Integrand size = 15, antiderivative size = 81 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=-\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}-\frac {(3 A-4 B) \sinh (x)}{35 (1-\cosh (x))^3}-\frac {2 (3 A-4 B) \sinh (x)}{105 (1-\cosh (x))^2}-\frac {2 (3 A-4 B) \sinh (x)}{105 (1-\cosh (x))} \] Output:
-1/7*(A+B)*sinh(x)/(1-cosh(x))^4-1/35*(3*A-4*B)*sinh(x)/(1-cosh(x))^3-2/10 5*(3*A-4*B)*sinh(x)/(1-cosh(x))^2-2*(3*A-4*B)*sinh(x)/(105-105*cosh(x))
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=\frac {\left (-36 A+13 B+13 (3 A-4 B) \cosh (x)-8 (3 A-4 B) \cosh ^2(x)+(6 A-8 B) \cosh ^3(x)\right ) \sinh (x)}{105 (-1+\cosh (x))^4} \] Input:
Integrate[(A + B*Cosh[x])/(1 - Cosh[x])^4,x]
Output:
((-36*A + 13*B + 13*(3*A - 4*B)*Cosh[x] - 8*(3*A - 4*B)*Cosh[x]^2 + (6*A - 8*B)*Cosh[x]^3)*Sinh[x])/(105*(-1 + Cosh[x])^4)
Time = 0.48 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3229, 3042, 3129, 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.
\(\displaystyle \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (\frac {\pi }{2}+i x\right )}{\left (1-\sin \left (\frac {\pi }{2}+i x\right )\right )^4}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {1}{7} (3 A-4 B) \int \frac {1}{(1-\cosh (x))^3}dx-\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}+\frac {1}{7} (3 A-4 B) \int \frac {1}{\left (1-\sin \left (i x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{7} (3 A-4 B) \left (\frac {2}{5} \int \frac {1}{(1-\cosh (x))^2}dx-\frac {\sinh (x)}{5 (1-\cosh (x))^3}\right )-\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}+\frac {1}{7} (3 A-4 B) \left (-\frac {\sinh (x)}{5 (1-\cosh (x))^3}+\frac {2}{5} \int \frac {1}{\left (1-\sin \left (i x+\frac {\pi }{2}\right )\right )^2}dx\right )\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{7} (3 A-4 B) \left (\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{1-\cosh (x)}dx-\frac {\sinh (x)}{3 (1-\cosh (x))^2}\right )-\frac {\sinh (x)}{5 (1-\cosh (x))^3}\right )-\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}+\frac {1}{7} (3 A-4 B) \left (-\frac {\sinh (x)}{5 (1-\cosh (x))^3}+\frac {2}{5} \left (-\frac {\sinh (x)}{3 (1-\cosh (x))^2}+\frac {1}{3} \int \frac {1}{1-\sin \left (i x+\frac {\pi }{2}\right )}dx\right )\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {1}{7} (3 A-4 B) \left (\frac {2}{5} \left (-\frac {\sinh (x)}{3 (1-\cosh (x))}-\frac {\sinh (x)}{3 (1-\cosh (x))^2}\right )-\frac {\sinh (x)}{5 (1-\cosh (x))^3}\right )-\frac {(A+B) \sinh (x)}{7 (1-\cosh (x))^4}\) |
Input:
Int[(A + B*Cosh[x])/(1 - Cosh[x])^4,x]
Output:
-1/7*((A + B)*Sinh[x])/(1 - Cosh[x])^4 + ((3*A - 4*B)*(-1/5*Sinh[x]/(1 - C osh[x])^3 + (2*(-1/3*Sinh[x]/(1 - Cosh[x])^2 - Sinh[x]/(3*(1 - Cosh[x])))) /5))/7
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 = 0.40 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.68
method | result | size |
parallelrisch | \(\frac {\coth \left (\frac {x}{2}\right )^{7} \left (\left (A -B \right ) \tanh \left (\frac {x}{2}\right )^{6}+\left (-A +\frac {B}{3}\right ) \tanh \left (\frac {x}{2}\right )^{4}+\frac {\left (3 A +B \right ) \tanh \left (\frac {x}{2}\right )^{2}}{5}-\frac {A}{7}-\frac {B}{7}\right )}{8}\) | \(55\) |
default | \(-\frac {-A +B}{8 \tanh \left (\frac {x}{2}\right )}-\frac {A +B}{56 \tanh \left (\frac {x}{2}\right )^{7}}-\frac {3 A -B}{24 \tanh \left (\frac {x}{2}\right )^{3}}-\frac {-3 A -B}{40 \tanh \left (\frac {x}{2}\right )^{5}}\) | \(56\) |
risch | \(-\frac {4 \left (70 B \,{\mathrm e}^{4 x}+105 A \,{\mathrm e}^{3 x}-70 B \,{\mathrm e}^{3 x}-63 A \,{\mathrm e}^{2 x}+84 B \,{\mathrm e}^{2 x}+21 A \,{\mathrm e}^{x}-28 B \,{\mathrm e}^{x}-3 A +4 B \right )}{105 \left ({\mathrm e}^{x}-1\right )^{7}}\) | \(61\) |
Input:
int((A+B*cosh(x))/(1-cosh(x))^4,x,method=_RETURNVERBOSE)
Output:
1/8*coth(1/2*x)^7*((A-B)*tanh(1/2*x)^6+(-A+1/3*B)*tanh(1/2*x)^4+1/5*(3*A+B )*tanh(1/2*x)^2-1/7*A-1/7*B)
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (65) = 130\).
Time = 0.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=\frac {4 \, {\left ({\left (3 \, A - 74 \, B\right )} \cosh \left (x\right )^{2} + {\left (3 \, A - 74 \, B\right )} \sinh \left (x\right )^{2} - 14 \, {\left (9 \, A - 7 \, B\right )} \cosh \left (x\right ) - 6 \, {\left ({\left (A + 22 \, B\right )} \cosh \left (x\right ) + 14 \, A - 7 \, B\right )} \sinh \left (x\right ) + 63 \, A - 84 \, B\right )}}{105 \, {\left (\cosh \left (x\right )^{5} + {\left (5 \, \cosh \left (x\right ) - 7\right )} \sinh \left (x\right )^{4} + \sinh \left (x\right )^{5} - 7 \, \cosh \left (x\right )^{4} + {\left (10 \, \cosh \left (x\right )^{2} - 28 \, \cosh \left (x\right ) + 21\right )} \sinh \left (x\right )^{3} + 21 \, \cosh \left (x\right )^{3} + {\left (10 \, \cosh \left (x\right )^{3} - 42 \, \cosh \left (x\right )^{2} + 63 \, \cosh \left (x\right ) - 36\right )} \sinh \left (x\right )^{2} - 36 \, \cosh \left (x\right )^{2} + {\left (5 \, \cosh \left (x\right )^{4} - 28 \, \cosh \left (x\right )^{3} + 63 \, \cosh \left (x\right )^{2} - 68 \, \cosh \left (x\right ) + 28\right )} \sinh \left (x\right ) + 42 \, \cosh \left (x\right ) - 21\right )}} \] Input:
integrate((A+B*cosh(x))/(1-cosh(x))^4,x, algorithm="fricas")
Output:
4/105*((3*A - 74*B)*cosh(x)^2 + (3*A - 74*B)*sinh(x)^2 - 14*(9*A - 7*B)*co sh(x) - 6*((A + 22*B)*cosh(x) + 14*A - 7*B)*sinh(x) + 63*A - 84*B)/(cosh(x )^5 + (5*cosh(x) - 7)*sinh(x)^4 + sinh(x)^5 - 7*cosh(x)^4 + (10*cosh(x)^2 - 28*cosh(x) + 21)*sinh(x)^3 + 21*cosh(x)^3 + (10*cosh(x)^3 - 42*cosh(x)^2 + 63*cosh(x) - 36)*sinh(x)^2 - 36*cosh(x)^2 + (5*cosh(x)^4 - 28*cosh(x)^3 + 63*cosh(x)^2 - 68*cosh(x) + 28)*sinh(x) + 42*cosh(x) - 21)
Time = 1.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=\frac {A}{8 \tanh {\left (\frac {x}{2} \right )}} - \frac {A}{8 \tanh ^{3}{\left (\frac {x}{2} \right )}} + \frac {3 A}{40 \tanh ^{5}{\left (\frac {x}{2} \right )}} - \frac {A}{56 \tanh ^{7}{\left (\frac {x}{2} \right )}} - \frac {B}{8 \tanh {\left (\frac {x}{2} \right )}} + \frac {B}{24 \tanh ^{3}{\left (\frac {x}{2} \right )}} + \frac {B}{40 \tanh ^{5}{\left (\frac {x}{2} \right )}} - \frac {B}{56 \tanh ^{7}{\left (\frac {x}{2} \right )}} \] Input:
integrate((A+B*cosh(x))/(1-cosh(x))**4,x)
Output:
A/(8*tanh(x/2)) - A/(8*tanh(x/2)**3) + 3*A/(40*tanh(x/2)**5) - A/(56*tanh( x/2)**7) - B/(8*tanh(x/2)) + B/(24*tanh(x/2)**3) + B/(40*tanh(x/2)**5) - B /(56*tanh(x/2)**7)
Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (65) = 130\).
Time = 0.05 (sec) , antiderivative size = 451, normalized size of antiderivative = 5.57 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cosh(x))/(1-cosh(x))^4,x, algorithm="maxima")
Output:
-8/105*B*(14*e^(-x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1) - 42*e^(-2*x)/(7*e^(-x) - 21*e^(- 2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1 ) + 35*e^(-3*x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e ^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1) - 35*e^(-4*x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1) - 2/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^ (-6*x) + e^(-7*x) - 1)) + 4/35*A*(7*e^(-x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^ (-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1) - 21*e^(-2 *x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7* e^(-6*x) + e^(-7*x) - 1) + 35*e^(-3*x)/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3* x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^(-7*x) - 1) - 1/(7*e^(-x) - 21*e^(-2*x) + 35*e^(-3*x) - 35*e^(-4*x) + 21*e^(-5*x) - 7*e^(-6*x) + e^( -7*x) - 1))
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 105 \, A e^{\left (3 \, x\right )} - 70 \, B e^{\left (3 \, x\right )} - 63 \, A e^{\left (2 \, x\right )} + 84 \, B e^{\left (2 \, x\right )} + 21 \, A e^{x} - 28 \, B e^{x} - 3 \, A + 4 \, B\right )}}{105 \, {\left (e^{x} - 1\right )}^{7}} \] Input:
integrate((A+B*cosh(x))/(1-cosh(x))^4,x, algorithm="giac")
Output:
-4/105*(70*B*e^(4*x) + 105*A*e^(3*x) - 70*B*e^(3*x) - 63*A*e^(2*x) + 84*B* e^(2*x) + 21*A*e^x - 28*B*e^x - 3*A + 4*B)/(e^x - 1)^7
Time = 1.91 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.88 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=\frac {\frac {8\,B}{105}+\frac {16\,A\,{\mathrm {e}}^x}{35}+\frac {16\,B\,{\mathrm {e}}^{2\,x}}{35}}{10\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^{5\,x}-5\,{\mathrm {e}}^x+1}-\frac {\frac {4\,A}{35}+\frac {8\,B\,{\mathrm {e}}^x}{35}}{6\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^x+1}-\frac {\frac {8\,B\,{\mathrm {e}}^x}{21}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{7}+\frac {16\,B\,{\mathrm {e}}^{3\,x}}{21}}{15\,{\mathrm {e}}^{2\,x}-20\,{\mathrm {e}}^{3\,x}+15\,{\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{6\,x}-6\,{\mathrm {e}}^x+1}+\frac {8\,B}{105\,\left (3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x+1\right )}+\frac {\frac {16\,A\,{\mathrm {e}}^{3\,x}}{7}+\frac {8\,B\,{\mathrm {e}}^{2\,x}}{7}+\frac {8\,B\,{\mathrm {e}}^{4\,x}}{7}}{21\,{\mathrm {e}}^{2\,x}-35\,{\mathrm {e}}^{3\,x}+35\,{\mathrm {e}}^{4\,x}-21\,{\mathrm {e}}^{5\,x}+7\,{\mathrm {e}}^{6\,x}-{\mathrm {e}}^{7\,x}-7\,{\mathrm {e}}^x+1} \] Input:
int((A + B*cosh(x))/(cosh(x) - 1)^4,x)
Output:
((8*B)/105 + (16*A*exp(x))/35 + (16*B*exp(2*x))/35)/(10*exp(2*x) - 10*exp( 3*x) + 5*exp(4*x) - exp(5*x) - 5*exp(x) + 1) - ((4*A)/35 + (8*B*exp(x))/35 )/(6*exp(2*x) - 4*exp(3*x) + exp(4*x) - 4*exp(x) + 1) - ((8*B*exp(x))/21 + (8*A*exp(2*x))/7 + (16*B*exp(3*x))/21)/(15*exp(2*x) - 20*exp(3*x) + 15*ex p(4*x) - 6*exp(5*x) + exp(6*x) - 6*exp(x) + 1) + (8*B)/(105*(3*exp(2*x) - exp(3*x) - 3*exp(x) + 1)) + ((16*A*exp(3*x))/7 + (8*B*exp(2*x))/7 + (8*B*e xp(4*x))/7)/(21*exp(2*x) - 35*exp(3*x) + 35*exp(4*x) - 21*exp(5*x) + 7*exp (6*x) - exp(7*x) - 7*exp(x) + 1)
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \cosh (x)}{(1-\cosh (x))^4} \, dx=\frac {-280 e^{4 x} b -420 e^{3 x} a +280 e^{3 x} b +252 e^{2 x} a -336 e^{2 x} b -84 e^{x} a +112 e^{x} b +12 a -16 b}{105 e^{7 x}-735 e^{6 x}+2205 e^{5 x}-3675 e^{4 x}+3675 e^{3 x}-2205 e^{2 x}+735 e^{x}-105} \] Input:
int((A+B*cosh(x))/(1-cosh(x))^4,x)
Output:
(4*( - 70*e**(4*x)*b - 105*e**(3*x)*a + 70*e**(3*x)*b + 63*e**(2*x)*a - 84 *e**(2*x)*b - 21*e**x*a + 28*e**x*b + 3*a - 4*b))/(105*(e**(7*x) - 7*e**(6 *x) + 21*e**(5*x) - 35*e**(4*x) + 35*e**(3*x) - 21*e**(2*x) + 7*e**x - 1))