Integrand size = 13, antiderivative size = 81 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\frac {\left (a^2+b^2\right )^2 \log (a+b \sinh (x))}{b^5}-\frac {a \left (a^2+2 b^2\right ) \sinh (x)}{b^4}+\frac {\left (a^2+2 b^2\right ) \sinh ^2(x)}{2 b^3}-\frac {a \sinh ^3(x)}{3 b^2}+\frac {\sinh ^4(x)}{4 b} \] Output:
(a^2+b^2)^2*ln(a+b*sinh(x))/b^5-a*(a^2+2*b^2)*sinh(x)/b^4+1/2*(a^2+2*b^2)* sinh(x)^2/b^3-1/3*a*sinh(x)^3/b^2+1/4*sinh(x)^4/b
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\frac {3 b^4 \cosh ^4(x)+12 \left (a^2+b^2\right )^2 \log (a+b \sinh (x))-12 a b \left (a^2+2 b^2\right ) \sinh (x)+6 b^2 \left (a^2+b^2\right ) \sinh ^2(x)-4 a b^3 \sinh ^3(x)}{12 b^5} \] Input:
Integrate[Cosh[x]^5/(a + b*Sinh[x]),x]
Output:
(3*b^4*Cosh[x]^4 + 12*(a^2 + b^2)^2*Log[a + b*Sinh[x]] - 12*a*b*(a^2 + 2*b ^2)*Sinh[x] + 6*b^2*(a^2 + b^2)*Sinh[x]^2 - 4*a*b^3*Sinh[x]^3)/(12*b^5)
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3147, 476, 2009}
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.
\(\displaystyle \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i x)^5}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {\left (\sinh ^2(x) b^2+b^2\right )^2}{a+b \sinh (x)}d(b \sinh (x))}{b^5}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (b^3 \sinh ^3(x)-a b^2 \sinh ^2(x)+b \left (a^2+2 b^2\right ) \sinh (x)-a \left (a^2+2 b^2\right )+\frac {\left (a^2+b^2\right )^2}{a+b \sinh (x)}\right )d(b \sinh (x))}{b^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2+2 b^2\right ) \sinh ^2(x)-a b \left (a^2+2 b^2\right ) \sinh (x)+\left (a^2+b^2\right )^2 \log (a+b \sinh (x))-\frac {1}{3} a b^3 \sinh ^3(x)+\frac {1}{4} b^4 \sinh ^4(x)}{b^5}\) |
Input:
Int[Cosh[x]^5/(a + b*Sinh[x]),x]
Output:
((a^2 + b^2)^2*Log[a + b*Sinh[x]] - a*b*(a^2 + 2*b^2)*Sinh[x] + (b^2*(a^2 + 2*b^2)*Sinh[x]^2)/2 - (a*b^3*Sinh[x]^3)/3 + (b^4*Sinh[x]^4)/4)/b^5
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 22.77 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(-\frac {-\frac {\sinh \left (x \right )^{4} b^{3}}{4}+\frac {a \,b^{2} \sinh \left (x \right )^{3}}{3}-\frac {\left (a^{2}+2 b^{2}\right ) \sinh \left (x \right )^{2} b}{2}+a \left (a^{2}+2 b^{2}\right ) \sinh \left (x \right )}{b^{4}}+\frac {\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sinh \left (x \right )\right )}{b^{5}}\) | \(82\) |
default | \(-\frac {-\frac {\sinh \left (x \right )^{4} b^{3}}{4}+\frac {a \,b^{2} \sinh \left (x \right )^{3}}{3}-\frac {\left (a^{2}+2 b^{2}\right ) \sinh \left (x \right )^{2} b}{2}+a \left (a^{2}+2 b^{2}\right ) \sinh \left (x \right )}{b^{4}}+\frac {\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sinh \left (x \right )\right )}{b^{5}}\) | \(82\) |
risch | \(-\frac {a^{4} x}{b^{5}}-\frac {2 x \,a^{2}}{b^{3}}-\frac {x}{b}+\frac {{\mathrm e}^{4 x}}{64 b}-\frac {a \,{\mathrm e}^{3 x}}{24 b^{2}}+\frac {{\mathrm e}^{2 x} a^{2}}{8 b^{3}}+\frac {3 \,{\mathrm e}^{2 x}}{16 b}-\frac {a^{3} {\mathrm e}^{x}}{2 b^{4}}-\frac {7 a \,{\mathrm e}^{x}}{8 b^{2}}+\frac {a^{3} {\mathrm e}^{-x}}{2 b^{4}}+\frac {7 a \,{\mathrm e}^{-x}}{8 b^{2}}+\frac {{\mathrm e}^{-2 x} a^{2}}{8 b^{3}}+\frac {3 \,{\mathrm e}^{-2 x}}{16 b}+\frac {a \,{\mathrm e}^{-3 x}}{24 b^{2}}+\frac {{\mathrm e}^{-4 x}}{64 b}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right ) a^{4}}{b^{5}}+\frac {2 \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right ) a^{2}}{b^{3}}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{b}\) | \(210\) |
Input:
int(cosh(x)^5/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
-1/b^4*(-1/4*sinh(x)^4*b^3+1/3*a*b^2*sinh(x)^3-1/2*(a^2+2*b^2)*sinh(x)^2*b +a*(a^2+2*b^2)*sinh(x))+(a^4+2*a^2*b^2+b^4)/b^5*ln(a+b*sinh(x))
Leaf count of result is larger than twice the leaf count of optimal. 865 vs. \(2 (75) = 150\).
Time = 0.12 (sec) , antiderivative size = 865, normalized size of antiderivative = 10.68 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \] Input:
integrate(cosh(x)^5/(a+b*sinh(x)),x, algorithm="fricas")
Output:
1/192*(3*b^4*cosh(x)^8 + 3*b^4*sinh(x)^8 - 8*a*b^3*cosh(x)^7 + 8*(3*b^4*co sh(x) - a*b^3)*sinh(x)^7 + 12*(2*a^2*b^2 + 3*b^4)*cosh(x)^6 + 4*(21*b^4*co sh(x)^2 - 14*a*b^3*cosh(x) + 6*a^2*b^2 + 9*b^4)*sinh(x)^6 - 192*(a^4 + 2*a ^2*b^2 + b^4)*x*cosh(x)^4 - 24*(4*a^3*b + 7*a*b^3)*cosh(x)^5 + 24*(7*b^4*c osh(x)^3 - 7*a*b^3*cosh(x)^2 - 4*a^3*b - 7*a*b^3 + 3*(2*a^2*b^2 + 3*b^4)*c osh(x))*sinh(x)^5 + 8*a*b^3*cosh(x) + 2*(105*b^4*cosh(x)^4 - 140*a*b^3*cos h(x)^3 + 90*(2*a^2*b^2 + 3*b^4)*cosh(x)^2 - 96*(a^4 + 2*a^2*b^2 + b^4)*x - 60*(4*a^3*b + 7*a*b^3)*cosh(x))*sinh(x)^4 + 3*b^4 + 24*(4*a^3*b + 7*a*b^3 )*cosh(x)^3 + 8*(21*b^4*cosh(x)^5 - 35*a*b^3*cosh(x)^4 + 12*a^3*b + 21*a*b ^3 + 30*(2*a^2*b^2 + 3*b^4)*cosh(x)^3 - 96*(a^4 + 2*a^2*b^2 + b^4)*x*cosh( x) - 30*(4*a^3*b + 7*a*b^3)*cosh(x)^2)*sinh(x)^3 + 12*(2*a^2*b^2 + 3*b^4)* cosh(x)^2 + 12*(7*b^4*cosh(x)^6 - 14*a*b^3*cosh(x)^5 + 15*(2*a^2*b^2 + 3*b ^4)*cosh(x)^4 + 2*a^2*b^2 + 3*b^4 - 96*(a^4 + 2*a^2*b^2 + b^4)*x*cosh(x)^2 - 20*(4*a^3*b + 7*a*b^3)*cosh(x)^3 + 6*(4*a^3*b + 7*a*b^3)*cosh(x))*sinh( x)^2 + 192*((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^4 + 4*(a^4 + 2*a^2*b^2 + b^4)* cosh(x)^3*sinh(x) + 6*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2*sinh(x)^2 + 4*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^4) *log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + 8*(3*b^4*cosh(x)^7 - 7*a*b^3 *cosh(x)^6 + 9*(2*a^2*b^2 + 3*b^4)*cosh(x)^5 - 96*(a^4 + 2*a^2*b^2 + b^4)* x*cosh(x)^3 - 15*(4*a^3*b + 7*a*b^3)*cosh(x)^4 + a*b^3 + 9*(4*a^3*b + 7...
Timed out. \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \] Input:
integrate(cosh(x)**5/(a+b*sinh(x)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (75) = 150\).
Time = 0.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.22 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=-\frac {{\left (8 \, a b^{2} e^{\left (-x\right )} - 3 \, b^{3} - 12 \, {\left (2 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 24 \, {\left (4 \, a^{3} + 7 \, a b^{2}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, b^{4}} + \frac {8 \, a b^{2} e^{\left (-3 \, x\right )} + 3 \, b^{3} e^{\left (-4 \, x\right )} + 24 \, {\left (4 \, a^{3} + 7 \, a b^{2}\right )} e^{\left (-x\right )} + 12 \, {\left (2 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (-2 \, x\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} x}{b^{5}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{b^{5}} \] Input:
integrate(cosh(x)^5/(a+b*sinh(x)),x, algorithm="maxima")
Output:
-1/192*(8*a*b^2*e^(-x) - 3*b^3 - 12*(2*a^2*b + 3*b^3)*e^(-2*x) + 24*(4*a^3 + 7*a*b^2)*e^(-3*x))*e^(4*x)/b^4 + 1/192*(8*a*b^2*e^(-3*x) + 3*b^3*e^(-4* x) + 24*(4*a^3 + 7*a*b^2)*e^(-x) + 12*(2*a^2*b + 3*b^3)*e^(-2*x))/b^4 + (a ^4 + 2*a^2*b^2 + b^4)*x/b^5 + (a^4 + 2*a^2*b^2 + b^4)*log(-2*a*e^(-x) + b* e^(-2*x) - b)/b^5
Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.72 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\frac {3 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{4} + 8 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 24 \, a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 48 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 96 \, a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} + 192 \, a b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{b^{5}} \] Input:
integrate(cosh(x)^5/(a+b*sinh(x)),x, algorithm="giac")
Output:
1/192*(3*b^3*(e^(-x) - e^x)^4 + 8*a*b^2*(e^(-x) - e^x)^3 + 24*a^2*b*(e^(-x ) - e^x)^2 + 48*b^3*(e^(-x) - e^x)^2 + 96*a^3*(e^(-x) - e^x) + 192*a*b^2*( e^(-x) - e^x))/b^4 + (a^4 + 2*a^2*b^2 + b^4)*log(abs(-b*(e^(-x) - e^x) + 2 *a))/b^5
Time = 1.72 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.09 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {{\mathrm {e}}^{4\,x}}{64\,b}+\frac {\ln \left (2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{b^5}+\frac {{\mathrm {e}}^{-x}\,\left (4\,a^3+7\,a\,b^2\right )}{8\,b^4}+\frac {a\,{\mathrm {e}}^{-3\,x}}{24\,b^2}-\frac {a\,{\mathrm {e}}^{3\,x}}{24\,b^2}-\frac {x\,{\left (a^2+b^2\right )}^2}{b^5}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a^2+3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a^2+3\,b^2\right )}{16\,b^3}-\frac {{\mathrm {e}}^x\,\left (4\,a^3+7\,a\,b^2\right )}{8\,b^4} \] Input:
int(cosh(x)^5/(a + b*sinh(x)),x)
Output:
exp(-4*x)/(64*b) + exp(4*x)/(64*b) + (log(2*a*exp(x) - b + b*exp(2*x))*(a^ 4 + b^4 + 2*a^2*b^2))/b^5 + (exp(-x)*(7*a*b^2 + 4*a^3))/(8*b^4) + (a*exp(- 3*x))/(24*b^2) - (a*exp(3*x))/(24*b^2) - (x*(a^2 + b^2)^2)/b^5 + (exp(-2*x )*(2*a^2 + 3*b^2))/(16*b^3) + (exp(2*x)*(2*a^2 + 3*b^2))/(16*b^3) - (exp(x )*(7*a*b^2 + 4*a^3))/(8*b^4)
Time = 0.16 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.22 \[ \int \frac {\cosh ^5(x)}{a+b \sinh (x)} \, dx=\frac {3 e^{8 x} b^{4}-8 e^{7 x} a \,b^{3}+24 e^{6 x} a^{2} b^{2}+36 e^{6 x} b^{4}-96 e^{5 x} a^{3} b -168 e^{5 x} a \,b^{3}+192 e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{4}+384 e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{2} b^{2}+192 e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{4}-192 e^{4 x} a^{4} x -384 e^{4 x} a^{2} b^{2} x -192 e^{4 x} b^{4} x +96 e^{3 x} a^{3} b +168 e^{3 x} a \,b^{3}+24 e^{2 x} a^{2} b^{2}+36 e^{2 x} b^{4}+8 e^{x} a \,b^{3}+3 b^{4}}{192 e^{4 x} b^{5}} \] Input:
int(cosh(x)^5/(a+b*sinh(x)),x)
Output:
(3*e**(8*x)*b**4 - 8*e**(7*x)*a*b**3 + 24*e**(6*x)*a**2*b**2 + 36*e**(6*x) *b**4 - 96*e**(5*x)*a**3*b - 168*e**(5*x)*a*b**3 + 192*e**(4*x)*log(e**(2* x)*b + 2*e**x*a - b)*a**4 + 384*e**(4*x)*log(e**(2*x)*b + 2*e**x*a - b)*a* *2*b**2 + 192*e**(4*x)*log(e**(2*x)*b + 2*e**x*a - b)*b**4 - 192*e**(4*x)* a**4*x - 384*e**(4*x)*a**2*b**2*x - 192*e**(4*x)*b**4*x + 96*e**(3*x)*a**3 *b + 168*e**(3*x)*a*b**3 + 24*e**(2*x)*a**2*b**2 + 36*e**(2*x)*b**4 + 8*e* *x*a*b**3 + 3*b**4)/(192*e**(4*x)*b**5)