Integrand size = 24, antiderivative size = 77 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=-\frac {e^{2 a+2 b x}}{2 b}+\frac {e^{2 (a+d)+4 b x}}{16 b}-\frac {1}{4} e^{2 a-2 d} x+\frac {e^{2 a-2 d} \log \left (1+e^{2 d+2 b x}\right )}{b} \] Output:
-1/2*exp(2*b*x+2*a)/b+1/16*exp(4*b*x+2*a+2*d)/b-1/4*exp(2*a-2*d)*x+exp(2*a -2*d)*ln(1+exp(2*b*x+2*d))/b
Time = 0.41 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=\frac {e^{2 a-2 d} \left (-4 d-8 e^{2 (d+b x)}+e^{4 (d+b x)}-4 b x+16 \log \left (1+e^{2 (d+b x)}\right )\right )}{16 b} \] Input:
Integrate[E^(2*(a + b*x))*Sinh[d + b*x]^2*Tanh[d + b*x],x]
Output:
(E^(2*a - 2*d)*(-4*d - 8*E^(2*(d + b*x)) + E^(4*(d + b*x)) - 4*b*x + 16*Lo g[1 + E^(2*(d + b*x))]))/(16*b)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.53, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2720, 27, 354, 93, 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.
\(\displaystyle \int e^{2 (a+b x)} \sinh ^2(b x+d) \tanh (b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^{2 a-b x} \left (1-e^{2 b x}\right )^3}{4 \left (1+e^{2 b x}\right )}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^{2 a} \int \frac {e^{-b x} \left (1-e^{2 b x}\right )^3}{1+e^{2 b x}}de^{b x}}{4 b}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {e^{2 a} \int \frac {e^{-b x} \left (1-e^{2 b x}\right )^3}{1+e^{2 b x}}de^{2 b x}}{8 b}\) |
\(\Big \downarrow \) 93 |
\(\displaystyle -\frac {e^{2 a} \int \left (e^{-b x}-e^{2 b x}+4-\frac {8}{1+e^{2 b x}}\right )de^{2 b x}}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} \left (\frac {7}{2} e^{2 b x}+\log \left (e^{2 b x}\right )-8 \log \left (e^{2 b x}+1\right )\right )}{8 b}\) |
Input:
Int[E^(2*(a + b*x))*Sinh[d + b*x]^2*Tanh[d + b*x],x]
Output:
-1/8*(E^(2*a)*((7*E^(2*b*x))/2 + Log[E^(2*b*x)] - 8*Log[1 + E^(2*b*x)]))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
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 = 0.98 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {{\mathrm e}^{2 a -2 d} x}{4}+\frac {{\mathrm e}^{4 b x +2 a +2 d}}{16 b}-\frac {{\mathrm e}^{2 b x +2 a}}{2 b}-\frac {2 \,{\mathrm e}^{2 a -2 d} a}{b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(89\) |
Input:
int(exp(2*b*x+2*a)*sinh(b*x+d)^2*tanh(b*x+d),x,method=_RETURNVERBOSE)
Output:
-1/4*exp(2*a-2*d)*x+1/16*exp(4*b*x+2*a+2*d)/b-1/2*exp(2*b*x+2*a)/b-2/b*exp (2*a-2*d)*a+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. 333 vs. \(2 (67) = 134\).
Time = 0.08 (sec) , antiderivative size = 333, normalized size of antiderivative = 4.32 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=\frac {\cosh \left (b x + d\right )^{4} \cosh \left (-2 \, a + 2 \, d\right ) + {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{4} + 4 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} - 4 \, b x \cosh \left (-2 \, a + 2 \, d\right ) - 8 \, \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) + 2 \, {\left (3 \, \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (3 \, \cosh \left (b x + d\right )^{2} - 4\right )} \sinh \left (-2 \, a + 2 \, d\right ) - 4 \, \cosh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} + 16 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \log \left (\frac {2 \, \cosh \left (b x + d\right )}{\cosh \left (b x + d\right ) - \sinh \left (b x + d\right )}\right ) + 4 \, {\left (\cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) - 4 \, \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (\cosh \left (b x + d\right )^{3} - 4 \, \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{4} - 4 \, b x - 8 \, \cosh \left (b x + d\right )^{2}\right )} \sinh \left (-2 \, a + 2 \, d\right )}{16 \, b} \] Input:
integrate(exp(2*b*x+2*a)*sinh(b*x+d)^2*tanh(b*x+d),x, algorithm="fricas")
Output:
1/16*(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 - 4*b*x*cosh(-2*a + 2*d) - 8*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 - 4)*sinh(-2*a + 2*d) - 4*cosh(-2*a + 2*d))*sinh(b*x + d)^2 + 16*( cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*log(2*cosh(b*x + d)/(cosh(b*x + d) - sinh(b*x + d))) + 4*(cosh(b*x + d)^3*cosh(-2*a + 2*d) - 4*cosh(b*x + d)*co sh(-2*a + 2*d) - (cosh(b*x + d)^3 - 4*cosh(b*x + d))*sinh(-2*a + 2*d))*sin h(b*x + d) - (cosh(b*x + d)^4 - 4*b*x - 8*cosh(b*x + d)^2)*sinh(-2*a + 2*d ))/b
\[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=e^{2 a} \int e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \tanh {\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*sinh(b*x+d)**2*tanh(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*b*x)*sinh(b*x + d)**2*tanh(b*x + d), x)
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=-\frac {{\left (8 \, e^{\left (-2 \, b x - 2 \, d\right )} - 1\right )} e^{\left (4 \, b x + 2 \, a + 2 \, d\right )}}{16 \, b} + \frac {7 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )}}{4 \, b} + \frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}{b} \] Input:
integrate(exp(2*b*x+2*a)*sinh(b*x+d)^2*tanh(b*x+d),x, algorithm="maxima")
Output:
-1/16*(8*e^(-2*b*x - 2*d) - 1)*e^(4*b*x + 2*a + 2*d)/b + 7/4*(b*x + d)*e^( 2*a - 2*d)/b + e^(2*a - 2*d)*log(e^(-2*b*x - 2*d) + 1)/b
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=-\frac {4 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )} - {\left (e^{\left (4 \, b x + 2 \, a + 6 \, d\right )} - 8 \, e^{\left (2 \, b x + 2 \, a + 4 \, d\right )}\right )} e^{\left (-4 \, d\right )} - 16 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}{16 \, b} \] Input:
integrate(exp(2*b*x+2*a)*sinh(b*x+d)^2*tanh(b*x+d),x, algorithm="giac")
Output:
-1/16*(4*(b*x + d)*e^(2*a - 2*d) - (e^(4*b*x + 2*a + 6*d) - 8*e^(2*b*x + 2 *a + 4*d))*e^(-4*d) - 16*e^(2*a - 2*d)*log(e^(2*b*x + 2*d) + 1))/b
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+2\,d+4\,b\,x}}{16\,b}-\frac {x\,{\mathrm {e}}^{2\,a-2\,d}}{4}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{2\,b}+\frac {{\mathrm {e}}^{2\,a-2\,d}\,\ln \left ({\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{2\,b\,x}+1\right )}{b} \] Input:
int(exp(2*a + 2*b*x)*sinh(d + b*x)^2*tanh(d + b*x),x)
Output:
exp(2*a + 2*d + 4*b*x)/(16*b) - (x*exp(2*a - 2*d))/4 - exp(2*a + 2*b*x)/(2 *b) + (exp(2*a - 2*d)*log(exp(2*d)*exp(2*b*x) + 1))/b
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.77 \[ \int e^{2 (a+b x)} \sinh ^2(d+b x) \tanh (d+b x) \, dx=\frac {e^{2 a} \left (e^{4 b x +4 d}-8 e^{2 b x +2 d}+16 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )-4 b x \right )}{16 e^{2 d} b} \] Input:
int(exp(2*b*x+2*a)*sinh(b*x+d)^2*tanh(b*x+d),x)
Output:
(e**(2*a)*(e**(4*b*x + 4*d) - 8*e**(2*b*x + 2*d) + 16*log(e**(2*b*x + 2*d) + 1) - 4*b*x))/(16*e**(2*d)*b)