Integrand size = 13, antiderivative size = 83 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=-\frac {a \left (a^2-2 b^2\right ) \cosh (x)}{b^4}+\frac {\left (a^2-2 b^2\right ) \cosh ^2(x)}{2 b^3}-\frac {a \cosh ^3(x)}{3 b^2}+\frac {\cosh ^4(x)}{4 b}+\frac {\left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{b^5} \]
-a*(a^2-2*b^2)*cosh(x)/b^4+1/2*(a^2-2*b^2)*cosh(x)^2/b^3-1/3*a*cosh(x)^3/b ^2+1/4*cosh(x)^4/b+(a^2-b^2)^2*ln(a+b*cosh(x))/b^5
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=\frac {-24 a b \left (4 a^2-7 b^2\right ) \cosh (x)-12 b^2 \left (-2 a^2+3 b^2\right ) \cosh (2 x)-8 a b^3 \cosh (3 x)+3 b^4 \cosh (4 x)+96 \left (a^2-b^2\right )^2 \log (a+b \cosh (x))}{96 b^5} \]
(-24*a*b*(4*a^2 - 7*b^2)*Cosh[x] - 12*b^2*(-2*a^2 + 3*b^2)*Cosh[2*x] - 8*a *b^3*Cosh[3*x] + 3*b^4*Cosh[4*x] + 96*(a^2 - b^2)^2*Log[a + b*Cosh[x]])/(9 6*b^5)
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 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.
\(\displaystyle \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \cos \left (-\frac {\pi }{2}+i x\right )^5}{a-b \sin \left (-\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\cos \left (i x-\frac {\pi }{2}\right )^5}{a-b \sin \left (i x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {\left (b^2-b^2 \cosh ^2(x)\right )^2}{a+b \cosh (x)}d(b \cosh (x))}{b^5}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left (-\left (\left (1-\frac {2 b^2}{a^2}\right ) a^3\right )-b^2 \cosh ^2(x) a+b^3 \cosh ^3(x)+b \left (a^2-2 b^2\right ) \cosh (x)+\frac {\left (a^2-b^2\right )^2}{a+b \cosh (x)}\right )d(b \cosh (x))}{b^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2-2 b^2\right ) \cosh ^2(x)-a b \left (a^2-2 b^2\right ) \cosh (x)+\left (a^2-b^2\right )^2 \log (a+b \cosh (x))-\frac {1}{3} a b^3 \cosh ^3(x)+\frac {1}{4} b^4 \cosh ^4(x)}{b^5}\) |
(-(a*b*(a^2 - 2*b^2)*Cosh[x]) + (b^2*(a^2 - 2*b^2)*Cosh[x]^2)/2 - (a*b^3*C osh[x]^3)/3 + (b^4*Cosh[x]^4)/4 + (a^2 - b^2)^2*Log[a + b*Cosh[x]])/b^5
3.2.67.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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 = 102.34 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(-\frac {-\frac {\cosh \left (x \right )^{4} b^{3}}{4}+\frac {a \cosh \left (x \right )^{3} b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \cosh \left (x \right )^{2} b}{2}+a \left (a^{2}-2 b^{2}\right ) \cosh \left (x \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \cosh \left (x \right )\right )}{b^{5}}\) | \(82\) |
default | \(-\frac {-\frac {\cosh \left (x \right )^{4} b^{3}}{4}+\frac {a \cosh \left (x \right )^{3} b^{2}}{3}-\frac {\left (a^{2}-2 b^{2}\right ) \cosh \left (x \right )^{2} b}{2}+a \left (a^{2}-2 b^{2}\right ) \cosh \left (x \right )}{b^{4}}+\frac {\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \cosh \left (x \right )\right )}{b^{5}}\) | \(82\) |
risch | \(-\frac {x \,a^{4}}{b^{5}}+\frac {2 x \,a^{2}}{b^{3}}-\frac {x}{b}+\frac {{\mathrm e}^{4 x}}{64 b}-\frac {a \,{\mathrm e}^{3 x}}{24 b^{2}}+\frac {{\mathrm e}^{2 x} a^{2}}{8 b^{3}}-\frac {3 \,{\mathrm e}^{2 x}}{16 b}-\frac {a^{3} {\mathrm e}^{x}}{2 b^{4}}+\frac {7 a \,{\mathrm e}^{x}}{8 b^{2}}-\frac {a^{3} {\mathrm e}^{-x}}{2 b^{4}}+\frac {7 a \,{\mathrm e}^{-x}}{8 b^{2}}+\frac {{\mathrm e}^{-2 x} a^{2}}{8 b^{3}}-\frac {3 \,{\mathrm e}^{-2 x}}{16 b}-\frac {a \,{\mathrm e}^{-3 x}}{24 b^{2}}+\frac {{\mathrm e}^{-4 x}}{64 b}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right ) a^{4}}{b^{5}}-\frac {2 \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right ) a^{2}}{b^{3}}+\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{b}\) | \(210\) |
-1/b^4*(-1/4*cosh(x)^4*b^3+1/3*a*cosh(x)^3*b^2-1/2*(a^2-2*b^2)*cosh(x)^2*b +a*(a^2-2*b^2)*cosh(x))+(a^4-2*a^2*b^2+b^4)/b^5*ln(a+b*cosh(x))
Leaf count of result is larger than twice the leaf count of optimal. 866 vs. \(2 (77) = 154\).
Time = 0.27 (sec) , antiderivative size = 866, normalized size of antiderivative = 10.43 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \]
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*cosh(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. \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (77) = 154\).
Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.14 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=-\frac {{\left (8 \, a b^{2} e^{\left (-x\right )} - 3 \, b^{3} - 12 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )} + 24 \, {\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-3 \, x\right )}\right )} e^{\left (4 \, x\right )}}{192 \, b^{4}} - \frac {8 \, a b^{2} e^{\left (-3 \, x\right )} - 3 \, b^{3} e^{\left (-4 \, x\right )} + 24 \, {\left (4 \, a^{3} - 7 \, a b^{2}\right )} e^{\left (-x\right )} - 12 \, {\left (2 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (-2 \, x\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x}{b^{5}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{b^{5}} \]
-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 = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.49 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=\frac {3 \, b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 8 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 24 \, a^{2} b {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 48 \, b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 96 \, a^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 192 \, a b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}}{192 \, b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{b^{5}} \]
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 = 1.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.04 \[ \int \frac {\sinh ^5(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{-4\,x}}{64\,b}+\frac {{\mathrm {e}}^{4\,x}}{64\,b}-\frac {x\,{\left (a^2-b^2\right )}^2}{b^5}+\frac {{\mathrm {e}}^{-x}\,\left (7\,a\,b^2-4\,a^3\right )}{8\,b^4}+\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{b^5}-\frac {a\,{\mathrm {e}}^{-3\,x}}{24\,b^2}-\frac {a\,{\mathrm {e}}^{3\,x}}{24\,b^2}+\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a^2-3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a^2-3\,b^2\right )}{16\,b^3}+\frac {{\mathrm {e}}^x\,\left (7\,a\,b^2-4\,a^3\right )}{8\,b^4} \]
exp(-4*x)/(64*b) + exp(4*x)/(64*b) - (x*(a^2 - b^2)^2)/b^5 + (exp(-x)*(7*a *b^2 - 4*a^3))/(8*b^4) + (log(b + 2*a*exp(x) + b*exp(2*x))*(a^4 + b^4 - 2* a^2*b^2))/b^5 - (a*exp(-3*x))/(24*b^2) - (a*exp(3*x))/(24*b^2) + (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)