Integrand size = 24, antiderivative size = 93 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=\frac {5 e^{2 a-d+b x}}{2 b}+\frac {e^{2 a+d+3 b x}}{6 b}+\frac {2 e^{2 a-d+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {4 e^{2 a-2 d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
5/2*exp(b*x+2*a-d)/b+1/6*exp(3*b*x+2*a+d)/b+2*exp(b*x+2*a-d)/b/(1-exp(2*b* x+2*d))-4*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 = 1.51 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.41 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=\frac {e^{2 a-7 d-5 b x} \left (-21 \left (36015+91925 e^{2 (d+b x)}+61158 e^{4 (d+b x)}-20166 e^{6 (d+b x)}-15061 e^{8 (d+b x)}+753 e^{10 (d+b x)}\right )-\frac {315 \left (-2401-5328 e^{2 (d+b x)}-1821 e^{4 (d+b x)}+3264 e^{6 (d+b x)}+1149 e^{8 (d+b x)}-240 e^{10 (d+b x)}+e^{12 (d+b x)}\right ) \text {arctanh}\left (\sqrt {e^{2 (d+b x)}}\right )}{\sqrt {e^{2 (d+b x)}}}+256 e^{8 (d+b x)} \left (1+e^{2 (d+b x)}\right )^3 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 (d+b x)}\right )\right )}{60480 b} \] Input:
Integrate[E^(2*(a + b*x))*Cosh[d + b*x]*Coth[d + b*x]^2,x]
Output:
(E^(2*a - 7*d - 5*b*x)*(-21*(36015 + 91925*E^(2*(d + b*x)) + 61158*E^(4*(d + b*x)) - 20166*E^(6*(d + b*x)) - 15061*E^(8*(d + b*x)) + 753*E^(10*(d + b*x))) - (315*(-2401 - 5328*E^(2*(d + b*x)) - 1821*E^(4*(d + b*x)) + 3264* E^(6*(d + b*x)) + 1149*E^(8*(d + b*x)) - 240*E^(10*(d + b*x)) + E^(12*(d + b*x)))*ArcTanh[Sqrt[E^(2*(d + b*x))]])/Sqrt[E^(2*(d + b*x))] + 256*E^(8*( d + b*x))*(1 + E^(2*(d + b*x)))^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*(d + b*x))]))/(60480*b)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2720, 27, 300, 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.
\(\displaystyle \int e^{2 (a+b x)} \cosh (b x+d) \coth ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {e^{2 a} \left (1+e^{2 b x}\right )^3}{2 \left (1-e^{2 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{2 a} \int \frac {\left (1+e^{2 b x}\right )^3}{\left (1-e^{2 b x}\right )^2}de^{b x}}{2 b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {e^{2 a} \int \left (-\frac {4 \left (1-3 e^{2 b x}\right )}{\left (1-e^{2 b x}\right )^2}+e^{2 b x}+5\right )de^{b x}}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 a} \left (-8 \text {arctanh}\left (e^{b x}\right )+5 e^{b x}+\frac {1}{3} e^{3 b x}+\frac {4 e^{b x}}{1-e^{2 b x}}\right )}{2 b}\) |
Input:
Int[E^(2*(a + b*x))*Cosh[d + b*x]*Coth[d + b*x]^2,x]
Output:
(E^(2*a)*(5*E^(b*x) + E^(3*b*x)/3 + (4*E^(b*x))/(1 - E^(2*b*x)) - 8*ArcTan h[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)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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 = 0.59 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.35
method | result | size |
risch | \(\frac {{\mathrm e}^{3 b x +2 a +d}}{6 b}+\frac {5 \,{\mathrm e}^{b x +2 a -d}}{2 b}+\frac {2 \,{\mathrm e}^{b x +4 a -d}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) b}-\frac {2 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{b}+\frac {2 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(126\) |
Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
1/6*exp(3*b*x+2*a+d)/b+5/2*exp(b*x+2*a-d)/b+2/(-exp(2*b*x+2*a+2*d)+exp(2*a ))/b*exp(b*x+4*a-d)-2*ln(exp(b*x+a)+exp(a-d))/b*exp(2*a-2*d)+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. 689 vs. \(2 (81) = 162\).
Time = 0.12 (sec) , antiderivative size = 689, normalized size of antiderivative = 7.41 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)^2,x, algorithm="fricas")
Output:
1/6*(cosh(b*x + d)^5*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2* d))*sinh(b*x + d)^5 + 5*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*si nh(-2*a + 2*d))*sinh(b*x + d)^4 + 14*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 2* (5*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (5*cosh(b*x + d)^2 + 7)*sinh(-2*a + 2*d) + 7*cosh(-2*a + 2*d))*sinh(b*x + d)^3 + 2*(5*cosh(b*x + d)^3*cosh(-2* a + 2*d) + 21*cosh(b*x + d)*cosh(-2*a + 2*d) - (5*cosh(b*x + d)^3 + 21*cos h(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - 27*cosh(b*x + d)*cosh(-2*a + 2*d) - 12*(cosh(b*x + d)^2*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh( -2*a + 2*d))*sinh(b*x + d)^2 + 2*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b* x + d)*sinh(-2*a + 2*d))*sinh(b*x + d) - (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) + 12*(cos h(b*x + d)^2*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh (b*x + d)^2 + 2*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d) - (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) + (5*cosh(b*x + d)^4*cosh( -2*a + 2*d) + 42*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (5*cosh(b*x + d)^4 + 4 2*cosh(b*x + d)^2 - 27)*sinh(-2*a + 2*d) - 27*cosh(-2*a + 2*d))*sinh(b*x + d) - (cosh(b*x + d)^5 + 14*cosh(b*x + d)^3 - 27*cosh(b*x + d))*sinh(-2*a + 2*d))/(b*cosh(b*x + d)^2 + 2*b*cosh(b*x + d)*sinh(b*x + d) + b*sinh(b*x + d)^2 - b)
\[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=e^{2 a} \int e^{2 b x} \cosh {\left (b x + d \right )} \coth ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)**2,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*cosh(b*x + d)*coth(b*x + d)**2, x)
Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=-\frac {2 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{b} + \frac {2 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{b} + \frac {{\left (14 \, e^{\left (-2 \, b x - 2 \, d\right )} - 27 \, e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )} e^{\left (2 \, a - 2 \, d\right )}}{6 \, b {\left (e^{\left (-3 \, b x - 3 \, d\right )} - e^{\left (-5 \, b x - 5 \, d\right )}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)^2,x, algorithm="maxima")
Output:
-2*e^(2*a - 2*d)*log(e^(-b*x - d) + 1)/b + 2*e^(2*a - 2*d)*log(e^(-b*x - d ) - 1)/b + 1/6*(14*e^(-2*b*x - 2*d) - 27*e^(-4*b*x - 4*d) + 1)*e^(2*a - 2* d)/(b*(e^(-3*b*x - 3*d) - e^(-5*b*x - 5*d)))
Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.10 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=\frac {{\left (e^{\left (3 \, b x + 2 \, a + 7 \, d\right )} + 15 \, e^{\left (b x + 2 \, a + 5 \, d\right )}\right )} e^{\left (-6 \, d\right )} - 12 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right ) + 12 \, e^{\left (2 \, a - 2 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right ) - \frac {12 \, e^{\left (b x + 2 \, a - d\right )}}{e^{\left (2 \, b x + 2 \, d\right )} - 1}}{6 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)^2,x, algorithm="giac")
Output:
1/6*((e^(3*b*x + 2*a + 7*d) + 15*e^(b*x + 2*a + 5*d))*e^(-6*d) - 12*e^(2*a - 2*d)*log(e^(b*x + d) + 1) + 12*e^(2*a - 2*d)*log(abs(e^(b*x + d) - 1)) - 12*e^(b*x + 2*a - d)/(e^(2*b*x + 2*d) - 1))/b
Time = 2.89 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.24 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+d+3\,b\,x}}{6\,b}+\frac {5\,{\mathrm {e}}^{2\,a-d+b\,x}}{2\,b}-\frac {2\,{\mathrm {e}}^{2\,a-d+b\,x}}{b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}-1\right )}-\frac {4\,\sqrt {{\mathrm {e}}^{4\,a-4\,d}}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )}{\sqrt {-b^2}} \] Input:
int(cosh(d + b*x)*coth(d + b*x)^2*exp(2*a + 2*b*x),x)
Output:
exp(2*a + d + 3*b*x)/(6*b) + (5*exp(2*a - d + b*x))/(2*b) - (2*exp(2*a - d + b*x))/(b*(exp(2*d + 2*b*x) - 1)) - (4*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 = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.41 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth ^2(d+b x) \, dx=\frac {e^{2 a} \left (e^{5 b x +5 d}+14 e^{3 b x +3 d}+12 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )-12 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )-27 e^{b x +d}-12 \,\mathrm {log}\left (e^{b x +d}-1\right )+12 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{6 e^{2 d} b \left (e^{2 b x +2 d}-1\right )} \] Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d)^2,x)
Output:
(e**(2*a)*(e**(5*b*x + 5*d) + 14*e**(3*b*x + 3*d) + 12*e**(2*b*x + 2*d)*lo g(e**(b*x + d) - 1) - 12*e**(2*b*x + 2*d)*log(e**(b*x + d) + 1) - 27*e**(b *x + d) - 12*log(e**(b*x + d) - 1) + 12*log(e**(b*x + d) + 1)))/(6*e**(2*d )*b*(e**(2*b*x + 2*d) - 1))