Integrand size = 13, antiderivative size = 67 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\frac {x}{b^2}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b}}-\frac {\sinh (x)}{b (a+b \cosh (x))} \] Output:
x/b^2-2*a*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/(a-b)^(1/2)/b^2/(a+ b)^(1/2)-sinh(x)/b/(a+b*cosh(x))
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\frac {x+\frac {2 a \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {b \sinh (x)}{a+b \cosh (x)}}{b^2} \] Input:
Integrate[Sinh[x]^2/(a + b*Cosh[x])^2,x]
Output:
(x + (2*a*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - (b*Sinh[x])/(a + b*Cosh[x]))/b^2
Time = 0.40 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 25, 3172, 25, 3042, 3214, 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.
\(\displaystyle \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (-\frac {\pi }{2}+i x\right )^2}{\left (a-b \sin \left (-\frac {\pi }{2}+i x\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (i x-\frac {\pi }{2}\right )^2}{\left (a-b \sin \left (i x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle -\frac {\int -\frac {\cosh (x)}{a+b \cosh (x)}dx}{b}-\frac {\sinh (x)}{b (a+b \cosh (x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\cosh (x)}{a+b \cosh (x)}dx}{b}-\frac {\sinh (x)}{b (a+b \cosh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sinh (x)}{b (a+b \cosh (x))}+\frac {\int \frac {\sin \left (i x+\frac {\pi }{2}\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \cosh (x)}dx}{b}}{b}-\frac {\sinh (x)}{b (a+b \cosh (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\sinh (x)}{b (a+b \cosh (x))}+\frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}}{b}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {x}{b}-\frac {2 a \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{b}}{b}-\frac {\sinh (x)}{b (a+b \cosh (x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {x}{b}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\sinh (x)}{b (a+b \cosh (x))}\) |
Input:
Int[Sinh[x]^2/(a + b*Cosh[x])^2,x]
Output:
(x/b - (2*a*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*S qrt[a + b]))/b - Sinh[x]/(b*(a + b*Cosh[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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 1.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.48
method | result | size |
default | \(\frac {\ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{b^{2}}+\frac {\frac {2 b \tanh \left (\frac {x}{2}\right )}{\tanh \left (\frac {x}{2}\right )^{2} a -b \tanh \left (\frac {x}{2}\right )^{2}-a -b}-\frac {2 a \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}\) | \(99\) |
risch | \(\frac {x}{b^{2}}+\frac {2 a \,{\mathrm e}^{x}+2 b}{b^{2} \left ({\mathrm e}^{2 x} b +2 a \,{\mathrm e}^{x}+b \right )}+\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{2}}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{2}}\) | \(148\) |
Input:
int(sinh(x)^2/(a+b*cosh(x))^2,x,method=_RETURNVERBOSE)
Output:
1/b^2*ln(1+tanh(1/2*x))+2/b^2*(b*tanh(1/2*x)/(tanh(1/2*x)^2*a-b*tanh(1/2*x )^2-a-b)-a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/ 2)))-1/b^2*ln(tanh(1/2*x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (57) = 114\).
Time = 0.10 (sec) , antiderivative size = 700, normalized size of antiderivative = 10.45 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx =\text {Too large to display} \] Input:
integrate(sinh(x)^2/(a+b*cosh(x))^2,x, algorithm="fricas")
Output:
[((a^2*b - b^3)*x*cosh(x)^2 + (a^2*b - b^3)*x*sinh(x)^2 + 2*a^2*b - 2*b^3 + (a*b*cosh(x)^2 + a*b*sinh(x)^2 + 2*a^2*cosh(x) + a*b + 2*(a*b*cosh(x) + a^2)*sinh(x))*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*c osh(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*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*( b*cosh(x) + a)*sinh(x) + b)) + (a^2*b - b^3)*x + 2*(a^3 - a*b^2 + (a^3 - a *b^2)*x)*cosh(x) + 2*(a^3 - a*b^2 + (a^2*b - b^3)*x*cosh(x) + (a^3 - a*b^2 )*x)*sinh(x))/(a^2*b^3 - b^5 + (a^2*b^3 - b^5)*cosh(x)^2 + (a^2*b^3 - b^5) *sinh(x)^2 + 2*(a^3*b^2 - a*b^4)*cosh(x) + 2*(a^3*b^2 - a*b^4 + (a^2*b^3 - b^5)*cosh(x))*sinh(x)), ((a^2*b - b^3)*x*cosh(x)^2 + (a^2*b - b^3)*x*sinh (x)^2 + 2*a^2*b - 2*b^3 + 2*(a*b*cosh(x)^2 + a*b*sinh(x)^2 + 2*a^2*cosh(x) + a*b + 2*(a*b*cosh(x) + a^2)*sinh(x))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (a^2*b - b^3)*x + 2*(a^ 3 - a*b^2 + (a^3 - a*b^2)*x)*cosh(x) + 2*(a^3 - a*b^2 + (a^2*b - b^3)*x*co sh(x) + (a^3 - a*b^2)*x)*sinh(x))/(a^2*b^3 - b^5 + (a^2*b^3 - b^5)*cosh(x) ^2 + (a^2*b^3 - b^5)*sinh(x)^2 + 2*(a^3*b^2 - a*b^4)*cosh(x) + 2*(a^3*b^2 - a*b^4 + (a^2*b^3 - b^5)*cosh(x))*sinh(x))]
Timed out. \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\text {Timed out} \] Input:
integrate(sinh(x)**2/(a+b*cosh(x))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sinh(x)^2/(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 = 68, normalized size of antiderivative = 1.01 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=-\frac {2 \, a \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} + \frac {2 \, {\left (a e^{x} + b\right )}}{{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} + b\right )} b^{2}} \] Input:
integrate(sinh(x)^2/(a+b*cosh(x))^2,x, algorithm="giac")
Output:
-2*a*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b^2) + x/b^2 + 2*(a*e^x + b)/((b*e^(2*x) + 2*a*e^x + b)*b^2)
Time = 2.11 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.07 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\frac {x}{b^2}+\frac {\frac {2}{b}+\frac {2\,a\,{\mathrm {e}}^x}{b^2}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}-\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}+\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}} \] Input:
int(sinh(x)^2/(a + b*cosh(x))^2,x)
Output:
x/b^2 + (2/b + (2*a*exp(x))/b^2)/(b + 2*a*exp(x) + b*exp(2*x)) + (a*log((2 *a*exp(x))/b^3 - (2*a*(b + a*exp(x)))/(b^3*(a + b)^(1/2)*(a - b)^(1/2))))/ (b^2*(a + b)^(1/2)*(a - b)^(1/2)) - (a*log((2*a*exp(x))/b^3 + (2*a*(b + a* exp(x)))/(b^3*(a + b)^(1/2)*(a - b)^(1/2))))/(b^2*(a + b)^(1/2)*(a - b)^(1 /2))
Time = 0.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.76 \[ \int \frac {\sinh ^2(x)}{(a+b \cosh (x))^2} \, dx=\frac {2 e^{2 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a b +4 e^{x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2}+2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a b +e^{2 x} a^{2} b x -e^{2 x} a^{2} b -e^{2 x} b^{3} x +e^{2 x} b^{3}+2 e^{x} a^{3} x -2 e^{x} a \,b^{2} x +a^{2} b x +a^{2} b -b^{3} x -b^{3}}{b^{2} \left (e^{2 x} a^{2} b -e^{2 x} b^{3}+2 e^{x} a^{3}-2 e^{x} a \,b^{2}+a^{2} b -b^{3}\right )} \] Input:
int(sinh(x)^2/(a+b*cosh(x))^2,x)
Output:
(2*e**(2*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2))*a *b + 4*e**x*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2))*a **2 + 2*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2))*a*b + e**(2*x)*a**2*b*x - e**(2*x)*a**2*b - e**(2*x)*b**3*x + e**(2*x)*b**3 + 2 *e**x*a**3*x - 2*e**x*a*b**2*x + a**2*b*x + a**2*b - b**3*x - b**3)/(b**2* (e**(2*x)*a**2*b - e**(2*x)*b**3 + 2*e**x*a**3 - 2*e**x*a*b**2 + a**2*b - b**3))