Integrand size = 15, antiderivative size = 40 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=-\frac {\cosh ^2(a+b x)}{b}+\frac {\cosh ^4(a+b x)}{4 b}+\frac {\log (\cosh (a+b x))}{b} \] Output:
-cosh(b*x+a)^2/b+1/4*cosh(b*x+a)^4/b+ln(cosh(b*x+a))/b
Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=\frac {-\cosh ^2(a+b x)+\frac {1}{4} \cosh ^4(a+b x)+\log (\cosh (a+b x))}{b} \] Input:
Integrate[Sinh[a + b*x]^4*Tanh[a + b*x],x]
Output:
(-Cosh[a + b*x]^2 + Cosh[a + b*x]^4/4 + Log[Cosh[a + b*x]])/b
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 26, 3070, 243, 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.
\(\displaystyle \int \sinh ^4(a+b x) \tanh (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i \sin (i a+i b x)^4 \tan (i a+i b x)dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \sin (i a+i b x)^4 \tan (i a+i b x)dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle \frac {\int \left (1-\cosh ^2(a+b x)\right )^2 \text {sech}(a+b x)d\cosh (a+b x)}{b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \left (1-\cosh ^2(a+b x)\right )^2 \text {sech}(a+b x)d\cosh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cosh ^2(a+b x)+\text {sech}(a+b x)-2\right )d\cosh ^2(a+b x)}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} \cosh ^4(a+b x)-2 \cosh ^2(a+b x)+\log \left (\cosh ^2(a+b x)\right )}{2 b}\) |
Input:
Int[Sinh[a + b*x]^4*Tanh[a + b*x],x]
Output:
(-2*Cosh[a + b*x]^2 + Cosh[a + b*x]^4/2 + Log[Cosh[a + b*x]^2])/(2*b)
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m + n - 1)/2)/x^n, x], x, Cos[e + f *x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
Time = 1.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{4}}{4}-\frac {\sinh \left (b x +a \right )^{2}}{2}+\ln \left (\cosh \left (b x +a \right )\right )}{b}\) | \(33\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{4}}{4}-\frac {\sinh \left (b x +a \right )^{2}}{2}+\ln \left (\cosh \left (b x +a \right )\right )}{b}\) | \(33\) |
risch | \(-x +\frac {{\mathrm e}^{4 b x +4 a}}{64 b}-\frac {3 \,{\mathrm e}^{2 b x +2 a}}{16 b}-\frac {3 \,{\mathrm e}^{-2 b x -2 a}}{16 b}+\frac {{\mathrm e}^{-4 b x -4 a}}{64 b}-\frac {2 a}{b}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b}\) | \(83\) |
Input:
int(sinh(b*x+a)^4*tanh(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/b*(1/4*sinh(b*x+a)^4-1/2*sinh(b*x+a)^2+ln(cosh(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (38) = 76\).
Time = 0.11 (sec) , antiderivative size = 457, normalized size of antiderivative = 11.42 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{8} + 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + \sinh \left (b x + a\right )^{8} + 4 \, {\left (7 \, \cosh \left (b x + a\right )^{2} - 3\right )} \sinh \left (b x + a\right )^{6} - 64 \, b x \cosh \left (b x + a\right )^{4} - 12 \, \cosh \left (b x + a\right )^{6} + 8 \, {\left (7 \, \cosh \left (b x + a\right )^{3} - 9 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 2 \, {\left (35 \, \cosh \left (b x + a\right )^{4} - 32 \, b x - 90 \, \cosh \left (b x + a\right )^{2}\right )} \sinh \left (b x + a\right )^{4} + 8 \, {\left (7 \, \cosh \left (b x + a\right )^{5} - 32 \, b x \cosh \left (b x + a\right ) - 30 \, \cosh \left (b x + a\right )^{3}\right )} \sinh \left (b x + a\right )^{3} + 4 \, {\left (7 \, \cosh \left (b x + a\right )^{6} - 96 \, b x \cosh \left (b x + a\right )^{2} - 45 \, \cosh \left (b x + a\right )^{4} - 3\right )} \sinh \left (b x + a\right )^{2} - 12 \, \cosh \left (b x + a\right )^{2} + 64 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4}\right )} \log \left (\frac {2 \, \cosh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 8 \, {\left (\cosh \left (b x + a\right )^{7} - 32 \, b x \cosh \left (b x + a\right )^{3} - 9 \, \cosh \left (b x + a\right )^{5} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{64 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4}\right )}} \] Input:
integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="fricas")
Output:
1/64*(cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 - 3)*sinh(b*x + a)^6 - 64*b*x*cosh(b*x + a)^4 - 12* cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 - 9*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 32*b*x - 90*cosh(b*x + a)^2)*sinh(b*x + a)^4 + 8 *(7*cosh(b*x + a)^5 - 32*b*x*cosh(b*x + a) - 30*cosh(b*x + a)^3)*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 96*b*x*cosh(b*x + a)^2 - 45*cosh(b*x + a)^ 4 - 3)*sinh(b*x + a)^2 - 12*cosh(b*x + a)^2 + 64*(cosh(b*x + a)^4 + 4*cosh (b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4)*log(2*cosh(b*x + a)/(cosh(b*x + a ) - sinh(b*x + a))) + 8*(cosh(b*x + a)^7 - 32*b*x*cosh(b*x + a)^3 - 9*cosh (b*x + a)^5 - 3*cosh(b*x + a))*sinh(b*x + a) + 1)/(b*cosh(b*x + a)^4 + 4*b *cosh(b*x + a)^3*sinh(b*x + a) + 6*b*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b *cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4)
\[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=\int \sinh ^{4}{\left (a + b x \right )} \tanh {\left (a + b x \right )}\, dx \] Input:
integrate(sinh(b*x+a)**4*tanh(b*x+a),x)
Output:
Integral(sinh(a + b*x)**4*tanh(a + b*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (38) = 76\).
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.02 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=-\frac {{\left (12 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac {b x + a}{b} - \frac {12 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} + \frac {\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} \] Input:
integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="maxima")
Output:
-1/64*(12*e^(-2*b*x - 2*a) - 1)*e^(4*b*x + 4*a)/b + (b*x + a)/b - 1/64*(12 *e^(-2*b*x - 2*a) - e^(-4*b*x - 4*a))/b + log(e^(-2*b*x - 2*a) + 1)/b
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (38) = 76\).
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.10 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=-\frac {64 \, b x - {\left (48 \, e^{\left (4 \, b x + 4 \, a\right )} - 12 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 64 \, a - e^{\left (4 \, b x + 4 \, a\right )} + 12 \, e^{\left (2 \, b x + 2 \, a\right )} - 64 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{64 \, b} \] Input:
integrate(sinh(b*x+a)^4*tanh(b*x+a),x, algorithm="giac")
Output:
-1/64*(64*b*x - (48*e^(4*b*x + 4*a) - 12*e^(2*b*x + 2*a) + 1)*e^(-4*b*x - 4*a) + 64*a - e^(4*b*x + 4*a) + 12*e^(2*b*x + 2*a) - 64*log(e^(2*b*x + 2*a ) + 1))/b
Time = 1.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.92 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1\right )}{b}-x-\frac {3\,{\mathrm {e}}^{-2\,a-2\,b\,x}}{16\,b}-\frac {3\,{\mathrm {e}}^{2\,a+2\,b\,x}}{16\,b}+\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}}{64\,b}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{64\,b} \] Input:
int(sinh(a + b*x)^4*tanh(a + b*x),x)
Output:
log(exp(2*a)*exp(2*b*x) + 1)/b - x - (3*exp(- 2*a - 2*b*x))/(16*b) - (3*ex p(2*a + 2*b*x))/(16*b) + exp(- 4*a - 4*b*x)/(64*b) + exp(4*a + 4*b*x)/(64* b)
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.30 \[ \int \sinh ^4(a+b x) \tanh (a+b x) \, dx=\frac {e^{8 b x +8 a}-12 e^{6 b x +6 a}+64 e^{4 b x +4 a} \mathrm {log}\left (e^{2 b x +2 a}+1\right )-64 e^{4 b x +4 a} b x -12 e^{2 b x +2 a}+1}{64 e^{4 b x +4 a} b} \] Input:
int(sinh(b*x+a)^4*tanh(b*x+a),x)
Output:
(e**(8*a + 8*b*x) - 12*e**(6*a + 6*b*x) + 64*e**(4*a + 4*b*x)*log(e**(2*a + 2*b*x) + 1) - 64*e**(4*a + 4*b*x)*b*x - 12*e**(2*a + 2*b*x) + 1)/(64*e** (4*a + 4*b*x)*b)