Integrand size = 15, antiderivative size = 82 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2}}-\frac {(A b-a B) \sinh (x)}{\left (a^2-b^2\right ) (a+b \cosh (x))} \] Output:
2*(A*a-B*b)*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(3/2)/(a+b) ^(3/2)-(A*b-B*a)*sinh(x)/(a^2-b^2)/(a+b*cosh(x))
Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {(-A b+a B) \sinh (x)}{(a-b) (a+b) (a+b \cosh (x))} \] Input:
Integrate[(A + B*Cosh[x])/(a + b*Cosh[x])^2,x]
Output:
(2*(a*A - b*B)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^ (3/2) + ((-(A*b) + a*B)*Sinh[x])/((a - b)*(a + b)*(a + b*Cosh[x]))
Time = 0.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3233, 25, 27, 3042, 3138, 221}
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.
\(\displaystyle \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (\frac {\pi }{2}+i x\right )}{\left (a+b \sin \left (\frac {\pi }{2}+i x\right )\right )^2}dx\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle -\frac {\int -\frac {a A-b B}{a+b \cosh (x)}dx}{a^2-b^2}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a A-b B}{a+b \cosh (x)}dx}{a^2-b^2}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a A-b B) \int \frac {1}{a+b \cosh (x)}dx}{a^2-b^2}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}+\frac {(a A-b B) \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {2 (a A-b B) \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a^2-b^2}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (a A-b B) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\sinh (x) (A b-a B)}{\left (a^2-b^2\right ) (a+b \cosh (x))}\) |
Input:
Int[(A + B*Cosh[x])/(a + b*Cosh[x])^2,x]
Output:
(2*(a*A - b*B)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]* Sqrt[a + b]*(a^2 - b^2)) - ((A*b - a*B)*Sinh[x])/((a^2 - b^2)*(a + b*Cosh[ x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 0.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32
method | result | size |
default | \(\frac {2 \left (A b -B a \right ) \tanh \left (\frac {x}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {x}{2}\right )^{2} a -b \tanh \left (\frac {x}{2}\right )^{2}-a -b \right )}+\frac {2 \left (A a -B b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\) | \(108\) |
risch | \(\frac {2 \left (A b -B a \right ) \left (a \,{\mathrm e}^{x}+b \right )}{b \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 x} b +2 a \,{\mathrm e}^{x}+b \right )}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) A a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) A a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right ) B b}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}\) | \(317\) |
Input:
int((A+B*cosh(x))/(a+b*cosh(x))^2,x,method=_RETURNVERBOSE)
Output:
2*(A*b-B*a)/(a^2-b^2)*tanh(1/2*x)/(tanh(1/2*x)^2*a-b*tanh(1/2*x)^2-a-b)+2* (A*a-B*b)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b) *(a-b))^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (71) = 142\).
Time = 0.10 (sec) , antiderivative size = 828, normalized size of antiderivative = 10.10 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cosh(x))/(a+b*cosh(x))^2,x, algorithm="fricas")
Output:
[-(2*B*a^3*b - 2*A*a^2*b^2 - 2*B*a*b^3 + 2*A*b^4 - (A*a*b^2 - B*b^3 + (A*a *b^2 - B*b^3)*cosh(x)^2 + (A*a*b^2 - B*b^3)*sinh(x)^2 + 2*(A*a^2*b - B*a*b ^2)*cosh(x) + 2*(A*a^2*b - B*a*b^2 + (A*a*b^2 - B*b^3)*cosh(x))*sinh(x))*s qrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*s inh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)* sinh(x) + b)) + 2*(B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*cosh(x) + 2*(B*a ^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*sinh(x))/(a^4*b^2 - 2*a^2*b^4 + b^6 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cosh(x)^2 + (a^4*b^2 - 2*a^2*b^4 + b^6)*sinh(x )^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cosh(x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cosh(x))*sinh(x)), -2*(B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4 + (A*a*b^2 - B*b^3 + (A*a*b^2 - B*b^3)*cosh(x)^2 + (A*a* b^2 - B*b^3)*sinh(x)^2 + 2*(A*a^2*b - B*a*b^2)*cosh(x) + 2*(A*a^2*b - B*a* b^2 + (A*a*b^2 - B*b^3)*cosh(x))*sinh(x))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a ^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (B*a^4 - A*a^3*b - B* a^2*b^2 + A*a*b^3)*cosh(x) + (B*a^4 - A*a^3*b - B*a^2*b^2 + A*a*b^3)*sinh( x))/(a^4*b^2 - 2*a^2*b^4 + b^6 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cosh(x)^2 + ( a^4*b^2 - 2*a^2*b^4 + b^6)*sinh(x)^2 + 2*(a^5*b - 2*a^3*b^3 + a*b^5)*cosh( x) + 2*(a^5*b - 2*a^3*b^3 + a*b^5 + (a^4*b^2 - 2*a^2*b^4 + b^6)*cosh(x))*s inh(x))]
Timed out. \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cosh(x))/(a+b*cosh(x))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+B*cosh(x))/(a+b*cosh(x))^2,x, algorithm="maxima")
Output:
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*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\frac {2 \, {\left (A a - B b\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (B a^{2} e^{x} - A a b e^{x} + B a b - A b^{2}\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )}} \] Input:
integrate((A+B*cosh(x))/(a+b*cosh(x))^2,x, algorithm="giac")
Output:
2*(A*a - B*b)*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/((a^2 - b^2)*sqrt(-a^2 + b^2)) - 2*(B*a^2*e^x - A*a*b*e^x + B*a*b - A*b^2)/((a^2*b - b^3)*(b*e^(2 *x) + 2*a*e^x + b))
Time = 2.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.00 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=\frac {\frac {2\,\left (A\,b^3-B\,a\,b^2\right )}{b\,\left (a^2\,b-b^3\right )}-\frac {2\,{\mathrm {e}}^x\,\left (B\,a^2\,b^2-A\,a\,b^3\right )}{b^2\,\left (a^2\,b-b^3\right )}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (A\,a-B\,b\right )}{b\,\left (a^2-b^2\right )}-\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )\,\left (A\,a-B\,b\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (A\,a-B\,b\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\ln \left (\frac {2\,\left (b+a\,{\mathrm {e}}^x\right )\,\left (A\,a-B\,b\right )}{b\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,{\mathrm {e}}^x\,\left (A\,a-B\,b\right )}{b\,\left (a^2-b^2\right )}\right )\,\left (A\,a-B\,b\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:
int((A + B*cosh(x))/(a + b*cosh(x))^2,x)
Output:
((2*(A*b^3 - B*a*b^2))/(b*(a^2*b - b^3)) - (2*exp(x)*(B*a^2*b^2 - A*a*b^3) )/(b^2*(a^2*b - b^3)))/(b + 2*a*exp(x) + b*exp(2*x)) + (log(- (2*exp(x)*(A *a - B*b))/(b*(a^2 - b^2)) - (2*(b + a*exp(x))*(A*a - B*b))/(b*(a + b)^(3/ 2)*(a - b)^(3/2)))*(A*a - B*b))/((a + b)^(3/2)*(a - b)^(3/2)) - (log((2*(b + a*exp(x))*(A*a - B*b))/(b*(a + b)^(3/2)*(a - b)^(3/2)) - (2*exp(x)*(A*a - B*b))/(b*(a^2 - b^2)))*(A*a - B*b))/((a + b)^(3/2)*(a - b)^(3/2))
Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54 \[ \int \frac {A+B \cosh (x)}{(a+b \cosh (x))^2} \, dx=-\frac {2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right )}{a^{2}-b^{2}} \] Input:
int((A+B*cosh(x))/(a+b*cosh(x))^2,x)
Output:
( - 2*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2)))/(a**2 - b**2)