b*(exp(2*d + 2*b*x) + 1))
Time = 0.24 (sec) , antiderivative size = 606, normalized size of antiderivative = 2.97 \[ \int e^{2 (a+b x)} \text {csch}^5(d+b x) \text {sech}^3(d+b x) \, dx=\frac {e^{2 a} \left (-35-3 \,\mathrm {log}\left (e^{b x +d}-1\right )-3 \,\mathrm {log}\left (e^{b x +d}+1\right )+6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )+6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )-3 e^{8 b x +8 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )+6 e^{10 b x +10 d} \mathrm {log}\left (e^{b x +d}-1\right )+6 e^{10 b x +10 d} \mathrm {log}\left (e^{b x +d}+1\right )-3 e^{4 b x +4 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-6 e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-12 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}-1\right )-12 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}+1\right )-3 e^{12 b x +12 d}+3 e^{8 b x +8 d} \mathrm {log}\left (e^{b x +d}-1\right )+3 e^{8 b x +8 d} \mathrm {log}\left (e^{b x +d}+1\right )+3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )+3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )+47 e^{4 b x +4 d}+64 e^{2 b x +2 d}+15 e^{8 b x +8 d}+12 e^{6 b x +6 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-136 e^{6 b x +6 d}+3 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )-3 e^{12 b x +12 d} \mathrm {log}\left (e^{b x +d}-1\right )-3 e^{12 b x +12 d} \mathrm {log}\left (e^{b x +d}+1\right )+3 e^{12 b x +12 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-6 e^{10 b x +10 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )\right )}{3 e^{2 d} b \left (e^{12 b x +12 d}-2 e^{10 b x +10 d}-e^{8 b x +8 d}+4 e^{6 b x +6 d}-e^{4 b x +4 d}-2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*csch(b*x+d)^5*sech(b*x+d)^3,x)
Output:
(e**(2*a)*( - 3*e**(12*b*x + 12*d)*log(e**(b*x + d) - 1) - 3*e**(12*b*x + 12*d)*log(e**(b*x + d) + 1) + 3*e**(12*b*x + 12*d)*log(e**(2*b*x + 2*d) + 1) - 3*e**(12*b*x + 12*d) + 6*e**(10*b*x + 10*d)*log(e**(b*x + d) - 1) + 6 *e**(10*b*x + 10*d)*log(e**(b*x + d) + 1) - 6*e**(10*b*x + 10*d)*log(e**(2 *b*x + 2*d) + 1) + 3*e**(8*b*x + 8*d)*log(e**(b*x + d) - 1) + 3*e**(8*b*x + 8*d)*log(e**(b*x + d) + 1) - 3*e**(8*b*x + 8*d)*log(e**(2*b*x + 2*d) + 1 ) + 15*e**(8*b*x + 8*d) - 12*e**(6*b*x + 6*d)*log(e**(b*x + d) - 1) - 12*e **(6*b*x + 6*d)*log(e**(b*x + d) + 1) + 12*e**(6*b*x + 6*d)*log(e**(2*b*x + 2*d) + 1) - 136*e**(6*b*x + 6*d) + 3*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) + 3*e**(4*b*x + 4*d)*log(e**(b*x + d) + 1) - 3*e**(4*b*x + 4*d)*log(e* *(2*b*x + 2*d) + 1) + 47*e**(4*b*x + 4*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**(2*b*x + 2*d) *log(e**(2*b*x + 2*d) + 1) + 64*e**(2*b*x + 2*d) - 3*log(e**(b*x + d) - 1) - 3*log(e**(b*x + d) + 1) + 3*log(e**(2*b*x + 2*d) + 1) - 35))/(3*e**(2*d )*b*(e**(12*b*x + 12*d) - 2*e**(10*b*x + 10*d) - e**(8*b*x + 8*d) + 4*e**( 6*b*x + 6*d) - e**(4*b*x + 4*d) - 2*e**(2*b*x + 2*d) + 1))