Integrand size = 13, antiderivative size = 75 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=\frac {(A-B) \sinh (x)}{7 (1+\cosh (x))^4}+\frac {(3 A+4 B) \sinh (x)}{35 (1+\cosh (x))^3}+\frac {2 (3 A+4 B) \sinh (x)}{105 (1+\cosh (x))^2}+\frac {2 (3 A+4 B) \sinh (x)}{105 (1+\cosh (x))} \] Output:
1/7*(A-B)*sinh(x)/(1+cosh(x))^4+1/35*(3*A+4*B)*sinh(x)/(1+cosh(x))^3+2/105 *(3*A+4*B)*sinh(x)/(1+cosh(x))^2+2*(3*A+4*B)*sinh(x)/(105+105*cosh(x))
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=\frac {\left (36 A+13 B+13 (3 A+4 B) \cosh (x)+8 (3 A+4 B) \cosh ^2(x)+(6 A+8 B) \cosh ^3(x)\right ) \sinh (x)}{105 (1+\cosh (x))^4} \] 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 = 0.43 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, 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.
\(\displaystyle \int \frac {A+B \cosh (x)}{(\cosh (x)+1)^4} \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {1}{7} (3 A+4 B) \int \frac {1}{(\cosh (x)+1)^3}dx+\frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}+\frac {1}{7} (3 A+4 B) \int \frac {1}{\left (\sin \left (i x+\frac {\pi }{2}\right )+1\right )^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{7} (3 A+4 B) \left (\frac {2}{5} \int \frac {1}{(\cosh (x)+1)^2}dx+\frac {\sinh (x)}{5 (\cosh (x)+1)^3}\right )+\frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}+\frac {1}{7} (3 A+4 B) \left (\frac {\sinh (x)}{5 (\cosh (x)+1)^3}+\frac {2}{5} \int \frac {1}{\left (\sin \left (i x+\frac {\pi }{2}\right )+1\right )^2}dx\right )\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {1}{7} (3 A+4 B) \left (\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{\cosh (x)+1}dx+\frac {\sinh (x)}{3 (\cosh (x)+1)^2}\right )+\frac {\sinh (x)}{5 (\cosh (x)+1)^3}\right )+\frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}+\frac {1}{7} (3 A+4 B) \left (\frac {\sinh (x)}{5 (\cosh (x)+1)^3}+\frac {2}{5} \left (\frac {\sinh (x)}{3 (\cosh (x)+1)^2}+\frac {1}{3} \int \frac {1}{\sin \left (i x+\frac {\pi }{2}\right )+1}dx\right )\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(A-B) \sinh (x)}{7 (\cosh (x)+1)^4}+\frac {1}{7} (3 A+4 B) \left (\frac {\sinh (x)}{5 (\cosh (x)+1)^3}+\frac {2}{5} \left (\frac {\sinh (x)}{3 (\cosh (x)+1)}+\frac {\sinh (x)}{3 (\cosh (x)+1)^2}\right )\right )\) |
Input:
Int[(A + B*Cosh[x])/(1 + Cosh[x])^4,x]
Output:
((A - B)*Sinh[x])/(7*(1 + Cosh[x])^4) + ((3*A + 4*B)*(Sinh[x]/(5*(1 + Cosh [x])^3) + (2*(Sinh[x]/(3*(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 = 0.41 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.69
method | result | size |
parallelrisch | \(-\frac {\tanh \left (\frac {x}{2}\right ) \left (\left (A -B \right ) \tanh \left (\frac {x}{2}\right )^{6}+\frac {7 \left (-3 A +B \right ) \tanh \left (\frac {x}{2}\right )^{4}}{5}+7 \left (A +\frac {B}{3}\right ) \tanh \left (\frac {x}{2}\right )^{2}-7 A -7 B \right )}{56}\) | \(52\) |
default | \(-\frac {\left (A -B \right ) \tanh \left (\frac {x}{2}\right )^{7}}{56}-\frac {\left (-3 A +B \right ) \tanh \left (\frac {x}{2}\right )^{5}}{40}-\frac {\left (3 A +B \right ) \tanh \left (\frac {x}{2}\right )^{3}}{24}+\frac {A \tanh \left (\frac {x}{2}\right )}{8}+\frac {B \tanh \left (\frac {x}{2}\right )}{8}\) | \(55\) |
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:
-1/56*tanh(1/2*x)*((A-B)*tanh(1/2*x)^6+7/5*(-3*A+B)*tanh(1/2*x)^4+7*(A+1/3 *B)*tanh(1/2*x)^2-7*A-7*B)
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (67) = 134\).
Time = 0.06 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.33 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=-\frac {4 \, {\left ({\left (3 \, A + 74 \, B\right )} \cosh \left (x\right )^{2} + {\left (3 \, A + 74 \, B\right )} \sinh \left (x\right )^{2} + 14 \, {\left (9 \, A + 7 \, B\right )} \cosh \left (x\right ) - 6 \, {\left ({\left (A - 22 \, B\right )} \cosh \left (x\right ) - 14 \, A - 7 \, B\right )} \sinh \left (x\right ) + 63 \, A + 84 \, B\right )}}{105 \, {\left (\cosh \left (x\right )^{5} + {\left (5 \, \cosh \left (x\right ) + 7\right )} \sinh \left (x\right )^{4} + \sinh \left (x\right )^{5} + 7 \, \cosh \left (x\right )^{4} + {\left (10 \, \cosh \left (x\right )^{2} + 28 \, \cosh \left (x\right ) + 21\right )} \sinh \left (x\right )^{3} + 21 \, \cosh \left (x\right )^{3} + {\left (10 \, \cosh \left (x\right )^{3} + 42 \, \cosh \left (x\right )^{2} + 63 \, \cosh \left (x\right ) + 36\right )} \sinh \left (x\right )^{2} + 36 \, \cosh \left (x\right )^{2} + {\left (5 \, \cosh \left (x\right )^{4} + 28 \, \cosh \left (x\right )^{3} + 63 \, \cosh \left (x\right )^{2} + 68 \, \cosh \left (x\right ) + 28\right )} \sinh \left (x\right ) + 42 \, \cosh \left (x\right ) + 21\right )}} \] 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)*c osh(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 = 0.87 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=- \frac {A \tanh ^{7}{\left (\frac {x}{2} \right )}}{56} + \frac {3 A \tanh ^{5}{\left (\frac {x}{2} \right )}}{40} - \frac {A \tanh ^{3}{\left (\frac {x}{2} \right )}}{8} + \frac {A \tanh {\left (\frac {x}{2} \right )}}{8} + \frac {B \tanh ^{7}{\left (\frac {x}{2} \right )}}{56} - \frac {B \tanh ^{5}{\left (\frac {x}{2} \right )}}{40} - \frac {B \tanh ^{3}{\left (\frac {x}{2} \right )}}{24} + \frac {B \tanh {\left (\frac {x}{2} \right )}}{8} \] Input:
integrate((A+B*cosh(x))/(1+cosh(x))**4,x)
Output:
-A*tanh(x/2)**7/56 + 3*A*tanh(x/2)**5/40 - A*tanh(x/2)**3/8 + A*tanh(x/2)/ 8 + B*tanh(x/2)**7/56 - B*tanh(x/2)**5/40 - B*tanh(x/2)**3/24 + B*tanh(x/2 )/8
Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (67) = 134\).
Time = 0.04 (sec) , antiderivative size = 449, normalized size of antiderivative = 5.99 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx =\text {Too large to display} \] 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) + 2 1*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 = 0.80 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 105 \, A e^{\left (3 \, x\right )} + 70 \, B e^{\left (3 \, x\right )} + 63 \, A e^{\left (2 \, x\right )} + 84 \, B e^{\left (2 \, x\right )} + 21 \, A e^{x} + 28 \, B e^{x} + 3 \, A + 4 \, B\right )}}{105 \, {\left (e^{x} + 1\right )}^{7}} \] 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 = 1.93 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.08 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=-\frac {\frac {4\,A}{35}+\frac {8\,B\,{\mathrm {e}}^x}{35}}{6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1}-\frac {8\,B}{105\,\left (3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1\right )}-\frac {\frac {16\,A\,{\mathrm {e}}^{3\,x}}{7}+\frac {8\,B\,{\mathrm {e}}^{2\,x}}{7}+\frac {8\,B\,{\mathrm {e}}^{4\,x}}{7}}{21\,{\mathrm {e}}^{2\,x}+35\,{\mathrm {e}}^{3\,x}+35\,{\mathrm {e}}^{4\,x}+21\,{\mathrm {e}}^{5\,x}+7\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{7\,x}+7\,{\mathrm {e}}^x+1}-\frac {\frac {8\,B\,{\mathrm {e}}^x}{21}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{7}+\frac {16\,B\,{\mathrm {e}}^{3\,x}}{21}}{15\,{\mathrm {e}}^{2\,x}+20\,{\mathrm {e}}^{3\,x}+15\,{\mathrm {e}}^{4\,x}+6\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{6\,x}+6\,{\mathrm {e}}^x+1}-\frac {\frac {8\,B}{105}+\frac {16\,A\,{\mathrm {e}}^x}{35}+\frac {16\,B\,{\mathrm {e}}^{2\,x}}{35}}{10\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^x+1} \] Input:
int((A + B*cosh(x))/(cosh(x) + 1)^4,x)
Output:
- ((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)/(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*exp(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) - ((8 *B*exp(x))/21 + (8*A*exp(2*x))/7 + (16*B*exp(3*x))/21)/(15*exp(2*x) + 20*e xp(3*x) + 15*exp(4*x) + 6*exp(5*x) + exp(6*x) + 6*exp(x) + 1) - ((8*B)/105 + (16*A*exp(x))/35 + (16*B*exp(2*x))/35)/(10*exp(2*x) + 10*exp(3*x) + 5*e xp(4*x) + exp(5*x) + 5*exp(x) + 1)
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \cosh (x)}{(1+\cosh (x))^4} \, dx=\frac {-280 e^{4 x} b -420 e^{3 x} a -280 e^{3 x} b -252 e^{2 x} a -336 e^{2 x} b -84 e^{x} a -112 e^{x} b -12 a -16 b}{105 e^{7 x}+735 e^{6 x}+2205 e^{5 x}+3675 e^{4 x}+3675 e^{3 x}+2205 e^{2 x}+735 e^{x}+105} \] 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))