Integrand size = 18, antiderivative size = 131 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=\frac {e^{2 a+2 b x}}{2 b}-\frac {8 e^{2 a-2 d}}{3 b \left (1+e^{2 d+2 b x}\right )^3}+\frac {8 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )^2}-\frac {12 e^{2 a-2 d}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {4 e^{2 a-2 d} \log \left (1+e^{2 d+2 b x}\right )}{b} \] Output:
1/2*exp(2*b*x+2*a)/b-8/3*exp(2*a-2*d)/b/(1+exp(2*b*x+2*d))^3+8*exp(2*a-2*d )/b/(1+exp(2*b*x+2*d))^2-12*exp(2*a-2*d)/b/(1+exp(2*b*x+2*d))-4*exp(2*a-2* d)*ln(1+exp(2*b*x+2*d))/b
Time = 0.35 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.44 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=\frac {e^{2 a} \left (3 e^{2 b x}-24 \cosh (2 d) \log \left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )-\frac {16 (\cosh (d)-\sinh (d))^5}{\left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )^3}+\frac {48 (\cosh (d)-\sinh (d))^4}{\left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {72 (\cosh (d)-\sinh (d))^3}{\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)}+24 \log \left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right ) \sinh (2 d)\right )}{6 b} \] Input:
Integrate[E^(2*(a + b*x))*Tanh[d + b*x]^4,x]
Output:
(E^(2*a)*(3*E^(2*b*x) - 24*Cosh[2*d]*Log[(1 + E^(2*b*x))*Cosh[d] + (-1 + E ^(2*b*x))*Sinh[d]] - (16*(Cosh[d] - Sinh[d])^5)/((1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2*b*x))*Sinh[d])^3 + (48*(Cosh[d] - Sinh[d])^4)/((1 + E^(2*b*x)) *Cosh[d] + (-1 + E^(2*b*x))*Sinh[d])^2 - (72*(Cosh[d] - Sinh[d])^3)/((1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2*b*x))*Sinh[d]) + 24*Log[(1 + E^(2*b*x))*Co sh[d] + (-1 + E^(2*b*x))*Sinh[d]]*Sinh[2*d]))/(6*b)
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2720, 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.
\(\displaystyle \int e^{2 (a+b x)} \tanh ^4(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {e^{2 a+b x} \left (1-e^{2 b x}\right )^4}{\left (1+e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{2 a} \int \frac {e^{b x} \left (1-e^{2 b x}\right )^4}{\left (1+e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {e^{2 a} \int \frac {\left (1-e^{2 b x}\right )^4}{\left (1+e^{2 b x}\right )^4}de^{2 b x}}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {e^{2 a} \int \left (1-\frac {8}{1+e^{2 b x}}+\frac {24}{\left (1+e^{2 b x}\right )^2}-\frac {32}{\left (1+e^{2 b x}\right )^3}+\frac {16}{\left (1+e^{2 b x}\right )^4}\right )de^{2 b x}}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 a} \left (e^{2 b x}-\frac {24}{e^{2 b x}+1}+\frac {16}{\left (e^{2 b x}+1\right )^2}-\frac {16}{3 \left (e^{2 b x}+1\right )^3}-8 \log \left (e^{2 b x}+1\right )\right )}{2 b}\) |
Input:
Int[E^(2*(a + b*x))*Tanh[d + b*x]^4,x]
Output:
(E^(2*a)*(E^(2*b*x) - 16/(3*(1 + E^(2*b*x))^3) + 16/(1 + E^(2*b*x))^2 - 24 /(1 + E^(2*b*x)) - 8*Log[1 + E^(2*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_))^(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 = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98
method | result | size |
risch | \(\frac {{\mathrm e}^{2 b x +2 a}}{2 b}+\frac {8 \,{\mathrm e}^{2 a -2 d} a}{b}-\frac {4 \left (9 \,{\mathrm e}^{4 b x +4 a +4 d}+12 \,{\mathrm e}^{2 b x +4 a +2 d}+5 \,{\mathrm e}^{4 a}\right ) {\mathrm e}^{4 a -2 d}}{3 \left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{3} b}-\frac {4 \ln \left ({\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 a -2 d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(129\) |
Input:
int(exp(2*b*x+2*a)*tanh(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
1/2*exp(2*b*x+2*a)/b+8/b*exp(2*a-2*d)*a-4/3/(exp(2*b*x+2*a+2*d)+exp(2*a))^ 3/b*(9*exp(4*b*x+4*a+4*d)+12*exp(2*b*x+4*a+2*d)+5*exp(4*a))*exp(4*a-2*d)-4 *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. 1482 vs. \(2 (118) = 236\).
Time = 0.10 (sec) , antiderivative size = 1482, normalized size of antiderivative = 11.31 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^4,x, algorithm="fricas")
Output:
1/6*(3*cosh(b*x + d)^8*cosh(-2*a + 2*d) + 3*(cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*sinh(b*x + d)^8 + 24*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^7 + 9*cosh(b*x + d)^6*cosh(-2*a + 2*d) + 3*(28*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (28*cosh(b*x + d)^2 + 3)*sinh(- 2*a + 2*d) + 3*cosh(-2*a + 2*d))*sinh(b*x + d)^6 + 6*(28*cosh(b*x + d)^3*c osh(-2*a + 2*d) + 9*cosh(b*x + d)*cosh(-2*a + 2*d) - (28*cosh(b*x + d)^3 + 9*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^5 - 63*cosh(b*x + d)^4*c osh(-2*a + 2*d) + 3*(70*cosh(b*x + d)^4*cosh(-2*a + 2*d) + 45*cosh(b*x + d )^2*cosh(-2*a + 2*d) - (70*cosh(b*x + d)^4 + 45*cosh(b*x + d)^2 - 21)*sinh (-2*a + 2*d) - 21*cosh(-2*a + 2*d))*sinh(b*x + d)^4 + 12*(14*cosh(b*x + d) ^5*cosh(-2*a + 2*d) + 15*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 21*cosh(b*x + d)*cosh(-2*a + 2*d) - (14*cosh(b*x + d)^5 + 15*cosh(b*x + d)^3 - 21*cosh(b *x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^3 - 93*cosh(b*x + d)^2*cosh(-2*a + 2*d) + 3*(28*cosh(b*x + d)^6*cosh(-2*a + 2*d) + 45*cosh(b*x + d)^4*cosh( -2*a + 2*d) - 126*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (28*cosh(b*x + d)^6 + 45*cosh(b*x + d)^4 - 126*cosh(b*x + d)^2 - 31)*sinh(-2*a + 2*d) - 31*cosh (-2*a + 2*d))*sinh(b*x + d)^2 - 24*(cosh(b*x + d)^6*cosh(-2*a + 2*d) + (co sh(-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)*sinh(-2*a + 2*d))*sinh(b*x + d)^5 + 3*cosh(b* x + d)^4*cosh(-2*a + 2*d) + 3*(5*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (5*...
\[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=e^{2 a} \int e^{2 b x} \tanh ^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)**4,x)
Output:
exp(2*a)*Integral(exp(2*b*x)*tanh(b*x + d)**4, x)
Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=-\frac {8 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )}}{b} - \frac {4 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )}{b} + \frac {{\left (49 \, e^{\left (-2 \, b x - 2 \, d\right )} + 57 \, e^{\left (-4 \, b x - 4 \, d\right )} + 27 \, e^{\left (-6 \, b x - 6 \, d\right )} + 3\right )} e^{\left (2 \, a - 2 \, d\right )}}{6 \, b {\left (e^{\left (-2 \, b x - 2 \, d\right )} + 3 \, e^{\left (-4 \, b x - 4 \, d\right )} + 3 \, e^{\left (-6 \, b x - 6 \, d\right )} + e^{\left (-8 \, b x - 8 \, d\right )}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^4,x, algorithm="maxima")
Output:
-8*(b*x + d)*e^(2*a - 2*d)/b - 4*e^(2*a - 2*d)*log(e^(-2*b*x - 2*d) + 1)/b + 1/6*(49*e^(-2*b*x - 2*d) + 57*e^(-4*b*x - 4*d) + 27*e^(-6*b*x - 6*d) + 3)*e^(2*a - 2*d)/(b*(e^(-2*b*x - 2*d) + 3*e^(-4*b*x - 4*d) + 3*e^(-6*b*x - 6*d) + e^(-8*b*x - 8*d)))
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.83 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=-\frac {4 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}{b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{2 \, b} + \frac {2 \, {\left (11 \, e^{\left (6 \, b x + 2 \, a + 6 \, d\right )} + 15 \, e^{\left (4 \, b x + 2 \, a + 4 \, d\right )} + 9 \, e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (-2 \, d\right )}}{3 \, b {\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{3}} \] Input:
integrate(exp(2*b*x+2*a)*tanh(b*x+d)^4,x, algorithm="giac")
Output:
-4*e^(2*a - 2*d)*log(e^(2*b*x + 2*d) + 1)/b + 1/2*e^(2*b*x + 2*a)/b + 2/3* (11*e^(6*b*x + 2*a + 6*d) + 15*e^(4*b*x + 2*a + 4*d) + 9*e^(2*b*x + 2*a + 2*d) + e^(2*a))*e^(-2*d)/(b*(e^(2*b*x + 2*d) + 1)^3)
Timed out. \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=\int {\mathrm {e}}^{2\,a+2\,b\,x}\,{\mathrm {tanh}\left (d+b\,x\right )}^4 \,d x \] Input:
int(exp(2*a + 2*b*x)*tanh(d + b*x)^4,x)
Output:
int(exp(2*a + 2*b*x)*tanh(d + b*x)^4, x)
Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.40 \[ \int e^{2 (a+b x)} \tanh ^4(d+b x) \, dx=\frac {e^{2 a} \left (3 e^{8 b x +8 d}-24 e^{6 b x +6 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )+30 e^{6 b x +6 d}-72 e^{4 b x +4 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-72 e^{2 b x +2 d} \mathrm {log}\left (e^{2 b x +2 d}+1\right )-30 e^{2 b x +2 d}-24 \,\mathrm {log}\left (e^{2 b x +2 d}+1\right )-19\right )}{6 e^{2 d} b \left (e^{6 b x +6 d}+3 e^{4 b x +4 d}+3 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(2*b*x+2*a)*tanh(b*x+d)^4,x)
Output:
(e**(2*a)*(3*e**(8*b*x + 8*d) - 24*e**(6*b*x + 6*d)*log(e**(2*b*x + 2*d) + 1) + 30*e**(6*b*x + 6*d) - 72*e**(4*b*x + 4*d)*log(e**(2*b*x + 2*d) + 1) - 72*e**(2*b*x + 2*d)*log(e**(2*b*x + 2*d) + 1) - 30*e**(2*b*x + 2*d) - 24 *log(e**(2*b*x + 2*d) + 1) - 19))/(6*e**(2*d)*b*(e**(6*b*x + 6*d) + 3*e**( 4*b*x + 4*d) + 3*e**(2*b*x + 2*d) + 1))