Integrand size = 15, antiderivative size = 60 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=-\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}-\frac {(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))^2}-\frac {(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))} \]
-1/5*(A+B)*sinh(x)/(1-cosh(x))^3-1/15*(2*A-3*B)*sinh(x)/(1-cosh(x))^2-1/15 *(2*A-3*B)*sinh(x)/(1-cosh(x))
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=\frac {\left (7 A-3 B+(-6 A+9 B) \cosh (x)+(2 A-3 B) \cosh ^2(x)\right ) \sinh (x)}{15 (-1+\cosh (x))^3} \]
Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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.
\(\displaystyle \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {1}{5} (2 A-3 B) \int \frac {1}{(1-\cosh (x))^2}dx-\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}+\frac {1}{5} (2 A-3 B) \int \frac {1}{\left (1-\sin \left (i x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{5} (2 A-3 B) \left (\frac {1}{3} \int \frac {1}{1-\cosh (x)}dx-\frac {\sinh (x)}{3 (1-\cosh (x))^2}\right )-\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}+\frac {1}{5} (2 A-3 B) \left (-\frac {\sinh (x)}{3 (1-\cosh (x))^2}+\frac {1}{3} \int \frac {1}{1-\sin \left (i x+\frac {\pi }{2}\right )}dx\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {1}{5} (2 A-3 B) \left (-\frac {\sinh (x)}{3 (1-\cosh (x))}-\frac {\sinh (x)}{3 (1-\cosh (x))^2}\right )-\frac {(A+B) \sinh (x)}{5 (1-\cosh (x))^3}\) |
-1/5*((A + B)*Sinh[x])/(1 - Cosh[x])^3 + ((2*A - 3*B)*(-1/3*Sinh[x]/(1 - C osh[x])^2 - Sinh[x]/(3*(1 - Cosh[x]))))/5
3.1.99.3.1 Defintions of rubi rules used
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 = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.65
method | result | size |
default | \(-\frac {-A +B}{4 \tanh \left (\frac {x}{2}\right )}-\frac {-A -B}{20 \tanh \left (\frac {x}{2}\right )^{5}}-\frac {A}{6 \tanh \left (\frac {x}{2}\right )^{3}}\) | \(39\) |
risch | \(\frac {2 B \,{\mathrm e}^{3 x}+\frac {8 A \,{\mathrm e}^{2 x}}{3}-2 B \,{\mathrm e}^{2 x}-\frac {4 A \,{\mathrm e}^{x}}{3}+2 B \,{\mathrm e}^{x}+\frac {4 A}{15}-\frac {2 B}{5}}{\left ({\mathrm e}^{x}-1\right )^{5}}\) | \(47\) |
parallelrisch | \(\frac {\left (-12 A +18 B \right ) \sinh \left (2 x \right )+\left (2 A -3 B \right ) \sinh \left (3 x \right )+30 \sinh \left (x \right ) \left (A -\frac {B}{2}\right )}{15 \cosh \left (3 x \right )+225 \cosh \left (x \right )-90 \cosh \left (2 x \right )-150}\) | \(56\) |
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (48) = 96\).
Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.12 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=\frac {2 \, {\left (15 \, B \cosh \left (x\right )^{2} + 15 \, B \sinh \left (x\right )^{2} + 2 \, {\left (11 \, A - 9 \, B\right )} \cosh \left (x\right ) + 6 \, {\left (5 \, B \cosh \left (x\right ) + 3 \, A - 2 \, B\right )} \sinh \left (x\right ) - 10 \, A + 15 \, B\right )}}{15 \, {\left (\cosh \left (x\right )^{4} + {\left (4 \, \cosh \left (x\right ) - 5\right )} \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 5 \, \cosh \left (x\right )^{3} + {\left (6 \, \cosh \left (x\right )^{2} - 15 \, \cosh \left (x\right ) + 10\right )} \sinh \left (x\right )^{2} + 10 \, \cosh \left (x\right )^{2} + {\left (4 \, \cosh \left (x\right )^{3} - 15 \, \cosh \left (x\right )^{2} + 20 \, \cosh \left (x\right ) - 9\right )} \sinh \left (x\right ) - 11 \, \cosh \left (x\right ) + 5\right )}} \]
2/15*(15*B*cosh(x)^2 + 15*B*sinh(x)^2 + 2*(11*A - 9*B)*cosh(x) + 6*(5*B*co sh(x) + 3*A - 2*B)*sinh(x) - 10*A + 15*B)/(cosh(x)^4 + (4*cosh(x) - 5)*sin h(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 = 0.55 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=\frac {A}{4 \tanh {\left (\frac {x}{2} \right )}} - \frac {A}{6 \tanh ^{3}{\left (\frac {x}{2} \right )}} + \frac {A}{20 \tanh ^{5}{\left (\frac {x}{2} \right )}} - \frac {B}{4 \tanh {\left (\frac {x}{2} \right )}} + \frac {B}{20 \tanh ^{5}{\left (\frac {x}{2} \right )}} \]
A/(4*tanh(x/2)) - A/(6*tanh(x/2)**3) + A/(20*tanh(x/2)**5) - B/(4*tanh(x/2 )) + B/(20*tanh(x/2)**5)
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (48) = 96\).
Time = 0.19 (sec) , antiderivative size = 267, normalized size of antiderivative = 4.45 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=-\frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {5 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} + \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} + \frac {4}{15} \, A {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} \]
-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)) + 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))
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} - 15 \, B e^{\left (2 \, x\right )} - 10 \, A e^{x} + 15 \, B e^{x} + 2 \, A - 3 \, B\right )}}{15 \, {\left (e^{x} - 1\right )}^{5}} \]
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 = 0.08 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.38 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^3} \, dx=\frac {\frac {B}{5}+\frac {4\,A\,{\mathrm {e}}^x}{5}+\frac {3\,B\,{\mathrm {e}}^{2\,x}}{5}}{6\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^x+1}-\frac {\frac {4\,A}{15}+\frac {2\,B\,{\mathrm {e}}^x}{5}}{3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x+1}-\frac {\frac {4\,B\,{\mathrm {e}}^x}{5}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{5}+\frac {4\,B\,{\mathrm {e}}^{3\,x}}{5}}{10\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^{5\,x}-5\,{\mathrm {e}}^x+1}+\frac {B}{5\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )} \]
(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) - ((4*B*exp(x))/5 + (8*A*exp(2*x))/5 + (4*B*exp(3*x))/5)/(1 0*exp(2*x) - 10*exp(3*x) + 5*exp(4*x) - exp(5*x) - 5*exp(x) + 1) + B/(5*(e xp(2*x) - 2*exp(x) + 1))