Integrand size = 22, antiderivative size = 80 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=\frac {e^{a+d+2 b x}}{4 b}-\frac {2 e^{a-d}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {1}{2} e^{a-d} x-\frac {e^{a-d} \log \left (1+e^{2 d+2 b x}\right )}{b} \] Output:
1/4*exp(2*b*x+a+d)/b-2*exp(a-d)/b/(1+exp(2*b*x+2*d))-1/2*exp(a-d)*x-exp(a- d)*ln(1+exp(2*b*x+2*d))/b
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=\frac {e^{a-d} \left (-d+\frac {1}{2} e^{2 (d+b x)}-\frac {4}{1+e^{2 (d+b x)}}-b x-2 \log \left (1+e^{2 (d+b x)}\right )\right )}{2 b} \] Input:
Integrate[E^(a + b*x)*Sinh[d + b*x]*Tanh[d + b*x]^2,x]
Output:
(E^(a - d)*(-d + E^(2*(d + b*x))/2 - 4/(1 + E^(2*(d + b*x))) - b*x - 2*Log [1 + E^(2*(d + b*x))]))/(2*b)
Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.61, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, 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.
\(\displaystyle \int e^{a+b x} \sinh (b x+d) \tanh ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^{a-b x} \left (1-e^{2 b x}\right )^3}{2 \left (1+e^{2 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^a \int \frac {e^{-b x} \left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^2}de^{b x}}{2 b}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {e^a \int \frac {e^{-b x} \left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^2}de^{2 b x}}{4 b}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {e^a \int \left (e^{-b x}-1+\frac {4}{1+e^{2 b x}}-\frac {8}{\left (1+e^{2 b x}\right )^2}\right )de^{2 b x}}{4 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^a \left (-e^{2 b x}+\frac {8}{e^{2 b x}+1}+\log \left (e^{2 b x}\right )+4 \log \left (e^{2 b x}+1\right )\right )}{4 b}\) |
Input:
Int[E^(a + b*x)*Sinh[d + b*x]*Tanh[d + b*x]^2,x]
Output:
-1/4*(E^a*(-E^(2*b*x) + 8/(1 + E^(2*b*x)) + Log[E^(2*b*x)] + 4*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[((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 = 0.57 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22
method | result | size |
risch | \(-\frac {{\mathrm e}^{a -d} x}{2}+\frac {{\mathrm e}^{2 b x +a +d}}{4 b}+\frac {2 \,{\mathrm e}^{a -d} a}{b}-\frac {2 \,{\mathrm e}^{3 a -d}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) b}-\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{a -d}}{b}\) | \(98\) |
Input:
int(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*exp(a-d)*x+1/4*exp(2*b*x+a+d)/b+2/b*exp(a-d)*a-2/(exp(2*b*x+2*a+2*d)+ exp(2*a))/b*exp(3*a-d)-ln(exp(2*b*x+2*a)+exp(2*a-2*d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (70) = 140\).
Time = 0.08 (sec) , antiderivative size = 461, normalized size of antiderivative = 5.76 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=\frac {\cosh \left (b x + d\right )^{4} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{4} - {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + 4 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} + {\left (6 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) - {\left (2 \, b x - 1\right )} \cosh \left (-a + d\right ) + {\left (2 \, b x - 6 \, \cosh \left (b x + d\right )^{2} - 1\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - 2 \, {\left (b x + 4\right )} \cosh \left (-a + d\right ) - 4 \, {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} + 2 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{2} + 1\right )} \sinh \left (-a + d\right ) + \cosh \left (-a + 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 ) + 2 \, {\left (2 \, \cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) - {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - {\left (2 \, \cosh \left (b x + d\right )^{3} - {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{4} - {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right )^{2} - 2 \, b x - 8\right )} \sinh \left (-a + d\right )}{4 \, {\left (b \cosh \left (b x + d\right )^{2} + 2 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) + b \sinh \left (b x + d\right )^{2} + b\right )}} \] Input:
integrate(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)^2,x, algorithm="fricas")
Output:
1/4*(cosh(b*x + d)^4*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^4 - (2*b*x - 1)*cosh(b*x + d)^2*cosh(-a + d) + 4*(cosh(b*x + d)*cosh (-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^3 + (6*cosh(b*x + d)^ 2*cosh(-a + d) - (2*b*x - 1)*cosh(-a + d) + (2*b*x - 6*cosh(b*x + d)^2 - 1 )*sinh(-a + d))*sinh(b*x + d)^2 - 2*(b*x + 4)*cosh(-a + d) - 4*(cosh(b*x + d)^2*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^2 + 2*(co sh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d) - (co sh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*log(2*cosh(b*x + d)/(cosh( b*x + d) - sinh(b*x + d))) + 2*(2*cosh(b*x + d)^3*cosh(-a + d) - (2*b*x - 1)*cosh(b*x + d)*cosh(-a + d) - (2*cosh(b*x + d)^3 - (2*b*x - 1)*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^4 - (2*b*x - 1)*cosh(b* x + d)^2 - 2*b*x - 8)*sinh(-a + d))/(b*cosh(b*x + d)^2 + 2*b*cosh(b*x + d) *sinh(b*x + d) + b*sinh(b*x + d)^2 + b)
\[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=e^{a} \int e^{b x} \sinh {\left (b x + d \right )} \tanh ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)**2,x)
Output:
exp(a)*Integral(exp(b*x)*sinh(b*x + d)*tanh(b*x + d)**2, x)
Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=-\frac {{\left (b x + a\right )} e^{\left (a - d\right )}}{2 \, b} - \frac {e^{\left (a - d\right )} \log \left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )}{b} + \frac {e^{\left (2 \, b x + a + d\right )}}{4 \, b} - \frac {2 \, e^{\left (3 \, a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a + 3 \, d\right )} + e^{\left (2 \, a + d\right )}\right )}} \] Input:
integrate(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)^2,x, algorithm="maxima")
Output:
-1/2*(b*x + a)*e^(a - d)/b - e^(a - d)*log(e^(2*b*x + 2*a + 2*d) + e^(2*a) )/b + 1/4*e^(2*b*x + a + d)/b - 2*e^(3*a)/(b*(e^(2*b*x + 2*a + 3*d) + e^(2 *a + d)))
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=-\frac {2 \, {\left (b x + d\right )} e^{\left (a - d\right )} + 4 \, e^{\left (a - d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right ) - \frac {4 \, {\left (e^{\left (2 \, b x + a + 2 \, d\right )} - e^{a}\right )} e^{\left (-d\right )}}{e^{\left (2 \, b x + 2 \, d\right )} + 1} - e^{\left (2 \, b x + a + d\right )}}{4 \, b} \] Input:
integrate(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)^2,x, algorithm="giac")
Output:
-1/4*(2*(b*x + d)*e^(a - d) + 4*e^(a - d)*log(e^(2*b*x + 2*d) + 1) - 4*(e^ (2*b*x + a + 2*d) - e^a)*e^(-d)/(e^(2*b*x + 2*d) + 1) - e^(2*b*x + a + d)) /b
Time = 3.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.10 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{a+d+2\,b\,x}}{4\,b}-\frac {x\,{\mathrm {e}}^{a-d}}{2}-\frac {2\,{\mathrm {e}}^{3\,a-3\,d}}{b\,\left ({\mathrm {e}}^{2\,a-2\,d}+{\mathrm {e}}^{2\,a+2\,b\,x}\right )}-\frac {{\mathrm {e}}^{a-d}\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,d}\right )}{b} \] Input:
int(exp(a + b*x)*sinh(d + b*x)*tanh(d + b*x)^2,x)
Output:
exp(a + d + 2*b*x)/(4*b) - (x*exp(a - d))/2 - (2*exp(3*a - 3*d))/(b*(exp(2 *a - 2*d) + exp(2*a + 2*b*x))) - (exp(a - d)*log(exp(2*a)*exp(2*b*x) + exp (2*a)*exp(-2*d)))/b
Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.35 \[ \int e^{a+b x} \sinh (d+b x) \tanh ^2(d+b x) \, dx=\frac {e^{a} \left (e^{4 b x +4 d}-4 e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-2 e^{2 b x +2 d} b x +9 e^{2 b x +2 d}-4 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )-2 b x \right )}{4 e^{d} b \left (e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*sinh(b*x+d)*tanh(b*x+d)^2,x)
Output:
(e**a*(e**(4*b*x + 4*d) - 4*e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d) + 1) - 2 *e**(2*b*x + 2*d)*b*x + 9*e**(2*b*x + 2*d) - 4*log(e**(2*b*x + 2*d) + 1) - 2*b*x))/(4*e**d*b*(e**(2*b*x + 2*d) + 1))