Integrand size = 24, antiderivative size = 106 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=\frac {2 e^{2 a-d+b x}}{b}-\frac {2 e^{2 a-d+b x}}{b \left (1-e^{2 d+2 b x}\right )^2}+\frac {5 e^{2 a-d+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {5 e^{2 a-2 d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
2*exp(b*x+2*a-d)/b-2*exp(b*x+2*a-d)/b/(1-exp(2*b*x+2*d))^2+5*exp(b*x+2*a-d )/b/(1-exp(2*b*x+2*d))-5*exp(2*a-2*d)*arctanh(exp(b*x+d))/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.77 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.37 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=-\frac {e^{2 a-5 d-3 b x} \left (-21 \left (56595+62725 e^{2 (d+b x)}-12071 e^{4 (d+b x)}-19353 e^{6 (d+b x)}+768 e^{8 (d+b x)}\right )+\frac {315 \left (3773+2924 e^{2 (d+b x)}-2534 e^{4 (d+b x)}-1548 e^{6 (d+b x)}+297 e^{8 (d+b x)}\right ) \text {arctanh}\left (\sqrt {e^{2 (d+b x)}}\right )}{\sqrt {e^{2 (d+b x)}}}+128 e^{8 (d+b x)} \left (9+16 e^{2 (d+b x)}+7 e^{4 (d+b x)}\right ) \, _4F_3\left (2,2,2,\frac {5}{2};1,1,\frac {11}{2};e^{2 (d+b x)}\right )+128 e^{8 (d+b x)} \left (1+e^{2 (d+b x)}\right )^2 \, _5F_4\left (2,2,2,2,\frac {5}{2};1,1,1,\frac {11}{2};e^{2 (d+b x)}\right )\right )}{10080 b} \] Input:
Integrate[E^(2*(a + b*x))*Coth[d + b*x]^2*Csch[d + b*x],x]
Output:
-1/10080*(E^(2*a - 5*d - 3*b*x)*(-21*(56595 + 62725*E^(2*(d + b*x)) - 1207 1*E^(4*(d + b*x)) - 19353*E^(6*(d + b*x)) + 768*E^(8*(d + b*x))) + (315*(3 773 + 2924*E^(2*(d + b*x)) - 2534*E^(4*(d + b*x)) - 1548*E^(6*(d + b*x)) + 297*E^(8*(d + b*x)))*ArcTanh[Sqrt[E^(2*(d + b*x))]])/Sqrt[E^(2*(d + b*x)) ] + 128*E^(8*(d + b*x))*(9 + 16*E^(2*(d + b*x)) + 7*E^(4*(d + b*x)))*Hyper geometricPFQ[{2, 2, 2, 5/2}, {1, 1, 11/2}, E^(2*(d + b*x))] + 128*E^(8*(d + b*x))*(1 + E^(2*(d + b*x)))^2*HypergeometricPFQ[{2, 2, 2, 2, 5/2}, {1, 1 , 1, 11/2}, E^(2*(d + b*x))]))/b
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2720, 27, 366, 27, 360, 299, 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.
\(\displaystyle \int e^{2 (a+b x)} \coth ^2(b x+d) \text {csch}(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {2 e^{2 a+2 b x} \left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 e^{2 a} \int \frac {e^{2 b x} \left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 366 |
\(\displaystyle -\frac {2 e^{2 a} \left (\frac {e^{3 b x}}{\left (1-e^{2 b x}\right )^2}-\frac {1}{4} \int \frac {4 e^{2 b x} \left (2+e^{2 b x}\right )}{\left (1-e^{2 b x}\right )^2}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2 e^{2 a} \left (\frac {e^{3 b x}}{\left (1-e^{2 b x}\right )^2}-\int \frac {e^{2 b x} \left (2+e^{2 b x}\right )}{\left (1-e^{2 b x}\right )^2}de^{b x}\right )}{b}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {2 e^{2 a} \left (\frac {1}{2} \int \frac {3+2 e^{2 b x}}{1-e^{2 b x}}de^{b x}-\frac {3 e^{b x}}{2 \left (1-e^{2 b x}\right )}+\frac {e^{3 b x}}{\left (1-e^{2 b x}\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle -\frac {2 e^{2 a} \left (\frac {1}{2} \left (5 \int \frac {1}{1-e^{2 b x}}de^{b x}-2 e^{b x}\right )-\frac {3 e^{b x}}{2 \left (1-e^{2 b x}\right )}+\frac {e^{3 b x}}{\left (1-e^{2 b x}\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 e^{2 a} \left (\frac {1}{2} \left (5 \text {arctanh}\left (e^{b x}\right )-2 e^{b x}\right )-\frac {3 e^{b x}}{2 \left (1-e^{2 b x}\right )}+\frac {e^{3 b x}}{\left (1-e^{2 b x}\right )^2}\right )}{b}\) |
Input:
Int[E^(2*(a + b*x))*Coth[d + b*x]^2*Csch[d + b*x],x]
Output:
(-2*E^(2*a)*(E^(3*b*x)/(1 - E^(2*b*x))^2 - (3*E^(b*x))/(2*(1 - E^(2*b*x))) + (-2*E^(b*x) + 5*ArcTanh[E^(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_)^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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
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 = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{b x +2 a -d}}{b}+\frac {\left (-5 \,{\mathrm e}^{2 b x +2 a +2 d}+3 \,{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +4 a -d}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}-\frac {5 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}+\frac {5 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{2 b}\) | \(131\) |
Input:
int(exp(2*b*x+2*a)*coth(b*x+d)^2*csch(b*x+d),x,method=_RETURNVERBOSE)
Output:
2*exp(b*x+2*a-d)/b+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))*exp(b*x+4*a-d)-5/2*ln(exp(b*x+a)+exp(a-d))/b*exp(2*a-2*d) +5/2*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. 1074 vs. \(2 (95) = 190\).
Time = 0.10 (sec) , antiderivative size = 1074, normalized size of antiderivative = 10.13 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*coth(b*x+d)^2*csch(b*x+d),x, algorithm="fricas")
Output:
1/2*(4*cosh(b*x + d)^5*cosh(-2*a + 2*d) + 4*(cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^5 + 20*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^4 - 18*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 2*(20*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (20*cosh(b*x + d)^2 - 9)*sinh( -2*a + 2*d) - 9*cosh(-2*a + 2*d))*sinh(b*x + d)^3 + 2*(20*cosh(b*x + d)^3* cosh(-2*a + 2*d) - 27*cosh(b*x + d)*cosh(-2*a + 2*d) - (20*cosh(b*x + d)^3 - 27*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 + 10*cosh(b*x + d)* cosh(-2*a + 2*d) - 5*(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*cos h(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) + 5*(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*co sh(-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...
\[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=e^{2 a} \int e^{2 b x} \coth ^{2}{\left (b x + d \right )} \operatorname {csch}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*coth(b*x+d)**2*csch(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*b*x)*coth(b*x + d)**2*csch(b*x + d), x)
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.13 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=-\frac {5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{2 \, b} + \frac {5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{2 \, b} - \frac {{\left (9 \, e^{\left (-2 \, b x - 2 \, d\right )} - 5 \, e^{\left (-4 \, b x - 4 \, d\right )} - 2\right )} e^{\left (2 \, a - 2 \, d\right )}}{b {\left (e^{\left (-b x - d\right )} - 2 \, e^{\left (-3 \, b x - 3 \, d\right )} + e^{\left (-5 \, b x - 5 \, d\right )}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*coth(b*x+d)^2*csch(b*x+d),x, algorithm="maxima")
Output:
-5/2*e^(2*a - 2*d)*log(e^(-b*x - d) + 1)/b + 5/2*e^(2*a - 2*d)*log(e^(-b*x - d) - 1)/b - (9*e^(-2*b*x - 2*d) - 5*e^(-4*b*x - 4*d) - 2)*e^(2*a - 2*d) /(b*(e^(-b*x - d) - 2*e^(-3*b*x - 3*d) + e^(-5*b*x - 5*d)))
Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=-\frac {5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right ) - 5 \, e^{\left (2 \, a - 2 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right ) + \frac {2 \, {\left (5 \, e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} - 3 \, e^{\left (b x + 2 \, a + d\right )}\right )} e^{\left (-2 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} - 1\right )}^{2}} - 4 \, e^{\left (b x + 2 \, a - d\right )}}{2 \, b} \] Input:
integrate(exp(2*b*x+2*a)*coth(b*x+d)^2*csch(b*x+d),x, algorithm="giac")
Output:
-1/2*(5*e^(2*a - 2*d)*log(e^(b*x + d) + 1) - 5*e^(2*a - 2*d)*log(abs(e^(b* x + d) - 1)) + 2*(5*e^(3*b*x + 2*a + 3*d) - 3*e^(b*x + 2*a + d))*e^(-2*d)/ (e^(2*b*x + 2*d) - 1)^2 - 4*e^(b*x + 2*a - d))/b
Time = 2.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.32 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=\frac {2\,{\mathrm {e}}^{2\,a-d+b\,x}}{b}-\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {2\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left ({\mathrm {e}}^{4\,d+4\,b\,x}-2\,{\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {5\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{\sqrt {-b^2}} \] Input:
int((coth(d + b*x)^2*exp(2*a + 2*b*x))/sinh(d + b*x),x)
Output:
(2*exp(2*a - d + b*x))/b - (5*exp(2*a - d + b*x))/(b*(exp(2*d + 2*b*x) - 1 )) - (2*exp(2*a - d + b*x))/(b*(exp(4*d + 4*b*x) - 2*exp(2*d + 2*b*x) + 1) ) - (5*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)
Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.78 \[ \int e^{2 (a+b x)} \coth ^2(d+b x) \text {csch}(d+b x) \, dx=\frac {e^{2 a} \left (4 e^{5 b x +5 d}+5 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )-5 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-18 e^{3 b x +3 d}-10 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )+10 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )+10 e^{b x +d}+5 \,\mathrm {log}\left (e^{b x +d}-1\right )-5 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{2 e^{2 d} b \left (e^{4 b x +4 d}-2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*coth(b*x+d)^2*csch(b*x+d),x)
Output:
(e**(2*a)*(4*e**(5*b*x + 5*d) + 5*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) - 5*e**(4*b*x + 4*d)*log(e**(b*x + d) + 1) - 18*e**(3*b*x + 3*d) - 10*e**(2 *b*x + 2*d)*log(e**(b*x + d) - 1) + 10*e**(2*b*x + 2*d)*log(e**(b*x + d) + 1) + 10*e**(b*x + d) + 5*log(e**(b*x + d) - 1) - 5*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))