Integrand size = 13, antiderivative size = 56 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=\frac {(A-B) \sinh (x)}{5 (1+\cosh (x))^3}+\frac {(2 A+3 B) \sinh (x)}{15 (1+\cosh (x))^2}+\frac {(2 A+3 B) \sinh (x)}{15 (1+\cosh (x))} \] Output:
1/5*(A-B)*sinh(x)/(1+cosh(x))^3+1/15*(2*A+3*B)*sinh(x)/(1+cosh(x))^2+(2*A+ 3*B)*sinh(x)/(15+15*cosh(x))
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=\frac {\left (7 A+3 B+(6 A+9 B) \cosh (x)+(2 A+3 B) \cosh ^2(x)\right ) \sinh (x)}{15 (1+\cosh (x))^3} \] Input:
Integrate[(A + B*Cosh[x])/(1 + Cosh[x])^3,x]
Output:
((7*A + 3*B + (6*A + 9*B)*Cosh[x] + (2*A + 3*B)*Cosh[x]^2)*Sinh[x])/(15*(1 + Cosh[x])^3)
Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3229, 3042, 3129, 3042, 3127}
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.
\(\displaystyle \int \frac {A+B \cosh (x)}{(\cosh (x)+1)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (\frac {\pi }{2}+i x\right )}{\left (1+\sin \left (\frac {\pi }{2}+i x\right )\right )^3}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {1}{5} (2 A+3 B) \int \frac {1}{(\cosh (x)+1)^2}dx+\frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3}+\frac {1}{5} (2 A+3 B) \int \frac {1}{\left (\sin \left (i x+\frac {\pi }{2}\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{5} (2 A+3 B) \left (\frac {1}{3} \int \frac {1}{\cosh (x)+1}dx+\frac {\sinh (x)}{3 (\cosh (x)+1)^2}\right )+\frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3}+\frac {1}{5} (2 A+3 B) \left (\frac {\sinh (x)}{3 (\cosh (x)+1)^2}+\frac {1}{3} \int \frac {1}{\sin \left (i x+\frac {\pi }{2}\right )+1}dx\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3}+\frac {1}{5} (2 A+3 B) \left (\frac {\sinh (x)}{3 (\cosh (x)+1)}+\frac {\sinh (x)}{3 (\cosh (x)+1)^2}\right )\) |
Input:
Int[(A + B*Cosh[x])/(1 + Cosh[x])^3,x]
Output:
((A - B)*Sinh[x])/(5*(1 + Cosh[x])^3) + ((2*A + 3*B)*(Sinh[x]/(3*(1 + Cosh [x])^2) + Sinh[x]/(3*(1 + Cosh[x]))))/5
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.62
method | result | size |
parallelrisch | \(\frac {\left (\left (A -B \right ) \tanh \left (\frac {x}{2}\right )^{4}-\frac {10 A \tanh \left (\frac {x}{2}\right )^{2}}{3}+5 A +5 B \right ) \tanh \left (\frac {x}{2}\right )}{20}\) | \(35\) |
default | \(\frac {\left (A -B \right ) \tanh \left (\frac {x}{2}\right )^{5}}{20}-\frac {A \tanh \left (\frac {x}{2}\right )^{3}}{6}+\frac {A \tanh \left (\frac {x}{2}\right )}{4}+\frac {B \tanh \left (\frac {x}{2}\right )}{4}\) | \(38\) |
risch | \(-\frac {2 \left (15 B \,{\mathrm e}^{3 x}+20 A \,{\mathrm e}^{2 x}+15 B \,{\mathrm e}^{2 x}+10 A \,{\mathrm e}^{x}+15 B \,{\mathrm e}^{x}+2 A +3 B \right )}{15 \left ({\mathrm e}^{x}+1\right )^{5}}\) | \(47\) |
Input:
int((A+B*cosh(x))/(1+cosh(x))^3,x,method=_RETURNVERBOSE)
Output:
1/20*((A-B)*tanh(1/2*x)^4-10/3*A*tanh(1/2*x)^2+5*A+5*B)*tanh(1/2*x)
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (50) = 100\).
Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.27 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B \cosh \left (x\right )^{2} + 15 \, B \sinh \left (x\right )^{2} + 2 \, {\left (11 \, A + 9 \, B\right )} \cosh \left (x\right ) + 6 \, {\left (5 \, B \cosh \left (x\right ) + 3 \, A + 2 \, B\right )} \sinh \left (x\right ) + 10 \, A + 15 \, B\right )}}{15 \, {\left (\cosh \left (x\right )^{4} + {\left (4 \, \cosh \left (x\right ) + 5\right )} \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 5 \, \cosh \left (x\right )^{3} + {\left (6 \, \cosh \left (x\right )^{2} + 15 \, \cosh \left (x\right ) + 10\right )} \sinh \left (x\right )^{2} + 10 \, \cosh \left (x\right )^{2} + {\left (4 \, \cosh \left (x\right )^{3} + 15 \, \cosh \left (x\right )^{2} + 20 \, \cosh \left (x\right ) + 9\right )} \sinh \left (x\right ) + 11 \, \cosh \left (x\right ) + 5\right )}} \] Input:
integrate((A+B*cosh(x))/(1+cosh(x))^3,x, algorithm="fricas")
Output:
-2/15*(15*B*cosh(x)^2 + 15*B*sinh(x)^2 + 2*(11*A + 9*B)*cosh(x) + 6*(5*B*c osh(x) + 3*A + 2*B)*sinh(x) + 10*A + 15*B)/(cosh(x)^4 + (4*cosh(x) + 5)*si nh(x)^3 + sinh(x)^4 + 5*cosh(x)^3 + (6*cosh(x)^2 + 15*cosh(x) + 10)*sinh(x )^2 + 10*cosh(x)^2 + (4*cosh(x)^3 + 15*cosh(x)^2 + 20*cosh(x) + 9)*sinh(x) + 11*cosh(x) + 5)
Time = 0.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=\frac {A \tanh ^{5}{\left (\frac {x}{2} \right )}}{20} - \frac {A \tanh ^{3}{\left (\frac {x}{2} \right )}}{6} + \frac {A \tanh {\left (\frac {x}{2} \right )}}{4} - \frac {B \tanh ^{5}{\left (\frac {x}{2} \right )}}{20} + \frac {B \tanh {\left (\frac {x}{2} \right )}}{4} \] Input:
integrate((A+B*cosh(x))/(1+cosh(x))**3,x)
Output:
A*tanh(x/2)**5/20 - A*tanh(x/2)**3/6 + A*tanh(x/2)/4 - B*tanh(x/2)**5/20 + B*tanh(x/2)/4
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (50) = 100\).
Time = 0.04 (sec) , antiderivative size = 263, normalized size of antiderivative = 4.70 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=\frac {4}{15} \, A {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {1}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1}\right )} + \frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {5 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {1}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1}\right )} \] Input:
integrate((A+B*cosh(x))/(1+cosh(x))^3,x, algorithm="maxima")
Output:
4/15*A*(5*e^(-x)/(5*e^(-x) + 10*e^(-2*x) + 10*e^(-3*x) + 5*e^(-4*x) + e^(- 5*x) + 1) + 10*e^(-2*x)/(5*e^(-x) + 10*e^(-2*x) + 10*e^(-3*x) + 5*e^(-4*x) + e^(-5*x) + 1) + 1/(5*e^(-x) + 10*e^(-2*x) + 10*e^(-3*x) + 5*e^(-4*x) + e^(-5*x) + 1)) + 2/5*B*(5*e^(-x)/(5*e^(-x) + 10*e^(-2*x) + 10*e^(-3*x) + 5 *e^(-4*x) + e^(-5*x) + 1) + 5*e^(-2*x)/(5*e^(-x) + 10*e^(-2*x) + 10*e^(-3* x) + 5*e^(-4*x) + e^(-5*x) + 1) + 5*e^(-3*x)/(5*e^(-x) + 10*e^(-2*x) + 10* e^(-3*x) + 5*e^(-4*x) + e^(-5*x) + 1) + 1/(5*e^(-x) + 10*e^(-2*x) + 10*e^( -3*x) + 5*e^(-4*x) + e^(-5*x) + 1))
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} + 15 \, B e^{\left (2 \, x\right )} + 10 \, A e^{x} + 15 \, B e^{x} + 2 \, A + 3 \, B\right )}}{15 \, {\left (e^{x} + 1\right )}^{5}} \] Input:
integrate((A+B*cosh(x))/(1+cosh(x))^3,x, algorithm="giac")
Output:
-2/15*(15*B*e^(3*x) + 20*A*e^(2*x) + 15*B*e^(2*x) + 10*A*e^x + 15*B*e^x + 2*A + 3*B)/(e^x + 1)^5
Time = 1.87 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.52 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=-\frac {\frac {4\,B\,{\mathrm {e}}^x}{5}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{5}+\frac {4\,B\,{\mathrm {e}}^{3\,x}}{5}}{10\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^x+1}-\frac {\frac {B}{5}+\frac {4\,A\,{\mathrm {e}}^x}{5}+\frac {3\,B\,{\mathrm {e}}^{2\,x}}{5}}{6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1}-\frac {\frac {4\,A}{15}+\frac {2\,B\,{\mathrm {e}}^x}{5}}{3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1}-\frac {B}{5\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )} \] Input:
int((A + B*cosh(x))/(cosh(x) + 1)^3,x)
Output:
- ((4*B*exp(x))/5 + (8*A*exp(2*x))/5 + (4*B*exp(3*x))/5)/(10*exp(2*x) + 10 *exp(3*x) + 5*exp(4*x) + exp(5*x) + 5*exp(x) + 1) - (B/5 + (4*A*exp(x))/5 + (3*B*exp(2*x))/5)/(6*exp(2*x) + 4*exp(3*x) + exp(4*x) + 4*exp(x) + 1) - ((4*A)/15 + (2*B*exp(x))/5)/(3*exp(2*x) + exp(3*x) + 3*exp(x) + 1) - B/(5* (exp(2*x) + 2*exp(x) + 1))
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.46 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^3} \, dx=\frac {-30 e^{3 x} b -40 e^{2 x} a -30 e^{2 x} b -20 e^{x} a -30 e^{x} b -4 a -6 b}{15 e^{5 x}+75 e^{4 x}+150 e^{3 x}+150 e^{2 x}+75 e^{x}+15} \] Input:
int((A+B*cosh(x))/(1+cosh(x))^3,x)
Output:
(2*( - 15*e**(3*x)*b - 20*e**(2*x)*a - 15*e**(2*x)*b - 10*e**x*a - 15*e**x *b - 2*a - 3*b))/(15*(e**(5*x) + 5*e**(4*x) + 10*e**(3*x) + 10*e**(2*x) + 5*e**x + 1))