Integrand size = 16, antiderivative size = 67 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=\frac {\left (a^2+b^2\right ) x}{\left (a^2-b^2\right )^2}-\frac {2 a b \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^2}+\frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))} \]
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=\frac {\left (a^2+b^2\right ) x-2 a b \log (a \cosh (x)+b \sinh (x))+\frac {b^2 \left (-a^2+b^2\right ) \sinh (x)}{a (a \cosh (x)+b \sinh (x))}}{(a-b)^2 (a+b)^2} \]
((a^2 + b^2)*x - 2*a*b*Log[a*Cosh[x] + b*Sinh[x]] + (b^2*(-a^2 + b^2)*Sinh [x])/(a*(a*Cosh[x] + b*Sinh[x])))/((a - b)^2*(a + b)^2)
Time = 0.51 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 3565, 3042, 3964, 3042, 4014, 26, 3042, 4013}
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.
\(\displaystyle \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i x)^2}{(a \cos (i x)-i b \sin (i x))^2}dx\) |
\(\Big \downarrow \) 3565 |
\(\displaystyle \int \frac {1}{(a+b \tanh (x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a-i b \tan (i x))^2}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle \frac {\int \frac {a-b \tanh (x)}{a+b \tanh (x)}dx}{a^2-b^2}+\frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}+\frac {\int \frac {a+i b \tan (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}+\frac {\frac {x \left (a^2+b^2\right )}{a^2-b^2}-\frac {2 i a b \int -\frac {i (b+a \tanh (x))}{a+b \tanh (x)}dx}{a^2-b^2}}{a^2-b^2}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\frac {x \left (a^2+b^2\right )}{a^2-b^2}-\frac {2 a b \int \frac {b+a \tanh (x)}{a+b \tanh (x)}dx}{a^2-b^2}}{a^2-b^2}+\frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}+\frac {\frac {x \left (a^2+b^2\right )}{a^2-b^2}-\frac {2 a b \int \frac {b-i a \tan (i x)}{a-i b \tan (i x)}dx}{a^2-b^2}}{a^2-b^2}\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle \frac {b}{\left (a^2-b^2\right ) (a+b \tanh (x))}+\frac {\frac {x \left (a^2+b^2\right )}{a^2-b^2}-\frac {2 a b \log (a \cosh (x)+b \sinh (x))}{a^2-b^2}}{a^2-b^2}\) |
(((a^2 + b^2)*x)/(a^2 - b^2) - (2*a*b*Log[a*Cosh[x] + b*Sinh[x]])/(a^2 - b ^2))/(a^2 - b^2) + b/((a^2 - b^2)*(a + b*Tanh[x]))
3.7.100.3.1 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[cos[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin [(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[(a + b*Tan[c + d*x])^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[m + n, 0] && IntegerQ[n] && NeQ[a^2 + b^2, 0 ]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* (x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si n[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.55
method | result | size |
default | \(-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{\left (a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{\left (a -b \right )^{2}}-\frac {2 b \left (\frac {b \left (a^{2}-b^{2}\right ) \tanh \left (\frac {x}{2}\right )}{a \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )}+a \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a +2 b \tanh \left (\frac {x}{2}\right )+a \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}\) | \(104\) |
parallelrisch | \(\frac {\left (-2 b^{2} \tanh \left (x \right ) a^{2}-2 a^{3} b \right ) \ln \left (a +b \tanh \left (x \right )\right )+\left (2 b^{2} \tanh \left (x \right ) a^{2}+2 a^{3} b \right ) \ln \left (1-\tanh \left (x \right )\right )+\left (b \left (a^{2} x +a \left (-1+x \right ) b +b^{2}\right ) \tanh \left (x \right )+a^{2} x \left (a +b \right )\right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (a +b \tanh \left (x \right )\right ) a}\) | \(108\) |
risch | \(\frac {x}{a^{2}+2 a b +b^{2}}+\frac {4 a b x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {2 b^{2}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right ) \left (a \,{\mathrm e}^{2 x}+b \,{\mathrm e}^{2 x}+a -b \right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(121\) |
-1/(a+b)^2*ln(tanh(1/2*x)-1)+1/(a-b)^2*ln(tanh(1/2*x)+1)-2*b/(a+b)^2/(a-b) ^2*(b*(a^2-b^2)/a*tanh(1/2*x)/(tanh(1/2*x)^2*a+2*b*tanh(1/2*x)+a)+a*ln(tan h(1/2*x)^2*a+2*b*tanh(1/2*x)+a))
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (67) = 134\).
Time = 0.26 (sec) , antiderivative size = 348, normalized size of antiderivative = 5.19 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} x \sinh \left (x\right )^{2} + 2 \, a b^{2} - 2 \, b^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} x - 2 \, {\left (a^{2} b - a b^{2} + {\left (a^{2} b + a b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{2} b + a b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{2} b + a b^{2}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \sinh \left (x\right )^{2}} \]
((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*x*cosh(x)^2 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*x*cosh(x)*sinh(x) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*x*sinh(x)^2 + 2 *a*b^2 - 2*b^3 + (a^3 + a^2*b - a*b^2 - b^3)*x - 2*(a^2*b - a*b^2 + (a^2*b + a*b^2)*cosh(x)^2 + 2*(a^2*b + a*b^2)*cosh(x)*sinh(x) + (a^2*b + a*b^2)* sinh(x)^2)*log(2*(a*cosh(x) + b*sinh(x))/(cosh(x) - sinh(x))))/(a^5 - a^4* b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5 + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2 *b^3 + a*b^4 + b^5)*cosh(x)^2 + 2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a *b^4 + b^5)*cosh(x)*sinh(x) + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*sinh(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 952 vs. \(2 (56) = 112\).
Time = 0.76 (sec) , antiderivative size = 952, normalized size of antiderivative = 14.21 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=\text {Too large to display} \]
Piecewise((zoo*(x - cosh(x)/sinh(x)), Eq(a, 0) & Eq(b, 0)), ((x - cosh(x)/ sinh(x))/b**2, Eq(a, 0)), (2*x*sinh(x)**2/(8*b**2*sinh(x)**2 - 16*b**2*sin h(x)*cosh(x) + 8*b**2*cosh(x)**2) - 4*x*sinh(x)*cosh(x)/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + 2*x*cosh(x)**2/(8*b**2*s inh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) - sinh(x)**2/(8*b **2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + 3*cosh(x)* *2/(8*b**2*sinh(x)**2 - 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2), Eq(a , -b)), (2*x*sinh(x)**2/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x)*cosh(x) + 8*b **2*cosh(x)**2) + 4*x*sinh(x)*cosh(x)/(8*b**2*sinh(x)**2 + 16*b**2*sinh(x) *cosh(x) + 8*b**2*cosh(x)**2) + 2*x*cosh(x)**2/(8*b**2*sinh(x)**2 + 16*b** 2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) + sinh(x)**2/(8*b**2*sinh(x)**2 + 1 6*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2) - 3*cosh(x)**2/(8*b**2*sinh(x) **2 + 16*b**2*sinh(x)*cosh(x) + 8*b**2*cosh(x)**2), Eq(a, b)), (a**3*x*cos h(x)/(a**5*cosh(x) + a**4*b*sinh(x) - 2*a**3*b**2*cosh(x) - 2*a**2*b**3*si nh(x) + a*b**4*cosh(x) + b**5*sinh(x)) + a**2*b*x*sinh(x)/(a**5*cosh(x) + a**4*b*sinh(x) - 2*a**3*b**2*cosh(x) - 2*a**2*b**3*sinh(x) + a*b**4*cosh(x ) + b**5*sinh(x)) - 2*a**2*b*log(cosh(x) + b*sinh(x)/a)*cosh(x)/(a**5*cosh (x) + a**4*b*sinh(x) - 2*a**3*b**2*cosh(x) - 2*a**2*b**3*sinh(x) + a*b**4* cosh(x) + b**5*sinh(x)) + a**2*b*cosh(x)/(a**5*cosh(x) + a**4*b*sinh(x) - 2*a**3*b**2*cosh(x) - 2*a**2*b**3*sinh(x) + a*b**4*cosh(x) + b**5*sinh(...
Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.55 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=-\frac {2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {2 \, b^{2}}{a^{4} - 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, x\right )}} + \frac {x}{a^{2} + 2 \, a b + b^{2}} \]
-2*a*b*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - 2*a^2*b^2 + b^4) - 2*b^2/(a^4 - 2*a^2*b^2 + b^4 + (a^4 - 2*a^3*b + 2*a*b^3 - b^4)*e^(-2*x)) + x/(a^2 + 2*a*b + b^2)
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=-\frac {2 \, a b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {x}{a^{2} - 2 \, a b + b^{2}} + \frac {2 \, {\left (a b e^{\left (2 \, x\right )} + a b - b^{2}\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3}\right )} {\left (a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b\right )}} \]
-2*a*b*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - 2*a^2*b^2 + b^4) + x /(a^2 - 2*a*b + b^2) + 2*(a*b*e^(2*x) + a*b - b^2)/((a^3 - a^2*b - a*b^2 + b^3)*(a*e^(2*x) + b*e^(2*x) + a - b))
Time = 0.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.55 \[ \int \frac {\cosh ^2(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx=\frac {\frac {b\,\mathrm {cosh}\left (x\right )}{a^2-b^2}+\frac {x\,\mathrm {sinh}\left (x\right )\,\left (a^2\,b+b^3\right )}{{\left (a^2-b^2\right )}^2}+\frac {a\,x\,\mathrm {cosh}\left (x\right )\,\left (a^2+b^2\right )}{{\left (a^2-b^2\right )}^2}}{a\,\mathrm {cosh}\left (x\right )+b\,\mathrm {sinh}\left (x\right )}+\ln \left (a\,\mathrm {cosh}\left (x\right )+b\,\mathrm {sinh}\left (x\right )\right )\,\left (\frac {1}{2\,{\left (a+b\right )}^2}-\frac {1}{2\,{\left (a-b\right )}^2}\right ) \]