Integrand size = 11, antiderivative size = 67 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=\frac {b \cosh (x)+a \sinh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}+\frac {2 \sinh (x)}{3 a \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))} \] Output:
1/3*(b*cosh(x)+a*sinh(x))/(a^2-b^2)/(a*cosh(x)+b*sinh(x))^3+2/3*sinh(x)/a/ (a^2-b^2)/(a*cosh(x)+b*sinh(x))
Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=\frac {a b \cosh (3 x)+\left (2 a^2-b^2+\left (a^2+b^2\right ) \cosh (2 x)\right ) \sinh (x)}{3 a (a-b) (a+b) (a \cosh (x)+b \sinh (x))^3} \] Input:
Integrate[(a*Cosh[x] + b*Sinh[x])^(-4),x]
Output:
(a*b*Cosh[3*x] + (2*a^2 - b^2 + (a^2 + b^2)*Cosh[2*x])*Sinh[x])/(3*a*(a - b)*(a + b)*(a*Cosh[x] + b*Sinh[x])^3)
Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3555, 3042, 3554}
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.
\(\displaystyle \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \cos (i x)-i b \sin (i x))^4}dx\) |
\(\Big \downarrow \) 3555 |
\(\displaystyle \frac {2 \int \frac {1}{(a \cosh (x)+b \sinh (x))^2}dx}{3 \left (a^2-b^2\right )}+\frac {a \sinh (x)+b \cosh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \sinh (x)+b \cosh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}+\frac {2 \int \frac {1}{(a \cos (i x)-i b \sin (i x))^2}dx}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3554 |
\(\displaystyle \frac {2 \sinh (x)}{3 a \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))}+\frac {a \sinh (x)+b \cosh (x)}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^3}\) |
Input:
Int[(a*Cosh[x] + b*Sinh[x])^(-4),x]
Output:
(b*Cosh[x] + a*Sinh[x])/(3*(a^2 - b^2)*(a*Cosh[x] + b*Sinh[x])^3) + (2*Sin h[x])/(3*a*(a^2 - b^2)*(a*Cosh[x] + b*Sinh[x]))
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-2), x _Symbol] :> Simp[Sin[c + d*x]/(a*d*(a*Cos[c + d*x] + b*Sin[c + d*x])), x] / ; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x _Symbol] :> Simp[(b*Cos[c + d*x] - a*Sin[c + d*x])*((a*Cos[c + d*x] + b*Sin [c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[(n + 2)/((n + 1)*(a^ 2 + b^2)) Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1] && NeQ[n, -2]
Time = 121.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {4 \left (3 a \,{\mathrm e}^{2 x}+3 b \,{\mathrm e}^{2 x}+a -b \right )}{3 \left (a \,{\mathrm e}^{2 x}+b \,{\mathrm e}^{2 x}+a -b \right )^{3} \left (a +b \right )^{2}}\) | \(46\) |
default | \(-\frac {2 \left (-\frac {\tanh \left (\frac {x}{2}\right )^{5}}{a}-\frac {2 b \tanh \left (\frac {x}{2}\right )^{4}}{a^{2}}-\frac {2 \left (a^{2}+2 b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{3}}{3 a^{3}}-\frac {2 b \tanh \left (\frac {x}{2}\right )^{2}}{a^{2}}-\frac {\tanh \left (\frac {x}{2}\right )}{a}\right )}{\left (a +2 b \tanh \left (\frac {x}{2}\right )+a \tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}\) | \(87\) |
Input:
int(1/(a*cosh(x)+b*sinh(x))^4,x,method=_RETURNVERBOSE)
Output:
-4/3*(3*a*exp(2*x)+3*b*exp(2*x)+a-b)/(a*exp(2*x)+b*exp(2*x)+a-b)^3/(a+b)^2
Leaf count of result is larger than twice the leaf count of optimal. 527 vs. \(2 (63) = 126\).
Time = 0.08 (sec) , antiderivative size = 527, normalized size of antiderivative = 7.87 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=-\frac {8 \, {\left ({\left (2 \, a + b\right )} \cosh \left (x\right ) + {\left (a + 2 \, b\right )} \sinh \left (x\right )\right )}}{3 \, {\left ({\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \left (x\right )^{5} + 5 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{4} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sinh \left (x\right )^{5} + 3 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )^{3} + {\left (3 \, a^{5} + 9 \, a^{4} b + 6 \, a^{3} b^{2} - 6 \, a^{2} b^{3} - 9 \, a b^{4} - 3 \, b^{5} + 10 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{3} + {\left (10 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \left (x\right )^{3} + 9 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 2 \, {\left (2 \, a^{5} + a^{4} b - 4 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \cosh \left (x\right ) + {\left (2 \, a^{5} + 4 \, a^{4} b - 4 \, a^{3} b^{2} - 8 \, a^{2} b^{3} + 2 \, a b^{4} + 4 \, b^{5} + 5 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \cosh \left (x\right )^{4} + 9 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )\right )}} \] Input:
integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="fricas")
Output:
-8/3*((2*a + b)*cosh(x) + (a + 2*b)*sinh(x))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^5 + 5*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)*sinh(x)^4 + (a^5 + 5*a^4*b + 10*a^3*b^ 2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sinh(x)^5 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)^3 + (3*a^5 + 9*a^4*b + 6*a^3*b^2 - 6*a ^2*b^3 - 9*a*b^4 - 3*b^5 + 10*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5 *a*b^4 + b^5)*cosh(x)^2)*sinh(x)^3 + (10*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10* a^2*b^3 + 5*a*b^4 + b^5)*cosh(x)^3 + 9*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2* b^3 - 3*a*b^4 - b^5)*cosh(x))*sinh(x)^2 + 2*(2*a^5 + a^4*b - 4*a^3*b^2 - 2 *a^2*b^3 + 2*a*b^4 + b^5)*cosh(x) + (2*a^5 + 4*a^4*b - 4*a^3*b^2 - 8*a^2*b ^3 + 2*a*b^4 + 4*b^5 + 5*(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^ 4 + b^5)*cosh(x)^4 + 9*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)^2)*sinh(x))
Timed out. \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=\text {Timed out} \] Input:
integrate(1/(a*cosh(x)+b*sinh(x))**4,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (63) = 126\).
Time = 0.07 (sec) , antiderivative size = 498, normalized size of antiderivative = 7.43 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=\frac {4 \, {\left (a - b\right )} e^{\left (-2 \, x\right )}}{a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}} + \frac {4 \, a}{3 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}\right )}} + \frac {4 \, b}{3 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{5} - 3 \, a^{4} b + 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 3 \, a b^{4} + b^{5}\right )} e^{\left (-4 \, x\right )} + {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, x\right )}\right )}} \] Input:
integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="maxima")
Output:
4*(a - b)*e^(-2*x)/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3* (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*e^(-2*x) + 3*(a^5 - 3* a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*e^(-4*x) + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*e^(-6*x)) + 4/3*a/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2* b^3 + a*b^4 - b^5)*e^(-2*x) + 3*(a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3 *a*b^4 + b^5)*e^(-4*x) + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^ 4 - b^5)*e^(-6*x)) + 4/3*b/(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*e^(-2*x) + 3*( a^5 - 3*a^4*b + 2*a^3*b^2 + 2*a^2*b^3 - 3*a*b^4 + b^5)*e^(-4*x) + (a^5 - 5 *a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*e^(-6*x))
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=-\frac {4 \, {\left (3 \, a e^{\left (2 \, x\right )} + 3 \, b e^{\left (2 \, x\right )} + a - b\right )}}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}^{3}} \] Input:
integrate(1/(a*cosh(x)+b*sinh(x))^4,x, algorithm="giac")
Output:
-4/3*(3*a*e^(2*x) + 3*b*e^(2*x) + a - b)/((a^2 + 2*a*b + b^2)*(a*e^(2*x) + b*e^(2*x) + a - b)^3)
Time = 0.91 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=-\frac {a\,\left (4\,{\mathrm {e}}^{2\,x}+\frac {4}{3}\right )+b\,\left (4\,{\mathrm {e}}^{2\,x}-\frac {4}{3}\right )}{{\left (a+b\right )}^2\,{\left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}^3} \] Input:
int(1/(a*cosh(x) + b*sinh(x))^4,x)
Output:
-(a*(4*exp(2*x) + 4/3) + b*(4*exp(2*x) - 4/3))/((a + b)^2*(a - b + a*exp(2 *x) + b*exp(2*x))^3)
Time = 0.22 (sec) , antiderivative size = 268, normalized size of antiderivative = 4.00 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^4} \, dx=\frac {-12 e^{2 x} a -12 e^{2 x} b -4 a +4 b}{3 e^{6 x} a^{5}+15 e^{6 x} a^{4} b +30 e^{6 x} a^{3} b^{2}+30 e^{6 x} a^{2} b^{3}+15 e^{6 x} a \,b^{4}+3 e^{6 x} b^{5}+9 e^{4 x} a^{5}+27 e^{4 x} a^{4} b +18 e^{4 x} a^{3} b^{2}-18 e^{4 x} a^{2} b^{3}-27 e^{4 x} a \,b^{4}-9 e^{4 x} b^{5}+9 e^{2 x} a^{5}+9 e^{2 x} a^{4} b -18 e^{2 x} a^{3} b^{2}-18 e^{2 x} a^{2} b^{3}+9 e^{2 x} a \,b^{4}+9 e^{2 x} b^{5}+3 a^{5}-3 a^{4} b -6 a^{3} b^{2}+6 a^{2} b^{3}+3 a \,b^{4}-3 b^{5}} \] Input:
int(1/(a*cosh(x)+b*sinh(x))^4,x)
Output:
(4*( - 3*e**(2*x)*a - 3*e**(2*x)*b - a + b))/(3*(e**(6*x)*a**5 + 5*e**(6*x )*a**4*b + 10*e**(6*x)*a**3*b**2 + 10*e**(6*x)*a**2*b**3 + 5*e**(6*x)*a*b* *4 + e**(6*x)*b**5 + 3*e**(4*x)*a**5 + 9*e**(4*x)*a**4*b + 6*e**(4*x)*a**3 *b**2 - 6*e**(4*x)*a**2*b**3 - 9*e**(4*x)*a*b**4 - 3*e**(4*x)*b**5 + 3*e** (2*x)*a**5 + 3*e**(2*x)*a**4*b - 6*e**(2*x)*a**3*b**2 - 6*e**(2*x)*a**2*b* *3 + 3*e**(2*x)*a*b**4 + 3*e**(2*x)*b**5 + a**5 - a**4*b - 2*a**3*b**2 + 2 *a**2*b**3 + a*b**4 - b**5))