Integrand size = 13, antiderivative size = 85 \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\frac {\left (a^2+2 b^2\right ) x}{2 a^3}-\frac {2 b^3 \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}-\frac {b \sinh (x)}{a^2}+\frac {\cosh (x) \sinh (x)}{2 a} \]
1/2*(a^2+2*b^2)*x/a^3-b*sinh(x)/a^2+1/2*cosh(x)*sinh(x)/a-2*b^3*arctan((a- b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/a^3/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\frac {2 a^2 x+4 b^2 x+\frac {8 b^3 \arctan \left (\frac {(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 a b \sinh (x)+a^2 \sinh (2 x)}{4 a^3} \]
(2*a^2*x + 4*b^2*x + (8*b^3*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/ Sqrt[a^2 - b^2] - 4*a*b*Sinh[x] + a^2*Sinh[2*x])/(4*a^3)
Time = 0.73 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 4340, 25, 3042, 4592, 3042, 4407, 3042, 4318, 3042, 3138, 218}
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.
\(\displaystyle \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\csc \left (\frac {\pi }{2}+i x\right )^2 \left (a+b \csc \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 4340 |
\(\displaystyle \frac {\int -\frac {\cosh (x) \left (-b \text {sech}^2(x)-a \text {sech}(x)+2 b\right )}{a+b \text {sech}(x)}dx}{2 a}+\frac {\sinh (x) \cosh (x)}{2 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\int \frac {\cosh (x) \left (-b \text {sech}^2(x)-a \text {sech}(x)+2 b\right )}{a+b \text {sech}(x)}dx}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\int \frac {-b \csc \left (i x+\frac {\pi }{2}\right )^2-a \csc \left (i x+\frac {\pi }{2}\right )+2 b}{\csc \left (i x+\frac {\pi }{2}\right ) \left (a+b \csc \left (i x+\frac {\pi }{2}\right )\right )}dx}{2 a}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\int \frac {a^2+b \text {sech}(x) a+2 b^2}{a+b \text {sech}(x)}dx}{a}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\int \frac {a^2+b \csc \left (i x+\frac {\pi }{2}\right ) a+2 b^2}{a+b \csc \left (i x+\frac {\pi }{2}\right )}dx}{a}}{2 a}\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {\text {sech}(x)}{a+b \text {sech}(x)}dx}{a}}{a}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^3 \int \frac {\csc \left (i x+\frac {\pi }{2}\right )}{a+b \csc \left (i x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^2 \int \frac {1}{\frac {a \cosh (x)}{b}+1}dx}{a}}{a}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {2 b^2 \int \frac {1}{\frac {a \sin \left (i x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a}}{2 a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {4 b^2 \int \frac {1}{\frac {a+b}{b}-\left (1-\frac {a}{b}\right ) \tanh ^2\left (\frac {x}{2}\right )}d\tanh \left (\frac {x}{2}\right )}{a}}{a}}{2 a}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sinh (x) \cosh (x)}{2 a}-\frac {\frac {2 b \sinh (x)}{a}-\frac {\frac {x \left (a^2+2 b^2\right )}{a}-\frac {4 b^3 \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\) |
(Cosh[x]*Sinh[x])/(2*a) - (-((((a^2 + 2*b^2)*x)/a - (4*b^3*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]))/a) + (2*b*Sinh[ x])/a)/(2*a)
3.1.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n)), x] - Sim p[1/(a*d*n) Int[((d*Csc[e + f*x])^(n + 1)/(a + b*Csc[e + f*x]))*Simp[b*n - a*(n + 1)*Csc[e + f*x] - b*(n + 1)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(152\) vs. \(2(71)=142\).
Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.80
method | result | size |
default | \(\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a^{3}}-\frac {2 b^{3} \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a^{3}}\) | \(153\) |
risch | \(\frac {x}{2 a}+\frac {x \,b^{2}}{a^{3}}+\frac {{\mathrm e}^{2 x}}{8 a}-\frac {b \,{\mathrm e}^{x}}{2 a^{2}}+\frac {b \,{\mathrm e}^{-x}}{2 a^{2}}-\frac {{\mathrm e}^{-2 x}}{8 a}-\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{3}}\) | \(171\) |
1/2/a/(tanh(1/2*x)-1)^2-1/2*(-a-2*b)/a^2/(tanh(1/2*x)-1)+1/2/a^3*(-a^2-2*b ^2)*ln(tanh(1/2*x)-1)-2*b^3/a^3/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2* x)/((a+b)*(a-b))^(1/2))-1/2/a/(tanh(1/2*x)+1)^2-1/2*(-a-2*b)/a^2/(tanh(1/2 *x)+1)+1/2*(a^2+2*b^2)/a^3*ln(tanh(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 860, normalized size of antiderivative = 10.12 \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\text {Too large to display} \]
[1/8*((a^4 - a^2*b^2)*cosh(x)^4 + (a^4 - a^2*b^2)*sinh(x)^4 - a^4 + a^2*b^ 2 + 4*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^4 - a^2*b^2)*cosh(x))*sinh(x)^3 + 2*(3*(a^4 - a^2*b^ 2)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*si nh(x)^2 - 8*(b^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2)*sqrt(- a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^ 2 + 2*(a^2*cosh(x) + a*b)*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh (x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cosh(x) + b)*sin h(x) + a)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + (a^4 - a^2*b^2 )*cosh(x)^3 + 2*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh (x)^2)*sinh(x))/((a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*b^2)*cosh(x)*sin h(x) + (a^5 - a^3*b^2)*sinh(x)^2), 1/8*((a^4 - a^2*b^2)*cosh(x)^4 + (a^4 - a^2*b^2)*sinh(x)^4 - a^4 + a^2*b^2 + 4*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x)^ 2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^4 - a^2*b^2)*cosh( x))*sinh(x)^3 + 2*(3*(a^4 - a^2*b^2)*cosh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4) *x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 16*(b^3*cosh(x)^2 + 2*b^3*cosh (x)*sinh(x) + b^3*sinh(x)^2)*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x ) + b)/sqrt(a^2 - b^2)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + ( a^4 - a^2*b^2)*cosh(x)^3 + 2*(a^4 + a^2*b^2 - 2*b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x)^2)*sinh(x))/((a^5 - a^3*b^2)*cosh(x)^2 + 2*(a^5 - a^3*...
\[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\cosh ^{2}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.08 \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a^{3}} + \frac {a e^{\left (2 \, x\right )} - 4 \, b e^{x}}{8 \, a^{2}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} + \frac {{\left (4 \, a b e^{x} - a^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, a^{3}} \]
-2*b^3*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*a^3) + 1/8*(a* e^(2*x) - 4*b*e^x)/a^2 + 1/2*(a^2 + 2*b^2)*x/a^3 + 1/8*(4*a*b*e^x - a^2)*e ^(-2*x)/a^3
Time = 2.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.96 \[ \int \frac {\cosh ^2(x)}{a+b \text {sech}(x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,a}-\frac {{\mathrm {e}}^{-2\,x}}{8\,a}-\frac {b\,{\mathrm {e}}^x}{2\,a^2}+\frac {b\,{\mathrm {e}}^{-x}}{2\,a^2}+\frac {x\,\left (a^2+2\,b^2\right )}{2\,a^3}+\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}-\frac {2\,b^3\,\left (a+b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a^3\,\sqrt {a+b}\,\sqrt {b-a}}-\frac {b^3\,\ln \left (\frac {2\,b^3\,{\mathrm {e}}^x}{a^4}+\frac {2\,b^3\,\left (a+b\,{\mathrm {e}}^x\right )}{a^4\,\sqrt {a+b}\,\sqrt {b-a}}\right )}{a^3\,\sqrt {a+b}\,\sqrt {b-a}} \]
exp(2*x)/(8*a) - exp(-2*x)/(8*a) - (b*exp(x))/(2*a^2) + (b*exp(-x))/(2*a^2 ) + (x*(a^2 + 2*b^2))/(2*a^3) + (b^3*log((2*b^3*exp(x))/a^4 - (2*b^3*(a + b*exp(x)))/(a^4*(a + b)^(1/2)*(b - a)^(1/2))))/(a^3*(a + b)^(1/2)*(b - a)^ (1/2)) - (b^3*log((2*b^3*exp(x))/a^4 + (2*b^3*(a + b*exp(x)))/(a^4*(a + b) ^(1/2)*(b - a)^(1/2))))/(a^3*(a + b)^(1/2)*(b - a)^(1/2))