Integrand size = 18, antiderivative size = 101 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=\frac {e^{2 a+2 b x}}{2 b}+\frac {2 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )^2}-\frac {6 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {3 e^{2 a-2 d} \log \left (1+e^{2 d+2 b x}\right )}{b} \] Output:
1/2*exp(2*b*x+2*a)/b+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))-3*exp(2*a-2*d)*ln(1+exp(2*b*x+2*d))/b
Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=\frac {e^{2 a} \left (e^{2 b x}-6 \cosh (2 d) \log \left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )+\frac {4 (\cosh (d)-\sinh (d))^4}{\left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {12 (\cosh (d)-\sinh (d))^3}{\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)}+6 \log \left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right ) \sinh (2 d)\right )}{2 b} \] Input:
Integrate[E^(2*(a + b*x))*Tanh[d + b*x]^3,x]
Output:
(E^(2*a)*(E^(2*b*x) - 6*Cosh[2*d]*Log[(1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2 *b*x))*Sinh[d]] + (4*(Cosh[d] - Sinh[d])^4)/((1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2*b*x))*Sinh[d])^2 - (12*(Cosh[d] - Sinh[d])^3)/((1 + E^(2*b*x))*Cos h[d] + (-1 + E^(2*b*x))*Sinh[d]) + 6*Log[(1 + E^(2*b*x))*Cosh[d] + (-1 + E ^(2*b*x))*Sinh[d]]*Sinh[2*d]))/(2*b)
Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2720, 25, 27, 353, 49, 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.
\(\displaystyle \int e^{2 (a+b x)} \tanh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^{2 a+b x} \left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{2 a+b x} \left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^{2 a} \int \frac {e^{b x} \left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {e^{2 a} \int \frac {\left (1-e^{2 b x}\right )^3}{\left (1+e^{2 b x}\right )^3}de^{2 b x}}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {e^{2 a} \int \left (-1+\frac {6}{1+e^{2 b x}}-\frac {12}{\left (1+e^{2 b x}\right )^2}+\frac {8}{\left (1+e^{2 b x}\right )^3}\right )de^{2 b x}}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} \left (-e^{2 b x}+\frac {12}{e^{2 b x}+1}-\frac {4}{\left (e^{2 b x}+1\right )^2}+6 \log \left (e^{2 b x}+1\right )\right )}{2 b}\) |
Input:
Int[E^(2*(a + b*x))*Tanh[d + b*x]^3,x]
Output:
-1/2*(E^(2*a)*(-E^(2*b*x) - 4/(1 + E^(2*b*x))^2 + 12/(1 + E^(2*b*x)) + 6*L og[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_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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 = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {{\mathrm e}^{2 b x +2 a}}{2 b}+\frac {6 \,{\mathrm e}^{2 a -2 d} a}{b}-\frac {2 \left (3 \,{\mathrm e}^{2 b x +2 a +2 d}+2 \,{\mathrm e}^{2 a}\right ) {\mathrm e}^{4 a -2 d}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}-\frac {3 \ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(115\) |
Input:
int(exp(2*b*x+2*a)*tanh(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/2*exp(2*b*x+2*a)/b+6/b*exp(2*a-2*d)*a-2/(exp(2*b*x+2*a+2*d)+exp(2*a))^2/ b*(3*exp(2*b*x+2*a+2*d)+2*exp(2*a))*exp(4*a-2*d)-3*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. 897 vs. \(2 (92) = 184\).
Time = 0.10 (sec) , antiderivative size = 897, normalized size of antiderivative = 8.88 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^3,x, algorithm="fricas")
Output:
1/2*(cosh(b*x + d)^6*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2* d))*sinh(b*x + d)^6 + 6*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*si nh(-2*a + 2*d))*sinh(b*x + d)^5 + 2*cosh(b*x + d)^4*cosh(-2*a + 2*d) + (15 *cosh(b*x + d)^2*cosh(-2*a + 2*d) - (15*cosh(b*x + d)^2 + 2)*sinh(-2*a + 2 *d) + 2*cosh(-2*a + 2*d))*sinh(b*x + d)^4 + 4*(5*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 2*cosh(b*x + d)*cosh(-2*a + 2*d) - (5*cosh(b*x + d)^3 + 2*cosh(b *x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^3 - 11*cosh(b*x + d)^2*cosh(-2*a + 2*d) + (15*cosh(b*x + d)^4*cosh(-2*a + 2*d) + 12*cosh(b*x + d)^2*cosh(-2 *a + 2*d) - (15*cosh(b*x + d)^4 + 12*cosh(b*x + d)^2 - 11)*sinh(-2*a + 2*d ) - 11*cosh(-2*a + 2*d))*sinh(b*x + d)^2 - 6*(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*cosh(b*x + d)^2 + 1)*sinh(-2*a + 2*d) + cosh(-2*a + 2 *d))*log(2*cosh(b*x + d)/(cosh(b*x + d) - sinh(b*x + d))) + 2*(3*cosh(b*x + d)^5*cosh(-2*a + 2*d) + 4*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 11*cosh(b*x + d)*cosh(-2*a + 2*d) - (3*cosh(b*x + d)^5 + 4*cosh(b*x + d)^3 - 11*co...
\[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=e^{2 a} \int e^{2 b x} \tanh ^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)**3,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*tanh(b*x + d)**3, x)
Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=-\frac {6 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )}}{b} - \frac {3 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}{b} + \frac {{\left (10 \, e^{\left (-2 \, b x - 2 \, d\right )} + 5 \, e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )} e^{\left (2 \, a - 2 \, d\right )}}{2 \, b {\left (e^{\left (-2 \, b x - 2 \, d\right )} + 2 \, e^{\left (-4 \, b x - 4 \, d\right )} + e^{\left (-6 \, b x - 6 \, d\right )}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^3,x, algorithm="maxima")
Output:
-6*(b*x + d)*e^(2*a - 2*d)/b - 3*e^(2*a - 2*d)*log(e^(-2*b*x - 2*d) + 1)/b + 1/2*(10*e^(-2*b*x - 2*d) + 5*e^(-4*b*x - 4*d) + 1)*e^(2*a - 2*d)/(b*(e^ (-2*b*x - 2*d) + 2*e^(-4*b*x - 4*d) + e^(-6*b*x - 6*d)))
Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=-\frac {3 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}{b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{2 \, b} + \frac {{\left (9 \, e^{\left (4 \, b x + 2 \, a + 4 \, d\right )} + 6 \, e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, d\right )}}{2 \, b {\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^3,x, algorithm="giac")
Output:
-3*e^(2*a - 2*d)*log(e^(2*b*x + 2*d) + 1)/b + 1/2*e^(2*b*x + 2*a)/b + 1/2* (9*e^(4*b*x + 2*a + 4*d) + 6*e^(2*b*x + 2*a + 2*d) + e^(2*a))*e^(-2*d)/(b* (e^(2*b*x + 2*d) + 1)^2)
Timed out. \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=\int {\mathrm {e}}^{2\,a+2\,b\,x}\,{\mathrm {tanh}\left (d+b\,x\right )}^3 \,d x \] Input:
int(exp(2*a + 2*b*x)*tanh(d + b*x)^3,x)
Output:
int(exp(2*a + 2*b*x)*tanh(d + b*x)^3, x)
Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.33 \[ \int e^{2 (a+b x)} \tanh ^3(d+b x) \, dx=\frac {e^{2 a} \left (2 e^{6 b x +6 d}-12 e^{4 b x +4 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )+15 e^{4 b x +4 d}-24 e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-12 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )-5\right )}{4 e^{2 d} b \left (e^{4 b x +4 d}+2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*tanh(b*x+d)^3,x)
Output:
(e**(2*a)*(2*e**(6*b*x + 6*d) - 12*e**(4*b*x + 4*d)*log(e**(2*b*x + 2*d) + 1) + 15*e**(4*b*x + 4*d) - 24*e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d) + 1) - 12*log(e**(2*b*x + 2*d) + 1) - 5))/(4*e**(2*d)*b*(e**(4*b*x + 4*d) + 2*e **(2*b*x + 2*d) + 1))